Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
UNIVERSITY OF EDINBURGH
EARTH MATERIALS
PRACTICAL BOOKLET
PRACTICAL HAND-OUTS FOR MINERAL SCIENCE AND COMPOSITION OF
THE EARTH SECTIONS OF THE EARTH MATERIALS COURSE
DR. GEOFFREY BROMILEY
PLEASE BRING THIS BOOKLET TO ALL PRACTICAL SESSIONS
Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
EARTH M ATERIALS WEEK-BY-WEEK SCHEDULE
Week/slot
Mon
lecture
1
2
3
4
Short
practical
Thur
lecture
Long
practical
Mon
lecture
Short
practical
Thur
lecture
Long
practical
Mon
lecture
Short
practical
Thur
lecture
Long
practical
Mon
lecture
Short
practical
Thur
lecture
Long
practical
Mineral Science (GDB)
Crystallography 1. Introduction to Crystallography: crystal faces
and interfacial angles, zones, form, habit, symmetry operators,
Miller indices and axial ratios; stereographic projection.
Crys P1. Stereographic projection of a tetragonal crystal
Crystallography 2. The 7 Crystal systems and 32 Crystal classes;
Hermann-Mauguin notation and sketch stereograms; crystal
structure and lattices; the 14 Bravais lattices
Crys P2. Stereographic projection of an orthorhombic crystal
Crystallography 3. Space groups, screw axes and glide planes,
lattice planes and X-ray diffraction, twinning
Crys P3. Indexing lattice planes and calculating d spacings
Optics 1. The theory of light and how light interacts with
minerals; introduction to the polarising microscope; mineral
colour, plechroism, relief and cleavage; introduction to
birefringence
Mineral Science Assessment 1.
Submitted at the end of the practical class
Optics 2. Anisotropy and the calcite experiment; birefringence,
retardation and interference colours
Optics P1. Recognition and use of interference colours
Mon
lecture
Short
practical
10
Optics 3. The uniaxial and the biaxial indicatrix; optic sign
Optics P2. The calcite rhomb experiment explained using the
indicatrix
Optics 4. Conoscopic observations; uniaxial and biaxial
interference figures
Optics P3. Interference figures in convergent light; uniaxial
minerals
Mineral Chemistry 1. Composition of the Earth; chemical
analysis of minerals; the Electron microprobe; recalculating
mineral compositions; graphically representing mineral
compositions
Optics P4. Interference figures in convergent light; biaxial
minerals
Mineral Science Assessment 2 handed out (due in end of week
8)
Composition of the Earth (GDB)
5
Mark
Composition 1. The atom; types of bonding; atomic/ionic radii;
close packing; classification of silicates
Chem P1. Plotting mineral compositions in 2- and 3- component
systems. Determining mineral formulae from microprobe data
20
Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
Thur
lecture
Long
practical
Mon
lecture
Short
practical
6
7
8
Thur
lecture
Long
practical
Mon
lecture
Short
practical
Thur
lecture
Long
practical
Mon
lecture
Short
practical
Composition 2. Ortho-, Dimer and Ring silicates; stability of
olivine; introduction to chain silicates (pyroxenes)
Comp P1. Close-packed minerals: spinel structures and olivine
Composition 3. Chain silicates continued; pyroxenes and
amphiboles; exchange vectors
Comp P2. Pyroxenes: structure, hand specimens and thin
sections
Submitted at the end of the practical class
Composition 4.Sheet silicates; structures and relationships
between talc, pyrophyllite and micas; serpentine minerals and
the asbestos problem; clay mineralogy
Comp P3. Amphiboles; structure, hand specimens and thin
sections
Composition 5. Framework silicates; feldspar structure,
composition, ordering and twinning
Comp P4. Sheet silicates: structure, hand specimens, thin
sections and X-ray determination.
Composition 6.Framework silicates cont. Feldspathoids; silica
polymorphs; mineral structures under extreme conditions
Comp P5. Feldspar structures, hand specimens and thin sections.
Determining feldspar compositions.
Submitted at the end of the practical class
Composition 7. Mineralogy of Earth’s deep interior continued.
Structure, composition and stability of carbonates
Mineralogy multiple choice test (30 minutes, closed book)
Comp_P6. Framework silicates.
15
15
10
Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
Crystallography Practical 1:
Stereographic projection of a
tetragonal crystal
Part 1: Manipulating the stereographic net
In the first part of the exercise we are going to practice the basic manipulations
which are necessary to plot a stereogram using data from a real crystal. We will run
through these together as a class; then you can have a go yourself.
a.
Carefully push a drawing pin through the exact centre of the cardboard net,
and then through the tracing paper. With a sharp, hard pencil, draw in the
primitive circle on the tracing paper.
b.
Mark a face pole (use an x) anyway on the primitive circle. Label it A.
c.
What angle does face A make with the plane of the paper? …………………………………
d.
Another face, B, has a horizontal face normal. It is 42˚ from A in a clockwise
direction. Plot B on your stereogram.
e.
A third face, C, is parallel to the plane of the paper, and plots in the upper
hemisphere. Plot C on your stereogram.
f.
A fourth face, D, is inclined with respect to both faces A and B. Its face
normal is neither vertical nor horizontal, but points obliquely out of the paper.
To fix its position we need two angles, from faces we have already plotted:
The interfacial angles are:
A^D 47˚
B^D 32˚
Turn your tracing paper so that the face A is at one of the two origins of the
small circles (the ‘North’ or ‘South’ pole if you were looking at a conventional
map of the whole Earth). The small circles represent lines of constant angle
from this point. Lightly sketch in the small circle which represents all points
47˚ from A.
Now turn your net so that B is at the origin of the small circles. Sketch in the
32˚ small circle. Where it intersects the small circle you drew previously is the
point where D will plot.
g.
Measure the interfacial angle C^D using the appropriate great circle, which you
should sketch in.
Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
h.
Close your eyes and mark another face pole, E, somewhere else on the upper
hemisphere of the stereogram (not where C is, and not on the primitive). Don’t
forget to mark it with an x.
i.
Measure the interfacial angle D^E (find the common circle and sketch it in).
………………………………………………………………………
j.
Now repeat stage h, but insert your new face pole, F, in the lower hemisphere.
Make sure you mark it with a neat O to show it is on the lower hemisphere.
k.
Now measure the angle D^F sketching the great circle you use. This is more
difficult than you may think at first sight, because D is on the upper
hemisphere and F on the lower hemisphere. Sketch in the common great circle
(which will be a rugby ball shape!) using a continuous line for the part on the
upper hemisphere and a broken line for the part in the lower hemisphere.
Hint: remember to visualise the stereogram as a sphere which you are looking
down upon (rather than a flat circle). For finding a great circle, you may find it
helpful to imagine that you are an ant, standing on the surface of the sphere at
D, looking for the quickest way to get to F, which is out of sight.
D^F = ……………………………………………………………
Part 2:
Plotting data from a tetragonal crystal
Now that you are familiar with some of
the basic stereogram manipulations, we
can try plotting interfacial angles from
an actual crystal. The crystal to the right
has 14 faces. Obviously, some of these
are related by symmetry.
Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
The following interfacial angles were measured:
Zone 1
Zone 2
Zone 3
m^m1 = 90˚
m^m2 = 180˚
m^m3 = 270˚
m^m = 360˚
c^p = 37˚
c^m = 90˚
c^p4 = 143˚
c^c1 = 180˚
c^p6 = 217˚
c^m2 = 270˚
c^p2 = 323˚
c^c = 360˚
c^p3 = 37˚
c^m3 = 90˚
c^p7 = 153˚
c^c1 = 180˚
c^p5 = 217˚
c^m1 = 270˚
c^p1 = 323˚
c^c = 360˚
a.
To help of understand the different zones and zone axes (the mutually parallel
edges in each zone) you may find it helpful to colour the sets of edges which
define each zone.
b.
Imagine that the crystal (much reduced in size) is sitting in the centre of your
stereogram as you see it. All the face poles for the prism zone (m, m1, m2, m3)
will plot on the primitive circle because their face normals are horizontal.
Plot face m on the right of your stereogram (at 3 o’clock). Face m1 is also on
the primitive circle and 90˚ away. Count the angle clockwise and mark in and
label m1. Continue clockwise, marking in and labelling the faces of zone 1.
c.
Where will face c plot? Continue with zone 2 starting with face c. Then plots
faces of zone 3, again starting with face c.
d.
Insert all elements of symmetry on the stereogram.
e.
Insert and label the crystallographic axes on the stereogram. Begin with the z
axis. Note that there are 2 possible orientations for the x and y axes. In this
case the convention is to plot the x and y axis bisecting the m faces (this
means that there is a face which intersects all three axes).
f.
Insert the miller indices for all the faces on the stereogram.
g.
Where would a (101) face be if it were developed? Plot it on the stereogram.
h.
Calculate the axial ratio between c and a.
i.
Draw a sketch stereogram showing all the symmetry elements and axes
Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
Part 3: symmetry operations and sketch stereograms
Complete each sketch stereogram below, showing all faces and all elements of
symmetry. Remember, the face poles in these sketch stereograms are only there to
make the stereogram easier to visualise.
a)
Reflection across mirror planes, symbolised m or as a bold line.
Insert all face poles.
m
m
A vertical mirror plane
b)
A horizontal mirror plane
Rotation about rotational axes, which may be 1-, 2-, 3-, 4- or 6-fold
Insert face poles and the correct symbols for the axes
(there is no symbol for 1)
1
2
4
3
6
Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
Crystallography Practical 2:
Stereographic projection of an
orthorhombic crystal
This exercise is similar to that in Practical 1. You will plot the faces (as poles),
identify the symmetry elements, choose the axes, determine the crystal class,
choose (111), index the faces and then calculate axial ratios.
The following goniometric measurements were made on an orthorhombic crystal.
Each column lists the angles between faces in a single zone.
Prism zone
Zone 1
Zone 2
a Λ b = 58.47°
b Λ c = 63.06°
c Λ d = 58.47°
d Λ e = 58.47
e Λ f = 63.06
f Λ a = 58.47°
a Λ g = 31.89°
g Λ h = 58.1°
h Λ i = 58.1°
i Λ d = 31.89°
d Λ j = 31.89°
j Λ k = 58.1°
k Λ l = 58.1°
l Λ a = 31.89°
b Λ m = 33.05°
m Λ h = 56.95°
h Λ o = 56.95°
o Λ e = 33.05°
e Λ s = 33.05°
s Λ k = 56.95°
k Λ q = 56.95°
q Λ b = 33.05°
(Note that the printed Wulff Net may not be accurately circular and that 45˚ near
the top may not be identical to 45˚ measured at the right or left edge. When you
plot the prism zone, work round from the top without moving the tracing paper very
far each time. Just put the last plotted point on top of the nearest 10˚ division and
measure on from there).
a)
Look for successive angles within each zone which add up to 90˚ or 180˚.
These will identify the faces at right angles or opposite to one another.
b)
Plot the Prism Zone first, anticlockwise round the primitive, starting with a at
the top of your stereogram. Extra marks will be given for accuracy.
c)
To plot the other two zones, look for faces in both zones 1 and 2 that you have
already plotted in the Prism Zone. Choose one face that occurs in both Zone 1
and Zone 2, and for which you now have the angle between that face and two
different faces in the prism zone, already plotted. The face that you have
chosen must lie at the intersection of two small circles of known radius centred
on two known faces. (Hint: h could be useful).
Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
d)
A fourth zone (zone 3):
fphncrktf
bmhoeskqb
is equivalent to the zone
Plot it.
e)
Identify and mark in the symmetry elements shown by the faces. Take care.
Look for mirror planes and then diads. Give the correct Hermann-Maughuin
symbol.
f)
Mark in the crystallographic axes. Note: z always goes in the middle and +y
always goes horizontally out to the right with +x to the south. If the axes are
all at 90˚ as here (the crystal is orthorhombic), x is due south on the primitive
and should coincide with one or more symmetry elements, as should y and z, in
this case.
g)
Choose an appropriate face as (111). Choose a face that cuts all three axes as
near to equally as possible and which also lies at the intersection of as many
zones as possible.
h)
Now draw in all the zones you can identify, as great circles, i.e. by rotating the
tracing paper, find individual great circles that link up as many poles as possible
(two or three non-parallel faces), across the diagram. The zones you have
already plotted are straight lines. Draw them in too.
i)
Index all of the faces. For some this can be done by simple visual inspection.
For other faces you will need to use the Zone Law and the Addition Rule.
Remember that you must have two pairs of known faces in two different zones
that intersect at the unknown face before you can index it. A zone must be
defined by two non-parallel faces. Consult the Handbook “Crystallography for
Geologists” pages 24-25. All zones used in indexing should be drawn in. The
zones you identified in h) should help you. If not, go back and see if you can
find some more zones!
j)
How many face forms are there, and what are they?
Additional advice:
Take care using the Addition Rule and always remember that you must have two zones
with known faces intersecting at your un-indexed face, before you can be sure of its
indices. For example, in the figure below, say you have identified a face p which lies
on a zone between (001) and (111). This means it could be (112) but it could also be
(113) or (223) or any linear combination of (001) and (111). You need another zone
which crosses the first at face p and has known faces, as does the zone (101) to (011).
Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
Because you can combine (101) and (011) to give (112) and (001) and (111), face P must
be (112) in this case.
k)
Calculate the axial ratio c/a. There are 2 different ways to do this. Firstly,
calculate the ratio using the goniometric measurements themselves. Now, mark
on the pole to a (101) face on your stereogram. Measure the angle (101) Λ (001),
and calculate c/a again. Comment on any differences between the results.
Part 2: more symmetry operators
If you have not done so already, complete part 3 from practical 1. Now we can
consider rotation and translation about inversion axes, which are symbolized as
follows:
_
1
No symbol
_
_
_
_
2 No symbol
3
4
6
1
2
4
3
6
Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
Crystallography Practical 3:
Indexing lattice planes and calculating
d spacings
In this practical we will use Miller indices to describe lattice planes in crystal
structures. The Practical begins with two-dimensional examples of indexing traces of
planes in a structure which has orthogonal crystal axes. We then consider the
indexing of planes in three dimensions.
Lattice planes are indexed using the Miller system in exactly the same way that
crystal faces are indexed. Remember that the indices are defined as (hkl) if the first
plane away from the origin makes intercepts on the +x, +y, and +z axes of a/h, b/k,
and c/l respectively.
Part 1: A two-dimensional example of indexing lattice planes
Rutile (TiO2) has a unit cell of dimensions a = b = 4.59 Å and c = 2.96 Å with
orthogonal axes. A projection on (100), i.e. looking down the x axis at the y–z plane, of
the rutile lattice is shown in Figure 1 below . The intersections (traces) of portions of
seven sets of lattice planes normal to the paper have been drawn on this plan.
a)
What is the crystal class of rutile?
b)
What is the h index for the planes shown in figure 1?
c)
For each set determine the indices k and l. (Draw the +y and +z axes from any
convenient lattice point on one of the planes as origin, but be consistent with
the signs of k and l.)
d)
What is the distinction between (hkl) and (h k l ) ? How does this differ from the
use of the Miller system for indexing crystal faces?
Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
Part 2: Your turn to draw lattice planes
A projection of part of the rutile lattice on (001) is shown in Figure 2. Use it to make
a drawing similar to Figure 1 showing the traces of sets of lattice planes (110), (010),
(210), (2 1 0) and (2 3 0) .
Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
Part 3: A three-dimensional example
a)
Figure 3 shows a view of a crystal lattice. Using the axes shown as your
reference axes, index the sets of lattice planes I to IV. Remember that in
each case you may choose any convenient lattice point as your origin.
c)
Near the numeral IV, add to the diagram the traces of a few planes of the set
(112 ).
Part 4: X-ray diffraction and calculating d spacings
Previously we have been considering lattice planes in the tetragonal mineral rutile.
Remember that the lattice is simply a representation of the crystal structure. The
lattice is an arbitrary concept, but a very useful one nevertheless, because it tells us
a lot about the actual crystal structure. We typically obtain information about the
lattice from X-ray diffraction measurements of actual crystals.
Below is an powder X-ray diffraction (XRD) pattern of rutile obtained using Cu Kα
radiation (λ=1.54).
Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
As geoscientists we often use powder XRD to quickly identify and characterise
material. Data are obtained from powdered samples (rather than from single
crystals). A powder can be thought of as a collection of thousands of very small
crystals with different orientations. When we obtain XRD patterns from a powder we
get reflection from all possible lattice planes simultaneously.
As you can see, the pattern consists of several peaks. These correspond to
reflections from lattice planes, and are distinctive of the mineral in question (the
pattern can be thought of as a fingerprint for rutile).
a)
the most prominent peak in the rutile spectra occurs at 2θ=27.47˚ Using
Bragg’s law, calculate the d spacing for this peak.
Bragg’s law: n λ = 2dsin θ
(remember to use θ and not 2 θ!!!)
Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
b)
You look up the ICDD entry for rutile and discover that the most prominent
peak is (110). The same entry allows you to assign all of the peaks in the XRD
pattern to (hkl) values. You discover that the peak at 39.23˚ is the (200)
reflection. Calculate the d spacing for this peak. Bearing in mind that this is a
(200) reflection, calculate the crystallographic a repeat distance.
Of course, you could have calculated this using the (110) d-spacing. How?
c)
If you managed that, then calculate the c repeat distance from the second
most prominent peak in the rutile pattern, which is the (101) peak at 36.12˚.
Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
Minerals science assessment 1
You will be given an exercise based on material covered in the first 3 practicals. This
exercise will be completed during the scheduled practical sessions and submitted at
the end. The session will be open book. You are free to work with each other and to
ask for assistance.
This assessment will be worth 10% of the overall class mark for the Earth Materials
course (i.e. 5% of the final course mark).
Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
Optics Practical 1:
Recognition and use of interference colours
Part 1: Interference colours and the quartz wedge
a)
Examine the quartz wedge under the microscope, using low power and cross
polars. Focus very carefully using dust or bubbles and avoid at all costs winding
the objective down on to the glass surface!
b)
Study the sequence of colours carefully and check them against against a
colour chart. Notice that a distinctive bright purplish red colour (magenta)
repeats at regular intervals along the wedge, but changes colour subtly as it
does.
The magenta colour is known as sensitive tint and the colours on the edge are
by convention divided into orders from the thin end. Thus, first order extends
from the grey to the first order sensitive tint band, the second order
extends from the blue colour just beyond first order sensitive tint up to
second order sensitive tint, and so on.
c)
Check how many orders can be observed on the wedge and mark the sensitive
tints as lines on the diagram below, representing the full birefringent part of
the wedge, numbering from the thin end.
Colour in (or label) the first two orders on your diagram to fix in your mind the
colour sequence (add more orders if you wish), and indicate the colour of the
high orders.
Thin end
d)
Check that the vibration directions of the wedge are perpendicular and parallel
to its length by rotating it to extinction and observing its orientation on the
stage.
Part 2: The sensitive tint plate
The sensitive tint plate provided with the microscopes have a quartz plate ground to
a thickness corresponding to first order sensitive tint or a retardation of 550nm.
They are fixed with the slow vibration direction cross-wise on to the length of the
Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
metal holder. When they are installed in the accessory plate slot, the slow direction
is then NE-SW. Not the mark indicating the slow direction
The quartz wedges are of two kinds. The newer owns designed for these microscopes
have their slow direction transverse to their length, like the sensitive tint plates. The
older black wedges have their slow direction along their length.
a)
Place the quartz wedge on the stage with its marked slow direction NE-SW.
position the wedge so that the first-order sensitive tint colour band lies under
the cross wire. Place the sensitive tint plate window directly on top of the
wedge with its slow direction line also NE-SW, parallel to that in the wedge.
What colour do you observe on the cross-wire (you may need to refocus)?
…………………………………………………………………………………
b)
Rotate the sensitive tint plate through 90˚ so that its slow direction is now at
right angles to the slow direction of the wedge. What colour do you observe
now?
…………………………………………………………………………………
EXPLANATION…make sure you understand this!
The petrographic microscope produces polarised light (light which vibrates in one
direction only). When this light passes through an anisotropic mineral such as quartz,
it is split into two components (or directions). These two components vibrate at right
angles to each other, and travel through the crystals at different speeds. When this
light emerges from the other side of the crystal, one component has been slowed
down more than the other.
The quartz wedge and sensitive tint plate are both pieces of quartz, and both of
them split light into 2 vibration directions at right angles to each other. In the 45˚
position used, the light energy is divided equally between the 2 directions and passes
through each quartz device as two rays. Each device slows down one ray relative to
the other by 550nm exactly, in the position adopted. When the slowed-down ray
emerging from the wedge enters the plate parallel to the slow direction in the plate
it is slowed down by a further 550nm. This is called addition. 550nm of retardation
is added making a total retardation of 1100nm. This corresponds to second order
sensitive tint.
Now consider what happens when we rotate the tint plate through 90˚. As the slow
ray emerges from the wedge and enters the plate parallel to the plate’s fast
vibration direction, it is the other ray that is now slowed down, by exactly the same
Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
amount that it was ahead on entering the plate (i.e. the fast ray becomes the new
slow ray, and the previous slow ray becomes the new fast ray). The net result is an
exact balancing of the retardations or compensation. The dark grey observed should
theoretically be black –no retardation- but rotation of the plane of polarisation at
lens surfaces gives an imperfect result.
The next part of the practical explores addition and compensation when the
retardations are not equal in the two devices. Remember that the colour seen is a
function of the final net retardation only, and is indifferent to which ray is finally
ahead or behind the other.
c)
The key observations concern the case when the initial retardation is less than
first order ST. If the slow ray in this situation leaves the wedge and enters
the plate above, parallel to the plate’s fast ray, and is speeded up by a whole
order, the initial retardation is completely overcome and the ray will end up
ahead (fast) but by less than a full 550nm. To investigate this:
Repeat the whole sequence of operations above, using the sensitive tint plate
superimposed on the following colour bands on the wedge, with slow directions
first parallel and then perpendicular, usually referred to as “slow parallel to
slow” and “fast parallel to slow”. Record your results below:
Initial colour on the Wedge
Colour Slow || Slow
Colour Slow || Fast
1st order grey
2nd order blue
2nd order red
In the parallel position the new colour observed is always exactly one order higher
than the original because the sensitive tint plate has a retardation equivalent to one
order. In the parallel position therefore, the retardation caused by the plate is
added to the retardation caused by the part of the wedge.
In the crossed or perpendicular position (slow parallel to fast) the new colour is not
always one order lower than the original.
Under what conditions is the new colour not one order lower than the original?
Answer……………………………………………………….……………………………………………………….……………………………
………………………….……………………………………………………….……………………………………………………………………….
Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
This is an important principle to get clearly understood. One useful way of thinking
about the problem is to envisage two Newton’s scales or interference colours joined
back to back, i.e. facing right and left joined at zero retardation. You can then take
any initial retardation on the right hand scale and add or subtract any additional
retardation and arrive at an answer, if necessary going back through zero and onto
the left hand scale –equivalent to ‘over-compensation’.
For example, an initial retardation of 300nm from the wedge becomes 300+550nm =
850nm with slow || slow. If slow || fast, it becomes 300-550nm = -250nm, which the
eye sees the same as +250 nm (as grey!). Your eye detects the magnitude of the final
result, but not which is the fast or slow ray!
Why is this important? We often use the tint plate to determine the directions of
the fast and slow rays in a mineral in thin section. This can be a very powerful tool
for quickly identifying minerals in thin section.
Part 3: Colour recognition
Go back to the wedge in the 45˚ position by itself, either on the stage or in the slot,
and note that the 1st, 2nd and 3rd sensitive tint colours are very similar. The whole of
the second order range is actually difficult to tell apart from the third order (both
are bright and intense).First order colours, greys through yellow, orange and red are
distinctive and unlike colours elsewhere. In practice there are several ways of
identifying a colour in a grain in thin section.
i.
The first and most important is by observation and association with a series of
colours from neighbouring grains of the same mineral. These will generally be
grains with the same thickness but different birefringence because of
differing orientation. A range of colours will be seen rather than one single
colour: birefringence will vary from zero up to a maximum value.
ii.
A development of this method is the technique of fringe counting. Look for a
grain that has a sloping edge, ideally at the edge of the slide. This edge will
reveal a sequence of colours down from the body colour of the grain, like a
miniature quartz wedge.
iii.
A third technique is to add and then subtract one order, using the sensitive
tint plate, first in the parallel position and thin in the crossed position. It is
usually possible to deduce what the original colour is by viewing colours which
you know to be order above or below (with the exception of what we learnt in
section 2c!). to do this, select your grain and turn it to the brightest position
under crossed polars. It will then be in the 45˚ position, but you will not know
Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
whether its fast or slow direction is parallel to the tint plate’s slow direction.
Insert the plate in the slot and not the colour. Rotate the stage through 90˚
which will bring the other vibration direction parallel to the plate’s slow
direction. Observe the colour. You now have 3 colours: original, original plus one
order, and original minus one order (with proviso as in 2c).
The sensitive tint plate method works best for isolated grains up to third
order.
iv.
Finally, one can slide in the quartz wedge to compensate the grain colour
exactly and achieve a dark grey, and then identify the position on the wedge
where this is achieved. This will be the grain colour.
This is often the only way to determine birefringence in isolated grains with
very high retardation.
a)
The sooner you learn to identify birefringence colours, the better. There are 4
thin sections of a sandstone cut to different thicknesses (3.1, 3.2, 3.3 and 3.4).
In addition there is a section labelled 3W which is a wedge-cut section from a
sandstone varying in retardation from low first order up to a highest order.
Use at least 3 techniques to confirm the identity of the highest order colour
visible (it is suggested that you use 3.2 of the wedged slice to start with).
As a general rule, please only take one section at a time, and return sections to
the tray when you have finished with them.
Given that the birefringence of quartz is 0.009, calculate the thickness of the slide
at the thickest point from the retardation equation and the colour chart.
retardation = thickness x birefringence
Slide
Maximum Birefringence colour
Thickness
3.1
……………………………………………………………………..
………………………..
3.2
……………………………………………………………………..
………………………..
3.3
……………………………………………………………………..
………………………….
3.4
………………………………………………………………………
………………………….
3W
(max)……………………………….(min)……………………………
……………………….(range)
Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
Optics Practical 2:
The calcite rhomb experiment explained using
the indicatrix
Objectives
To observe the 2 images of a spot under a calcite rhomb and so make the key
observations relating to double refraction in calcite:
• Position of the 2 rays relative to the crystallographic axes
• The directions of polarisation of the 2 rays
• The behaviour of the 2 rays on rotating the rhomb
Then to interpret and account for these observations using the concept of the
uniaxial indicatrix. The final end point will be a diagrammatic cross section through
the rhomb, with included section through the indicatrix that will tie all the observed
features together, and confirm the optic sign of calcite.
Part 1: Sorting out the crystal symmetry of the calcite rhomb.
Calcite is trigonal and has a single 3-fold axis of symmetry. It is important to locate
the direction in the crystal parallel to the axis because it controls the symmetry of
all calcite’s physical properties, including refractive index. Calcite has a perfect
rhombohedral cleavage, which means that it parts along 3 planes all at an equal angle
to the 3-fold axis, and structurally equivalent to each other.
a)
examine the cleavage rhomb. Find a corner that has the crystal edges all at
equal angles to each other. This is where the 3-fold symmetry axis (the triad)
emerges. There is another class of corner which has two angles equal but the
third unequal (this is not a symmetry axis). In addition to the 3-fold axis there
are 2-fold axes going through pairs of opposite edges. However, we are only
interested in the highest order symmetry axis, the 3-fold axis.
When you have found the 3-fold axis stick a pin on the corner with blue tac,
making the pin stick out at an angle equal to the 3 faces which meet at that
corner. This is the direction in the crystal of the 3-fold axis and is
present throughout the lattice, not just at that corner.
Part 2: A first encounter with the indicatrix
Calcite is a trigonal mineral, so it is uniaxial. This means that the refractive index of
light varies with direction in the crystal in such a way that one can represent it by a
spheroid (a squashed or an elongated sphere). It may either be represented as an
elongate (or prolate) spheroid (as our spaghetti squash) or as a flattened (or oblate)
Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
spheroid (as our tangerine), depending on whether it is uniaxially positive or uniaxially
negative.
a)
The axis of the spheroid must coincide with the 3-fold axis. Why do you think
this is?
………………………………………………………………………………………………………………………………………………………..
………………………………………………………………………………………………………………………………………………………...
…………………………………………………………………………………………………………………………………………………………
Remember, the distance from the centre of the spheroid to the outer surface
represents the magnitude of the refractive index for light vibrating in that direction.
It is useful at this point to imagine the spheroid sitting in the centre of the crystal,
its axis aligned along the 3-fold axis. At this stage we do not know whether it is
oblate or prolate. Look at your lecture notes to remind yourself of the 2 shapes.
To proceed further we are going to make use of the fact that light passing through
anisotropic minerals splits into 2 rays which vibrate at right angles to each other.
For a given wave-normal (the direction perpendicular to the wave front) these
two directions are the axes of the elliptical section through the indicatrix
perpendicular to the wave normal. This statement is key to using the indicatrix to
understand the behaviour of light in crystals……..do not worry; it is much easier to
understand that to write down!
For the moment we are only interested in confirming that light does split into 2 rays
going through calcite. We will then determine the directions of vibration for these 2
rays, and how these directions relate to the indicatrix.
Part 3: observations
In this part you are going to use a dot on paper as an object and treat it as if light
rays leave the dot and travel up through the rhomb to your eye. This is not strictly
true, but geometrically works fine.
a)
Place the rhomb over a dot an observe the two images from above. Rotate the
rhomb through 360º, keeping it flat on the paper and centred on the dot as far
as possible, with your eye directly above the dots.
Are both images at the same depth? …………………………………………………………………………….
Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
b)
Do they both move when the rhomb is rotated about an axis through the spot?
………………………………………………………………………………………………………………………………………………
c)
Which one appears to move, the upper or the lower? …………………………………………….
d)
Place your eye over the fixed image, rotate the rhomb back and forth about
30º each way, and at the same time note where the 3-fold axis projects out of
the rhomb at the obtuse-angled corner. Does the moving image lie between the
fixed image and the 3-fold axis, or does it lie on the opposite side?
