Fermat’s Principle and Its Applications
A
S
B
S
3.1 INTRODUCTION
The propagation of light in the realm of geometrical optics
employs the concept of rays. Consider a circular aperture in front of
a point source P as shown in Fig. 3.1. When the diameter of the
aperture is quite large (~1 cm), we can see a patch of light with welldefined boundaries on the screen SS‘. As we reduce the size of the
aperture to be very small (<0.1 mm), then the pattern obtained on SS'
ceases to have well-defined boundaries. This phenomenon is known
as diffraction and is a direct consequence of the finiteness of the
wavelength (which is denoted by ). In Chapters 18 and 20, we will
show that in the limit of
0 the diffraction effects will be absent.
Consequently we can obtain an infinitesimally thin pencil of light;
this is called a ray. The field of optics neglecting the finiteness of the
wavelength is called geometrical optics.
Light has a wavelength of the order of 10 5 cm, which is small
compared to the dimensions of normal optical instruments like
lenses, mirrors, etc., one can, in many applications, neglect the
finiteness of the wavelength.
The path of the rays an be studied by using Fermat's principle:
the ray will correspond to that Path for which the time taken is an
extremum in comparison to nearby paths. The time taken to
traverse the geometric path ds in a medium of refractive index n(x,
y, z) is given by
ds
c/n
nds
c
where n(x, y, z) is the position dependent refractive index and c
is the speed of light in free space.
P
Fig. 3.1 The light emitted by the point source P is allowed to
pass through a circular hole and if the diameter of the hole is
very large compared to the wavelength of light then the light
patch on the screen SS has well defined boundaries.
If the ray to traverse the path AB along the curve C (Fig. 3.2) then
the total elapsed time is
1
c
1
cA
ni dsi
i
nds.
(1)
B
C
A
B
The symbol C below the integral represents the fact that the
integration is from the point A to B through the curve C. Where dsi
represents the ith arc length and ni the corresponding refractive
index. If ACB indeed represents the path of a ray, then will be
either less than, greater than or equal to ’ for all nearby paths like
AC’B. Since c is a constant, one can alternatively define a ray as
the path for which optical path
nds
A
is an extremum.
B
C
(2)
Mathematically, the ray would follow the path for which
nds
B
A
ds
C
C
A
Fig. 3.2 If the path ACB represents the actual ray path
then the time taken in traversing the path ACT will be an
extremum in comparison to any nearby path AC B.
(3)
B
where the left-hand side represents the change in the value of
the integral due to an infinitesimal variation of the ray path. The
actual ray path between two points is the one for which the
optical path length is stationary with respect to variations of the
path.
The Fermat principle tells that in a homogenous medium,
the rays will be straight lines because a straight line will
correspond to a minimum value of the optical path connecting
two points in the medium (referring to Fig. 3.3).
3.2 LAWS OF REFLECTION AND REFRACTION
FROM FERMA'T'S PRINCIPLE
E
B
C
D
A
Fig. 3.3 Since the shortest distance between two
points is along a straight line, light rays in a
homogeneous medium are straight lines; all nearby
paths like AEB or ADB will take longer times.
Consider a reflecting surface as shown in Figure 3.4. Light
from point A is deflected to point B by this surface, forming the
angle of incidence i (= APS) and the angle of reflection r (= SPB)
measured from the normal to the surface. The time required for the
ray of light to travel the path AP+PB is given by =(AP+PB)/v,
where v is the velocity of light in the medium containing the points
APB. We find
1
v
( z)
hi2 (d
z )2
hr2
z2 .
To extremize (z) with respect to variations in z, we set d (z)/dz=0
to get
d
hi2 (d
z
z
z)2
hr2
z2
i
r.
We thus obtain the laws of reflection from the Fermat's principle.
Let us now use Fermat‘s principle to analyze refraction as
illustrated in Figure 3.5. 1 and 2 are the angles of incidence and
transmission, respectively, measured once again from the normal to
the interface. The optical path length taken by the light to travel the
distance ARB is
Lop
n1 AR n2 RB
n1 x
2
2
1
h
n2
2
L x
2
2
h .
Thus Eq. (5) becomes
n1 sin
n2 sin 2 .
1
(6)
which is the Snell's law of refraction.
(4)
The laws of reflection and refraction form the basic laws
for tracing light rays through simple optical systems, like a
system of lenses and mirrors, etc.
