Wave Optics
1.1 INTRODUCTION
The field of optics (study of light) is divided into three kinds:
(a) Geometrical optics (Macroscopic optics) which is concerned with the behaviour of light on
a large scale (macro) and is treated by the method of light rays.
(b) Physical optics (Microscopic optics) which is concerned with the study of the finer details
of light and involves theory of waves.
(c) Quantum optics which is concerned with the interaction of light with the matter and its
treated by the method of quantum mechanics.
Light is propagated (transported) in a straight line as long as there is no change of the medium or
no change in the properties of the same medium and is called a ray. Actually it is not possible to
seperate a ray from a light wave. As we are concerned with the physical optics, which is required to
explain the finer details of light, we have to consider the wave picture of light.
The light is considered to be transverse wave which is defined as: when a wave passes in a
direction the particles move at right angles to the direction of the propagation of the wave. When
particles move up and down in simple harmonic motion, a series of waves will be produced in which
the displacement of the wave is proportional to the sine of the S.H.M. generating angle ‘q’
i.e.,
y µ sin q
= a sin q
= a sin
2π
λ
Such waves are called sine waves and is shown in the Fig. 1.1.
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Fig. 1.1
Here ‘a’ is the amplitude (the maximum displacement of the wave),‘l’ is the wavelength (length
of complete wave from crest to crest or from trough to trough)
The wavelengths of visible light is in the range of 4000 Å to 7200 Å (i.e., 4 to 7.2 × 10–5 cm).
Faraday and Maxwell showed that light is composed of electric and magnetic fields at right angles to
each other and again these fields are at right angles to the direction of propagation of light. This can be
explained as follows:
H
When a point charge is at rest, it produces electric field (E ) and when it is in motion, the
H
magnetic field (B) is also produced is addition to the electric field. Further, if charge oscillates with
simple periodic motion an electromagnetic wave is produced like shown in the Fig. 1.2.
Fig. 1.2
Hertz demonstrated the existance of such electromagnetic waves.
Interference
Newton could not observe the interference effect, so he held that light must be particle like in nature
whereas Huygen was able to observe the interference effect so he held that light must be wave-like in
nature. To explain the wave propagation, Huygen established a principle saying that “every point along
a wavefront serves as a source of secondary spherical wavelets. He also assured that the intensity of the
spherical wavelts is not uniform in all directions but varies continuously from a maximum in the
forward direction to a minimum of zero in the backward direction.
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For light waves, due to various process of emission, one cannot observe interference between the
waves from two independent sources although the interference does takes place. Thus, one tries to
derive the interfering waves from a single wave so that the phase relationship is maintained. The
methods to achieve this can be classified under two broad categories.
Under the first category, a beam is allowed to fall on two closely-spaced holes/slits and the two
beams emerging from the holes interfere. Division of wavefront (E.g. : Young’s double slit experiment
(or) Bi-Prism.
In the second category, a beam is allowed to fall on two or more reflecting surfaces/mediums and
the divided reflected beams interfere.
Division of amplitude (E.g. : Newton's rings experiment).
1.2 INTERFERENCE OF LIGHT
If two waves of the same frequency travel in approximately the same direction and have a phase
difference that remains constant with time, they may combine so that their energy is not distributed
uniformly in space but is a maximum at certain points and a minimum (or zero) at other points. This
effect is called the Interference.
This was first demonstrated by Thomas Young in 1801 and wave theory of light was experimentally
proved. He was the first person to deduce the wavelength value from this experiment.
Young in his original experiment allowed sunlight to fall on a pinhole So punched in a screen A
as shown in Fig. 1.3.
Fig. 1.3 Young’s double slit experiment.
At a considerable distance away, the light is allowed to pass through two pinholes S1 and S2. The
two sets of spherical waves emerging from the two holes interfered with each other in such a way as to
form a symmetrical pattern of varying intensity on the screen C.