Keeping your eye, as before, over the fixed image, confirm that the directions of the
z axis (i.e. the 3-fold axis) and the two rays which are apparently coming from the
dot images to your eye, are all in the same vertical plane. Remember that the z-axis is
a direction everywhere in the crystal parallel to your pin.
e)
Figure 1, overleaf, is a sketch of a rhomb from above with two images. Label
them as fixed and moving, consistent with your observations, given that the zaxis is coming out of the top corner. Label them fast and slow, bearing in mind
that fast means lower refractive index and slow, higher refractive index, and
that
HINT: Remember, the faster the velocity, the farther away the light seems to come
from; if you look down into a swimming pool, the bottom appears to be shallower than
it really is, because light travels slower in water than in air.
f)
Refractive Index (R.I. or n) is inversely proportional to the velocity of light in
a mineral (or any other substance):
nmineral / nvacuum = velocityvacuum / velocitymineral
Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
Therefore, a direction in a mineral with a high refractive index is better at slowing
down light. For comparison, here are some indices of refraction:
1.000 in vacuum (by definition)
1.003 in air
1.333 in water
1.4 to 2.0 in most non-opaque minerals
In a unixaxial mineral, the o-ray or e-ray can be the fast ray (the one with the lower
refractive index); this depends on whether the mineral is optically positive or
negative.
Reference books on mineralogy give 2 indices of refraction for calcite: 1.486 and
1.658. Thus, for calcite:
The R.I. for the ordinary ray is ………………………………………………………………………. And,
The R.I for the extraordinary ray is ………………………………………………………………
Figure 2 is an E-W cross section (in the plane of the optic axis/triad) showing a ray
travelling up from the real dot, but which appears to the eye to come from the image
of the fixed dot at shallower depth. You have observed that the image of the moving
spot is at one of the four possible positions shown (A to D), behind or in front of the
fixed spot, and above or below it.
g)
Draw you own version of figure 2, marking on the correct apparent position of
the moving dot (A, B, C or D). Then Sketch on the path of a ray which must
actually come from the dot itself, travel up through the rhomb and then bend
Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
back to the vertical when it enters the air so as to appear to have originated at
your chosen image. When your eye is directly over the correct moving image
the ray must be travelling vertically up from the rhomb surface to your eye.
Draw in that ‘in-air’ part of the path. The ray in the crystal has to get from the
real object, the dot, to that point on the surface of the rhomb below your eye.
You should be able to draw in that part of the path. Projecting the vertical air
part of the path straight back down leads to the apparent position of the dot,
the moving image itself. Your ray cannot change direction in the rhomb, only at
the crystal-air interface.
Part 4: Determining the vibration directions of the 2 rays
Polaroid acts like a grid, allowing light to pass through vibrating in one direction only.
Light is partially polarised by reflection so that glare off the sea or snow is much
reduced by Polaroid sunglasses set to cut out horizontal polarised light. They also cut
the total amount of light down. The Polaroid provided has its vibration direction
marked on it. You can now go on to make the observations with the rhomb using that
information.
a)
Use your marked Polaroid to confirm that the 2 rays form the dot are both
polarised, and then to determine the direction of polarisation. Using the figure
below, mark the directions of vibration with short lines on the 2 dots. Note
how the vibration directions relate to the plane containing the rays, the normal
to the surface on which the rhomb sits, and the z axis. Make the slow ray
have a longer line than the fast ray. Write m and f beside the appropriate
image for moving and fixed.
Note that for a single wave normal the 2 vibration directions you have drawn actually
belong to one elliptical section. If you have correctly drawn the relative lengths of
Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
the 2 lines you can tell at this stage whether the elliptical section is longer in the
direction of the z axis or shorter (A or B).
The observations may be
completed (with some
difficulty), by tilting the
rhomb and observing that the
two images tend to converge on
each other s the z-axis
approaches the vertical. If one
had ground faces perpendicular
to the axis to sit the rhomb on
it would exhibit no double
refraction.
Part 5: Using the indicatrix (vibration directions)
As noted before, the indicatrix can tell us where the vibration directions are for a
given wave normal. The section of the indicatrix perpendicular to the wave normal is
an ellipse (the exception is the sectional normal to the optic axis, which is a circle). In
our case, we are going to make use of the fact that our observations are all made
looking vertically down through the rhomb, perpendicular to its top and bottom
surface. The rays that enter the rhomb from the dot going to your eye have a single,
vertical wave normal. Thus we know where the indicatrix axis is (the z axis) and we
know where the wave normal is. We don’t know which shape the indicatrix applies, but
we do know that it must be either of the configurations below:
Positive indicatrix
Negative indicatrix
Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
a)
Take careful not of the elliptical sections that apply in both cases. Which
corresponds to your observations so far? Make sure that you understand why
they differ in orientation and which is the slow ray and which is the fast ray.
Also make sure you appreciate which one has the fast ray parallel to the trace
of the optic axis.
Write ‘correct’ next to the figure above which corresponds with your
observations.
e)
Is calcite positive or negative? ……………………………………………………………..
Part 6: Birefringence
Birefringence, or double refraction, is the difference between the index of
refraction of the slow ray and the fast ray.
a)
Birefringence for calcite is ……………………….. - ……………………….. = …………………………..
In the calcite rhomb experiment we eliminate one variable: orientation of the optical
directions. In thin section, the grains typically have random orientation, so that each
can have a different optical orientation. Calcite is a uniaxial mineral. There is a fixed
value of the o-ray’s R.I. but the e-ray’s index (as seen from one grain to the next) will
range from a maximum value (noted above) to that equalling the value of the ordinary
ray. We can visualise this by considering how we section the optical indicatrix (what
shape ellipse we end up with) when we section a mineral, as shown in the figure
overleaf. In this figure, the indicatrix for an optically negative mineral (such as
calcite) is shown, with the optic axis vertical. This indicatrix has been sectioned in 4
different directions, with corresponding sections shown in plan view. Remember, the
2 light rays in the mineral will vibrate perpendicular to the radii of these sections. In
(a) it is sectioned horizontally (perpendicular to the optic axis). The angle at which
the indicatrix is cut increases from (a) to (d), with the indicatrix section parallel to
the optic axis in (d).
b)
How do the 2 radii of the elliptical sections vary from (a) to (d)?
………………………………………………………………………………………………………………………………………………..
c)
………………………………………………………………………………………………………………………………………………..
What would be the appearance of a calcite grain in crossed polars if sectioned
in the orientation shown in (a)?
……………………………………………………………………………………………………………………………………………….
Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
d)
In which orientation would a grain need to be sectioned to produce the
maximum birefringence in crossed polars, and why?
………………………………………………………………………………………………………………………………………………..
………………………………………………………………………………………………………………………………………………..
You may wish to label the appearance of grains in cross polars which have been
sectioned as shown in the figure above (i.e. in terms of relative birefringence).
Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
e)
Approximate values for the maximum birefringence of other anisotropic
minerals are:
quartz and feldspar ~ 0.01
amphiboles and pyroxenes ~ 0.02 to 0.03
biotite and some hornblende ~ 0.04-0.10
Compare these values to your answer for 6a above. What does this tell you
about calcite and its appearance under crossed polars? You may wish to look at
a Michel-Levy chart to quantify this.
…………………………………………………………………………………………………………………………………………………
…………………………………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………………………………..
Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
Optics Practical 3: Interference figures in convergent light 1:
uniaxial minerals
Objectives
The aims of this practical are:
• To become proficient at obtaining an interference figure.
• To appreciate the two components of a centred interference figure, the black
cross and the radially increasing sequence of interference colours , and to
understand why they form in convergent light and crossed polars.
• To recognise and understand off-centred uniaxial figures and the flash figure.
• To understand why the sensitive tint plate produces second order blue in the
upper right quadrant for a centred positive figure and yellow for a negative
figure.
Procedure for obtaining an interference figure
1.
Identify and centre your target grain under low magnification and check
that it is in focus. In general the target grain will be the one with the
lowest birefringence –the darkest grey interference colour, but some test
sections with a single mounted grain (as in this practical) are not the lowest.
Close the diaphragm down to partially improve the image.
2.
With the polars uncrossed, move up through the magnification stages to the
highest power available. Focus carefully using the fine focus control but be
careful to avoid impacting the slide with the lens.
3.
Check the centring of the objective lens by rotating the stage and noting
the path taken by any inclusions/dust particles. The centre of rotation
should coincide with the cross hairs. If there is a discrepancy of more than
a few per cent, ask a demonstrator for help or use the Allen keys in the two
socket-headed adjustment screws at the top of the lens mounting in the NE
and NW positions to steer the centre of rotation into the centre of the
field.
4.
Switch in the substage condensing lens and open the substage diaphragm
fully. Cross the polars. Lamp power should be at the mormal bright level,
about 4.5V.
5.
Either insert the Bertrand lens or take out one eyepiece. You should see an
interference figure with a black isogyre against a background grading from a
low order interference colour upwards.
Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
Part 1: The centred uniaxial figure
a)
Slide 5.1 is cut from a single quartz crystal, normal to its optic axis (the optic
axis, which coincides with the crystallographic z axis, is vertical and you are
looking down it when the slide is on the stage. There is just one crystal slice.
Hold the slide up to the light and note where the crystal is. Place it on the
microscope slide and follow the instructions above to obtain a goof figure. You
should eventually see a black cross, slightly fuzzy at the edges, set against a 1st
order grey background. This is a centred uniaxial interefence figure.
If the grain were thicker or the mineral were not quartz, but a uniaxial mineral
with a higher birefringence, you would see one or more concentric rings of
Newton’s scale of interference colours, increasing in order outwards.
b)
Insert the sensitive tint plate in the accessory slot and not the colours that
result in the four quadrants of the figure. Note also the cross itself, which was
black, becomes sensitive tint in colour. Add to the sketch of the figure in the
circle below the colours seen when the sensitive tiny plate is inserted. Note
that in this thin quartz section the critical colours are only seen at the very
edge of the four quadrants.
Now we need to understand how the black cross and the interference colours in the
background form. Firstly, we need to think about the vibration directions of light in
different parts of the figure. Remember, we obtain interference figures using
condensed light……the condenser lens produces a cone of light into the base of the
crystal. The Bertrand lens takes this cone of light after it has passed through the
mineral and converts it back to parallel light rays.
Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
Figure 7.36 Nesse “Intro to min”
a) Cross section showing light passing through the quartz
crystal (note shape of the uniaxial indicatrix)
b) Perspective view of the same quartz crystal. Inclined
rays pass through increasingly thick sections of the
crystal.
The key point here is that different light rays pass through the section at different
angles to the optic axis. Using convergent light is a clever way of observing how light
passes through at crystal at lots of different orientations. This is equivalent to
cutting (or sectioning) the uniaxial indicatrix in different directions:
light
light
Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
This figure shows how sections of the indicatrix vary when
a uniaxial crystal such as quartz is cut at different angles.
Remember that when we cut an indicatrix we form a circle
or an ellipse. The axes of the ellipse are the directions in
which light vibrates, and the length of the axes (radii)
represent the magnitude of the refractive indices of light
vibrating in those directions.
Light passing through the centre of the uniaxial
interference figure passes along the optic axis. The
section of the uniaxial indicatrix in this direction is a
circle.
For all other parts of the interference figure, light has
passed through at an angle to the optic axes,
corresponding to elliptical sections through the indicatrix.
Why does this matter? Well, the important observation
here is in all of the elliptical sections, the radial axes
are extraordinary (e or e’) rays, and the tangential
axes are all o rays. In other words, once we have
obtained a centred uniaxial interference figure we know, for each of the quadrants
of the figure, which directions the o and e rays are vibrating in. This is the most
important step in determining which is faster….i.e. to determine the optic sign. This
is a useful thing to now when you are trying to identify an unknown mineral!
c) On the interference figure to the right,
mark on the o, e and e’ rays on the
elliptical sections in the four quadrants
of the indicatrix.
Looking at this figure, can you explain
how the black cross is formed ?
…………………………………………………………………………..
…………………………………………………………………………..
……………………………………………………………………………
………………………………………………………………………………………………………………………………………………………..
Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
d)
Quartz is a uniaxial positive mineral. Which ray is faster (o or e)?
…………………………………………………………………………………………………………………………………..
e)
Now explain why putting the sensitive tint plate with its slow direction NE-SW
makes the colour in the NE and SW quadrants go from 1st order grey to second
order blue, and the colour in the NW and SE quadrants go from grey to 1st
order yellow. Write your explanation in your own words, draw a diagram or
both.
f)
Now draw a sketch of a centred cross for a
negative figure (it will look identical), and
show some correctly oriented ellipses to
represent the vibrations and relative speeds
of the rays making up the figure along the NE
SW line in the NE quadrant. Remember, for a
negative mineral, the sensitive tine plate gives
yellow in the upper right quadrant of the
figure.
Remember: Blue Upper Right Positive (BURP)
Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
Part 2: Off-centred figures and flash figures
a)
Sections 5.3 and 5.4 are cut from quartz single
crystals at 64º and 38º to the optics axes,
respectively. Using 5.3 for preference, obtain an
interference figure following the standard
procedure. Watch how the figure moves as you
rotate the stage. In essence the centre of the
figure lies outside the field of view, but rotates
around the field of view as you rotate the stage.
The arms of the cross stay NS and EW. You can
work out where the centre of the cross lies by
visualising where the arms that you see in the
sequence must meet.
The sign can be determined by positioning the
cross down to the SW of the field of view so that you are looking at the NE
quadrant of the figure, even though no arms are visible. Check that it goes to
the correct blue colour with the ST plate.
c) The Flash Figure is obtained from a section cut perpendicular to the optic
axes, as in section 5.2. Set up 5.2 to obtain a figure and observe that on
rotation the field becomes a broad dark cross very briefly four times in a
complete revolution and is brightly illuminated the rest of the time. These dark
instances correspond to the positions of extinction in ordinary crossed polars.
The flash figure cannot be used to determine optic sign, so is of little use to
us!
Part 3: Obtaining figures from grains in a normal thin-section
a)
Using section 3.2, the thick quartzite from practical 1 (if unobtainable use the
thick end of 3W) select a grain under low power and crossed polars which will
give a centred figure, i.e. one with the lowest birefringence visible (darkest
grey) and follow the procedure carefully. You should now see a figure with
coloured rings because the retardation for the rays making up the outer part
of the figure is the product of birefringence and thickness.
Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
Optics Practical 4:
Interference figures in convergent light 2:
biaxial minerals
Note: The mineral science assessment 2 will be handed out at the end of the
practical class. This is to be completed in your own time and is due into the
undergrad TO (Grant Institute) by 4pm Friday, week8.
Part 1: Observing an acute biaxial indicatrix figure
a)
Set up your microscope for interference figure observation using muscovite
flake 6.1, following the same procedure outlined in practical 3 (high power,
focus, max. illumination, condenser in, Bertrand Lens in). Turn the stage so that
the figure is in the conventional 45º postion with two isogyres in the top
right and bottom left quadrants. You should observe that as you rotate the
stage the curved isogyres come together and then part. In the 45º position the
figure is symmetrical about the NE-SW line. The points of maximum curvature
of the isogyres represent the points of emergence of the optic axes (i.e. the
perpendiculars to the two circular sections through the indicatrix). Remove the
Bertrand lens to confirm that in this position, the crystal is in extinction.