(5)
Example 3.1 (see Fig. 3.6)
Example 3.2 (see Fig. 3.9)
Example 3.3 (see Fig. 3.10)
Example 3.4 (see Fig. 3.11)
To extremize this, we must have
dLop
dx
n2 L x
n1 x
0
x 2 h12
L x
2
0.
h22
From Fig. 3.5, we see that
sin
A
x
1
x 2 h12
and sin
n2 L x
2
L x
2
.
2
2
h
A
a S
z
hi
B
1
h1
hr
1
R
N
M
P
Q
P
L -x
N
M
R
n1
Q
n2
x
2
h2
2
A
Fig. 3.4 The shortest path connecting the two points A
and B via the mirror is along the path APB where the
point P is such that AP, PS and PB are in the same plane
and APS = SPB; PS being the normal to the plane of
the mirror. The straight line path AB is also a ray.
B
L
Fig. 3.5 A and B are two points in media of refractive indices
n1 and n2. The ray path connecting A and B will be such that
n1 sin 1 = n2 sin 2 .
A
Q
L
P
Q
S
L
C
B
Fig. 3.6 All rays parallel to the axis of a paraboloidal
reflector pass through the focus after reflection (the
line ACB is the directrix). It is for this reason that
antennas (for collecting electromagnetic waves) or solar
collectors are often paraboloidal in shape.
Fig. 3.7: A paraboloidal satellite dish. Photograph
courtesy McGraw Hill Digital Access Library
P
S1
Fig. 3.8 Fully steerable 45m paraboloidal dishes of the Giant Metrewave Radio
Telescope (GMRT) in Pune, India. The GMRT consists of 30 dishes of 45m
diameter with 14 antennas in the Central Array. Photograph courtesy: Professor
Govind Swarup, GMRT, Pune.
Q
S2
Fig. 3.9 All rays emanating from one of the foci of an
ellipsoidal reflector will pass through the other focus.
S
n1
n2
n1
P
S
P
O
x
C
Q
I
r
O
I
y0
y
n2
M
Fig. 3.10
SPM is a spherical refracting surface
separating two media of refractive indices n1 and n2. C
represents the center of the spherical surface.
3.3 RAY PATHS IN AN INHOMOGENEOUS MEDIUM
In an inhomogeneous medium, the refractive index varies in
a continuous manner and, in general, the ray paths are curved. For
example, on a hot day, the air near the ground has a higher
temperature than the air which is much above the surface. Since the
density of air decreases with increase in temperature, the refractive
index increases continuously as we go above the ground. This leads
to the phenomenon known as mirage.
We will use Snell's law (or Fermat‘s principle) to study the
ray paths in an inhomogeneous medium. We further assume the
refractive index changes continuously along one direction only; the
x axis.
The inhomogeneous medium can be thought of as a limiting
case of a medium consisting of a continuous set of thin slices of
media of different refractive indices [see Fig. 3.12(a)].
Fig. 3.11 : The refracted ray is assumed to diverge away
from the principal axis.
At each interface, the light ray satisfies Snell‘s law and one
obtains
n1 sin
1
n2 sin
2
n3 sin
.
3
(13)
Thus, we may state that the product
n x cos
x
n x sin
x
(14)
is an invariant of the ray path; we will denote this invariant by .
If the ray initially makes an angle 1 (with the z axis) at a
point where the refractive index is n1, then the value of
is
n1cos 1. The piecewise straight lines shown in Fig. 3.12(a)
form a continuous curve which is determined from the
equation
n x cos
x
n1 cos 1 = .
(15)
3.3.1 The Phenomenon of Mirage
x
n4
x
On a hot day the refractive index continuously decreases as
we go near the ground. Indeed, the refractive index variation can
be approximately assumed to be of the form
4
n3
3
ds
3
n2
dx
2
n1
n x
1
z
1
z
(a)
(b)
Fig. 3.12 (a) In a layered structure, the ray bends in such a
way that the product ni cos i remains constant. (b) For a
medium with continuously varying refractive index, the ray
path bends in such a way that the product n(x) cos (x)
remains constant.
3
x(m)
x few metres
n0
n0
ne
C
P
0
(16)
n0: the refractive index of air at x=0 (just above the ground)
k : a constant. The calculated ray paths (see Example 3.8) are
shown in Fig. 3.13.
Consider a ray which touches horizontal at x = 0.