To be convenient, at a later stage, the pin holes were replaced by narrow slits and a source giving
monochromatic light (light of a single wavelength l ) was used. And the condition a £ l is fulfilled,
where ‘a’ is the slit width. If the circular line (see Fig. 1.2) represent crests of waves, the intersections
of any two lines represent the arrival at those points of two waves with the same phase or with phases
differing by a multiple of 2p radians. Such points are therefore those of maximum disturbance or
brightness.
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That means if crest falls on crest and trough falls on trough ® disturbance add up
But if crest falls on trough or trough falls on crest ® disturbances cancels out
In interference, we get equispaced fringes (dark and bright bands). This interference phenomenon
is not limited to light waves but applicable to all other kinds of waves like sound waves and other
electromagnetic waves.
1.3
YOUNG'S DOUBLE SLIT EXPERIMENT AND MAXIMA AND MINIMA
CONDITIONS
Let us now analyse Young’s experiment quantitatively assuming that the incident light consists of a
single wavelength only. In the figure shown here P is an arbitrary point on the screen. It is at a distance
of r1 and r2 from the narrow slits S1 and S2, respectively. Draw a line from S2 to b in such a way that
the lines PS2 and Pb are equal. If d, the distance between the two slits, is much smaller than the distance
D between the slit system and the screen S2, b is almost perpendicular to both r1 and r2. This means that
angle S Sˆ b is almost equal to angle PAˆ O and is marked as q (d << D and r || r ).
1
1 2
2
The rays arriving at O from S1 and S2 have same path lengths and there is no phase difference.
Fig. 1.4
The two rays arriving at P from S1 and S2 are in phase at the source slits, both being derived from
the same wavefront in the incident plane wave. Because the rays have different optical path lengths,
theyarrive at P with a phase difference. The number of wavelengths contained in S1b (path difference),
determines the nature of the interference at P.
To have maximum (brightness) at P, S, b must contain an integral number of wavelengths, or
S1b = m l where m = 0, 1, 2, . . . . .
but S1b = d sin q (from the triangle S1bS2S1)
d sin q = ml
gives maxima.
...(1.1)
∴
The central maximum is described by m = 0 and in between two maxima there should be a
minima.
1

S1b = d sin q =  m + λ gives minima
...(1.2)
∴
2

where m = 0, 1, 2, . . . . .
Wave Optics
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Now let us define the terms, phase and coherence. The term phase tells us what fraction of a
complete vibration (oscillation) the particle has executed at a given instant.
And phase difference µ path difference
f µ (x2 – x1)
or
f = K (x2 – x1)
2
= π( x 2 − x1 )
...(1.3)
λ
Fig. 1.5
1.4
COHERENCE
If the slits S1and S2 are illuminated by two completely independent light sources, no interference
fringes are observed. The phase difference between the two beams arriving at P varies with time in a
random way. At one instant the wave from source S2 may be in phase with that from source S1 ; but in
a very short time (10–8sec), this phase relation may change from reinforcement to cancellation. The
same random phase behaviour holds for points on screen, with the result that the screen is uniformly
illuminated. If instead, the light waves that travel from S2 and S1 to P have a phase difference (f) that
remains constant with time, the two beams are said to be coherent.
Fig. 1.6
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In the case of coherent waves, combine amplitudes vectorially and square the resulting amplitude
to obtain a quantity proportional to Luminous Intensity.
In the case of incoherent waves, square the individual amplitudes to obtain quantity proportional
to individual intensity and add the individual intensities.
1.5 INTENSITY IN YOUNG'S DOUBLE SLIT EXPERIMENT
We know that light wave is electromagnetic in nature and is represented by electric and magnetic field
H
H
vectors ( E and B ). Let us assume that the electric field components of the two waves from slits S1 and
S2 vary with time at point P (Fig. 1.7) as
H
...(1.4)
E 1 = Eo sin wt
H
E2 = Eo sin (wt + f)
and
...(1.5)
where w is the angular frequency and is equal to 2pn and f is the phase difference between the two
waves which depends on location of P which is described by q.