You are always observing the acute biaxial indicatrix when you see a figure
like this. You will also be looking down one of the Principal Vibration Directions
–either X (for a negative mineral) or Z (for a positive mineral).
b)
Insert the sensitive tint plate in the slot and note the new colour in the centre
of the field. Also note the colour in the upper right quadrant on the inside of
the curve of the isogyre.
Colour in centre:
………………………………………………………………………………………………………………………..
Colour in upper right:
…………………………………………………………………………………………………………….
Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
For completeness, sketch the appearance with the sensitive tint plate inserted:
You should have established that the colour goes yellow in the Upper Right
Quadrant. Just as in the Uniaxial case, which means that muscovite is negative
The figure to the right is a perspective
view of the biaxial indicatrix. We
observe the acute biaxial indicatrix by
viewing a mineral down the indicatrix Z
axis. The Z axis bisects the optic axes.
The angle between the optic axes is 2V,
and is a diagnostic characteristic for
mineral species.
By definition, an anisotropic mineral
appears to be isotropic when viewed
along an optic axis. This is shown in the
figure: both sections perpendicular to
the optic axes are circular (i.e. the RI is
equal in all directions). These two
circular sections intersect at (and are
hinged about) the Y axis.
There are 3 RI:
nα along X
nβ along Y
nγ along Z where nγ > nβ > nα
Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
Biaxial minerals can also be optically positive and optically negative. This depends on
whether the β refractive index is closer to that of α or γ. This is easiest to explain if
we draw a cross section through the figure shown above.
Biaxial Positive
Angle between optic axes (2V) is bisected
by γ (γ is the acute bisectrix, BXA).
c)
Biaxial Negative
Acute angle, 2V, between the
optic axes is bisected by the
α refractive index (α is the acute
bisectrix).
The angle between the optic axes is an important feature. There are 2 special
cases. If 2V = 90º the mineral has no optic sign. Why?
………………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………………………………..
d)
What about the case where 2V=0? What is this equivalent to?
…………………………………………………………………………………………………………………………………………………
…………………………………………………………………………………………………………………………………………………
.
Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
The following figure shows the directions of vibration of fast and slow rays in the
acute indicatrix for both optically positive and negative biaxial minerals. OP = optic
plane (the plane through the indictrix which contains both optic axes.
e)
Examine again the acute biaxial interference figure for muscovite (in the 45º
position), and verify the optic sign. Observe what happens to the colours inside
the curve of the isogyre in the upper right quadrant. The same rule applies as
for uniaxial minerals:
Blue Upper Right Positive
…………………………………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………………..
Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
f)
Draw two confirmatory sketches of a positive and a negative acute biaxial
figure (or acute bisectrix figure) with the colours seen on sensitive tint
insertion added, or labelled.
Part 2: Rotating the acute bisectrix figure
a)
Note that when you rotate the 6.1 slide the figure transforms to a black cross
with one arm thicker and one slightly thiner. The thinner arm is the trace of
the optic axial plane. At the point where the figure becomes a cross, remove
the Bertrand lens and confirm that the cross occurs in the normal extinction
position for the grain. This makes sense: the optic axial plane has been rotated
to be either NS or EW. At either position the vinration directions are the axes
of a principle section through the indicatrix which you will have turned to be
NS and EW. The crystal is thus in extinction.
Part 3: Centred Optic axis Figures and Estimating the Size of 2V
In this part of the practical we will practice obtaining centred optic axis figures
from suitable grains in rocks, and using such figures to determine optic sign and the
size of 2V, both of which can be diagnostic features of minerals.
For a centred optic axis figure you need to obtain a grain which is oriented with one
of its optic axes vertical; this grain will present itself as an isotropic section to an
oncoming wave normal. This grain will, therefore, have the lowest available
retardation –ideally black or very dark grey. It is always worth searching hard for
the best available grain, but note in passing how, for a given mineral, the
birefringence colour varies according to how well centred the figure is.
Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
A centred figure should rotate about a fixed point on the isogyre and this should
be the point of maximum curvature when the isogyre is in the 45º position. If the
isogyre is centred and curved, you can tell in which direction the acute bisectrix
lies and thus whether you are indeed looking at the upper right quadrant of the
figure (as shown in the figure below).
For the following minerals, obtain and draw centred optic axis figures, using the
circles provided, which are labelled with increasing 2V size. Add the optic sign of the
mineral concerned.
a)
In the thin section of anhydrite 11.3, find a suitable grain and obtain as near
a centred figure as you can. Avoid grains with abundant twin lamellae
(birefringent bands). Determine the optic sign. Anhydrite has a 2V of 44º.
b)
Obtain a figure from titanaugite (pinkish pyroxene with a good cleavage) in
section GG (4.2). Determine its sign and not the curvature. This has been
measured at 48º.
c)
Use an olivine in either 4.3 or 10136 or Rhöne Spinel Lherzolite G7 and take
care to find the lowest birefringence grain. This olivine has a 2V close to 90º
and cannot clearly be resolved in sign. Olivine as a mineral series shows a range
of 2V.
d)
4.4 (JMZ1) has crystals of the K-feldspar sanidine, with 2V close to 30º
(Quartz, without cleavage and uniaxial, and plagioclase, with multiple twins and
large 2V are also present). Note that the other isogyre is visible for some of
the time during rotation of the optic axis figure but that it is outside the field
Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
of view in the 45º position. This is a good example of a small 2V mineral. You
may encounter, by chance, and acute bisectrix section of sanidine in 4.4.
e)
Section C228 contains the pyroxene augite, with 2V approx. 60 º.
sanidine
anhydrite
Ti-augite
2V=30º
2V=44º
2V=48º
augite (C322)
olivine
2V=60º
2V=90º
Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
Mineral Chemistry practical 1: plotting mineral compositions and
determining mineral formulae
Part 1: solid solutions and plotting compositional data
Most rock-forming minerals do not have an exact chemical composition. Instead, we
find that the composition of many minerals varies between certain end-member
compositions, representing extremes in the compositional range. A simple example of
such a mineral is olivine. We commonly write the chemical formula of olivine as
(Mg,Fe)2SiO4. This means that we can have variable amounts of Mg and Fe in olivine,
as long as the sum of all Mg and Fe in the formula adds up to 2. Note that the
amounts of Si and O are fixed and do not vary.
The reason that the relative amounts of Mg and Fe in olivine can vary is that both
these cations can freely substitute for each other in the olivine crystal structure,
without unduly affecting the structure. All olivine, regardless of Fe/Mg content, has
the same basic crystal structure (space group Pbnm). If we substitute increasing
amounts of Fe2+ for Mg2+ we do, however, observe some changes in optical properties,
particularly in 2V angle and birefringence, and some changes in crystal structure,
such as cell parameters. It is important to note, however, that the underlying crystal
symmetry does not change.
a)
If olivine has the space group Pbnm, what is the crystal class (expressed using
the Hermann-Maughin notation), and what is the crystal system?
……………………………………………………………………………………………………………………………………………..
………………………………………………………………………………………………………………………………………………
b)
Draw a sketch stereogram for olivine showing
the combination of symmetry elements, and
show a face pole affected by these elements
Olivine has a range of compositions due to Fe2+-Mg2+
solid solution. We can, therefore, consider the
composition of most olivines to lie between those of
2 extreme end-members, representing the pure Fe
and pure Mg olivine compositions:
The Mg end-member, Mg2SiO4, is known as forsterite
The Fe end-member, Fe2SiO4, is known as fayallite
Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
Most natural olivines actually have a composition closer to that of forsterite. The
olivine structure is, compared to many other minerals, stable at very high pressure
and high temperature conditions, and olivine is one of the main constituents of
Earth’s upper mantle. The typical composition of olivine in the upper mantle is close to
Mg0.9Fe0.1SiO4 (this highlights the fact that the mantle contains much more Mg than
Fe). We will learn more about olivine later in the course.
Pyroxenes are a group of minerals which share a common structural unit, long chains
made up of silica tetrahedral. Compositions of pyroxenes are much more varied than
olivine, with a number of different end-member compositions and solid solution series.
However, much of the compositional range of pyroxenes can be expressed using the
pyroxene quadrilateral:
The pyroxene ‘quadrilateral’
Pyroxenes plot within the
shaded area of this triangular
diagram. No pyroxenes are
stable within the unshaded
part of the diagram.
This diagram has 3 end-member compositions. There is simple Fe2+-Mg2+ solid solution
along the base of the triangle between enstatite and ferrosilite.
c)
Write a general formula, comparable to the formula of olivine given above, to
express the composition of solid solutions between enstatite and ferrosilite.
…………………………………………………………………………………………………………………………………………
Pyroxenes along this solid solution series are orthorhombic, and are referred to as
orthopyroxenes. In addition to Fe-Mg exchange, pyroxenes can also contain large
amounts of Ca2+. The 2 other end-member pyroxene compositions in the quadrilateral
contain Ca, either with Mg (as in diopside) or with Fe (as in hedenbergite).
Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
d)
Note that there is no end-member composition plotted at the top apex of the
triangular diagram. What would be the composition of a pyroxene that did plot
here?
………………………………………………………………………………………………………………………………………………
This hypothetical end-member is not shown because pyroxenes which have this
composition do not exist (either in nature or in the laboratory). The pyroxene
structure is not stable in very Ca-rich compositions. Instead, a related mineral, known
as wollastonite, is stable. We will explore the reasons for this later in the course
when we look at pyroxenes in more detail.
e)
We do again observe complete solid solution involving Mg-Fe exchange between
the end-members diopside and hedenbergite. Write another general formula to
describe the composition of pyroxenes lying on this join.
……………………………………………………………………………………………………………………………………………..
Pyroxenes with this composition belong to the monoclinic system. This is because
substitution of Ca2+ into the structure results in distortion of the crystal structure
and a ‘loss’ or lowering of the symmetry (Ca2+ is significantly larger than Mg2+ and
Fe2+). Monoclinic pyroxenes are known as clinopyroxenes.
Now, lets start plotting the compositions of some actual pyroxenes on the
quadrilateral.
f)
Plot the following orthopyroxene composition (expressed as percentage endmembers to make it easier to plot onto the triangular graph opposite):
enstatite 63.6%, ferrosilite 33.5%, wollastonite 2.9%
You should find that this composition plots close to the enstatite-ferrosilite
join.
Now plot the following clinopyroxene composition
Enstatite 39.1%, ferrosilite 18.9%, wollastonite 42.0%
g)
These 2 pyroxenes are found co-existing together in the same rock. To signify
this, draw a straight line between the two points representing their
compositions on your diagram.
Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
h)
Why do you think that 2 pyroxenes coexist in this rock, rather than one
pyroxene with an intermediate composition? (The 2 pyroxenes formed together
as part of a stable assemblage of minerals).
………………………………………………………………………………………………………………………………….
………………………………………………………………………………………………………………………………….
………………………………………………………………………………………………………………………………….
Part 2: Determining mineral formulae from electron microprobe data
Mineral chemistry is inextricably linked with crystal structure, as revealed in a
mineral’s optical properties. The composition of individual minerals depends on the
bulk composition of the rock in which it is found, the pressures and temperatures
under which it formed, as well as a host of other conditions. The composition of a
mineral may also vary from the core to the rim of a grain, providing a snap shot of
chemical processes ‘frozen’ into a rock. As such, mineral chemistry can give us a lot of
information regarding how a rock formed and what changes it has undergone.
One of the most important tools used in modern Earth Science is the electon
microprobe. As described in the lectures, this machine can be used to provide
accurate and detailed chemical information on earth materials. However, in order to
interpret and make use of electron microprobe data we need to know something about
the mineralogy of the material we are investigating. This is why electron microprobe
analysis (EMPA) and petrological examination using an optical microscope work so well
in tandem.
Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
You will need to use a calculator to help you with the following calculations
(alternatively, use a spreadsheet program!)
Recalculation of an olivine microprobe data
The following EMPA data was obtained from a crystal of olivine (which you identified
in thin section). Results have been outputted as oxide weight percent.
1
2
3
Mol weight
SiO2
Al2O3
FeO1
MnO
MgO
CaO
Analysis
wt%
38.48
0.13
20.99
0.33
40.45
0.02
Total
ideal
100.4
100.0
oxide
4
5
6
7
8
60.08
101.96
71.85
70.94
40.30
56.08
wt% = percent oxide by weight
1
Amount of iron is expressed as FeO. The microprobe cannot differentiate oxidation
states, but for olivine it is reasonable to assume that all iron is divalent.
The first column in the table above gives oxide formula, and the second column the
oxide weight percent data determined from EMPA (i.e. the olivine grain contains
38.48 weight percent SiO2). The first step in calculating the mineral formula from
this data is to convert these weight percent values to Molar proportions. We do this
by dividing by the molecular weight of the oxide.
a)
Create a new heading in the next empty column (4), entitled Mol Prop (for
molecular proportion). For each oxide calculate this value by dividing the weight
percent value (column 2) by the molecular weight (column 3).
b)
Entitle the next column (5) Mol Prop Cations, short for molecular proportion of
cations. Here, we need to calculate the proportion of cations in the oxide
formula. Geologists usually express amounts in oxides, but it is important to
remember that the proportions of cations present in an oxide varies; for
Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
instance, 1 mole of SiO2 contains 1 mole of Si, but 1 mole of Al2O3 contains 2
moles of Al. To calculate the molecular proportion of cations, simple multiply
your values in column 4 by the proportion of cations present in the
corresponding oxide formula (i.e. for SiO2 multiple by 1, for Al2O3 multiply by
2).
c)
Next, we need to calculate the molar proportion of oxygens. As you will see, we
need to renormalize the data to the number of oxygens which are present in
the olivine formula. Use the title Mol Prop O for column 6. As you might
suspect, we calculate this by multiplying the molecular proportion (column 4) by
the number of oxygen atoms in the oxide formula (i.e. x2 for SiO2, x3 for
Al2O3 etc).
d)
We can now renormalize the O proportions. First, sum (add up) all the values in
column 6. Enter this value in an appropriate cell in the table. Now, divide this
number by the number of O atoms in the olivine formula. Again, write this value
somewhere in the table. This is going to be out correction factor.
e)
Finally, divide the molecular proportions for each cation (column 5) by the
correction factor. Enter the value in column 7, which should have the heading
‘Cations pfu’ (cations per formula unit…the number of cations in the formula as
it is normally written).
What we have done, in essence, is to recalculate the proportions of each cation with
respect to oxygen (or oxygen units) in our mineral of choice. We know that the
mineral is olivine, so we already no something about the formula. The calculation
simply allows us to re-express the olivine probe data in a way in which we are more
familiar in dealing with.
The final thing to do is write down a consistent olivine formula based on the EMP
data. This is actually fairly simple to do for olivine, because there are a very limited
number of end-members, and there is mixing on a limited number of crystallographic
sites. For some minerals (such as amphibole) this calculation can be quite complex, and
it is of critical importance that you now the ranges of solid solutions possible (and
which sites in the structure certain cations prefer to occupy)……more about this later
in the course!
