26.4 Polarization and the Reflection and Refraction of Light
Brewster’s law
When the incident
angle is equal to
θ B = tan −1
n2
n1
(a) The reflected ray is perpendicular to the
refracted ray
(b) The reflected ray is completely polarized
in the horizontal plane.
If the incident ray is linearly polarized in the
incident plane  100% transmission
−1
For air to glass: θ B = tan
1.5
= 56°
1.0
n1
n2
Dispersion The index of refraction of visible light
changes slightly with wavelength. In glass n is an
decreasing function of λ (and an increasing
function of f )
1
26.5 The Dispersion of Light: Prisms and Rainbows
In seeing a rainbow, we are looking at sunlight that
has gone through a refraction-reflection-refraction
interaction in the water droplets suspended in air
The reflection is
NOT total internal
reflection
Common
Misconception
2
Fermat’s Principle
C
30.0 cm
θ2
x
Glass has refractive index n1=1.50
n2=1.00
n1=1.50
B
θ1
Find x, where the ray crosses the boundary
30.0 cm
A
c 30.0
c 30.0
=
= 20.0 cm/ns, v2 =
=
= 30.0 cm/ns
n1 1.50
n2 1.00
c = 30.0 cm/ns, v1 =
(50 − x ) 2 + (30) 2
x 2 + (30) 2
+
20
30
AB BC
+
=
t=
v1
v2
sinθ 2 =
An interesting solution:
Fermat’s Principle: The ray from A to C
takes the path of least time
AB = x 2 + (30) 2 , BC = (50 − x ) 2 + (30) 2
50.0 cm
sinθ1 =
A ray originates from point A (0,0) inside a piece
of glass, and crosses the glass-air boundary at
point B (x, 30.0 cm), and is detected by a sensor
at point C (50.0 cm, 60.0 cm).
17
(17) + (30)
2
2
33
(33) + (30)
2
2
= 0.493
Minimum at
x=17.0 cm
= 0.740
sinθ 2 0.740
n
=
= 1.500 = 1
sinθ1 0.493
n2
!!!
3
26.6 Lenses
Refractive Lenses are typically formed by grinding and polishing the
surface of glass or special plastics to spherical and/or planar shapes.
They come in two general types:
With a converging lens,
paraxial rays that are
parallel to the principal axis
converge to the focal point.
With a diverging lens,
paraxial rays that are
parallel to the principal
axis appear to originate
from the (virtual) focal
point.
−
4
26.6 Lenses
For lenses, the object and the viewer are on opposite
sides (for mirrors they were on the same side) :
Converging lenses are analogous to concave mirrors.
For ray tracing we commonly use three rays.
[1] from tip of object,
parallel to the principle
axis on the object
sides, through the lens
and then through the
real focal point on the
viewer side
[2] from tip of object,
through the real focal
point on the object
side, then through the
lens and emerge
parallel to the
principle axis on the
viewer side
4
[3] from tip of object
straight through the
middle of the lens
5
26.7 The Formation of Images by Lenses
Image Formation by a Converging Lens
Case 1: (Example: ordinary camera taking a picture of a distant object)
d0 > 2 f
f < di < 2 f
Object is placed beyond 2F (object side): 2F for a converging lens is analogous to C
 Shrunken, inverted REAL image located between F and 2F on observer side.
Real images are always inverted and are always on the viewer side of the lens
The same equations that applies to mirrors also apply to lenses
1 1 1
+ =
do di f
do > 2 f →
do > 0 →
1
1
<
do 2 f
1
>0
do
and
1
1 1
= −
di
f do
1
1 1
= −
di
f do
di
hi
m= =−
ho
do
→
→
1
1
1
> −
di
f 2f
1 1
<
di
f
→
1
1
>
di 2 f
→ di > f
→ di < 2 f
→ f < di < 2 f
6
26.7 The Formation of Images by Lenses
Case 2: (Example: projection of a downward nearby object onto a
upward REAL image on screen)
f < d0 < 2 f
di > 2 f
Object placed between F (object side) and 2F (analogous to C for mirrors)
 enlarged, inverted REAL image beyond 2F on observer side.
do < 2 f →
1
1
>
do 2 f
1
1 1
= −
di
f do
di < 0
→
1 1
1
< −
di
f 2f
→
1
1
<
di 2 f
→ di > 2 f
Case 3: (Example: magnifying glass)
d0 < f
Object placed inside F (object side)
 enlarged, upright VIRTUAL image on object side
0 < do < f
→
1
1
>
do
f
→
1
1 1
= −
< 0 → di < 0
di
f do
m=−
di
f
=
>1
do
f − do
7
26.8 The Thin-Lens Equation and the Magnification Equation
Example:
The Real Image Formed by a Camera Lens
A 1.70-m tall person is standing 2.50 m in front of a camera. The
camera uses a converging lens whose focal length is 0.0500 m.