Assume the refractive index is ne at the eye position E (x=xe)
where the ray makes an angle e with the horizontal then
Usually
2
n0 kx
dz
e<<
ne cos e .
(17)
1 so that
cos
e
1
1
2
2
e
2 1
n0
.
ne
(18)
At constant air pressure, the dependence of n on temperature T
gives
E
W W
n0 1 T0
1
ne 1 Te .
(19)
Eq. (19) tells that
0
0
B
B
R1
R2
M
200
400
600
800
1000
1200
1
n0 1
T0
ne 1
1
2 1
1
ne
n n
Te
or e 0
T0
ne
ne 1
T
1 e
ne
T0
z (m)
Fig. 3.13 Ray paths in a medium characterized by a linear variation of
refractive index [see Eq.(16)] with k 1.234x10-5 m-1 . The object point is at a
height of 1.5 m and the curves correspond to + 0.2 , 0 , -0.2 , -0.28 ,–0. 3486
and –0.5 . The shading shows that the refractive index increases with x.
e
1
Te
T0
(20)
On a typical hot day the temperature near road surface
T0 323 K (=50 C) and, about 1.5 m above the ground, Te 303 K
(= 30 C). Now, at 30 C, ne 1.00026 giving e 5.67 10
radians 0.325 . In Fig. 3.13 we have shown rays emanating (at
different angles) from a point P which is 1.5 m above the ground;
each ray has a specified value of the invariant
n1 cos 1 . The figure shows that when the object point P and
the observation point E are close to the ground, the only ray path
connecting points P and E will be along the curve PME and that a
ray emanating horizontally from the point P will propagate in the
upward direction as PC as shown in figure. The eye at E will see
the mirage and not see the object directly at P. There is a region
R2 where none of the rays (emanating from the point P) reaches
where neither the object nor its image can be seen the shadow
region. There is also a region R1 where only the object is directly
visible and the virtual image is not seen.
In Fig. 3.13, we see that the ray bending up after it becomes
parallel to the z axis (the point at which =0 is known as the
turning point). The bending up of the cannot be directly inferred
from Eq. (15) because at such a point one may expect the ray to
proceed horizontally beyond the turning point as shown by a
dotted line. However this behaviour may be explained from
considerations of symmetry and from the reversibility of ray
paths; the ray path should be symmetrical about the turning point.
Physically, the bending of the ray can be understood by
considering a small portion of a wave front such as W (see Fig.
3.13); the upper edge will travel with a smaller speed in
comparison with the lower edge, and this will cause the wave
front to tilt (see W') making the ray to bend. Furthermore, a
straight line path like BB‘ does not correspond to an extremum
value of the optical path.
We next consider a refractive index variation which saturates to a
constant value as x
:
n2 x
n02
n22 1 e
x
, x
0
2
(21)
R
where n0, n2 and
are constants and x represents the height
above the ground. The exact ray paths are obtained by solving
the ray equation (see Example 3.10) and are shown in Figs 3.14
and 3.15; they correspond to the following values of various
parameters:
n0=1.000233, n2= 0.45836 and
= 2.303 m
(22)
The values of equation (22) are not actually realistic,
nevertheless, it allows us to understand qualitatively the ray
paths in a graded index medium. Figures 3.14 and 3.15 show the
ray paths emanating from points that are 0.43 m and 2.8 m
above the ground respectively.
1
C
E
P
R2
1.10
1.05
1.00
0
R1 M
0
3
6
z (m)
Fig. 3.14 Ray paths in a medium characterized by Eqs. (21) and (22).
The object point is at a height of 1/ ( 0.43m) and the curves
correspond to 1 = + /10, 0, - /60, - /30, - /15 and - /10. The
shading shows that the refractive index increases with x.
3
In Fig. 3.14, the point P corresponds to a value of the refractive
index equal to 1.06455 (=n1) and different rays correspond to
different values of 1, the angle that the ray makes with the z
axis at the point P. We see that when the object point P and the
observation point E are close to the ground, the only ray path
connecting points P and E will be along the curve PME and that
a ray emanating horizontally from the point P will bend up in
the upward direction, shown as PC. The eye at E will see the
mirage and not see the object directly at P. If points P and E are
much above the ground as in Fig. 3.15, the eye will see the
object almost directly (because of rays like PCE) and will also
receive rays appearing to emanate from points like P'. Since
different rays do not appear to come from the same point and
hence the reflected image seen will have considerable
aberrations.