For simplicities sake, we can take a bit of a short cut here. Let us ignore everything
other than Si, Mg and Fe (and of course O!). These are what we would term the major
elements in olivine. They are, of course, the elements which make up the 2 end
members, forsterite and fayalite. Some other elements are present, but the amount
of these in our olivine sample are very small. We would term these other elements
minor elements. If we did a much more detailed analysis using an even more
Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
sophisticate machine such as an Ion Microprobe (and being in Edinburgh you are in
one of the few Geoscience departments worldwide where you could do this!) we might
find that there were other oxides present at the level of parts per million by
weight…..we would term these trace elements. We do not worry too much about minor
and trace elements when quoting mineralogical formaule, because they have
insignificant effects on mineral/crystal properties. Most minerals will contain trace
amounts of dozens of different elements because geological systems (i.e. rocks!) are
what chemists would term ‘very dirty systems’. In other course, however, you will
learn to appreciate how the study of trace elements in minerals can reveal huge
amounts of geological information!
f)
Back to the problem at hand! Write a formula for our olivine using only Fe, Mg
and Si proportions calculated from your table. By definition, you will set
O=6.00 exactly.
………………………………………………………………………………………………………………………………………………..
………………………………………………………………………………………………………………………………………………..
g)
Now, rewrite this as proportions of the 2 end-members, forsterite and
fayalite.
…………………………………………………………………………………………………………………………………………………
…………………………………………………………………………………………………………………………………………………
Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
‘Composition’ Practical 1: close-packed minerals (spinels and olivine)
Practicals in the ‘composition’ section of the course typically have three components:
structural models, hand specimens and thin sections. Work on material that is least in
demand at any time, following the recommendations as to the relative length of time
to spend on each component.
Part 1: structures
These structures have close-packed oxygen layers in ABABABABA sequences
(hexagonal close-packing) or ABCABCABCA sequences (cubic close-packing). The
cations sit in the interstitial tetrahedral [4] and octahedral [6] sites. The family
embraces spinels (AB2O4), various HCP oxides (e.g. hematite, Fe2O3), olivine
(Fe,Mg)2SiO4 as well as several sulphides. Close-packed structures are also important
in high-pressure minerals science as well as broad areas of materials science,
condensed matter physics an solid-state chemistry.
The spinel structure
a)
Examine a model of spinel (either normal or inverse), which is of one unit cell.
Identify the close-packed layers of oxygens (red). Spinels are cubic, so there
are triads axes along the body diagonals of the cubic cell. Identify these.
What is the orientation of the close-packed oxygen layers relative to the triad
axes?
…………………………………………………………………………………………………………………………………………………
b)
What is the packing sequence; is it abcabcabc or abababab? Is spinel hcp or
ccp?
………………………………………………………………………………………………………………………………………………..
c)
Identify the tetrahedral A-sites and the octahedral B-sites, which in any one
cation layer will be partially occupied by silver divalent ions or gold trivalent
ions. In the cell (i.e. the whole model) there are 64 potential A-sites and 32
potential B sites. Can you find any ‘unoccupied’ octahedral sites? What about
unoccupied tetrahedral sites?
Confirm that there are a total of 8 A-sites and 16 B-sites occupied per unit
cell, giving a formula 8x XY2O4.
Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
Observe that there are 2 different types of cation layer, one with 3 out of 4 B sites
occupied, alternating with a layer in which 2 A sites and 1 B site, per unit mesh, are
occupied.
d)
Look at the co-ordinating polyhedra of oxygens around the cation sites and
note down whether they share corners, edges, faces or a combination of these.
Bear in mind that positive cations tend to avoid each other and this is best
achieved by sharing as few corners of your polyhedron with neighbours as
possible.
e)
Look at both spinel models, which represent end-members of the normal and
inverse structures, and note that the difference arises simply from the
occupancy of the A and B sites (in this case, the distribution of silver an gold
ions).
If silver ions are divalent (2+) and gold ions are trivalent (3+), complete the
following:
In normal spinels X2+ are on the ……… site, and Y3+ and on the ……….. site.
In inverse spinels, 8 Y3+ are on the ……… site, and 8 Y3+ and 8 X2+ are on the
..……….. site.
The corundum/hematite structure
f)
Examine the model of the hematite structure (this is the same structure as
corundum and ilmenite). Identify the close-packed oxygen layers. Is the
packing sequence ababababa (hcp) or abcabcabca (ccp)?
………………………………………………………………………………………………………………………………………………..
g)
Confirm that the co-ordination of the cations is [6]. The [4] sites are not
occupied. Note the each octahedron shares a face with one in the layer above
and below. What effect does this have on the Fe or Al cations which occupy
these octahedral (note, you might normally expected cations to be in the
centre of co-ordination polyhedra), and why?
………………………………………………………………………………………………………………………………………………..
………………………………………………………………………………………………………………………………………………..
………………………………………………………………………………………………………………………………………………..
Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
The olivine structure
Examine the olivine model, and identify the sheets of oxygen ions. These sheets are
parallel to (100). Confirm that they have an ABABABABA packing sequence.
h)
Now identify the (SiO4) groups. You should note that these point up and down
relative to the oxygen sheet. What is the co-ordination of Si? What about the
other cation?
…………………………………………………………………………………………………………………………………………………
i)
Identify the unit cell. Confirm by eye that it has orthorhombic symmetry. If
the cell dimensions are a=0.4755 nm, b=1.021 nm and c=0.5985 nm, what plane
are the oxygen sheets parallel to?
j)
If Fe-rich olivines are heated in air, parallel plates of hematite grow in olivine.
How would you expect these to be orientated?
Part 2: hand specimens
Spinel-group minerals and HCP oxides
In igneous and metamorphic rocks, small, black, opaque grains are probably
oxides. If they are equidimensional (octahedral or cubes, or rounded) they are
probably spinel-group. If they are elongate or platy they are probably hematite
or ilmentite.
a)
Examine the large crystals of spinel and magnetite in speciments M163, 164,
153 and 148 from various metamorphic and high-temperature vein settings.
Note the octahedral and distorted octahedral forms.
b)
Examine the more common examples of spinel concentrations in basic plutonic
rocks, in which layers from mm to m thickness, occasionally traceable for km(!),
can occur. More commonly, spinel concentrations are irregular pods and
stringers.
221 is a typical Cr-spinel layer lying in this case between Ca-plagioclase and
olivine layers in the Rhum intrusion
220a is a typical chromite accumulation in serpentinised ultramafic rock, in this
case from Vavdos, N. Greece.
Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
c)
Hematite variations are shown in 143 and 144 kidney ore, and 142 and 149
specular hematite.
d)
Corundum are single crystals is rare. 140 and 145 show the characteristic
barrel-shape. 3-fold symmetry is preserved even in abraded crystals.
Olivine
Fresh olivine in perdiotite nodules or as phenocrysts in basalts is usually a
distinctive light green colour and has a glassy, conchoidal fracture because of its
poor cleavage. Weathered olivine-rich rocks are often a distinctive sandy-brown
colour. Serpentine, the common alteration product of olivine, as a variety of
greenish, reddish, black or brownish colours and often has shiny shear surfaces.
Olivine is a major constituent of the upper mantle, and nodules rich in magnesian
olivine can be found in some varieties of basaltic lava flows (as in San Carlos
specimens).
What is the name of the Mg end-member olivine? ………………………………………………………………
a)
Examine in olivine-rich nodules in G2/3 M2.22, from the Navajo Reservation,
San Carlos, Arizona. These are interpreted as residual mantle from which
basalt has been extracted , into which a small amount of another basaltic liquid
has percolated. All of this happened at a depth of 60-70 km. The nodules were
then ripped off the walls of a melt conduit by this basalt (the matrix) and
incorporated in a basalt flow, through which, in turn, they have sunk to the
base. The other dark nodules are pyroxene-bearing mantle rocks (pyroxenites
and lherzolites).
b)
G2/3 M3 is a spinel lherzolite nodule in basalt. Olivine here is weathers to a
yellow-brown colour. The black crystals are orthopyroxene.
c)
At high levels in the crust, olivine is often the first silicate mineral to
crystallise from a basaltic magma relatively poor in SiO2. Being heavier than
the magma, it may sink and accumulate at the base of the magma chamber.
G2/3 M221 is a 3-layer plagioclase/spinel/olivine rock from Rhum. The
yellowish-brown colour of the olivine is typical of a northern latitude layered
olivine cumulate.
d)
G2/3 M4 is the unusual Ca-rich olivine monticellite, CaMgSiO4. This is from a
contact skarn, Skye.
Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
Part 3: Thin sections
Oxides
Oxides in thin section are usually opaque, although some may have deep, diagnostic
colours. The most important identifier is the shape: cubic spinel-group minerals
have equidimensional cubic shapes and the HCP oxides have platy or prismatic
shapes.
a)
Examine sections 57 or 1210, typical basalts with groundmass magnetite
(spinel-group). The rock has phenocrysts of clinopyroxene (high relief, good
cleavage), patches of green sheet silicate after olivine, and plagioclase feldspar
(low birefringence, twinning and zoning). Switch to high power and look at the
groundmass. Sketch a range of shapes of the opaque spinel and satisfy yourself
that they are consistent with sections through octahedral. The other
groundmass phases are plagioclase feldspar, clinopyroxene, green clay after
olivine and brown glass.
Sketches:
b)
Section S is a Cr-spinel layer is a
speciment from Rhum (similar to 221).
Note that if you put in the condenser
and observe under high power with the
diaphragm open you can see the deep
red colour indicative of chromite. Note
also the rounded shapes of the grains,
possibly indicating some partial
dissolution or original octahedral by
percolating melt. Sketch a
representative area. GD1 and GD2 also
contain Cr-spinel with serpentine
Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
c)
Section 79.36 contains an opaque phase. Note that it is in elongate grains
typical of hematite and ilmenite, and also shows the red colour of hematite
when then enough. The rock is a calcite-dolomite-chlorite-actinolite schist.
e) Sections DTB49 and 12192 contain corundum. DTB49 is a corundummagnetite-plagioclase rock from Tievebulliagh, Northern Ireland. It is a
thermally metamorphosed laterite (a weathered lava-flow top heated up by a
magma passing through a feeder channel to higher, younger flows). Corundum is
blue (sapphire). Magnetite is again well-formed octahedral. 12192 is an
aluminous metasediment which also contains corundum as a green spinel.
Olivine
Olivine is a distinctive mineral, usually colourless, high birefringence, with
characteristic curving fractures, and in phenocrystals, a distinctive shape. The
alteration along cleavage planes to sheet silicates like serpentine is rarely absent
and may have gone to completion, leaving a pseudomorph. Large 2V, straight
extinction and a faint cleavage may support identification.
a)
GG is a silica undersaturated dolerite with olivine, titanaugite, spinels and
plagioclase. Examine the slide under p.p.l. taking care to open the lower
diaphragm only just far enough to give a sharp picture. The olivine should be
clearly distinguishable from the pinkish-brown pyroxene by its higher relief,
faint greenish colour and cracking patterns.
Crystal outline, cleavage, fracture, alteration, birefringence and extinction
angle are all very characteristic. Make a couple of careful sketch of at least
two, diagnostic, olivine grains. Add a scale. Next to your sketches, describe the
diagnostic properties of olivine, and how you would distinguish it from
clinopyroxene.
Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
Hydration by low temperate aqueous solution is the commonest olivine alteration
process. The effect is often the progressive replacement of olivine crystals along
fractures by fibrous or platy, low-relief, low-birefringence serpentine, a sheet
silicate Mg3Si2O5.(OH)4. The fayalite component (Fe end-member) may be oxidised to
granules of magnetite at the same time. Dunite may be completely transformed to
serpentinite.
b)
To make serpentine by the hydration of olivine requires water and some extra
SiO2 because the Si:Mg ratio is not the same in forsterite and serpentine.
Write a balanced equation for the hydration of forsterite:
Forsterite + silica + water = serpentine
By assigning arbitrary coefficients, a-d, to the reactants and products and
solving simultaneous equations (e.g. for Mg, 2a=3d).
Alteration frequently produces yellowish or brownish minerals which appear to be
related to chlorite or serpentine but are often intimate mixtures of two or more
substances, not always with perfectly ordered crystal structures. These, and other
products, are best described as ‘hydrous alteration products’.
c)
Section 86, an analcite basanite, illustrates patch pseudomorphing of olivine
crystals by such minerals. All stages from intact olivines to euhedral
pseudomorphs are present. Sketch a partially altered grain (use F or GG if
there are no sections available). Remember to add a scale and highlight
diagnostic features.
86
F
Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
d)
Examine thin section F, a troctolite, or olivine-plagioclase rock, in which this
process has partially transformed the olivine crystals. Note the radiating
cracks, which probably result from stresses set up in the crystals due to the
volume increase accompanying serpentinisation. Make a sketch to illustrate the
principle features of the texture. How are the serpentine fibres oriented in
the cracks in the olivine. Draw a labelled sketch (at high magnification) to show
this texture.
F
Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
‘Composition’ practical 2: Pyroxenes
Note: your script for this practical should be handed in at the end of the
session. It accounts of 15% of the overall class work mark for the course.
Pyroxenes have a structure based on single chains of SiO4 tetrahedra. The chains
are parallel to the z-axis and are arranged with the tetrahedral pointing one way
in each chain with the chains face-to-face andback-to-back. Co-ordinating
cations sit between the chains in 2 types of site, M1 and M2, which are [6] and [8]
in Diopside, CaMgSi2O6.
Part 1: Structure
a)
Examine the diopside model. Notice the chains of SiO4 tetrahedra parallel to
the z-axis. The chains are linked together by Ca and Mg ions. The Mg ions are
in 6-fold co-ordination by oxygens linked to 1 Si only. This is the M1 site. The
larger Ca are in [8] co-ordination by oxygens, two sets of two which are shared
between neighbouring tetrahedral in the chain. This is the M2 site.
Some people refer to the M2 site in diopside (and other clinopyroxenes) as
having a co-ordination of [6+2]. Looking at the structure, why do you think this
might be?
…………………………………………………………………………………………………………………………………………………
…………………………………………………………………………………………………………………………………………………
b)
Notice that the
chains have a
narrow face and a
broader back. The
chains are face-toface and back-toback (i.e. not faceto-back). Compare
the ball and stick
model with the
pyroxene I-beam
cartoon which we
spoke about in the
lecture.
Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
c)
The planes have planes of relative weakness at the sides of the chains where
there are no bonds and the spaces between the backs occupied by long and
relatively weak Ca-O bonds. Try to identify these on the large model by looking
at it in the same orientation as the cartoon above. Now, trace out these zig-zag
planes on the figure above, observing that they intersect at 90º to each other.
These are the 2 prismatic cleavages {110} common to all pyroxenes.
Going back to the large model, notice that neighbouring chains are displaced along z
relative to each other, making the cell monoclinic. Now, examine the orthopyroxene
model and note that in orthorhombic pyroxenes M1 and M2 are both occupied by Mg
and Fe2+. Both of these cations are smaller than Ca2+. Notice that the chains are
stacked together differently. M1 and M2 become similar, but not identical, and the
unit cell is orthorhombic.
It is the larger ionic radius of Ca (compared to Mg and Fe) which distorts the
structure and changes the symmetry.
Part 2:
Hand specimens
Pyroxenes are not always easy to identify in hand specimen. Short green or darkgreen/black prisms in basic igneous rocks are quite likely to be pyroxenes. Shape
as single crystals, and cleavage angle and cleavage traces can also sometimes be
seen. Amphiboles (which we will look at in the next lab) tend to be more elongate
with more prominent cleavage.
