optics interference of light: many theories were put forward to … · 2019. 6. 18. · distances...
TRANSCRIPT
1
OPTICS
Interference of Light:
Many theories were put forward to explain the nature of light:
Newton’s corpuscular theory
Huygens’ wave theory
Electromagnetic theory
Quantum theory
The above-mentioned theories explain that some of the
properties like refraction, reflection, interference etc. can be
explained by considering the light as a wave while the
phenomenon like photoelectric effect, Compton Effect etc. may
be described by considering the light as particle.
Interference:
It is the phenomenon of superposition of two coherent*
waves in the region of superposition.
At some points in the medium, the intensity of light is
maximum while at some other points the intensity is
minimum.
The positions of maximum intensity are called maxima,
while those of minimum intensity are called minima.
Principle of Superposition:
In a medium, the resultant displacement of a particle acted upon
by two or more waves simultaneously is equal to the algebraic
sum of the displacements of the particle due to individual
waves.
Thus if the displacement due to a single wave train at a point be
y1 and y2 be the displacement due to the another wave train in
the same direction, then the resultant displacement y may be
written as,
2
21 yyy
If, tay sin11 and tay sin22
Where δ is the phase difference between two waves.
tatay sinsin 21
tataa cossinsincos 221
Let coscos21 Aaa and sinsin2 aa ; we have,
tAtAy cossinsincos
tAsin
Where, 2/1
21
2
2
2
1 cos2 aaaaA
and
cos
sintan
21
2
aa
a
The intensity at any point is proportional to the square of the
amplitude, i.e. 2AI , then
cos2 21
2
1
2 aacaAI …………(1)
Since 2
1a and 2
2a are the intensities of the two interfering waves,
the resultant intensity at any point is not just the sum of the
intensities due to the separate waves 2
2
2
1 aa .
Condition for maximum and minimum intensities:
From (1) it is clear that I will be maximum at points where
.........2,1,0;21cos nnor
Therefore, the intensity will be maximum when phase
difference is even multiple of , but we know that,
Path difference
2phase difference
3
n2
2 or 2
2
n
Therefore, 2
21max aaI
Similarly I will be minimum at points where cos δ=-1and
therefore,
12 n
2
12
n n=0,1,2……….
And, 221min aaI
*Coherent Sources: The light waves of practically the same
intensity, of exactly the same wavelength and of exactly a
constant initial phase difference are known as Coherent
Sources. For Interference of light to take place the two light
waves must have the same plane of polarization. Two light
waves from independent sources can never be coherent. Thus a
pair of coherent waves may be obtained from a single source of
light.
Young’s Double Slit experiment: In 1801, Young first
demonstrated experimentally the phenomenon of interference.
4
In his experiment light from a monochromatic source is
allowed to fall on a slit S and then through a double slit S1
and S2.
The interference pattern was observed on a screen XY
where few dark and bright bands or fringes are observed.
These fringes are of equal distances.
The bold lines show the amplitude in positive direction
(crest) where the dotted lines show the amplitude in the
negative direction (troughs).
At points where crests fall on crests or troughs fall on
troughs the amplitudes add up and the resultant intensity
increases (I α A2) and thus we obtain constructive
interference.
At points where crests fall on the troughs or the troughs
fall on the crests, the amplitudes are reduced (the resultant
amplitude will be zero if the two lights from S1 and S2 have
equal amplitudes and so the intensity is also zero). This is
known as destructive interference.
S1
S2
S
I
X
Y
5
Thus on the screen alternate bright and dark regions of
equal width, called interference fringes are observed.
Analytical treatment for Young’s double Experiment:
Let S1and S2 be two coherent sources separated by a
distance d and made from a monochromatic source S.
Let a screen be placed at a distance D from the coherent
sources.
O is a point on the screen which is equidistant from S1and
S2.
Therefore, path difference at O will be zero and the
intensity at O will be maximum.
Let P be a point on the screen such that,
OP= y
Therefore,
2,
2
dyPR
dyPQ
S1
S2
d
P
R
C O
d/2
d/2
D
Q y
6
D
dy
Dd
yDPS2
2,
2
2
2/12
2
1
and D
dy
DPS2
2
2
2
The path difference of two light rays emerging from S1and S2
and reaching to a point P on the screen is,
D
ydPSPS 12
Therefore the phase difference,
2path difference=
D
yd
2
Position of bright fringes: As mentioned earlier that for
bright fringes path difference is even multiple of λ/2, i.e.
nnD
yd
22
Thus the distance of nth bright fringe from point O is given by:
d
Dnyn
.
Position of dark fringes: For dark fringe or the minimum
intensity at P, the path difference must be an odd multiple
of half wavelength, i.e.
2
12
nD
yd
7
Thus the distance of the nth dark fringe from point O is given
by,
d
Dnyn
2
12
.
Fringe Width: The distance between any two consecutive
dark fringes or bright fringes is same and known as fringe
width. It is given by:
d
Dyy nn
1
Conclusions: 1. There is a bright fringe at the centre of the screen.
2. Alternatively dark and bright fringes on both the sides
of the sides of the central maximum occur.
3. Fringe widths of dark and bright fringes are same.
4. Fringe width depends upon the wavelength of light,
separation of sources and source to screen separation.
Condition for Sustained Interference: To obtain a well
defined and observable interference the condition for
interference of light are given as under:
The beams from two sources must propagate along the same
line; otherwise the vibrations cannot be along a common line.
The two sources must be very narrow because a broad source
is equivalent to a no. of sources and so positions of darkness
and brightness cannot be well defined.
The two sources must have equal intensities.
The two sources should have equal amplitudes and
frequencies. In case of unequal frequencies the phase
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difference between the interfering waves will vary
continuously and as a result, intensity at any point will not be
constant but vary with time. If the amplitudes are different
then a clear cut distinction between maximum (a1+ a2) and
minimum (a1- a2) is not possible.
The common original source must be monochromatic, i.e.,
emitting light of single wavelength.
The light from the two sources must have either zero phases
or a constant difference of phase. This condition is very well
observed in the case of coherent sources.
The separation between source and screen should be large to
get wide fringes.
If the interference pattern has to be obtained by polarised
light then the polarised waves must be in the same state of
polarization.
Two Classes of Interference The interference is divided into
two classes on the basis of the way of obtaining the two
coherent sources:
1. Division of wavefront:
In this case the wavefront originating from one source is
divided into two parts. This division can be made with the
help of any optical system like Fresnel’s biprism, Fresnel’s
S1
S
S2
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mirror etc. The two wavefronts after traveling certain
distances are brought together to give interference pattern.
2. Division of amplitude:
In this process the amplitude is divided into two or more parts
either by partial reflection or refraction and which after
traveling different paths produce interference at a point.
Newton’s rings, Michelson interferometer are the examples of
this class.