Fig. 3.16 A typical mirage as seen on a hot road on a warm day;
photograph adapted from
http://fizyka.phys.put.poznan.pl/~pieransk/Physics%20Around%20Us/
Air%20mirror.jpg. The photograph was taken by Professor Piotr
Pieranski of Poznan University of Technology in Poland; used with
permission from Professor Pieranski.
CC
P
E
E
RR1
2
1
1
MM
0
1.10
1.05
1.00
0
5
RR22
10
15
20
z (m)
n
P
Fig. 3.15 Ray paths in a medium characterized by Eqs. (21)
and (22). The object point is at a height of 2.8m and the
curves correspond to 1 (the initial launch angle) = 0, /60, - /30, - /16 , - /11 , - /10 and - /8. The shading
shows that the refractive index increases with x.
Fig. 3.17 This is actually not a reflection in the ocean, but the miraged
(inverted) image of the Sun's lower edge. A few seconds later (notice
the motion of the bird to the left of the Sun!), the reflection fuses with
the erect image. The photographs were taken by Dr. George Kaplan of
the U. S. Naval Observatory and are on the website
http://mintaka.sdsu.edu/GF/explain/simulations/infmir/Kaplan_photos
.html created by Dr. A Young; photographs used with permissions
from Dr Kaplan and Dr. A. Young.
There is also a shadow region R2 where none of the rays (from
the point P) will reache; an eye in this region can neither see the
object nor its image. The actual formation of mirage is shown in
Figs. 3.16 and 3.17.
Refer to
(1) Example 3.5
(2) Example 3.6
2
P
1
P
E
0
1.00
1.05
3.3.2 The Phenomenon of Looming
0
5
10
15
z (m)
Fig. 3.18 Ray paths corresponding to the refractive index
distribution given by Eq.(23) for an object at a height of 0.5 m;
the values of n0, n2 and are given by Eq. (22).
The equation describing the ray path is discussed in Problem
3.13. We assume the values of n0, n1 and to be given by Eq.(22).
For an object point P at a height of 0.5 m, the ray paths are shown
in Fig. 3.18. If the eye is at E, then it will receive rays appearing
to emanate from P'. Such a phenomenon in which the object
appears to be above its actual position is known as looming; it is
commonly observed in viewing ships over cold sea waters (see
Figs 3.19 and 3.20). Since no other rays emanating from P reach
E, the object cannot be observed directly.
3.3.3 The Graded Index Atmosphere
The noncircular shape of the setting or the rising sun (see Fig.
3.21 and Fig. 8 in the prelim pages) is due to the graded index
medium Atmosphere. The refractive index of the air gradually
decreases as we move outwards.
Above cold sea water, the air near the water surface is
colder than the air above it and hence there is an opposite
temperature gradient. A suitable refractive index variation for
such a case an be written as:
n2 x
n02
n22e
x
, x
0
(23)
If we approximate the continuous refractive index gradient by a
finite number of layers (each layer having a specific refractive
index) then the ray will bend in a way similar to that shown in
Fig. 3.22. Thus the sun (which is actually at S) appears to be in
the direction of S‘. It is for this reason that the setting sun
appears flattened and also leads to the fact that the days are
usually about 5 minutes longer than they would have been in the
absence of the atmosphere.
Obviously, if we were on the surface of the moon, the rising
or the setting sun would not only look white but also circular in
shape!
Fig. 3.19 The superior mirage occurs under reverse atmospheric conditions from the
inferior mirage. For it to be seen, the air close to the surface must be much colder than
the air above it. This condition is common over snow, ice and cold water surfaces. When
very cold air lies below warm air, light rays are bent downward toward the surface, thus
tricking our eyes into thinking an object is located higher or is taller in appearance than
it actually is. Figure adapted from
http://www.islandnet.com/~see/weather/elements/mirage1.htm
Fig. 3.20 A house in the archipelago with a superior mirage. Figure adapted
from
http://virtual.finland.fi/netcomm/news/showarticle.asp?intNWSAID=25722.
The photograph was taken by Dr. Pekka Parviainen in Turku, Finland; used
with permission from Dr. Parviainen.
(b)
(a)
(c)
Fig. 3.21 The non circular shape of the setting sun.
Additional Figure: Photographs on the
moon. Because moon does not have any
atmosphere, the sky and shadows are very
dark. In (a) and (b) we can also see the
earth.Photographs courtesy McGraw Hill
Digital Access Library.