Eo is the initial amplitude.
Let us assume that the slits are narrow (a < < l) and illuminates the central portion uniformly.
But the resultant wave disturbance at P on the screen is found and is given by
H
E = E1 + E2 , and using eqns. (1.4) and (1.5), we have
= Eo sin wt + Eo sin (wt + f)
= Eo [sin wt + sin (wt + f)]
Fig. 1.7
 B+C 
C − B
 we have
 cos 
Using the trigonometric relation, sin B + sin C = 2 sin 
 2 
 2 
H

 ωt + ωt + φ 
 ωt + φ − ωt 
E = Eo 2 sin

 cos 
2
2





Wave Optics
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= 2Eo [sin (wt +f/2) cos f/2]
= 2Eo cos f/2 sin f/2)
(put f/2 = b and 2 Eo = Em)
= Em cos b sin (wt + b)
H
\
E = E θ sin (ωt + β)
\
Eq = Em cos b
...(1.6)
= 2Eo cos b
where Eq is the amplitude of resultant wave disturbance which determines the I of interference fringes.
Em is the maximum possible amplitude.
We knew in the case of coherent waves, the intensity is proportional to square of the amplitude
i.e., I µ E2 where E is the electric field strength.
If Iq is the intensity of resultant wave at P and Io is the intensity that a single wave acting alone
would produce, then
Iq µ Eq2 and Io µ Eo2
I θ = kEθ2
I o = kEθ2
\
\
or
I θ  Eθ 

=
I o  Eo 
2
Iθ
= (2 cos b)2 from equation (1.6)
Io
= 4 cos2 b
(or)
I q = 4I o cos2 β
(or)
I q = Im cos2 b
To compute Iq as a function of q, we substitute b =
...(1.7)
φ
value in terms of q from the relation
Phase difference µ Path difference (eqn. 1.3)
i.e.,
 2π 
f =   d sin q
 λ 
or
b =
φ
=
π
d sin q
λ
The intensity pattern for the double slit interference is shown in the Fig. 1.8.
Fig. 1.8 Energy distribution in Young’s experiment.
...(1.8)
...(1.9)
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1.6 BIPRISM - FRINGEWIDTH
After Young’s double slit experiment, objection was raised that the bright fringes may be due to some
modification of the light by the edges of the slits and not true interference. Thus wave theory of light
was still questioned.
Fresnel brought forward several new experiments in which the interference of two beams of light
was proved. One of them is the Fresnel bi-prism experiment.
The arrangement is as shown in the Figure 1.9. The bi-Prism consists of two prisms attached
back to back. The angle at the edges of the bi-prism is of the order of 30' and the other angle in the biprism is of the order of 179º. If a monochromatic source is placed in front of the bi-prism, the light
spreads out in the form of two beams which superimpose one over the other. Here the interference
condition has been fulfilled. The two beams looks like originating from the imaginary source positions
S1 and S2. The location of S1 and S2 can be obtained by extending the two beam backwards. These are
the two virtual images of the source S and act as two slit sources in Young’s double slit experiment and
produce stationary interference pattern.
If D x is the distance between successive fringes and d is the distance between S1 and S2, then the
wavelength value can be obtained from the equation.
∆x d
d
l=
= ∆x
...(1.10)
a+c
D
Fig. 1.9 A Typical biprism arrangement.
Fig. 1.10 Determination of slit separation (d).
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The distances d and D can easily be determined by placing a convex lens between the biprism and
the screen/eyepiece. For a fixed position of the eyepiece, there will be two positions of the lens
(L1 or L2) where the images S1 and S2 can be seen at the eyepiece. Let d1 be the distance between the
two images where the lens is at L1 (b1 from eyepiece to L1). Similarly d2 × b2 for L2 position of the
lens.