Fresnel’s Biprism: However Young demonstrated the
phenomenon of interference, but it was doubted that the fringes
are not due to interference of light waves but due to some
modification of light at the slits. These doubts are removed by
the Fresnel’s biprism experiment. By using biprism, Fresnel
obtained two coherent sources from a single source by
refraction.
Construction: It consists of two acute angle prisms with their
bases in contact forming a single obtuse angle prism ABC. In
actual practice the prism is grounded from a single optically
S Reflected
System
Transmitted
system
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true glass plate. The obtuse angle of the prism is nearly 1790, so
that each of acute angles is nearly 30’.
Working:
A narrow vertical slit S is illuminated by a source of
monochromatic light.
The prism is placed with its refracting edge parallel to the
line joining the sources.
The light from the slit S is allowed to fall symmetrically on
the biprism ABC with its refracting edge vertical (i.e.,
parallel to the line source S).
When the light from S falls on the upper half of the prism,
after refraction it appears to come from virtual source S1.
Similarly after refraction from the lower half of the prism
it appears to be coming from virtual source S2.
The distance between the biprism and the source S is so
adjusted that the sources S1 and S2 are very close to each
other.
The interference fringes are observed in the region EF of
the screen.
S
S2
S1
d
a b
D
B
O’
A
C
G
E
H
O
F
Screen
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The analytical treatment is same as for Young’s Double
Slit experiment.
If d be the distance between the sources S1 and S2 and D
between the source and the screen, then the fringe width is
given by,
d
D
And thus
ba
d
D
d
Moreover the distance of n th bright fringe from O.
,)(d
Dny brightn
Similarly
d
Dny darkn
2
12)(
.
Where )1(2 ad
Applications: a) Determination of wavelength of light:
D
d
b) Determination of thickness of a thin sheet of transparent
material:
a) Biprism can be used to determine the thickness of a
given thin sheet of transparent material e.g., glass,
mica.
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b) Suppose A and B are two virtual coherent sources.
c) The point C is equidistant from A and B.
d) When a transparent plate G of thickness t and
refractive index µ is introduced in the path of one of
the beams, the fringe which was originally at C shifts
to P.
e) The time taken by the wave from B to P in air is the
same as the time taken by the wave from A to P partly
through air and partly through the plate.
f) Suppose c0 is the velocity of light in air and c its
velocity in medium then,
c
t
c
tAP
c
BP
00
t
c
ctAPBP 0
, But
c
c0
tttAPBP 1
If P is the point originally occupied by the n th fringe, then the
path difference
nAPBP
nt 1 ……..(i)
P t
d
x
D
C
A
B
G
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Also the distance x through which the fringe shifted
d
Dn
Where
d
D, the fringe width.
d
Dnx
Also, D
xdn
Or,
D
xdt 1
g) Therefore, knowing x, the distance through which the
central fringe is shifted, D, d and µ, the thickness of
the transparent plate can be calculated.
h) If a monochromatic source is used, the fringes will be
similar and it is difficult to locate the position where
the central fringe shifts after the introduction of the
transparent plate. Hence white light is used. The
fringes will be coloured but the central fringe is white.
Interference in Thin films: It has been observed that
interference in the case of thin films takes place due to (1)
reflected light and (2) transmitted light. Newton was able to
show the interference rings when a convex lens was placed on a
plane glass-plate.
Interference Due To Reflected Light(Thin films):
Consider a transparent film of thickness t and refractive
index µ.
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A ray SA incident on the upper surface of the film is partly
reflected along AT and partly refracted along AB.
At B part of it is reflected along BC and finally emerges
out along CQ.
The difference in path between the two rays AT and CQ
can be calculated.
CN is normal to AT and AM is normal to BC.
The angle of incidence is i and the angle of refraction is r.
CB and AE are extended to meet at P.
rAPC The optical path difference
ANBCABx
Here, CM
AN
r
i
sin
sin
CMAN .
CMBCABx .
CMPCCMBCABx .
PM.
In the ΔAPM,AP
PMr cos
or rEPAErAPPM coscos.
rt cos2 tEPAE
rtPMx cos.2. ……………(1)
This eq. (1), in the case of reflected light does not represent the
correct path difference but only the apparent. It has been
established on the basis of electromagnetic theory that, when
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light is reflected from the surface of an optically denser
medium (air-medium interface) a phase change equivalent to
a path difference 2
occurs.
Therefore, the correct path difference in this case,
2cos2
rtx ………..(2)
1) If the path difference x = n where n = 0,1,2,3….etc.,
constructive interference takes place and the film appears
bright.
nrt 2
cos2 or, 2
12cos2
nrt …..(3)
2) If the path difference is 2
12
nx where n = 0,1, 2,
3….etc., destructive interference takes place and the film
appears dark. 2
122
cos2
nrt
or, 1cos2 nrt ………….(4)
Here n is an integer only, therefore (n+1) can also be taken
as n.
nrt cos2 ………….(5)
Where n = 0,1, 2, 3……………etc.
i
i
r
S T Q
C A
E
P F
M
N
B
r
AIR
AIR
µ t
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Interference Due To Transmitted light (Thin Film):
Let there be a thin transparent film of thickness r and
refractive index μ.
A ray SA after refraction goes along AB.
At B it is partly reflected along BC and partly refracted
along BR.
The ray BC after reflection at C, finally emerges along DQ.
Here at B and C reflection takes place at the rarer medium
(medium-air interface).
Therefore, no phase change occurs.
BM is normal to CD and DN is normal to BR.
The optical path difference between DQ and BR is given
by
BNCDBCx
also MD
BN
r
i
sin
sin or MDBN .
from fig. rBPC and CP = BC = CD
t
B
R Q
i N
D M
C
P
A
AIR
AIR
μ r
i
r
r r r
i
S
17
PDCDBC
PMMDPDMDPDx
In the BP
PMrBPM cos, or rBPPM cos.
But, tBP 2
rtPM cos2 rtPMx cos2.
1) For bright fringes, the pat difference nx
nrt cos2
where n = 0, 1, 2, 3...etc.
2) For dark fringes, the path difference 2
12
nx
2
12cos2
nrt
where n = 0, 1, 2, 3...etc.
In the case of transmitted light, the interference fringes
obtained are less distinct because the difference in amplitude
between BR and DQ is very large.
Important points: When the film is exposed to white light
then only those wavelengths will be present for which the
condition of maxima is satisfied or which have the path
difference satisfying the condition for bright fringe.
Newton’s Rings: When a Plano convex lens of long focal
length is placed on a plane glass plate, a thin film of air is
enclosed between the lower surface of the lens and the upper
surface of the plate. The thickness of the air film is very small
at the point of contact and gradually increases from the centre
outwards. The fringes produced with the monochromatic light
are circular. The fringes are concentric circles, uniform in
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thickness and with the point of contact as the centre. These
fringes are known as Newton’s Rings.
Construction:
S is a source of monochromatic light at the focus of the
lens L1.