S
Sun
Sun
S
Earth
3.4 THE RAY EQUATION AND ITS SOLUTIONS
We may derive the ray equation now, the solution of
which will give the precise ray paths in an inhomogeneous
medium. We assume the refractive index changes along the x
axis. It can be considered as the limiting case of a medium
comprising of a continuous set of thin slices of media of different
refractive indices. For a continuously varying refractive index,
the product n(x)cos (x) is an invariant of the ray path:
n x cos
Fig. 3.22 Because of refraction, light from S
appears to come from S .
dz
ds
cos
n x
.
(26)
2
n2 x
2
(24)
2
dx
2
2
ds
dz
2
dz or
dx
dz
2
1.
(25)
Both Eqs (27) and (28) represent rigorously correct ray
equations when the refractive index depends only on the x
coordinate. Refer to Example 3.7 and Example 3.8.
3.4.1 Ray Paths in Parabolic Index Media
Consider a parabolic index medium with the following
Plug Eq. (26) into Eq. (25), we have
dx
dz
.
For a continuous variation of refractive index, the piecewise
straight lines shown in Fig. 3.12(a) forms a continuous curve
as in Fig. 3.12(b). Let ds represents the infinitesimal arc
length along the curve, then
ds
Refer to Fig. 3.12(b), we find that
x
1.
(27)
refractive index distribution:
n2 x
n12
2
x2
(36)
Equation (27) can be written as
For a given n(x) variation, Eq. (27) can be integrated to give the
ray path x(z). We may also differentiate it with respect to z to
obtain:
dx d 2 x
2
dz dz 2
2
1 dn x dx
2
dx dz
d 2x
2 2
dz
2
1 dn x
. (28)
2
dx
1
n
2
2
x
dx
1
dz.
(37)
dz.
(38)
Substituting dx leads to
1
2
0
x
x
dx
where
x0
1
n12
2
Writing x=x0sin we get
(39) and
x
x0 sin
(40)
z
z0
n2 x
n12 1 2
n22
n12 1 2
x0 sin z
(42)
n1=1.5,
= 0.01 ,
a=20 m.
n2 1.485, =1.0607 104 m
Typical ray paths for different values of 1 are shown in Fig. 3.23.
We may write
n2
n1
Guided rays
n2
,
x
a core
,
x
a cladding
(43)
(45)
We obtain
2
2
(44)
In a typical parabolic index fiber,
We could have also used Eq. (28) to obtain the ray path.
In an practical optical wave-guide the refractive index
distribution is usually written in the form:
x
a
n1 2
.
a
(41)
Choose the origin such that z0= 0, we obtain
x
The region |x|<a is known as the core of the waveguide and the
region |x|>a is usually referred to as the cladding. Thus
Refracting rays
(46)
In Fig. 3.23, the ray paths shown correspond to z0=0 and 1=4,
8.130 and 20 ; the corresponding values of are pproximately 1.496
(>n2), 1.485 (=n2) and 1.410 (<n2)—the last ray undergoes
refraction at the core-cladding interface.
the periodical length zp of the sinusoidal path is given by
zp
1 = 20
Cladding
0.02
0.02
1
Core
= 8.13
0
1
=4
zp
-0.02
-0.02
1
0
z (mm)
Fig. 3.23 Typical ray paths in a parabolic index medium
for parameters given by Eq.(45) for 1=4 , 8.13 and 20 .
2
2 a cos
2
1
.
(47)
In Fig. 3.23, the values of zp are 0.8864 mm and 0.8796 mm
with 1=4 and 8.13 , respectively. In the paraxial approximation,
cos 1 1 and all rays have the same periodic length. In Fig. 3.25,
typical paraxial ray paths are plotted for rays launched along the
z axis. Different rays (shown in the figure) correspond to
different values of .
Remarks on the Features:
(i) In the paraxial approximation (
n1) all rays launched
horizontally come to a focus at a particular point. Thus the
medium acts as a converging lens of focal length given by:
f
2 a cos
1
lim
0
4 1
2
1
a
.
2 2
(48)
(iii) Rays launched at different angles with the axis (see, for
instance, the rays emerging from point P) get trapped in the medium
and hence the medium acts like a 'guide'. Indeed such media are
referred to as optical waveguides and their study forms a subject of
great contemporary interest.