Clinopyroxenes
a)
Diopside. Fe-free diopside is colourless. Examine G2/3M63, G2/3M69,
G2/3M73 and G2/3M71 (all from metamorphosed impure dolomites) with a
hand lens and look for monoclinic terminations to the prisms and the 90º
(actually 86º!) angle between prism faces. Fe-bearing diopsides and
hedenbergite are green in various shades.
b)
G2/3M65,67 and 70 are augite (Ca,Mg,Fe2+,Fe3+,Ti,Al)2(Al,Si)2O6 phenocrysts
from lava. Well-formed crystals show {110}, {100}, {010} and {111}.
c)
G2/3M72 are xenocrysts of Cr-bearing diopside. Note the bright green colour
indicative of Cr. Over recent years, deposits of this material have been
commercially exploited (Russian Cr-diopside, as seen on QVC)>
Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
d)
G2/3M64 and G2/3M68 are fassaite (diopside with substantial substitution of
Al for Si and Al for Mg….the CaAlAlSiO6 molecule). This is quite common in
alpine veins (as here) but otherwise quite rare. Coupled substitution of 2Al for
Mg and Si is important in a number of minerals, and involved Al substituting
onto 2 different cation sites to preserve charge neutrality.
Orthopyroxenes
e)
G2/3M209. Bronzite. The bronze colour is very characteristic of massive
orthopyroxene rocks of moderate Fe content. Not the kink bands due to
deformation.
f)
G2/3M66. Basanite pseudomorphs after orthopyroxene. Orthopyroxene can be
transformed to serpentinite in a similar way to olivine. The characteristic platy
form of serpentine showing optical interference effects or schiller structure
is termed bastite after BAST in the Harz where specimens identical to M66
occur.
g)
G2/3M62. Hypersthene (term for enstatite-ferrosilite solid solution commonly
used in older texts). Large crystals from a pegmatitic facies of the Cuillin
Gabbro-Peroditite complex, Skye.
Sodic pyroxenes
h)
G2/3M43. Aegerine (NaFe3+Si2O6) pegmatite, with nepheline (Ilimaussaq,
Greenland).
i)
G2M31. Jadeite (NaAlSi2O6) from a block of serpentinite surrounded by
glaucophane schist. Syros, Greece.
Relative rarities
j)
G2/3M74. Spodumene (LiAlSi2O6) single crystal from a Li-rich pegmatite.
k)
Wollastonite (CaSiO3). Not a pyroxene, but a pyroxenoid with a single chain
structure repeating every three tetrahedral. This sample is from a
metamorphosed siliceous limestone.
Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
Part 3: Thin sections
Pyroxenes have 2 cleavages {110} at 86º (essentially at right angles). The cleavages
are parallel to z, so that the 2 cleavages and the characteristic angle are seen in
the cross sections of prisms. Well-formed crystals may have 8 sides with {100},
{010} and {110} as faces. More elongate sections have a single cleavage trace.
Monoclinic pyroxenes have inclined extinction (augites commonly at 40-50º),
moderate relief and moderate birefringence (high in aegerine-rich pyroxenes);
orthorhombic pyroxenes have a low birefringence and generally straight
extinction in prism sections.
Random sections through augite prisms as in the DHZ illustration look like this:
In order to accurately measure extinction angle we need to find a section which will
give the true angle γ^z, the angle between an extinction position (vibration direction)
and a cleavage trace. Such a section will:
1) Have the y axis (=β) perpendicular to it.
2) Show one cleavage trace –the trace of the {110} cleavage planes. This cleavage
trace is, by definition, parallel to the z axis.
3) Possess maximum birefringence (i.e. highest order colour) because it is the α-γ
plane.
Other sections will show oblique extinction to a single cleavage (smaller extinction
angle than the maximum), straight extinction to one cleavage trace, symmetrical
extinction to 2 cleavage traces or unsymmetrical extinction to unequal cleavages,
depending upon the orientation of the section. These sections are of no diagnostic
use.
a)
Sections GG (=4.2) or 10.5 have clinopyroxene as the only pyroxene present.
Identify is by its pinkish brown colour in GG (greenish in 10.5), 2nd order
interference colour and visible cleavage. Use the following procedure to
determine extinction angle. Measure angles in at least 3 different grains,
taking the maximum value. Draw a sketch of the grain which gave you the
maximum extinction angle (showing or labelling optical properties!).
Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
1)
2)
3)
4)
5)
6)
Under crossed polars, select the maximum birefringence pyroxene grain,
which should show one cleavage.
Set the cleavage parallel to the N-S cross-hair and record the stage
reading.
Rotate the crystal to the nearest extinction position and record the stage
reading.
Rotate the crystal 45º clockwise so that the vibration direction you have
just measured the angle to is now NE-SW.
Insert the S.T. plate and determine whether the colour goes up an order,
indicating γ, or down an order, indicating α.
Record the answer as γ^z (subtract the angle from 90º if you measured the
angle to α).
Stage
Stage
reading reading,
N-S
nearest
extinction
position
b)
Extinction
angle
Colour with Result
ST after
γ^z
45º
rotation..up
or down?
Grain showing max. extinction angle:
Now chose one of the 2-pyroxene-andesite sections, 12.1, Casc-1 or J. In
this section you will need to locate and distinguish the two pyroxenes, augite
(clinopyroxene) and hypersthene (orthopyroxene). The groundmass is likely to
have feldspar, one or more pyroxenes, an opaque oxide, possibly some glass
(isotropic) ± some alteration.
Prepare a labelled drawing showing the following:
1.
2.
3.
4.
5.
Texture of the section showing the coexistence of the 2 pyroxenes.
The distinction between the 2 pyroxenes (show this in you section and
list/label the diagnostic features of each separately).
An indication of other phenocryst mineral present in the section .
Part of the groundmass illustrated at the same scale or as an insert.
Slide number and scale!
Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
Remember, you can draw a representative sketch of your section (you do not need to
exactly reproduce what you see in any one field of view; your drawing might have
features observed in different parts of the slide!). Clarity and scientific content are
more important than artistry.
Using the same procedure as before, determine the extinction angle in at least 3
clinopyroxene grains
Grain showing max. extinction angle:
Stage
Stage
reading reading,
N-S
nearest
extinction
position
Extinction
angle
Colour with Result
ST after
γ^z
45º
rotation..up
or down?
Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
‘Composition’ practical 3: Amphiboles
Part 1: Structure
Amphiboles have a double chain structure, with 2 pyroxene-type chains linked
across as a band with a trapezium cross section, the narrow side as face and the
broad side as back. These bands are arranged face-to-face and back-to-back,
with three [6] cations sites, M1-3, between the faces, creating a broad “I-beam”,
and two larger cations sites, M4 (approx. [8]) and A between the backs. The Ibeams are cross-linked by the M2 and M4 cations. The A-site is 12-fold coordinated and is large. It may be vacant (as in tremolite) or occupied by large
monovalent cations such as Na an K. OH (and F and Cl when present) sits in the
centre of the ring of tetrahedral on the face-side of the band at the level of the
apical oxygens. Cleavage is presumed to follow an analogous path to that in
pyroxenes, passing along the sides of the I-beams and crossing between the backs
through the M4 and A sites.
The model has Ca (gold), Mg (silver), Si (black), O (red) and OH (green). The cell
dimensions of tremolite are a=0.985nm, b=1.81nm, c=0.53nm and β=105º. Thus c is the
shortest side of the model, which is best inspected by looking down c.
a)
On the model locate the Si4O11 chains and note the face-to-face, back-to-back
arrangement.
b)
Locate and note the co-ordination of the M1, M2 and M3 sites, which contain
Mg, and M4 site which contains Ca.
M1=
………………….. M2= ……..…………… M3= …………………. M4= …………………….
Note that the two (OH) ions take part in co-ordinating M3 on opposite sides of
the co-ordination polyhedron, two adjacent (OH) ions are involved in M1 coordination but that M2 is co-ordinated entirely by oxygen, as is the M4 site.
The A site is a recognisable empty hole between opposing SiO4 rings.
c)
Note the probably location of the cleavage planes. Convention suggests that
they pass along the sides of the chains and zig-zag through the M4 and A sites.
Label the cations sites on the following I-beam cartoon of the tremolite
structure and then show the zig-zag traces of the cleavage planes.
Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
Part 2: Hand specimens
Amphiboles are commonly more elongate as prisms and more striated than
pyroxenes, with better cleavage. If crystals are separate from one another, then
the cross section or even the 120º angle between cleavages may be seen. Colour in
amphiboles is usually diagnostic of composition
The main objective of studying the hand specimens here is to note common colours of
different amphiboles and the crystal form, which make amphiboles more easily
identified in igneous and metamorphic rocks.
Green amphiboles are most likely to be in the hornblende/actinolite family (but could
be a member of the Fe-Mg group). Hornblends can be virtually black, various shades
of brown or deep red if oxidised. In a volcanic rock with dark-green amphibole shaped
phencrysts, or in a plutonic, granitic-looking rock with similar green crystals, or in a
metamorphic rock rich in green amphibole, hornblende is going to be the correct
diagnosis 99% of the time.
Blue or purple amphiboles are likely to be alkali and the identity is going to depend on
the host rock composition, or in metamorphics, the metamorphic grade.
Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
Look over the specimens and convince yourself that you would recognise the dominant
coloured mineral as an amphibole and could give it a name.
Calcic amphiboles
Tremolite: G2/3M81, from a meta-dolomite. The white colour is unusual because
this rock is virtually F-free. The form and cleavage still point to
amphibole.
Actinolite-talc rock: G2/3M80. Formed at a serpentinised/siliceous-calcareous
country rock contact during metamorphism. Note that the prism crosssection shape can be seen with the naked eye and confirmed as having
the amphibole 124º angle between adjacent faces and the symmetrical 6sided shape.
Actinolite/hornblende rock. G2/3M78 and G2/3M79. Each is a mass of intergrown
green crystals with strong preferred orientation, typical of a
metamorphic setting. Composition somewhere in the actinolite to
hornblende range (no way of being more definite without EMPA).
Hornblende would be an acceptable guess.
Hornblende. SOP. Hornblende phencrysts in a volcanic dacite, SE Spain. Typical
colour and crystal shape, with identifiable amphibole cross section.
G2/3M217. Hornlende as the distinctive dark prismatic mafic mineral
along with biotite (perfect platy cleavage). Granodiorite with a mafic
clot. Garabal Hill, Argyll. More difficult to pin down as hornblende but
the cleavage and shape is more amphibole than pyroxene, and the rock is
quartz- and feldspar-rich and contains visible biotite, all of which would
be consistent with the accompanying dark mineral being hornblende.
Alkali amphiboles
Riebeckite. G2/3. Na2Fe2+3Fe2+2Si8O22(OH)2. Alisa Craig microgranite from which
traditional curling stones are made. Its granitic and with the eye of
faith the deep purple colour can be detected, pointing to alkali granite as
the rock and riebeckite as the amphibole.
Glaucophane. Na2(Mg,Fe)3(Al,Fe)2Si8O22(OH)2, a monomineralic glaucophanite from a
high-P/low-T metamorphic (blueschist) terrain, Syros, Greece. Bright
blue in good sunlight. Unmistakeable.
Fe-Mg amphiboles
Anthophyllite. G2/3M59. Mg7Si8O22(OH)2. The more obviously prismatic crystals are
the anthophyllite, set in a matrix of orthopyroxene, but you would be
doing well to identify the prismatic mineral as an amphibole, let alone an
orthorhombic one. This rock is made of magnesian silicates and is the
Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
product of metamorphism of an original ultrabasic protolith. Niddister,
Hillswick, Shetland.
Part 3: Thin-sections
Amphiboles have {110} cleavages parallel to prominent prism faces at 124º. Most,
but not all, are coloured and pleochroic, commonly in shades of green and broan, or
blues and purples for alkali amphiboles. Extinction angles are usually <25º γ^z which
distinguishes them from most pyroxenes which have larger angles, typically 40-50º.
Ortho-amphiboles (like orthopyroxenes) have straight extinction.
a)
Look at section GAR9 or 5078 which both contain typical hornblende. Note the
characteristic cleavage and extinction, and the pleochroism in ppl.
For the main part of the exercise use section SOP1 or TAM. We will now determine
the extinction angle (as we did for pyroxenes in the previous practical) and pleochroic
scheme of the amphiboles in these sections.
b)
Use the procedure opposite to determine the extinction angle γ^z on suitable
grains. Record your answers in the table below and present your final estimate.
Note that amphibole may have high birefringence and may be deeply coloured.
Then determine the pleochroic scheme.
Slide number:
……………………………………..
Stage reading Stage, nearest
Extinction
N-S
extinction position
angle
Answers: 1…………………….
2…………………………..
Colour up or
down with ST?
Result γ^z
3………………… Final answer……………………..
Pleochroic scheme: α ………………………………………………………
β ………………………………………………………
γ ………………………………………………………
Relative absorption: ………………………………………………………………………………..
Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
i.
1.
2.
3.
4.
5.
6.
7.
8.
ii.
iii.
Select a grain as in (ii) which shows the maximum retardation (highest
colour). It should be close to being parallel to (010) which is the Optic Axial
Plane in most amphiboles, so contains α and γ, the smallest and largest
principal refractive indices. You are trying to measure the true angle
between γ and the trace of the z axis, also in the (010) plane. The trace of
the {110} cleavage planes are parallel to the z axis in the (010) section.
Set the cleavage trace N-S and record the stage setting.
Rotate the stage with polars crossed, to the nearest extinction position.
Record the stage setting and calculate the extinction angle.
You now have a vibration direction N-S which should be γ. Rotate the stage
clockwise by 45º and insert the S.T. plate. The colour should go up an order
(if not, and the colour goes down, you have measured the angle to α not γ,
you will need to subtract the extinction angle from 90º).
All being well, you will have a grain with 2 principal vibration directions (α
and γ). Remove the S.T. plate and return the grain to the original position of
extinction with γ=N-S and α=E-W. Uncross the polars. The grain now shows
the α colour since it is illuminated with E-W polarised light. Describe the
colour.
Cross the polars and rotate the stage 90º so that the other vibration
direction γ is now E-W. Uncross the polars and record the γ absorption
colour.
We now need to determine the absorption direction in the β direction. Β will
lie in the plane of the slide and E-W in a grain which shows both {110}
cleavages sharp and at 124º, as in Fig. (iii). In this position you are looking
straight down the z axis. Find such a grain (it should also show symmetrical
extinction) and record the β absorption colour.
The remaining task, which is best done alongside the pleochroism
determination, is to record the relative absorption in the three principal
vibration directions. This is simply how dark the grain goes in the three
positions, and is expressed as an inequality (e.g. α ≤ β >> γ implies that the α
colour is slightly less absorbing than the β colour but that both are much
darker than the γ colour).
Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
c)
Section X615 or X615A contain the blue amphibole glaucophane. Determine
the extinction position by finding the highest birefringence grain and using the
grain long edges as the z-axis trace in the cleavage is hard to see.
Stage reading Stage, nearest
N-S
extinction position
Extinction
angle
Colour up or
down with ST?