A horizontal beam of light falls on the glass plate B at 450.
The glass plate B reflects a part of the incident light
towards the air film enclosed by the lens L and the plane
glass plate G.
The reflected beam from the air film is viewed with a
microscope.
Working:
Interference takes place and dark and bright circular
fringes are produced.
L
G
AIR FILM
M
S
L1
B 45
0
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This is due to the interference between the light reflected
from the lower surface of the lens and the upper surface of
the glass plate G.
Theory: 1. Newton’s ring by reflected light: Let the radius of curvature
of the lens is R and the air film is of thickness t at a distance
OQ= r, from the point of contact O.
Here interference is due to reflected light. Therefore for the
bright rings
2
12cos2
nrt ….(i) where n = 1,2,3……….etc.
For normal incidence r = 00 therefore, cos r = 1
For air, μ = 1
Hence 2
122
nt …..(ii)
For the dark rings, nrt cos2
Or, nt 2 ……(iii) where n = 0,1,2,3…….etc.
From fig. in ΔCEP 222 EPCECP
Or 222 EPOECOCP
L
G
REFLECTED
LIGHT
AIR FILM
C
R
O Q
H E P
rn
t
20
or 222
nrtRR
or Rtrn 22 (ignoring t2)
hence R
rt n
2
2
substituting the value of t in (ii) and (iii) we get,
for bright rings
2
122 Rnrn
……..(iv)
or
2
12 Rnrn
………..(v)
for dark rings Rnrn 2……..(vi)
or Rnrn ………..(vii)
Result:
The radius of nth dark ring is proportional to
i. n
ii.
iii. R
Similarly the radius of nth bright ring is proportional to
i. 2
)12( n
ii.
iii. R
If Dn is the diameter of the nth dark ring,
RnrD nn 42 ……..(viii)
If Dn is the diameter of the nth bright ring,
21
2
1222
RnrD nn
…..(ix)
2. Newton’s ring by transmitted light:
In case of transmitted light the interference fringes are
produced such that for bright rings, nrt cos2 ………(x)
For air, μ = 1and cos r = 1 then nt 2 ……….(xi)
For dark rings,
2
122
nt ………(xii)
Hence in this case the radius of nth bright is
Rnrn …….(xiii)
And the radius of the nth dark ring is
2
12 Rnrn
…….(xiv)
Applications of Newton’s Ring:
L
G
AIR FILM
TRANSMITTED
LIGHT
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Determination of wavelength of light: For nth dark ring we
have
RnDn 42 ……..(i)
or nR
Dn 4
2
…….(ii)
again for (n+p)th dark ring,
pnR
D pn
4
2
……(iii)
pR
DD npn
4
22
……(iv)
Determination of Refractive Index of a liquid: The
refractive index of a liquid, forming film between lens and
glass plate, can be obtained from the following eq.
pR
DDliquidnpn
4
22
………..(i)
pR
DDairnpn
4
22
Since for air μ= 1
Hence
pR
DDairnpn
4
22
……..(ii)
Dividing (ii) by (i) we get
liquidnpn
airnpn
DD
DD
22
22
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Newton’s Rings with white light: In case of white light the
diameter of the rings of different colours will be different and
there will be an overlapping of the rings of different colours
over each other. The only first few rings will be clear while
other rings cannot be viewed.
Michelson’s Interferometer: Principle: The amplitude of light beam from a source is
divided into two parts of equal intensities by partial reflection
and transmission. These beams are sent in two directions at
right angles and are brought after they suffer reflection from
plane mirrors to produce interference fringes.
Construction:
A
G1 G2
M1
M2
T
S
C
B Transmitted
Reflected
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It consists of two highly polished plane mirrors M1 and M2
two plane parallel glass plates G1 and G2 of exactly the same
thickness.
The plate G1 is semi silvered at the back so that the
incident beam is divided into reflected and transmitted
beams of equal intensities.
M1 and M2 are mutually at right angles to each other.
The plates G1 and G2 are parallel to each other and are at an
angle of 450 to M1 M2.
The mirrors M1 and M2 can be adjusted exactly perpendicular
to each other with the help of the screws on their backs.
The mirror M1 can be moved exactly parallel to itself with a
carriage on which it is moved.
Working:
The light falling on the glass plate G1 is partly reflected and
partly transmitted.
The reflected ray AC travels normally towards plane mirror
M1 reflected back to the same path and comes out along AT.
The transmitted ray after reflection from mirror M2 follows
the same path and then moves along AT after reflection from
the back surface of G1.
So the rays received at T are produced from a single source
by the division of amplitude and we get an interference
pattern.
The reflected ray AC travels through glass plate G1 twice and
in the absence of plate G2 the transmitted ray does not cross
plate G1.
To compensate it a second plate G2 is introduced in the path
of transmitted ray.
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Adjustment:
Distance of the mirrors M1 and M2 are adjusted to be nearly
equal from glass plate G1.
In order to obtain a parallel ray of light a lens is adjusted
between the source S and plate G1.
Formation of fringes:
1. Circular fringes:
If eye is placed to see mirror M1 directly then in addition it
will see virtual image of M2 formed by reflection in glass
plate G1.
Thus one of the interfering beams cones from M1 and other
appears to come from the virtual image of M2 i.e. M2’.
If M2 is exactly at right angles to M1 and the plate G1 is at
450 to each of then, the air film formed between M1 and
M2’ will have uniform thickness.
Let the two interfering beams appear to come to the eye
from two virtual images S1 and S2 of the original source
and let 2d be the distance between them.
Now since reflected ray from M2 suffers reflection at the
silvered surface of plate G1, an additional path difference
λ/2 is introduced.
Therefore, the total path difference 2/cos2 d .
Therefore, for brightness,
.....2,1,0,2/cos2 etcnnd
If M2’coinsides with M1 then total path difference will be
λ/2 which is the condition for destructive interference.
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If the light is obtained from an extended monochromatic
light source then for a given value of n, the value of α is
constant and so the locus of the fringes is circle.
2. Localized fringes:
If M1 and M2’ are not exactly parallel the path difference is
very small; we observe fringes as in the case of a wedge
shaped film.
In this case the locus of points of equal thickness is a
straight line parallel to the edge of the wedge.
M1
M2’
2d α
S
S1
S2
2d cos α
27
For small path difference, the fringes are nearly straight but
for larger path difference the fringes are generally curved
with the convex surface of the film towards the edge of the
wedge.
Important points: 1. As the circular fringes are formed due to two parallel
interfering waves, hence they are formed at infinity.
2. Here the circular fringes are formed at same inclination
hence they are called equal inclination fringes or Haidinger
fringes.
Applications:
1. Determination of wavelength of Monochromatic light:
Mirror M1 and M2 are adjusted so that circular fringes are
visible in the field of view.
If M1 and M2 are equidistant from the glass plate G1, the
field of view will be perfectly dark.