(iii) Ray paths would be allowed only in the region where is less
than or equal to n(x) [see Eq. (27)]. Further, dx/dz would be zero
(i.e., the ray would become parallel to the z axis) when n(x)
equals .
(iv) The rays periodically focus and defocus as shown in Fig. 3.25.
Physically, although the ray PLQ traverses a larger path in
comparison to PMQ, it does so in a medium of ‘lower’ average
refractive index—thus the greater path length is compensated for by
a greater ‘average speed’ and hence all rays take the same time to
propagate through a certain distance of the waveguide. It is for this
reason that parabolic index waveguides are extensively used in
fiber-optic communication.
Gradient-Index (GRIN) lenses, characterized by parabolic
variation of refractive index in the transverse direction, are now
commercially available and find many applications (see Fig.
3.24). For example a GRIN lens can be used to couple the output
of a laser diode to an optical fiber; the length of such a GRIN lens
would be zp/4 (see Fig. 3.25); typically zp few cm and the
diameter of the lens would be few millimeters. Such small size
lenses find many applications. Similarly, a GRIN lens of length
zp/2 can be used to transfer collimated light from one end of the
lens to the other.
x ( m)
0.02
20
L
M
0
Q
P
-0.02
-20
0
Fig. 3.24 A gradient-index lens with a parabolic variation of
refractive index. The lens focuses light in a way similar to a
conventional lens; figure adapted from
http://en.wikipedia.org/wiki/Refractive_index.
1
zp
z (mm)
0
1
Fig. 3.25 Paraxial ray paths in a square law medium. Notice the periodic
focussing and defocussing of the beam.
3.4.2 Transit Time Calculations in a Parabolic Index Waveguide
As shown in Sec. 3.4.1, the ray path through a parabolic
index waveguide as described by Eq. (36) (inside the core) is
given by
x
x0 sin z ,
ds
,
c/n x
n12
c
z
,
1 2
n x dz
c
1
d
n12
c
1 2
n1
=
c
z
When
=n1,
z
z
.
c / n1
(53)
as expected. For large values of z, the second term on the RHS of
Eq. (52) would make a negligible contribution to z(z) and we
may write
n12
1
(54)
z
z
2c
If a pulse of light is incident on one end of the waveguide, it
would in general excite all rays and since different rays take
different amounts of time, the pulse will get temporally bro
adened. For a parabolic index waveguide, this broadening will be
given by
n2
n1
z n1 n2
2c
n2
2
zn2
2c
2
. (55)
1
n12
c
2
2
x 2 dz
x02 sin z dz
(50)
The time taken by the ray to traverse a distance z along the
waveguide is given as
(50)
where c is the speed of light in free space. Since
dz
n x
ds
d
(49)
where x0 and have been defined through Eqs (39) and (40).
Let d represent the time taken by a ray to traverse the arc
length ds [see Fig. 3.12 (b)]:
d
we may write Eq. (50) as
We have assumed
z
0
2
dz
x02
1 cos 2 z
dz
0
2
z
c
2
2
1
2c
2
0
x
x02 1
sin 2 z
c 2
z
2
1
n
2
n1 n2
2n12
z
2
n12
4c
2
sin 2 z
n1 n2
.
n2
(52)
(56)
For the fiber parameters given by Eq. (45), we get
0.25 ns/km.
Refer to Example 3.9 [p. 3.16]
3.4.3 Reflections from the Ionosphere
The ultraviolet rays in the solar radiation results in the ionization
of the constituent gases in the atmosphere resulting in the
formation of what is known as the ionosphere (The ionization is
almost negligible below a height of about 60 km).
2
1
Because of the presence of the free electrons (in the ionosphere),
the refractive index is given by (Eq. (76), Chapter 7):
= 60
x (m)
1
1
= 45
n2 x
= 30
Ne x q2
.
m 0 2
= 20
cos
-2
-2
0
2
(63)
Thus as the electron density starts increasing from 0 the
refractive index starts decreasing and the ray paths would be
similar to that described in Example 3.9. If nT represents the
refractive index at the turning point (see Fig. 3.27) then
0
1
1
4
1
nT .
(64)
6
z (m)
Fig. 3.26 Parabolic ray paths (corresponding to 1 = 20 , 30 ,45 and 60 ) in
a medium characterized by linear refractive index variation in the region
x > 0 [see Eq.(29)]. The ray paths in the region x < 0 are straight lines.
Therefore the short wave radio broadcasts (
20m) sent at a
particular angle from a particular city (say London) would reach
another city (say New Delhi) after undergoing reflection from the
ionosphere.