\
d =
and
D = b1 + b 2
= a + c = distance from source to screen or eyepiece.
d1d 2
Interference in Thin Films - Reflected Light
Fig. 1.11 Interference in thin films.
Let XY and X'Y' be the two surfaces of a transparent film of uniform thickness t and refractive
index µ as shown in figure 1.11. Suppose S is a monochromatic source of light. Suppose a ray SA is
incident on the upper surface XY at an angle i. This ray is partly reflected along AR and refracted along
AB at an angle r. At B it is incident at angle r. Here it undergoes reflection along BC at an angle r and
refraction along BT at an angle i. At C also it undergoes refraction along CR1 and reflection along CD.
This process will continue for a number of times until the intensity becomes very very small. The rays
BR and CR1 are derived from the same ray SA and travel in the same direction they interfere. To find
out the effective path difference between the rays AR and CR1 draw a normal CE on AR and normal
AF on BC. Produce the normal at A and the ray CB in the backward direction until they meet. Suppose
they meet at Q. From the geometry of the figure, ∠ACE = i and ∠CAF = r. The optical path difference
between the two reflected light rays (AR and CR1) is given by
D = Path (AB + BC) in film; Path AE in air
= µ (AB + BC) – AE
...(1.11)
10
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From triangles ACE and ACF we know that
µ = sin i/sin r = AE/AC ¸ CF/AC = AE/CF
AE = µ CF
...(1.12)
From equations (1.11) and (1.12), we can write
D = µ (AB + BC) – µ (CF)
= µ (AB + BF + FC) – µ (CF)
= µ (AB + BF)
= µ (QF)
...(1.13)
From triangle AQF, cos r = QF/AQ
or QF = AQ cos r = 2t = cos r
...(1.14)
(Since AQ = AP + PQ = t + t = 2t)
Substituting the value of QE from equation (1.14) in equation (1.13), we have
D = µ × 2t cos r = 2 µ t cos r
...(1.15)
It should be remembered that a ray reflected at a surface backed by a denser medium suffers an
abrupt phase change of p which is equivalent to a path difference l/2.
Thus the effective path difference between the two reflected rays is (2 µ t cos r ± l/2).
We know that maxima occur when effective path difference = n l
For interference maximum 2 µ t cos r ± l/2 = n l
Or
2 µ t cos r = (2n ± 1) l/2
...(1.16)
If this condition is fulfilled, the film will appear bright in the reflected light.
The minima occur when the effective path difference is (2n ±1) l/2 i.e.,
2 µ t cos r ± l/2 = (2n ±) l/2 or 2 µ t cos r = (2n ± 1) l/2 ± l/2 = n l
...(1.17)
because (n + 1) or (n – 1) can also be taken as integer. Here n = 1, 2, 3... etc.
When this condition is fulfilled the film will appear dark in the reflected light.
Looking at the same point as we move our eye the angle of incidence and the corresponding angle
of refraction changes. Therefore the conditions of maxima and minima are changed alternately. Hence
we observe a number of bright and dark regions.
Keeping the eye fixed if we change the point of observation then also we observe bright and dark
regions. If the film is illuminated with white light, the maxima of different colours are observed at
different angles. Hence the film appears coloured.
1.7 NEWTON’S RINGS
When a plano-convex lens with its convex surface is placed on a plane glass plate, an air film of
gradually increasing thickness is formed between the two. The thickness of the film at the point of
contact is zero. If monochromatic light is allowed to fall normally, and the film is viewed in reflected
light, alternate dark and bright rings concentric around the point of contact between the lens and glass
plates are seen.