The mirror M2 is kept fixed and the mirror M1 is moved
with the help of the handle of the micrometer screw and
the number of fringes that cross the field of view is
counted.
M1
M2’ M1
M2’
M1
M2’
28
If mirror is moved through a distance of d for a
monochromatic light of wavelength λ and no. of fringes
that cross the centre of the field of view is n then 2
nd ,
because for one fringe shift the mirror moves through a
distance equal the half the wavelength. Hence λ can be
determined.
2. Determination of the difference in wavelength between
two neighbouring spectral lines:
Set the apparatus for circular fringes.
Let the source has two wavelengths λ1 and λ2 (λ1> λ2)
which are very close to each other.
The two wavelengths form their separate fringe patterns,
but because of very small change in wavelength, the two
patterns overlap.
As the mirror M1 is moved slowly the two patterns separate
out slowly and when the path difference is such that the
dark fringe of λ1 falls on the bright fringe of λ2 maximum
indistinctness occurs.
Again when the path difference is such that the bright
fringe of λ1 falls on bright fringe of λ2 maximum two
successive positions of maximum distinctness occurs.
Let the mirror M1 is moved through a distance of d
between distinctness.
This will happen only when nth
fringe of λ1 coincides with
n+1 th
fringe of λ2 then
21 )1(2 nnd
1
2
dn and
2
21
dn
29
Subtracting we get,
12
221
dd =
21
212
d
Or d2
2121
21 hence 2
21
hence d2
2
21
Thus by measuring the distance d moved by M1 the
difference between two wavelengths can be determined.
3. Determination of thickness of a thin plate:
The apparatus is set for parallel fringes.
White light source is used instead of monochromatic
source.
The cross wire is set on the central fringe.
Now the thin plate whose thickness is to be determined is
introduced in the path of one of the interfering beams.
Due to the insertion of the plate path difference is changed
to 2(μ-1)t.
Thus a shift in the fringe system occurs.
Now the mirror M1 is moved till the central fringe
coincides with the cross wire.
The distance x moved by mirror M1 is measured using
micrometer screw. Hence we have,
tx )1(22
30
1
xt
Using above equation t can be determined.
Diffraction: The bending of light beam round the corners or
edge and spreading of light into the geometrical shadow of an
opaque obstacle placed in its path is known as diffraction. In
other words the diffraction may be defined as the encroachment
of light into the geometrical shadow region of small opaque
obstacle or aperture placed in the path of light.
Types of diffraction: There are two classes of diffraction:
1. Fresnel Diffraction
2. Fraunhofer Diffraction
1. Fresnel Diffraction:
In this case diffraction is considered to take place from
different parts of the aperture.
Here the source or screen or both are at finite distance from
obstacle.
In this class the incident wave front is divergent, either
spherical or cylindrical.
S
A
B
A’
B’
31
Here the observed pattern is a projection of diffracting
element modified by diffracting effects and the geometry
of the source.
Here the diffraction centre of the pattern may be either
bright or dark.
2. Fraunhofer Diffraction:
Here the source of light and screen are at infinite distance
from the obstacle.
Here the inclination is important.
Here the incident wave front is generally plane and the
centre of diffraction pattern is always bright.
Fraunhofer Diffraction At A Single Slit:
The light from a monochromatic point source S is
converted into parallel beams of light by a convex lens
L1L1’.
Now this beam is incident normally on a slit AB of width e.
θ
θ θ
θ
θ
e
A
θ
B
θ
B’
θ
A’
θ
P
θ
L2
θ
L2’
θ
W1
θ
W1’ θ
W2
θ
W2’
θ
L1’
L1
W
W’
O
Y
Y’
K
32
The incident light wave fronts are shown by W1W1’ and
W2W2’.
Every point on the wave front incident on the slit and lying
within the width of the slit emits secondary waves which
superimpose to give diffraction pattern on the screen YY’.
In this diffraction pattern a central bright band is obtained.
On either side of this band alternate dark and bright bands
of decreasing intensity are observed.
Analysis: AK is perpendicular from A on BB’. The path
difference of the rays meeting at P is clearly BK. Here,
sinsin eABBK Therefore, the corresponding phase difference
sin
22eencepathdiffer
If the slit is assumed to be made up of n equal parts, the
amplitude of the secondary waves originating from these parts
will be equal. Let the amplitude of each secondary wave be ‘a’
and then the phase difference between any two successive
waves will be
)(,sin21
sayden
Thus the resultant in a direction θ, due to the superposition of
these n secondary waves can be calculated in the following
way:
Let there be n-harmonic waves having equal amplitudes ‘a’ and
common phase difference ‘δ’ between successive vibrations. To
find the resultant we construct a polygon, the closing side of
which gives the resultant amplitude R and a resultant phase Ф.
33
Now resolving the amplitudes parallel and perpendicular, we
get,
)1....(..........)1cos(.........2coscoscos naaaaR
)2....(..........)1sin(.........2sinsinsin naaaR
Multiplying eq. (1) by 2sin
2
,we
)1cos(
2sin2.........2cos
2sin2cos
2sin2
2sin2
2sincos2 naR
From the trigonometrical relation )sin()sin(sincos2 BABABA we have,
]2
3sin
2
1sin...
2
3sin
2
5sin
2sin
2
3sin
2sin2[
2sincos2
nn
aR
2
1cos
2sin2
2
1sin
2sin
nna
na
R
a
a
a
1δ
2δ
3δ
a
Ф
O
34
or
2sin
2
1cos
2sin
cos
nna
R …………(3)
similarly, multiplying (2) by 2sin2
and applying
)cos()cos(sinsin2 BABABA
and 2
sin2
sin2coscosCDDC
DC
we get,
2sin
2
1sin
2sin
sin
nna
R……..(4)
from (3) and (4) we get
2sin
2sin
na
Rand
2
1
n
Hence for the present case
n
e
ea
en
en
n
a
d
nda
Rsin
sin
sinsin
2
sin21
sin
2
sin21
sin
2sin
2sin
Let
sinethen,
35
n
a
n
aR
/
sin
/sin
sin
As n/ is small,
nn
sin
Again since ‘n’ is infinitely large and ‘a’ is negligibly small,
Let na =A, a finite no.
Hence,
sinAR ...........(5)
The secondary waves diffracted by the slit in a direction θ are
focused by the lens L2L2’ at a point P on the screen. Therefore,
the resultant intensity at P, 2RI . i.e.
2
22 sin
AI ........(6)
Conditions for maxima and minima:
0d
dI
or 0sin
2
22
A
d
d
or 0sincossin2
2
2
A .....(7)
Case I: When 0sin
or sin α =0 or n ( o , if α
= 0 then (sin α)/α = indeterminate)
Therefore, .......3,2,1,0 nn
or
n
e
sin
or ne sin ..........(8)
36
which gives the condition for nth minima.