Result γ^z
Now determine the pleochroic scheme by noting the two colours shown in the grain
you have just measured the extinction angle on and the β colour from a cross section
showing 2 sharp cleavages and set with is acute bisector E-W (these grains are not
abundant because of the strong preferred orientation!).
Pleochroic scheme: α ………………………………………………………
β ………………………………………………………
γ ………………………………………………………
Relative absorption: ………………………………………………………………………………..
d)
Locate the amphibole anthophyllite in M59 confirm that the highest
birefringence grains have straight extinction. Make a sketch showing such a
grain with its immediately surrounding grains drawn so as to be identifiable as
orthopyroxene or anthophyllite.
Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
‘Composition’ practical 4: Sheet silicates
Note: your script for this practical should be handed in at the end of the
session. It accounts of 15% of the overall class work mark for the course.
Part 1: Structures
The main objectives of this part of the practical are to study the crystal models
available and learn to identify different sheet silicate structures based on their
defining characteristics:
1)
2)
3)
2-layer structures (open sandwiches of a tetrahedral layer backing an
octahedral layer, e.g. kaolinite and serpentine) vs 3-layer structures (closed
sandwiches of two inward facing tetrahedral layers and an octahedral layer
between them as filling, e.g. talc, pyrophyllite and micas).
Dioctahedral minerals (2 trivalent ions in the 3 available sites, e.g. kaolinite,
pyrophyllite, muscovite) vs trioctahedral minerals (3 divalent ions in the 3
available sites, e.g. serpentine, talc, phlogopite).
Presence of interlayer cations due to the net charge on the 2- or 3-layer
sandwich leads to the distinction between: minerals with no interlayer cations,
minerals (e.g. talc) with regular interlayer cations (e.g. micas) and minerals
with variable interlayer cations and water molecules (e.g. clays).
At the same time, not the offset of the two tetrahedral layers in all three layer,
filled sandwich structures, and how this layer stagger gives rise to stacking
polytypes.
We will then consider how X-ray diffraction may be used to help identify sheet
silicate structures which may be difficult to determine optically.
Basic structures without interlayer cations
a)
In PYROPHYLLITE, Al2Si4O10(OH)2, note the double sandwich, or 3-layer
structure of inward pointing Si4O10 layers, with 2 out of 3 octahedral sites
occupied by Al3+ in the octahedral layer between them. All the tetrahedral in
any one layer point in the same direction, and share 3 out of 4 O with adjacent
tetrahedral. The apical oxygens together with the OH ions in the centres of
the rings of apices are available to co-ordinate the cations in the octahedral
layer.
b)
In KAOLINITE, Al2Si2O5(OH)4,note the single tetrahedral layer, and the
layer of OH ions which provides the octahedral layer of aluminiums with
coordinating anions. Note that the thickness of the repeating unit is less in
Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
kaolinite than in pyrophyllite (or in micas). We will examine the importance of
this when using XRD to determine sheet silicate structures later in the
practical.
Structures with Al-Si substitution and charge-balancing interlayer cations
c)
Examine the model of MUSCOVITE, KAl2AlSi3O10(OH)2, and confirm the 3layer structure. Can you identify a ‘pyrophyllite layer’ in this structure (in
phlogopite this would be a talc layer)? These layers are separated by layers of
12-coordinated K ions in the centres of the rings of basal oxygens.
d)
Note that there is no offset between the sandwiches at the level of the K ions,
as the O rings on either side of the K are exactly above one another.
Substitution of Al for Si in the tetrahedral layers in the mica models is not shown as
it is essentially random (there is a 1 in 3 probability of any one T site being occupied
by Al).
Clay minerals (sheet silicates with excess charge <1 per formula unit)
e)
Examine MONTMORILLONITE, (Na,Ca)0.33(Al,Mg)2(Si4O10)(OH)2.nH2O (an
example of a smectite group mineral). Smectites have a charge excess on the
sandwich of 0.26-0.6, which is balanced by an interlayer sandwich of water
molecules and cations. Smectites can be di-octahedral or tri-octahedral; which
is montmorillonite?
……………………………………………………………………………………………………………………………………………….
Chlorite (Mg,Fe,Al)6[(Si,Al)4O10](OH)8
f)
Examine the CHLORITE model and note the composite structure of alternating
talc and brucite (magnesium hydroxide, Mg(OH)2) layers. The charge excess on
the talc sandwich is variable.
X-ray diffraction as a tool for identifying sheet silicates.
Some sheet silicates, as we shall discover in part 3, are readily identifiable in thin
section. Others are not, and we must use different tools to distinguish them. X-ray
diffraction offers perhaps the most powerful tool for identifying sheet silicates,
especially clays, because it is highly sensitive to small changes in unit cell parameters.
X-ray diffraction patterns obtained from 3 different sheet structures are shown on
the next page. The most important peaks in the patterns have been identified for
you.
Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
g)
These XRD patterns have several obvious similarities, the most obvious being
that the first large peak in all patterns is due to reflection from the (001)
plane. Why is this an important reflection for sheet silicate minerals, and how
might we use this?
…………………………………………………………………………………………………………………………………………………
…………………………………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………………………………..
h)
The positions of these peaks for the 3 structures are: 2θ001 pyrophyllite =9.62,
2θ001 talc= 9.46, and 2θ001 kaolinite=12.37. Use the Bragg equation and the
wavelength for CuKα radiation (1.5418Ǻ) to determine the interlayer spacing in
the 3 structures.
Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
i)
How do you answers compare with observations you made on the ball and stick
models?
This is somewhat of a simplified example in differentiating between different sheet
structures. In some instances, notably with clay minerals, simple XRD is not
sufficient, and samples of material have to be prepared in several different ways to
aid identification. One useful method applicable to clay identification is to allow
samples to absorb (basically to ‘suck up’) different solvents. This causes different
structures to swell, resulting in predictable changes in inter-layer distances. This
technique can be used to differentiate between clay samples which may have very
similar inter-layer distances or to help identify several ifferent clays in a mixed
sample.
Part 2: Hand specimens
Sheet silicates have a predominantly platy character. There is a broad correlation
in 3-layer minerals between the size and excess charge, and the strength, size and
flatness of individual crystal plates. Neutral sheets result in weak charges between
sheets, producing soft minerals with small crystals. Large excess charges result in
stronger interlayer bonding, producing brittle, stiff minerals. 2-layer structures
produce minerals which are usually finely crystalline or fibrous and soft. Depth of
colour towards green or brown generally correlates with increasing Fe content.
Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
Pyrophyllite and talc:
No excess chare-very soft. Samples have a ‘silky’ feel to
them. Pale colour indicates little or no Fr.
Micas (muscovite, phlogopite, biotite):
High charge excess-flexible plates with
a perfect (basal) cleavage. Colour varies from Fe-free muscovite (colourless),
through magnesian phlogopite (pale green) to Fe and Ti-rich dark biotite.
Exotic colours: rose pink (lithium mica), bright green chromium muscovite mica
(fuchsite).
Chlorite:
Variable in appearance, commonly dark green, softer, finer-grained and
less platy than micas, but harder than talc.
Serpentine: Fine-grained, massive antigorite. Colours of serpentine minerals can vary
considerably (green, blue-green, yellow-green to black). Fibrous serpentine is
crysotile (one of the major asbestos minerals)
a)
Identify the 2 minerals in sample MIN-4A (one from this practical, one
technically from the previous practical):
……………………………………………………………………………………………………………………………………………….
b)
Identify the green mineral that surrounds the black octahedral of magnetite
(Fe3O4) in the specimen MIN-4B and give a reasonable formula.
……………………………………………………………………………………………………………………………………………….
Part 3. Thin sections
Sheet silicate minerals are difficult to identify in thin section beyond broad family
categories. Key properties are birefringence (high vs low), colour in ppl and whether
cleavage and crystallinity are prominent or not. All sheet silicates are flaky or platy,
with a single cleavage and rarely have good crystal forms (with the rare exception of
hexagonal biotite plates). Look at one example of each group at least.
White Mica
Colourless, high birefringence, well cleaved flakes are identified as “white mica”.
Depending on the ‘context’ of the rock, this may be muscovite, paragonite, talc or Fepoor pholopite. Some clay minerals can also be deceptively mica-like.
10
is a quartz-muscovite rock, a genisenised (H-metasomatised) granite from
Cornwall. The original feldspar has been converted to white mica (probably a
Li-bearing muscovite) by acidic watery fluids at high temperature. It also
Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
contains fluorite (CaF2) [isotropic, high negative relief, purplish in places] from
HF in the metasomatising conditions.
Biotite and phlogopite
Coloured, high-birefringence, well-cleaved flakes are almost certainly biotite or
phlogopite (but beware of very high birefringence, grotty brown flakes in weathered
rocks which may be the clay mineral vermiculite, altered from chlorite at low T).
4.1 contains biotite partially altered to chlorite in granite
30
contains biotite in syenite. Syenite is an alkali feldspar+mafics plutonic igneous
rock. This one contains augite, green and brown hornblende, biotite, sphene
and alkali feldspar. There are at least 2 generations of alkali feldspar, one in
large ragged plates, the second as small euhedral tablets. The latter are
enclosed in hornblende crystals (poikilitic texture) which implies that the
hornblende crystallised later. This same feldspar generation appears to be
interstitial (in the gaps) yet the large plates enclose all three ferromagnesian
minerals. The feldspar may therefore have crystallised last, and swept up the
other minerals into the gaps, or enveloped them, as the liquid disappeared.
Serpentine, chlorite etc (low birefringence group)
Colourless, low birefringence, platy or flaky minerals can be a number of things. If
well cleaved, chlorite is most likely. Many chlorites have anomalous grey-blueish
interference colours. Serpentine can usually be identified as a product of olivine or
opx alteration. Various other clays are also possible. Identification of fine-grained
alteration products or diagenetic clays is usually not possible.
GD2 serpentinite
TS12/TS10 serpentinite. All serpentine but in 2 forms, possibly different structural
polymorphs. One showing characteristic mesh texture, with square ‘windows’
with sub-microscopic fibers arranged cross-wise on to the bars of the window,
and a second form in large plates, more or less in optical continuity, but which
are also composite aggregates of parallel fibres. These plates me be after opx.
This rock is probably a fragment of an upper mantle peridotite (olivine-opx
rock) serpentinised at high level in the crust. The low-birefringence, low relief,
mesh and platy textures and very fine crystallite size are all characteristic.
11.5 serpentinite. An altered picrite from Inchcolm, Firth of Forth. The primary
minerals in this very Fe-rich, Mg-rich, feldspar-poor igneous rock are euhedral
olivine, titanaugiote rimmed by a brown hornblende with green hornblende
fringes, phlogopite and a more Fe-rich biotite. The low temperature alteration
minerals are mostly unidentifiable optically. The olivine is altered to at least 3
different minerals; yellow cores of clay(?), serpentine outside that with a pale
blue-green serpentine/chlorite mineral at the rim, probably reflecting
successive stages of alteration. A chlorite with anomalous bright blue
interference colours is abundant, possible after phlogopite. This rock may have
reacted with water more or less continuously from the melt stage to the initial
Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
cooling stage (amphibole rims on pyroxene) all the way down to temperatures
less than 100˚C
a)
Draw one grain of one of the sheet silicates which you have identified in one of
the previous thin sections, with its immediately adjacent mineral grains. Choose
a grain showing its typical and preferable diagnostic properties, label all
minerals and give a scale, section number and note whether it is illustrated in
ppl or crossed polars. THIS IS NOT A FULL PETROGRAPHIC ILLUSTRATION
TASK.
Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
‘Composition’ practical 5: Feldspars
Note: your script for this practical should be handed in at the end of the
session. It accounts of 15% of the overall class work mark for the course.
Part 1: Structure
Feldspars have framework structures based on cross-linked crankshaft-like
tunnels of SiO4 tetrahedra. The sites for Na, K and Ca ions lie between these
tunnels. Collapse of the framework around these sites through Al-Si ordering in K
feldspars and when the structure contains the smaller Na and Ca ions lowers the
symmetry from monoclinic to triclinic.
a)
Examine the model of orthoclase. Note the distorted 4-tetrahedra rings,
arranged in chains parallel to x. Notes also how the remaining free oxygens link
adjacent chains so that all oxygens are shared by two Si/Al ions.
b)
The (SiAl)-O framework of orthoclase has large interstices occupied by K ions.
These lie on mirror planes parallel to (010) edges and the centre of the cell.
Can you identify the mirror planes? Co-ordination of the K ions is irregular and
at lest 9-fold.
If you are feeling brave, try the following:
Identify the glide planes which are parallel to (010) and midway between the
mirror planes. These involve reflection and translation through a/2 along x.
Note then which Si atoms in a chain are related by this symmetry operation.
Identify the diad axes parallel to Y and again note which Si atoms are thus
equivalent.
Identify the screw diads with two non-equivalent types of Si/Al site in the cell
(T1 and T2). It is on these that ordering of Si and Al can occur.
c)
In microcline and albite the large intercises change shape slightly and the
mirror planes and diads disappear. Compare one of these structures to the
structure of orthoclase to verify this. This distortion doubles the number of
non-equivalent sites and lowers the symmetry to triclinic. Notice how the
structure has collapsed around the Na site in a unsymmetrical fashion.
Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
d)
The figure above shows ball-and-stick representations of 2 feldspar
structures. Large spheres are X-site cations (Na, K, Ca) and small dark sphere
are T site cations (Al, Si). Anions (O) are not shown. Orientation of both
structures is the same. Using your observations from above, decide which
structure is sanidine, and which is albite. Give 2 reasons.
LHS: ………………………………………..
RHS: …………………………………………..
Reason 1: …………………………………………………………………………………………………………………………….
Reason 2: ……………………………………………………………………………………………………………………………
Part 2: Hand specimens
Well-formed feldspars are generally roughly rectangular tablets, light in colour with
visible cleavage and often exsolution. Weathering tends to turn most feldspars milky
white, but reddish tinges are quite common. Fresh basic rocks can often have dark
broan or green glassy feldspars, if free of microscopic pores. The cleavage will
generally rule out quartz as a possibility.
a)
Read these general descriptive notes and then look at the specimens on display
with them in mind.
Sanidine in volcanic rocks is commonly flattened on (010) and is occasionally in large
enough phenocrysts for the 6-edged tablet form to be evident and for the common
Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
interpenetrant Carlsbad twinning to be seen. In plutonic rocks, orthoclase and
microcline may be pinkish due to exsolved iron oxide, a colour rarely seen in
plagioclases. Adularia, a K-rich vein feldspar, occurs in distinctive rhombohedra
(distorted). K-feldspar in pegmatites frequently contains coarse perthitic exsolution
lamellae or blebs visible to the naked eye.
Lamellar twinning absent from alkali feldspars (exept albite) may occasionally be seen
with a hand lens in plagioclase. Iridescence due to finely spaced lamellae acting as a
diffraction grating is most common in coarse plutonic plagioclases around An50 but
also occurs in plagioclases in the oligoclase range, and in some alkali feldspars
(microcline of the moonstone variety). The green colour of some microclines (Amazonstone) is distinctive but rare.
Part 3: Thin sections
Feldspars are all colourless, with low birefringence and two good, fine cleavages.