Experimental Arrangement
The experimental arrangement of obtaining Newton's rings is shown in figure. L is a plano convex lens
of large radius of curvature. This lens with its convex surface is placed on a plane glass plate G. The
lens makes contact with the plate at O. Light from an extended monochromatic source such as sodium
Wave Optics
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lamp falls on a glass plate G' held at an angle 45° with the horizontal. The glass plate G' reflects a part
of the incident light towards the air film enclosed by the lens L and the glass plate G. A part of the
incident light is reflected by the curved surface of the lens L and a part is transmitted which is reflected
back from the plane surface of the plate. These two reflected rays interfere and give rise to an interference
pattern in the form of circular rings. These rings are localised in the air film, and can be seen with a
microscope focussed on the film.
Fig. 1.12a Newton’s rings apparatus.
Fig. 1.12b
Fig. 1.12c Plano-convex lens.
Explanation of the Formation of Newton’s Rings
Newton’s rings are formed due to interference between the waves reflected from the top and bottom
surfaces of the air film formed between the plates. The formation of Newton’s rings can be explained
with the help of the Fig. 1.12c. AB is a monochromatic ray of light, which falls on the system. A part
is reflected at B (glass-air boundary) which goes out in the form of ray R1 without any phase reversal.
The other part is refracted along BC. At point C it is again reflected and goes out in the form of ray R2
with a phase reversal of p. The reflected rays R1 and R2 are in a position to produce interference fringes
as they have been derived from the same ray AB and hence fulfill the condition of interference. As the
rings are observed in the reflected light, the path difference between them is (2µt cos r + l/2). For air
film µ = 1 and for normal incidence r = 0. Hence in this case, path difference is (2 t + l/2). At the point
of contact t = 0, and the path difference is l/2, which is the condition of minimum intensity. Thus the
central spot is dark.
For nth maximum, we have
2 t + l/2 = nl
This expression shows that a maximum of a particular order will occur for a constant value of t.
In this system, ‘t’ remains constant along a circle Thus the maximum is in the form of a circle. For
different value of ‘t’, different maxima will occur. Hence we get a number of concentric bright circular
rings. In a similar way, this can be shown that minima are also in the circular form.
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Theory : Newton’s Rings by Reflected Light
Now we shall calculate the diameters of dark and bright rings. Let LOL' be the lens placed on a glass
plate G. The curved surface LOL' is the part of spherical surface with centre at C. Let R be the radius
of curvature and r be the radius of nth bright ring corresponding to the constant film of thickness t. As
discussed above,
2 t + l/2 = nl
Or 2t = (2n – 1) l/2 for the bright ring where n = 1, 2, 3, ...etc.
For the property of the circle
EP × PF = PQ × PQ
Substituting the values
r × r = t × (2R – t) = 2Rt – t2 » 2Rt (approximately)
r2 = 2 R t or t = r2/2 R.
Thus for a bright ring
2r2/2 R = (2n – 1) l/2
or
r2
=
(2n − 1)λ R
2
Replacing r by D/2, we get the diameter of nth bright ring as
(2n − 1)λR
D2
=
or
2
4
or
or
Dn2 = 2(2n – 1)lR
D =
2λR (2n − 1)
D ∝ A (2n − 1)
Thus the diameters of the bright rings are proportional to the square roots of odd natural numbers
as (2n –1) is an odd number.
This shows that the difference in the squares of the diameters of the rings is constant.
Similarly for a dark ring
or
2r2/2r = nlR
or
D 2 = 4 nlR
or
D = 2 nλR ∝ n
Thus diameters of dark rings are proportional to the square roots of natural numbers.
If Dm and Dn are the diameters of the mth and the nth rings we have
Dm2 − Dn2 = 4(m − n)λR
This shows that the difference in the squares of the diameters of the rings is constant.