Case II: The second bracketed term in eq.(7) will give the
condition for maxima.
Thus 0sincos
2
or tan ..............(9)
The intersection of the curves tany and y will give
the solution of eq.(9). The points of intersection will give the
values of α as
,.......2
7,
2
5,
2
3,0
Here α = 0 corresponds to central maximum.
Hence for central maximum the intensity is given by,
0
2
2
22 sinIA
AI
y=
tan
α
y=
tan
α
y=
tan
α
y = α
450
O
y
α
π/2 3π/2 5π/2
37
e e e e e e
38
Important Results:
1. The width of central maximum is directly proportional to
the wavelength of light used and as the wavelength of red
light is more than the wavelength of violet light hence the
width of the central is more for red light than for violet
light.
2. The width of the central maximum is inversely
proportional to the width of slit e hence width of central
maximum is greater for narrow slits.
Diffraction At Double Slit:
Let there be two slits AB and CD of equal width ‘e’,
separated by a distance‘d’.
The incident light gets diffracted from these two slits and
focused on the screen XY.
The diffraction by two parallel slits is a case of diffraction as
well as interference.
A
B
C
D
S1
S2
L L
P
O
P’
M
S θ
θ
θ
e+d
Y
Y’
39
Thus the pattern obtained on the screen consists of equally
spaced interference fringes due to both the slits and their
intensity being modulated by the diffraction phenomenon
occurring due to individual slits.
The two slits AB and CD may be considered equivalent to
two coherent sources S1 and S2.
Now due to individual slit, the resultant amplitude due to all
secondary waves diffracted in the direction θ is given by:
sinAR
..................(1)
Where,
sine
For simplicity, the resultant of all the secondary waves may
be taken as a single wave of amplitude R.
The resultant at point P on the screen will be the result of
interference between these two waves, of same
amplitudes
sinA, starting from S1 and S2.
The path difference between the wavelets from S1 and S2 in a
direction θ is, sin)2 deMS and therefore the phase
difference
sin
222 deMS .
The resultant amplitude at P may be given as,
cos2 21
2
2
2
1
2 RRRRRr
cos
sin2
sinsin222
AAA
40
2cos
sin4 2
2
22
A
Hence resultant intensity,
2cos
sin4 2
2
222
ARI rr
or
2
2
222 cos
sin4
ARr ...........(2)
where
sin
2de
From the above expression, it is clear that the resultant
intensity depends on the two factors.
(a). Diffraction term
2
22 sin
A :This factor is same as in the
case of single slit. Thus this term corresponds to
diffraction pattern due to secondary waves from the two
slits individually. This term gives a central maximum
having alternately minima and subsidiary maxima of
decreasing intensity on either side. The direction of
minima are given by,
0sin or m or
m
e
sin
........3,2,1,sin mme .............(3)
and the position of secondary maxima approach to
..........2/7,2/5,2/3
(b). Interference term 2cos :This term corresponds to
the two diffraction patterns coming out from the two slits.
This term gives a set of equidistant dark and bright
fringes. The directions of the maxima are given by
41
1cos 2 or n or
nde sin
or nde sin where n=0,1,2,3,.....(4)
Thus in a direction θ = 0, the central maxima due to
interference and diffraction coincide. The minima due to
interference term is given by,
0cos 2 or 2/12 n
or 2/12sin nde
Missing Orders: In the double slit arrangement, we find that
some of the interference maxima are missing. Since for the
same value of θ, the following two relations hold true,
nde sin , Interference Maxima.....(5)
me sin , Diffraction Minima....(6)
When both the above conditions are satisfied simultaneously,
then the interference maxima will be absent in the direction for
which θ is common. From eq. (5) and (6) we have,
m
n
e
de
Now the following conditions may be considered:
1. If d = e then n = 2m = 2,4,6,..... Hence the 2nd
,4th
,6th
etc.
order interference maxima will be absent.
2. If d = 2e, then n = 3m = 3,6,9.... Thus 3rd
, 6th
, 9th
order of the
interference maxima will be absent.
3. If e +d =e, i.e. d = 0, then the diffraction pattern will similar
to as observed to a single slit of width equal to 2e.
42
Effect of Increasing the slit width e: On increasing the slit
width e, the central peak will become sharper, but the fringe
spacing remains unchanged. Hence less interference maxima
fall within the central diffraction maximum.
Effect of increasing the distance between slits d: On
increasing the separation between the slit d keeping the slit
width constant, the fringes become closer together but the
I
θ
4A2sin
2αcos
2β/α
Cos2β
β π 2π -π -2π 0
43
envelope of the pattern remains unchanged. Hence more
interference maxima fall within the central envelope.
Effect of increasing the wavelength : When the wavelength
of monochromatic light falls on the slit increases, the envelope
becomes broader. As a result, the fringes move farther apart.
Diffraction Due to N-Slits (Plane Transmission Grating): A
diffraction grating consists of a large no. of parallel slits of
equal width and equal separation. It generally consists of 10000
to 15000 lines per inch.
Theory: The diffraction grating is equivalent to N-slits
arrangements and the diffraction pattern we obtain will be the
combined effect of all such slits. Let ‘e’ be the width of each
slit and‘d’ that of opaque part between them, then (e + d) is
known as grating element.
θ P
(e+d)
S1
S2
SN
M1
M2
MN-1
L
X
Y
44
The diffracted rays from each of the slits are allowed to fall on
a convex lens which focuses all of them at a point P on the
screen. As in the single slit, the waves diffracted from each slit
are equivalent to a single wave of amplitude,
sinAR
…….(1)
where
sine ………(2)
The path difference between any two consecutive waves from
two slits sin)( de . Therefore, the corresponding phase
difference will be
sin)(
2de
. Since the phase difference
is constant between any two consecutive waves it can be taken
as
2sin)(
2 de
………(3)
Thus we have to find out the resultant of N vibrations in a
direction θ and each vibration is of amplitude
sinA
.
Now similar to the resultant of n-harmonic waves, the resultant
of N-slits may be given as,
sin
sin.
sin
sin
sin
2
2sin
2
2sin
'NANR
NR
R
Thus the resultant intensity at P may be given as,
45
2
2
2
222
sin
sin.
sin'
NARI
………..(4)
Here the factor 2
22 sin
Ais the intensity factor due to a single
slit while
2
2
sin
sin N is due to the interference from all the N-slits.
Principle Maxima: for maximum intensity, we have,
0sin or n , .....2,1,0n
But in this condition 0
0
sin
sin
N is in indeterminate form. Thus
to find its value we adopt the differential calculus method and
then we get,
NN
n
sin
sinlim
. Thus ,
2
2
22 sinN
AI
………(5)
The intensity at these maxima is maximum and that is why it is
principal maxima. Therefore, the condition for principal
maxima is
0sin or n or
nde sin)( or
nde sin ……….(6)
Now for n = 0, we get θ = 0, which gives the direction of the
central order maxima.