(a) Find the image distance and determine whether the image is
real or virtual.
(b) Find the magnification and height of the image on the film.
1 1 1
+ =
do di f
(a)
and
1 1 1
1
1
= −
=
−
= 19.6 m −1
d i f d o 0.0500 m 2.50 m
d i = 0.0510 m
(b)
di
hi
m= =−
do
ho
m=−
real image
di
0.0510 m
=−
= −0.0204
2.50 m
do
hi = mho = (− 0.0204 )(1.70 m ) = −0.0347 m
8
26.6 Lenses
Diverging Lenses are analogous
to convex mirrors
Ray-tracing for diverging lenses
[1] from tip of object, parallel
to the principle axis on the
object sides, through the
lens and bending outward
as if it came from the virtual
focal point on the object side
[2] from tip of object,
toward the real focal
point on the viewer
side, then through the
lens and emerge
parallel to the principle
axis on the viewer side
[3] from tip of object
straight through the
middle of the lens
4
9
26.7 The Formation of Images by Lenses
Image Formation by a Diverging Lens
d0 > 0
f <0
di < 0
A diverging lens has a VIRTUAL focus on the object side of the lens
A diverging lens always forms a shrunken, upright, VIRTUAL, image.
Virtual images are always upright and are always on the object side of the lens
d > 0, f < 0
m=−
→
1
1 1
1
1
= −
=− −
<0
di
f do
f do
di
f
=
do
f − do
=
− f
− f − do
=
f
f + do
→ di < 0
→ m <1
10
26.8 The Thin-Lens Equation and the Magnification Equation
Textbook Summary of Sign Conventions for Lenses
f is + for a converging lens.
f is − for a diverging lens.
d o is + if the object is to the left of the lens.
d o is − if the object is to the right of the lens.
d i is + for an image formed to the right of the lens (real image).
d i is − for an image formed to the left of the lens (virtual image).
m is + for an upright image.
m is − for an inverted image.
11
26.8 The Thin-Lens Equation and the Magnification Equation
Prof. Jui’s Conventions for Thin Lenses
f , d o and d i are + for a
Real objects are on the object side
REAL focus, object and image
Real image are on the observer side
f , d o and d i are − for a
Virtual objects are on the observer side
VIRTUAL focus, object and image
Virtual image are on the object side
ho and hi are + for upward
Converging lenses have REAL foci on
both sides of the lens.
objects and images
ho and hi are − for downward
Diverging lenses have VIRTUAL foci
on both sides of the lens.
objects and images
m=
hi
is − for real images
ho
hi
is + for virtual images
m=
ho
12
26.9 Lenses in Combination
The image produced by one
lens can serves as the object
for the next lens.
The pair of lenses shown here is functionally identical
to a compound microscope
Example: The objective and eyepiece of the compound microscope in the above figure are
both converging lenses and have focal lengths of fo = 15.0 mm and fe = 25.5 mm. A distance of
61.0 mm separates the lenses. The microscope is being used to examine an object
placed do1 = 24.1 mm in front of the objective. Find the final image distance.
First look at the action of the objective
lens: both do1, di1 are relative to this lens,
and expressed in mm
1
1
1
1
1
=
−
=
−
= 0.02517 mm -1
d i1
f
d o1 15.0 24.1
o
d i1 =
1
= 39.7 mm
-1
0.02517 mm
13
d i1 = 39.7 mm
d o 2 = 61.0 mm − d i1
= 61.0 mm − 39.7 mm
= 21.3 mm
f = 25.5 mm
e
1
1
1
1
1
=
−
=
−
= −0.007733 mm -1
di2
f
d o 2 25.5 21.3
e
di2 =
1
= − 129 mm
-1
( −0.007733 mm )
14
26.10 The Human Eye
The human eye operates very much
like a modern electronic camera
Iris: controls the amount of light
energy entering the lens
Lens: focus light onto retina
(adjustable)…refraction also
provided by cornea + A.H.
Retina: Layer of electronic
(ok…neural) pixel elements
Monocular Vision
The lens focuses some of the rays (“emitted” in all
directions) from points on the pencil (the object) on to
individual points (the image) on the retina
The electrical impulses are carried by the optic nerve
into the brain for processing  shapes and colors
15
26.10 The Human Eye
Muscles in the eye changes the
shape (focal length) of the lens
in response to near and far
objects  depth perception
with just one eye
relaxed lens
This is a skill learned by a baby in the first few days after
birth. It is difficult to demonstrate—it is so automatic
(1) Cover one eye. Stare at this screen with other eye. Then
move a finger into field of view.