For normal incidence,
m 0 2
.
(65)
N e xT
q2
Thus if an electromagnetic signal is sent from the point A (at an
angle 1) is received at the point B, one can determine the
refractive index (and hence the electron density) of the
ionospheric layer where the beam has undergone the reflection.
F region
180 km
100 km
4 1011 m-3
(Ne) max
2.6 1011 m-3
T
In a typical experiment, an electromagnetic pulse (of frequency
between 0.5 to 20 MHz) is sent vertically upwards and if the echo
is received after a delay of t seconds, then
2h
t
.
(66)
c
h is the height at which it undergoes reflection. Thus if
electromagnetic pulse is reflected from the E layer of ionosphere
(which is at a height of about 100 km), the echo will be received
after about 670 s.
(Ne) max
E region
A
1
B
Fig. 3.27 Reflection from the E region of the ionosphere. The point T
represents the turning point. The shading shows the variation of the
electron density.
In Fig. 3.28 is a plot of the frequency dependence of the
equivalent height of reflection (as obtained from the delay time of
echo) from the E and F regions of the ionosphere. From the figure
we find that at =4.6 106 Hz, echoes suddenly disappear from the
100 km height. Thus,
N e 100 km
9.11 10
31
m
0
2
q2
2
8.854 10 12 (2
(1.6 10 19 ) 2
Equivalent height
(km)
Alternatively, by measuring the delay t, one can determine the
height (at which the pulse gets reflected) from the following
relation
2 t
h
.
(67)
2c
300
200
100
0
1
2
3
4
5
6
7
8
9
10
Frequency of the exploring waves in MHz
4.6 106 ) 2
Fig. 3.28 Frequency dependence of the equivalent height of reflection
from the E and F regions of the ionosphere.
2.6 1011 electrons/m3 .
If we further increase the frequency, the echoes appear from the
F region of the ionosphere.
Refer to Example 3.10 [p. 3.18]
3.5 Refraction of Rays at the Interface between an Isotropic
Medium and an Anisotropic Medium
In an isotropic medium the properties remain the same in all
directions; typical examples are glass, water and air. On the other
hand, in an anisotropic medium, some of the properties (such as
speed of light) may be different in different directions. When a
light ray is incident on a crystal like calcite, it (in general) splits
into two rays known as ordinary and extraordinary rays. The
velocity of the ordinary ray is the same in all directions. Thus the
ordinary ray obeys Snell's laws but the extraordinary ray does not.
In a uniaxial homogeneous medium, the refractive index
variation for the extraordinary ray is given by [see Eq. (121) of
Chapter 22]
n2
n02cos2 +ne2sin 2 .
(77)
where no and ne are constants of the crystal and
the angle that the ray makes with the optic axis.
represents
3.5.1 Optic Axis Normal to the Surface
Consider the particularly simple case of the optic axis being
normal to the surface. Referring to Fig. 3.29, the optical path
length from A and B is given by
Lop
n1 h12
L x
2
n
h22
x2 .
(78)
n1: the refractive index of medium I
*The incident ray, the refracted ray and the optic axis are coplanar.
Since
A
h2
cos
2
2
h
n1
h1
x
2
2
2
h
x
2
,
We obtain
i
I
n1 h12
Lop
x
P
L x
2
no2h22
For the actual ray path, we require
II
x
and sin
Optic Axis
L-x
h2
dx
h12
B
Fig. 3.29 The direction of the refracted extraordinary ray when the optic axis
(of the uniaxial crystal) is normal to the surface.
no n1 sin i
ne ne2
n12 sin i
.
(81)
As a simple example, let the first medium to be air so that n1= 1.
Then
no sin i
ne ne2
n12 sin i
.
(82)
2
no2h22
ne2 x 2
n2 sin r.
ne tan r
n1 sin i
2
o
(80)
ne2 tan 2 r
n
[Snells's Law]
Consider a more general case of the optic axis making an angle
with the normal. We further assume the optic axis lies in the
plane of incidence as shown in Fig. 3.30. In general, in an
anisotropic medium, the refracted ray does not lie in the plane of
incidence. It can be shown that if the optic axis lies in the plane
of incidence then the refracted ray also lies in the plane of
incidence. The optical path length from A to B (see Fig. 3.30) is
given by
Lop
Assume the second medium to be calcite, then no=1.65836, and
ne=1.48641. Thus for i=45°, we get r 31.1°. If no=ne= n2 then
Eq. (80) simplifies to
n1 sin i
L x
3.5.2 Optic Axis in the Plane of Incidence
tan r =x/h2.