The values n = 1,2,3…….. correspond to the first, second, third
….order maxima.
46
Minima: For sin N β =0. But 0sin . We get the minimum
intensity. A series of minima, thus, obtained for
N
mormN
or
mdeN sin)(
mdeN sin)( ……….(7)
Thus for all integral values of m except 0,N,2N,3N…….we get
a minima, because for these values sin β = 0 and this will give
the position of principal maxima. Between m = 0 and N, i.e.
two maxima the no. of minima exist for m= 1,2,3,……,(N-1).
Since maxima and minima are obtained alternately there will be
(N-2) other maxima. These (N-2) ,maxima are known as
secondary maxima and the position of these maxima can be
obtained by differentiating eq. (4) w.r.t. β and equating it with
zero. Thus,
0sin
cossinsincos
sin
sin2.
sin2
2
NNNNA
d
dI
or 0cossinsincos NNN
or NN tantan ……..(8)
Therefore the roots of this equation except for which n
(central maxima) correspond to the positions of secondary
maxima.
47
If we construct a right angle triangle with its sides as 1, N tan β
and 2tan1 N then we have
22 tan1
tansin
N
NN
……….(9)
22 tan1 N
Hence from eq. (9) the intensity of secondary maxima may be
given as
222
22
2
22
sintan1
tan.
sin'
N
NAI
2222
2
2
22
sinsincos.
sin'
N
NAI
22
2
2
22
sin11.
sin'
N
NAI
………(10)
Hence from (5) and (10) we obtain
imafprincipalIntensityo
imaondaryfIntensityo
N
N
I
I
max
maxsec
sin11
'22
2
1
N tan β
Nβ
48
Condition For Absent Spectra With a diffraction grating: For a grating the direction of n
th order principal maxima is
given by
nde sin)( ............(1)
And the direction of minima in the diffraction pattern due to a
single slit is given by the eq.
me sin ...............(2)
The resultant intensity in the direction θ is given by
2
2
2
22
sin
sin.
sin NAI .........(3)
Intensity Distribution Curve
Central Maxima
Intensity
49
Where,
sine and
sin)( de
In eq. (3), 2
22 sin
Ais the diffraction term and represents the
intensity due to diffraction at a single slit while
2
2
sin
sin N is
interference term representing the intensity of waves interfering
from all the N slits. Therefore, if in any direction, diffraction
term is zero and interference term has maximum value, the
principal maxima will not be present in that direction. Form (1)
and (3) we get,
m
n
e
de
or
m
e
den
..............(4)
Eq (4) represents the condition for the nth
order to be absent
from the grating spectra. Thus,
1. If d = e then n = 2m = 2,4,6,..... Hence the 2nd
,4th
,6th
etc.
order interference maxima will be absent.
2. If d = 2e, then n = 3m = 3,6,9.... Thus 3rd
, 6th
, 9th
order of
the interference maxima will be absent.
Maximum No. Of Orders Of The Spectra With A Grating: In a diffraction grating the direction of the nth order principal
maxima for wavelength λ is given by
nde sin)(
or
sin)( den
Thus the maximum no. of possible orders nmax. is given by
)(max
den
1)(sin max
50
Dispersive Power Of A Diffraction Grating: The dispersive
power of diffraction grating is defined as the rate of change of
the angle of diffraction with the wavelength of light. It is
expressed as
d
d.
For a grating we have nde sin)(
Differentiating it w.r.t. λ we get,
nd
dde
cos)(
or
cos)( de
n
d
d
Linear Dispersive Power: Linear dispersive power is defined
asd
dx, where dx is the linear separation between two
wavelengths differing each other by dλ. If the focal length of
the lens used is f then.
Linear dispersive power
d
df
d
dx
f Angular dispersive power
cos)( de
nf
Characteristics Of Grating Spectra: 1. Spectra of different orders are situated symmetrically on
both sides of zero order image.
2. Spectral lines are almost straight and quite sharp.
3. Spectral colours are in order from violet to red.
51
4. Most of the intensity goes to zero order and rest is
distributed among the other orders.
5. The lines are more and more dispersed as we go to higher
orders.
Resolving Power Of An Optical Instrument: The ability of
an optical instrument to form separate images of two objects,
placed very close to each other, is known as its resolving power.
The minimum separation between two objects up to which they
can be seen as distinct objects by an optical instrument is called
the limit of resolution of that instrument.
Rayleigh’s Criterion Of Resolving: According to Rayleigh
“Two nearby point objects are just resolved by an optical
instrument when the principal maxima in the diffraction pattern
of one falls over the first minimum in the diffraction pattern of
the other and vice versa. The same thing happens in case of
spectral lines of two different wavelengths, the lines are
resolvable when the principal maximum due to one wavelength
falls over the first minimum due to the other or vice versa.
52
λ1 λ2
Not Resolvable
Just Resolvable
DIP Resultant
Intensity curve
λ1 λ2
Easily Resolvable
λ2 λ1
Principle Maxima Principle Maxima
53
Resolving Power Of A Grating: The resolving power of a
grating is defined as its ability to form separate diffraction
maxima of two wavelengths which are close to each other. If d
λ is the smallest difference in two wavelengths, which can be
just resolved by a grating and λ is the wavelength of either of
them or mean wavelength, then λ /d λ is known as the resolving
power of the grating.
Expression for resolving power:
Let (e + d) be the grating element and N the total number of
slits. Let P1 be position of the nth principal maximum of
spectral line of wavelength λ while nth principal maximum due
to wavelength (λ + d λ) be at P2.
θn
dθn P1
P2
Central Image
A
B
X
Y
54
Now according to Rayleigh’s criterion of resolution the two
wavelengths can be resolved if the position of P2 coincides with
the first minimum due to wavelength λ.
In fig. the dotted line represents the diffraction pattern due to
wavelength (λ + d λ) and the solid line represents the diffraction
pattern due to wavelength λ.
Now the principal maximum of wavelength λ in a direction θn
is given by
nde n sin)( ...........(1)
And the equation of minimum is given by
mdeN n sin)( .........(2)
Where m has all integral values except 0, N, 2N..........nN,
because for these values of m the condition for maxima is
satisfied.
Therefore, the first minimum adjacent to nth principal
maximum in the direction )( nn d may be obtained by putting
m=nN +1.
Thus the first minimum in the direction )( nn d is given by
)1()sin()( nNddeN nn ..........(3)
And the principal maximum of (λ + d λ) in the direction
)( nn d may be given as
)()sin()( dndde nn ............(4)
or )()sin()( dnNddeN nn ..........(5)
comparing (3) and (5), we get
)()1( dnNnN
or nNd / ......(6)
also from eq. (1) we have
55
nden
sin
Hence (6) can be modified as
ndeN
dsin
/
.......(7)
Here N(e + d) is the total width of the grating.