(2) Cover one eye. Look down at one finger. Raise your head
until this screen comes into field of view
tensed lens
Binocular Vision
(a) Eyeballs rotate to center the object in each
eye (conscious but fairly automatic response by
the brain)  more depth perception
(1) Put one finger from each hand in front of you—
one at twice the distance of the other.
(2) Alternately focus on one finger—the other will be
seen in “double”
(b) The slightly different images seen in the
two eyes are interpreted by the brain to given
even more depth perception – 3D glasses!
16
26.10 The Human Eye
Distant
Object
Nearsightedness
(myopia)
Relaxed Eye Lens
DF (non-standard notation)
Far Point of
nearsighted eye
Image formed in
front of retina
Ideally, the lens of the eye
should be able to adjust to
objects at any distance.
But the nearsighted eye has a lens-retina combination that cannot relax itself enough to focus
objects out to infinity. A distant object focus to a real image in front of (but missing) the retina.
Usually there is a maximum object distance, called the far point, to which the eye can focus
Diverging
Lens
Distant
Object
Far Point of
nearsighted eye
Distant
Object
Corrective Lens
The patient is prescribed a
diverging lens to compensate
for the over-convergence
Image formed
on the retina
Prescription
Virtual Image formed
by diverging lens
Far Point of
nearsighted eye
DL
We want to put the virtual
image made by the diverging
lens of a distant object (i.e.
do = ∞) at the far point: DF.
Remember that the corrective lens is worn at a small distance DL in front of the eye (DL=0 for a contact lens)
do = ∞
d i = − (D F − D L )
1
1
1
=
+
f do di
=
1
1
+
∞ − ( DF − DL )
→ f = −( DF − DL )
17
26.10 The Human Eye
With this prescription, objects at finite, but far distances are mapped into virtual
images located between the corrective lens (at distance DL from the eye) and the
far point (at distance DF from the eye)
Example : Eyeglasses for the Nearsighted Person
A nearsighted person has a far point located only 521 cm from the eye. Assuming that
eyeglasses are to be worn 2 cm in front of the eye, find the focal length needed for the
diverging lens of the glasses so the person can see distant objects.
1
1
1
1
1
=
+
= −
f d o d i ∞ DF − DL
=
1
1
−
∞ DF − DL
=
−1
521 cm − 2 cm
→
f = −519 cm
THE REFRACTIVE POWER OF A LENS – THE DIOPTER
Optometrists who prescribe correctional lenses and the opticians who make the lenses do
not specify the focal length. Instead they use the concept of refractive power.
1
Refractive Power ( RP : in diopters) =
f (in meters )
RP is not a standard notation, and diopter is not an SI unit.
f = −519 cm
→ RP = −0.193 m -1
= −0.193 dpt
18
26.10 The Human Eye
Tensed Eye Lens
Near Point of
nearsighted eye
Sharp image
formed behind
the retina
Close-by object
DN (non-standard notation)
Farsightedness
(hyperopia)
Ideally, the lens of the eye
should be able to adjust to
objects at any distance.
But the Farsighted eye has a lens-retina combination that cannot tense itself enough to focus
objects close by. A close-by object focus to a sharp, real image behind (but missing) the retina.
Usually there is a minimum object distance, called the near point, to which the eye can focus
Converging
Lens
Near Point of
nearsighted eye
Sharp image
on retina
Close-by object
Virtual Image formed by
converging lens
Near Point of
nearsighted eye
Corrective Lens
Converging
Lens
Close-by object
DL
RP =
The patient is prescribed a
converging lens to
compensate for the underconvergence
Prescription
Put the virtual image made by the
converging lens of the nearest
object you want to see (typically at
DMIN = 25 cm) to the near point: DN.
1
1
1
1
1
=
−
=
+
f d o d i ( DMIN − DL ) ( DN − DL )
19
Example of corrective lens for farsightedness:
this is a pathology everyone gets as they get older – starting at ~40 yrs of age
(nearsightedness improves somewhat in combination with this)
Your professor wears reading glasses with refractive power of RP = 1.75 dpt = 1.75 m-1.
Where is his near point (inside of which he cannot see). Assume the glasses to correct
for objects as near as 25 cm, and that the glasses are worn 2 cm from the eyes.
RP =
1
1
1
=
+
f do di
1.75 m −1 =
=
1
1
−
( DMIN − DL ) ( DN − DL )
1
1
−
(0.25 m − 0.02 m) ( DN − 0.02 m)
1
1
=
− 1.75 m −1 = 2.60 m −1
( DN − 0.02 m) 0.23 m
DN − 0.02 m =
1
= 0.38 m
2.60 m −1
DN = 0.40 m
20