This gives the angle of refraction for a given angle of incidence.
tan r
0,
or
Q
tan r
(79)
ne2 x 2
n1 L x
thus
c / no
r: angle of refraction
We find
dLop
c / ne
ne2 x 2 .
(83)
n1 h12
Since =r
, we have
n2
n02cos2 r
L x
2
n
h22
x2 .
(84)
+ne2sin 2 r
=
n02 cos r cos
sin r sin
2
+ne2 sin r cos
cos r sin
2
2
2
0
h2
=n
2
2
h
x
2
x
cos +
2
2
h
x
For the actual ray path, we require
sin
2
dx
thus
2
2
e
x
n
2
2
h
x
2
h2
cos -
2
2
h
x
2
sin
L x
n02 h2 cos
2
x sin
no2 h2 cos
n
h22
x2
n02 h2 cos
x sin
2
ne2 x cos
h2 sin
2
n1 h12
n1 sin i
L x
n02 h2 cos
x sin
x sin
2
ne2 x cos
h2 sin
h2 sin
cos
2
n02 cos sin +ne2 sin cos
no2 cos2
ne2 sin 2
(87)
2
2
ne2 x cos
h2 sin
2
.
For given values of the angles i and , the above equation can be
solved to give the values of and hence the angle of refraction r
(= + ).
(86)
Remarks on Several Particular Cases
A
I
sin +ne2 x cos
or
(85)
and
Lop
0,
n1 L x
h12
Thus
1
dLop
h1
(i) When no=ne=n2, the anisotropic medium becomes isotropic
and Eq. (87) simplifies to
x
i
n1
P
S1
n( )
II
L-x
r
h2
B
S2
n1sin i =n2sin( + )=n2sin r
which is nothing but Snell's law.
(ii) When = 0, i.e., the optic axis is normal to the surface, Eq.
(87) becomes
ne2 sin
n1 sin i
no2 cos2
ne2 sin 2
Optic axis Q
L
Fig. 3.30 The direction of the refracted extraordinary ray when the optic axis (of
the uniaxial crystal) lies in the plane of incidence making an angle with the
normal to the interface.
=
ne2 sin r
no2 cos2 r ne2 sin 2 r
(88)
where we have used the fact that r = . The above equation is
identical to Eq. (80).
(iii) Finally, we consider normal incidence, i.e., i =0. Thus, Eq.
(87) gives us
n02 cos sin +ne2 sin cos
0
or
n02 cos r
sin +ne2 sin r
cos
0
I
or
II
2
0
cos r n cos sin
2
e
n sin cos
2
0
sin r n sin
2
2
e
n cos
2
0
or
tan r
ne2 n02 sin cos
n02sin 2
(89)
ne2 cos 2
Optic axis
Equation (89) shows that in general r
0 (see Fig. 3.31).
For normal incidence, the above analysis is valid for an
arbitrary orientation of the optic axis; the refracted (extraordinary) ray lies in the plane containing the normal and the optic
axis. Furthermore, for normal incidence, when the crystal is
rotated about the normal, the refracted ray also rotates on the
surface of a cone.
Eq. (89) tells that when the optic axis is normal to the surface
( =0) or when the optic axis is parallel to the surface but lying in
the plane of incidence ( = /2), r=0 and the ray goes un-deviated
(see Fig. 3.31).
Fig. 3.31 For normal incidence, in general, the refracted extraordinary ray
undergoes finite deviation. However, the ray proceeds undeviated when the
optic axis is parallel or normal to the surface.
S
Q
P
C
O
x
r
y
M
Fig. 3.32 Paraxial image formation by a concave mirror.
S
n1
n2
S
P
Q
C
y
O
x
Q
O
C
r
P
y
r
x
M
M
Fig. 3.33 Paraxial image formation by a convex mirror.
Fig. 3.34 Paraxial image formation by a concave refracting surface SPM.
O
n1
Q
B
R
n2
A
C
O
C
Fig. 3.35 All rays parallel to the major axis of the ellipsoid of
revolution will focus to one of the focal points of the ellipse provided
the eccentricity = n1/ n2 .
P1
C
P2
Fig. 3.36 A spherical reflector.