Resolving Power Of A Telescope: A telescope is used to see
distant objects. The ability of telescope to form separate images
of two distant point objects situated close to each other is
known as its resolving power. The resolving power of a
telescope is measured by means of the angle subtended by two
nearby point objects at its objective; it does not depend on the
linear separation between the objects. When the images of the
nearby distant objects are just resolved by the telescope then
the angle subtended by these two objects at the telescope
objective is called the limit of resolution of the telescope. Thus
the resolving power of telescope is defined as the reciprocal of
the smallest angle subtended at the objective by the two distant
objects, the images of which are just seen as separate ones
through a telescope.
Expression for resolving power:
dθ
O
dθ
A
C
P1
P2
56
Let a be the diameter of the objective of the telescope.
Consider the incident ray of light from two neighbouring
points of a distant object.
The image of each point is a Fraunhofer diffraction pattern.
According to Rayleigh, these two images are said to be
resolved if the position of the central maximum of the
second image coincides with the first minimum of the first
image and vice versa.
The path difference between the secondary waves traveling
in the directions AP1 and BP1 is zero and hence they
reinforce with one another at P1.
Similarly, all the secondary waves from the corresponding
points between A and B will have zero path difference.
Thus, P1 corresponds to the position of the central maxima
of the first image.
The secondary waves traveling in the directions AP1 and
BP2 will meet at P2 on the screen.
Let the angle P2AP1 be dθ. The path difference between the
secondary waves travelling in the directions BP2 and AP1
is BC.
From fig. ΔABC,
daABddABBC .sin
57
If this path difference da. , the position of P2
corresponds to the first minimum of the first image.
But P2 is also the position of the central maximum of the
second image.
Thus, Rayleigh’s condition of resolution is satisfied if
da.
or ad
............(1)
The whole aperture AB can be considered to be made up of
two halves AO and OB. The path difference between the
secondary waves from the corresponding points in the
halves will be2
.
All the secondary waves destructively interfere with one
another and hence P2 will be the first minimum of the first
image.
The eq. ad
holds good for rectangular apertures. For
circular aperture, this eq., according to Airy, can be written
as
ad
22.1 ............(2)
where λ is the wavelength of light and a is the aperture of
the telescope objective, which is equal to the diameter of
the metal ring in which the objective lens is mounted.
Here dθ refers to the limit of resolution of the telescope.
The reciprocal of dθ measures the resolving power of the
telescope.
58
22.1
1 a
d .........(3)
Resolving Power Of Microscope: The ability of a microscope
to form separate distinct images of two nearby small objects,
which cannot be seen by necked eye, is known as its resolving
power. The limit of resolution of a microscope is the smallest
separation between the two objects when their images are just
resolved.
Expression for resolving power:
In fig. MN is the aperture of the objective of a microscope
and A and B are two object points at a distance d apart.
A’ and B’ are the corresponding Fraunhofer diffraction
patterns of the two images.
A’ is the position of the central maximum of A and B’ is
the position of the central maximum of B.
A’ and B’ are surrounded by alternate dark and bright
diffraction rings.
The two images are said to be just resolved if the position
of the central maximum B’ also corresponds to the first
minimum of the image of A’.
The path difference between the extreme rays from the
point B and reaching A’ is given by
2α A
B
A’ A
B’
O
M
N
59
)'()'( MABMNABN
But NA’ = MA’
path MNBNdifference
In fig. AD is perpendicular to DM and AC is perpendicular
to BN
)()( DBDMCNBCBMBN
But DMAMANCN
path DBBCdifference
From Δs ACB and ADB
sinsin dABBC
and sinsin dABDB
Path difference sin2d
If the path difference 22.1sin2 d
Then limit of resolution
sin2
22.1d
Thus resolving power
22.1
.).(2
22.1
sin21..
AN
dPR , for self-
luminous object.
α
α
α
α
A
B
M
O C
D
60
If a medium of refractive index μ is introduced between the
object and objective then,
22.1
.).(2
22.1
sin21..
AN
dPR .
Resolving Power Of A Prism: Resolving power of prism is
defined as its ability to form separate images of two
neighbouring wavelengths.
Expression for resolving power:
Let S is a source of light; L1 is a collimating lens and L2 in
the telescope objective.
As the two wavelengths, λ and λ + d λ are very close, if the
prism is set in the minimum deviation position it would
hold good for both the wavelengths.
a A α
α α
δ dδ
S α
L1 α
L2 α
I1 α
I2 α t
61
The final image I1 corresponds to the principal maximum
for wavelength λ and image I2 corresponds to the principal
maximum for wavelength λ +d λ.
I1 and I2 are formed at the focal plane of the telescope
objective L2.
The face of the prism limits the incident beam to a
rectangular section of width a.
Hence, the Rayleigh criterion can be applied in the case of
a rectangular aperture.
In the case of diffraction at a rectangular aperture, the
position of I2 will corresponds to the first minimum of the
image I1 for wavelength λ provided
ad
or a
d
......................(1)
Here δ is the angle of minimum deviation for wavelength λ.
From fig. A
22
A
22sinsin
A
or
2cossin
A
But
l
asin
l
aA
2cos
............(2)
62
Also l
tA
22sin ............(3)
In the case of a prism
2sin
2sin
A
A
2sin
2sin
AA
...........(4)
Here μ and δ are dependent on wavelength of light λ.
Differentiating eq. (4) w.r.t. λ
2sin
2cos
2
1 A
d
d
d
dA
Substituting the values from eq. (2) and (4)
l
t
d
d
d
d
l
a
22
1
or
d
dt
d
da .. ...................(6)
substituting the values of dδ from eq. (1)
d
dt
d.
The expression
d measures the resolving power of the
prism.
Polarisation Of Light: In transverse waves there can be a no.
of directions perpendicular to the direction of propagation in
which the particles of the medium can execute periodic
vibrations. Transverse waves in which the vibrations are
restricted to one particular direction are known as polarised
waves and the phenomenon is as polarisation.
63
Unpolarised Light: The light having vibrations along all
possible straight lines perpendicular to the direction of
propagation of light is known as Unpolarised Light.
Polarised Light: The light, which has acquired the property of
one-sidedness, is called polarised light. Therefore, the polarised
light is not symmetrical about the direction of propagation but
its vibrations are confined only to a single direction
perpendicular to the direction of propagation. The crystal which
makes the light polarised is known as polariser and the crystal,
which analyses the incoming polarised light, is called analyser.
1. Plane Polarised Light: Light is said to be plane polarised
when vibrations take place only in one direction parallel to
the plane through the axis of a beam.
2. Circularly Polarised Light: Light is said to be circularly
polarised when the vibrations in transverse plane are circular
having constant period. Here the amplitude of vibrations
remains constant but the direction changes only.