Sanidine, being monoclinic, can only show simple twins or no twninning. Orthoclase
is monoclinic optically, but triclinic in sub-microscopic domains, and microcline is
triclinic. Only microcline shows visible multiple or cross-hatch twinning. Both may
be untwined. The triclinic plagioclases commonly show multiple lamellar twinning,
may show simple twinning in addition and occasionally show no twinning.
The objective of this part of the practical is to use R.I., 2V, optic sign, the nature of
twinning and exsolution phenomena to distinguish the alkali feldspars from each other
Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
and from plagioclase. We will use plagioclase extinction angles to estimate
composition.
Alkali feldspars: the K-feldspars (sanidine, orthoclase and microcline) and albite
a)
Section 4.112.7 contains microcline with clearly identifiable cross-hatch
twinning.
b)
Section CC (Mourne Mtn. granite) contains a K-feldspar. Locate this using
relief, straight extinction and the presence of faint perthitic exsolution. In
some cases it may prove difficult to distinguish orthoclase from microcline.
XRD provides a useful means of doing this. In thin section, microcline (if
lacking its characteristic cross-hatch twinning) and orthoclase may be
reasonably well distinguish on the basis of 2V angle. Microcline usually has 2V
greater than 65˚. A 2V angle significantly lower than this (but typically above
40˚) suggests orthoclase. Estimate 2V for the K-feldspar in this section (using
the procedure outlined in optics practical 4), and use this establish its probable
identity.
………………………………………………………………………………………………………………………………………………..
………………………………………………………………………………………………………………………………………………..
c)
Examples of sanidine can be found in the following sections. Study at least one
of these:
7
Devitrified porphyritic pitchstone with ferro-augite and fayalite, and
sanidine phenocrysts.
P125 Sanidine-bearing phonolite with nepheline, titanaugite zoned out to
aegerine-augite, and sphene. Island of Pricipe, Gulf of Guinea. This rock
has 3 phenocryst phases. Dark phenocrysts are pyroxenes, colourless
ones are nepheline (which we will look at in more detail in the final
practical) and sanidine. Nepheline is present as regular hexagonal short
prisms, giving rise to 6- and 4-sides crystals (also some 5-sided). In this
rock is looks rather like quartz (but has negative relief and is uniaxial
negative). It has curving cracks, a faint cleavage, no twinning and no
alteration (which is unusual for nepheline). It has low birefringence but
is generally a bright grey interference colour.
Sandine crystals are typically rectangular (2x to 4x longer than wide)
with simple twining. They show one or more cleavages (one along the
length, one transverse) with some alteration along the cleavage traces
and birefringence is a bit lower than nepheline.
Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
JMZ 1 (4.4) Rhyolitic ash-flow tuff. Fragments of sanidine, quartz and cpx in
a matrix of devitrified glass. Bandolier Tuff, Jemez Mountains, New
Mexico.
JMZ 9 Rhyolitic welded tuff (ignimbrite) with sanidine, plagioclase and quartz.
Note in passing the collapsed glassy pumice fragments. Bandolier Tuff,
Jemez Mountains, New Mexico.
Perthitic and anti-perthitic exsolution
Visible feldspar exsolution is virtually confined to alkali feldspars. If K-feldspar is
the host it may show microcline cross-hatch twinning or simple Carlsbad tins, and
often no twinning. If the texture is anti-perthite with albitic feldspar as the host,
the host may show fine-multiple twinning, although this is rare. Exsolution lamellae
are most commonly ragged-edges sub-parallel inclusions or completely irregular blebs
or blotches. The finer they are, the greater the tendency for them to be parallel
sided.
d)
Foyaite (No.8, γ1, or Gr8). This is a nepheline syenite with bladed crystals of
alkali feldspar showing exsolution. Colourless, low birefringence squarish
crystals are the feldspathoid nepheline. Striking green, pleochroic crystals are
the pyroxene aegerine. Interstitial, moderate birefringence grains are another
feldspathoid, cancrinite.
In the circle overleaf prepare a careful, labelled sketch showing a clear
example of perthitic of anti-perthitic exsolution. Show the context of the
feldspar grain (surrounding minerals identified and labelled), add a scale bar
and identify slide used, and make sure you clearly label the sodic and potassic
components in the exsolution texture. This can be confirmed using the Becke
Line test (see box overleaf); the bright line will move into the higher R.I.
material, which will be the sodic component (albitic). Your illustration should be
at a sufficient magnification to show the exsolution (shape of blebs, how they
are aligned etc rather than a simle impressionistic view).
Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
e)
For comparison with (d), look at section 4.14.1, a biotite granite. Orthoclase in
this rock has a much finer microperthitic exsolution, and the plagioclase has
broad twin lamellae and shows oscillatory zoning.
Plagioclase composition determination
Composition of plagioclase feldspars can be estimated using the Maximum
Symmetrical Extinction Method or Michel-Lévy Method. The extinction angles
between the fast vibration directions and the trace of the albite twin lamellae in
twinned plagioclase crystals set with the (010) plane vertical should be the same for
both sets of lamellae (i.e. symmetrical extinction). The angle varies with the
orientation of the crystal. The maximum angle is diagnostic of composition. It is
useful to be able to make a rapid assessment of whether a plagioclase is sodic,
intermediate or calcic when studying rocks in thin section. Use the following
procedure:
Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
Determining plagioclase composition using the Michel-Lévy method
1.
2.
3.
4.
5.
6.
Find a grain showing albite twinning, which must have the following features:
• When set N-S the twin lamellae should be equally illuminated (light or
dark).
• The twin boundaries should be sharp and stay sharp when the focus is
moved up and down (i.e. they are set vertical).
•
Set the grain so that it is vertical (as in the middle illustration above) and
record the stage setting.
Turn the crystal to one side to the nearest position of extinction for one
set of lamellae. Record the stage setting.
Rotate the stage back so that the sample is again N-S. Now, rotate in the
opposite direction so that the other set of lamellae are in the nearest
extinction position (stages 3 and 4 and shown in the figure above). Record
this stage setting.
Calculate the 2 extinction angles, If they differ by more than 5˚ reject
them and start again.
In most cases the extinction angle between the trace of {010} and the fast
ray vibration is less than 45˚, but for calcic plagioclase it may be larger.
Therefore, it is important to use the S.T. plate to identify the fast
vibration direction within one set of lamellae. Begin by rotating the stage
until a set of lamellae come into extinction. Then rotate 45˚clockwise from
the extinction position and insert the S.T. plate. If retardations in the
lamellae in question subtract then the fast ray in those lamellae was
correctly placed N-S at extinction. If the colours increase the slow ray was
in the N-S position. In this case, subtract your angles from 90˚.
Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
7.
8.
f)
Take several readings and determine an average extinction angle. Then read
off the composition on the chart below.
Below An20 there is a change in trend line. It may be necessary to
determine the optic sign (using the procedure outlined in optics practical 4)
of the plagioclase to verify which part of the chart gives the correct
composition is maximum extinction angles are low (do not worry about this in
the following exercise).
Select one of the slides ARD-1, 5456, P, 64, Ex50. Select a suitable grain,
as detailed in the procedure above and assess whether the maximum observed
extinction angle is generally:
Sodic (<15˚), typical of granodiorites and diorites (and some volcanic
equivalents).
Intermediate (15-27˚), typical of diorites and some very evolved
ferrogabbros, and most equivalent volcanics.
Calcic (27˚+) as in gabbros and basalts.
Try to determine the extinction angle in at least 3 grains.
Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
Slide
number
Part 4:
a)
Max. extinction angle
Composition (An%)
Sodic/intermed. or
calcic?
Plagioclase composition determined by electron microprobe
analysis
The following oxide data was obtained from a plagioclase feldspar phenocryst.
Using the procedure outlined in the mineral chemistry practical, determine the
number of cations per formula unit based on a unit cell with 8 oxygens. In
column 7, round this value to the first decimal place.
a mineralogical formula for the sample of the form (Na,Ca)(Al)(Si)O8. C
1
2
3
oxide
Analysis Mol weight
SiO2
Al2O3
Fe2O31
CaO
Na2O
K2O
wt%
52.42
29.70
0.46
12.65
4.01
0.31
Total
ideal
99.55
100
b)
4
5
Mol
prop
Cation
prop
6
7
8
Oxygen Cations
prop
pfu
Using data from this calculation, write a mineralogical formula for this
plagioclase of the form (Na,Ca)(Al)(Si)O8.
………………………………………………………………………………………………………………………………………………..
Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
c)
What is the mole% Anorthite for this plagioclase solid solution. With
reference to the feldspar composition ternary diagram shown at the end of
section 2, give this composition a mineralogical name.
………………………………………………………………………………………………………………………………………………..
Part 5: Plagioclase crystallisation behaviour
Plagioclases exhibit nearly ideal solid solution at high temperature and have melting
and crystallisation behaviour similar to the olivines which we covered in the lectures.
The diagram below shows the crystal-liquid equilibrium diagram for plagioclase
feldspars. A liquid of composition L1 (i.e. a magma) is cooled under equilibrium.
Indicate the following with a labelled dot on the appropriate axes:
a)
b)
c)
d)
The temperature at which the first solid appears
The composition of the first solid to appear
The temperature at which the last liquid disappears
The composition of the last liquid present before crystallisation is complete.
Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
‘Composition’ practical 6: Other framework silicates
In the first part of the practical you will be given a mineralogy multiple choice test.
The duration of this test is 30 minutes. In the remaining time we will look briefly at
some other minerals with framework structures.
Part 1: Hand specimens
The irregular hexagonal pyramid topping a hexagonal prism is almost always present
in crystalline quartz. Otherwise there are few universal diagnostic features for
amorphous or cryptocrystalline forms, except for an absence of cleavage.
Nepheline is generally identifiable by its grey, greasy appearance and its
association with alkali pyroxenes and amphiboles. In volcanic or dyke rocks,
phenocrysts of leucite and the sodalite group minerals are characteristically
rounded, multi-faceted crystals reflecting their cubic character. Most are light
coloured, occasionally blue (haüyne).
Crystalline quartz
G2.3M200 single crystal showing the common {1010} and {0110} prisms and the
{1011} and {0111} rhombohedra combining to give the familiar 6-sided
prism topped by 6-faced pyramid.
Coloured variation in crystalline quartz
G2/3M201
G2/3M202
G2/3M203
G2/3M204
G2/3M205
Milky white prismatic quartz with pyrite (FeS2).
Red-stained stumpy prisms (pyramid dominant).
Smoky quartz, single crystal.
Amethystine quartz with calcite.
White vein quartz, intergrown and polycrystalline.
Cryptocrystalline or chalcedonic silica
G2/3M206 Agate (a term usually given to the most compact, often translucent,
banded or otherwise ornamented forms).
G2/3M207 Chalcedony (usually uniform in colour, often bluish or greenish and milky.
Commonly botryoidal –globular form resembling a bunch of grapes-and
ornamental, but not as fine as agate).
G2/3M208 Chert (opaque, dull, commonly black, grey or brown, but may be red,
green, cream or buff in colour). This specimen is flint, the name given to
Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
nodular chert found in chalk. Deep red, very compact chalcedonic silica is
commonly called jasper.
Opalescent silica
Precious opal is composed of sub-microscopic SiO2 spheres in close-packed sheets
with interstitial water films. This structure gives rise to the distinctive diffraction
colours. Semi-opal and other low density more or less opalescent forms have less
packing perfection. Interstitial air renders the mineral opaque.
G2/3M209 Semi-opal (unknown locality).
G2/3M210 Siliceous sinter. Deposited from hot-springs.
G2/3M211 Silicified wood, here chert rather than opal which also commonly
replaces wood.
Nepheline
In coarse grained plutonic rocks (e.g. syenites) nepheline often has a distinctive greygreen greasy appearance. In volcanic and hypabyssal rocks it often occurs in short
hexagonal prisms with approximately square section.
G2/3M212 and G2/3M213 are pegmatitic nepheline syenites (with alkali feldspar
and ferromagneisam minerals)
Leucite
The commonest manifestation is as white or glass-clear icositetrahedral phenocrysts,
often in great abundance, in volcanic rocks rich in K2O and poor in SiO2.
G2/3M214 Leucitite. Roccamonfina, S. Italy
G2/3M215 Leucite tephrite (=olivine-poor basalt +leucite) from Vesuvius.
Analcite occasionally occurs in an identical manner (this may represent chemical
transformation of primary leucites).
Sodalite group
Sodalite, haüyne and nosean cannot be differentiated in hand specimen or thin
section with certainty. All can be colourless. Sodalite and haüyne are often bright
blue in hand specimen and nosean greyish.
G2/3M216 Sodalite syenite
G2/3M217 Haüyne crystal in phonolite
Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
G2/3M218 Haüyne nephelinite. Mt. Etinde, Cameroon.
Zeolites
Examine the specimens on display. They all come from cavities in lava flows where
they have crystallised from hot circulating waters as the flow cools. Further
specimens are on display in the mineral cabinets in the Department. Spectacular large
specimens feature in the Royal Museum of Scotland collections. Zeolites are generally
colourless or white and exhibit a very wide range of habits.
Part 2: Thin sections
Nepheline has low birefringence, is unixaxial negative and has straight extinction
to a cleavage or rectangular crystal outlines, when you can see them, and
occasionally has good hexagonal cross section.
Leucite only occurs in volcanic rocks. It is unmistakable as it occurs as
icositetrahedra (Greek for 24 faces) which are almost spherical, almost isotropic
and frequently show complex multiple twinning because of a structural inversion
from cubic to tetragonal on cooling.
The Sodalite Group minerals, Sodalite, Haüyne and Nosean are virtually identical
in thin section: hexagonal or octagonal outlines (they are cubic) and completely
isotropic. All have lower refractive indices than feldspar. None ever coexists with
quartz.
Feldspathoid minerals can also be recognised by their characteristic association
with alkali ferromagnesian minerals (bright green pyroxenes and/or blue/purple
amphiboles).
FF
Ijolite (a mafic nepheline-ferromags rock with minor alkali feldspar) from
Lobau, Saxony. Euhedral, large colourless nephelines with smaller, turbid
isotropic sodalite.
Foyalite (Nepheline Syenite).You looked at this rock in the previous practical (see
detailed notes). Find the nepheline in between the large alkali feldspars and
aegerine. Note also cancrinite as a nepheline alteration product. It shares the
low refractive index but is otherwise very different from the other
feldspathoids. It looks like muscovite mica, with moderate birefringence,
perfect cleavage and straight extinction. It is, however, uniaxial negative. A
muscovite-looking mineral in an igneous rock with identifiable nepheline is
almost certainly cancrinite.
Generated by Foxit PDF Creator © Foxit Software
http://www.foxitsoftware.com For evaluation only.
81
Leucite phonolite, Reiden, Eifel. Leucite, haüyne, nosean in addition to
aegirine.
9.2 or 87 or 11413 or 1709. Leucite-bearing basaltic rocks. The last 3 probably all
from Vesuvius. Leucite is distinct as icositetrahedra.
a)
Select one of the feldspathoid-bearing thin sections to illustrate the
appearance of as many different feldspathoids as possible, together with
other minerals also present, to bring our the general context of the
feldspathoids. Remember to include a scale bar and label the section studied.
Expand
groundmass
as
an
insert
if
need
be.