64
3. Elliptically Polarised Light: Light is said to be elliptically
polarised when the vibrations are elliptical and having a
constant period and takes place in a plane perpendicular to the
direction of propagation. Here the amplitude of vibrations
changes in magnitude as well as in direction.
Plane of Polarisation and Plane of Vibration: The plane in
which the vibrations occurs known as plane of vibration and a
plane perpendicular to plane of vibration in which no vibration
occurs is known as plane of polarisation.
B C
G H
A
B C
D
E F
G H
65
Plane of Vibration
Plane of Polarisation
Brewster’s law: In 1811, Brewster discovered a simple relation
between angle of polarisation and refractive index μ of the
medium. He found that the refractive index of the material
medium is equal to tangent of the angle of polarisation. i.e.
pitan ........(1)
Thus when angle of incidence becomes equal to the angle of
polarisation, then from Snell’s law,
r
i p
sin
sin ........(2)
Thus from (1) and (2), rip sincos
or rip sin)90sin(
ABCD EFGH
ip ip
r
θ
X Y O
A
C
B
66
or 090 pir
Also from fig. if θ be the angle between reflected and refracted
rays then, 09090180)(180 pir
Therefore, the reflected and refracted rays are at right angle to
each to other.
Double Refraction: When an ordinary light beam after passing
through a crystal splits up into ordinary and extraordinary light
the crystal is known as double refracting crystal and the
phenomenon is known as Double Refraction.
Ordinary and Extraordinary Rays: When ordinary light
beam when passed through a uniaxial crystal splits up into two
refracted rays. One of them obeys laws of refraction known as
(o-Ray) ordinary ray while the other does not obey laws of
refraction known as (E-ray) extraordinary ray.
These two rays have the following characteristics:
1. The plane of polarisation for O-ray is same as the principle
plane. Therefore, it has vibrations perpendicular to the
principal plane. For E-ray the case is opposite.
2. The two rays emerge from the crystal along the parallel
directions.
re
ro
E-Ray
O-Ray
Incident
Ray
67
3. The refractive index of the material for the two rays is given
as:
o
or
i
sin
sin and
e
er
i
sin
sin
The value of μo is constant while μe varies with the angle of
incidence.
4. In a crystal if μo >μe the crystal is known as negative crystal
in which the velocity of O-ray is less than E- ray.
5. In a crystal if μo <μe the crystal is known as positive crystal in
which the velocity of E- ray is less than O-ray.
6. Since the refractive index of O- ray is constant, hence it
travels with same speed in all directions.
7. Unpolarised light incident along optic axis does not split into
O- ray and E- ray.
8. The difference between the refractive indices of O- ray and
E- ray is known as birefringence i.e. birefringence
= eo .
NICOL PRISM: It was invented by William Nicol in 1828. It
is an excellent optical device used for producing and analyzing
the plane polarised light.
Principle: 1. It works on the phenomenon of double refraction.
2. When Unpolarised light is passed through uniaxial crystal, it
splits up into ordinary and extraordinary light.
3. If by some optical method one of the two rays is eliminated
the ray, emerging through the crystal will be plane polarised.
4. In Nicol prism O- ray is eliminated by total internal reflection
and E- ray is transmitted through the crystal.
68
Construction: It is most commonly used device to get plane
polarised light. It is made from a calcite crystal in the following
manner:
1. The length of the calcite crystal is taken three times its width.
2. The end faces in the crystal are cut in such a manner that the
acute angle (710) reduces to 68
0.
3. The crystal is then cut into two parts along a plane which
passes through the blunt corners A’ and D’ and is
perpendicular to the principal section.
4. Now the two cut surfaces are polished and cemented back
with a transparent substance known as Canada balsam, the
refractive index of which is midway for O and E ray.
5. The two end faces of the crystal are left open while the other
faces are coated with lampblack.
6. Finally, the crystal is enclosed in a brass tube.
Action: When Nicol crystal is used for, producing polarised
light is known as polariser and when it is used for the analysis
of polarised light is known as analyser.
(a). Nicol as a Polariser: 1. A beam of Unpolarised light when falls on the face of the
crystal it splits up into O- and E- rays.
2. Now after traversing some distances into Nicol prism
both the rays reach to the Canada balsam layer.
680 71
0
140
140
O O’ A’ A
D’ P
S0
SE
S
69
3. Since this layer offers a rarer medium to O-ray coming
from a denser medium of calcite, the O- ray suffers total
internal reflection.
4. This totally reflected ray is absorbed by the lampblack
coating.
5. Here the E-ray emerges out of the crystal and in this way
the crystal produces plane polarised E-ray and thus known
as polariser.
(b). Nicol as a Analyser:
POLARISAER ANALYSER
E- Ray
O- Ray
E- Ray
E- Ray
O- Ray
E- Ray
POLARISAER ANALYSER
70
1. When the two Nicol prisms are arranged coaxially then
the first Nicol act as a polariser and produces plane
polarised E-ray.
2. The ray emerging from polariser falls on the second nicol
prism known as analyser.
3. When the principal section of the two nicols are parallel
to each other, the plane polarised ray coming out from the
first nicol is easily transmitted through the second nicol.
4. Now if we rotate the second nicol gradually, the intensity
of transmitted ray from the second nicol decreases.
5. The intensity becomes zero when principal axis of both
the nicol become perpendicular to each other. Hence the
second nicol works as an analyser.
Quarter wave plate: A plate of doubly refracting crystal viz,
calcite or quartz whose refracting faces are cut parallel to the
direction of optic axis and its thickness is such that it produces
a phase difference of π/2 and path difference of λ/4 between
emerging ordinary and extraordinary rays is known as Quarter
Wave Plate.
If μ0 and μe be the refractive indices for ordinary and
extraordinary rays then, for normal incidence, the path
difference introduced between emerging O- and E- rays for
negative crystal (calcite) 4)(
teo
or )(4 eo
t
(for negative crystal eo )
and for positive crystal (quartz) 4)(
toe
71
or )(4 oe
t
(for positive crystal oe ).
This plate is used to produce circularly and elliptically
polarised light. When a plane polarised light is allowed to fall
on a quarter wave plate such that its vibrations make an angle
450 with the optic axis then the emergent beam will be
circularly polarised. When the vibrations of incident light make
an angle other than 450 with the optic axis the emergent light
will be elliptically polarised.
Half Wave Plate: A plate of doubly refracting crystal viz,
calcite or quartz whose refracting faces are cut parallel to the
direction of optic axis and its thickness ‘t’ is such that it
produces a phase difference of π and path difference of λ/2
between emerging ordinary and extraordinary rays is known as
Half Wave Plate.
Now for negative crystals like calcite ( eo ), path difference
is
2)(
teo
or )(2 eo
t
and for negative crystals like quartz ( eo )
)(2 oe
t
When a plane polarised is allowed to pass through half wave
plate then the emergent light is also plane polarised.
These two plates are also known as retardation plates as they
retard the motion of one of the beam.
72
END