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1.8 DETERMINATION OF WAVELENGTH OF SODIUM LIGHT
USING NEWTON'S RING
Experimental Arrangement
The experimental arrangement of obtaining Newton's rings is shown in the Fig. 1.12. L is a plano
convex lens of large radius of curvature placed with its convex surface on a plane glass plate P. The lens
makes contact with the plate at O. Light from an extended monochromatic source such as sodium lamp
falls on at glass plate G' held at an angle 45° with the horizontal. The glass plate G' reflects a part of the
incident light towards the air film enclosed by the lens L and the glass plate P. A part of the incident
light is reflected by the curved surface of the lens L and a part is transmitted which is reflected back
from the plane surface of the plate. These two reflected rays interfere and give rise to an interference
pattern in the form of circular rings. These rings are localised in the air film, and can be seen with a
microscope focussed on the film.
Procedure
First of all the eyepiece of the microscope is adjusted on its crosswires. Now the distance of the
microscope from the film is adjusted such that the rings with dark centre are in focus.
The centre of the crosswires is adjusted at the centre of the rings pattern. The microscope is
moved to the extreme left of the pattern and the crosswire is adjusted tangentially in the middle of a
clearly nth bright or dark ring. The reading of micrometer screw is noted. The microscope is now
moved to the right and the reading of micrometer screw are noted at successive rings etc., till we are
very near to the central dark spot.
Again crossing the central dark spot in the same direction, the readings corresponding to successive
rings are noted on other side. Now a graph is plotted between number of rings n and the square of the
corresponding diameter. The graph is shown in Fig. 1.13. If Dm and Dn are the diameters of the mth and
nth rings and R is the radius of curvature of curved surface of the lens the wavelength of the sodium
light is given by
l=
Dm2 − Dn2
Slope
=
4(m − n) R
4R
D2
S lo p e
n
Fig. 1.13
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The radius R of the plano-convex lens can be obtained with the help of spherometer using the
l2 h
+ . Here l is the distance between the two legs of the spherometer and h is
6h 2
the difference of the readings of the spherometer when it is placed on the lens as well as when placed
on lens surface.
Let R be the radius of curvature of the surface in contact with the plate, l the wavelength of light
used and Dm and Dn be the diameters of mth and nth bright rings respectively, then
following formula R =
Dm2 = 2(2m – 1) lR
Dn2 = 2(2n – 1) lR
and
Dm2 – Dn2 = 4 (m – n)lR
or
l = Dm2 – Dn2 / 4(m – n)R
or
...(1.18)
Using this formula, l can be determined.
To find the refractive index of a liquid, it is introduced between the lens and glass plate and the
experiment is repeated as before. If D'm and D'n are the diameters of mth and nth rings in liquid then the
refractive index µ can be calculated using µ =
Dm2 − Dn2 .
2
D'm
−D'2n
...(1.19)
NUMERICAL EXAMPLES
1. A thin sheet of plastic of refractive index 1.6 is placed in the path of one of the interfering
beams in Young’s experiment using light of wavelength 5890 Å. If the central fringe shifts
through 12 fringes, calculate the thickness of the sheet.
(OU, 2003)
Solution: Given Data
l = 5890 × 10–8 cm
refractive index (µ) = 1.6
order of the fringe (m) = 12
optical path of beam with plastic sheet = x – t + µt
and the path of another beam
=x
\ path difference
= x – t + nt – x
= (µ – 1) t = nl for maxima
\
l=
( µ − 1)t
or
m
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λn
λ × 12
5890 × 10 −8 × 12
=
=
(µ − 1)
(1.6 − 1)
0.6
–8
= 1178 ×10 cm.
t=
2. The path of one of the interfering beams in biprism experiment, a thin sheet of mica of
refractive index 1.55 is placed. A light of wavelength 5893 Å is incident on it. Calculate the
thickness of the sheet if the central fringe shifts through 10 fringes.
(OU, 2000)
Solution: Given Data
l = 5893 × 10–8 cm
refractive index (n) = 1.55
order of the fringe (m) = 10
optical path of beam with plastic sheet = x – t + µt
and the path of another beam = x
\ path difference is = x – t + µt – x
= (µ – 1) t = n l
10 × 5893× 10 −8
5893 × 10 −7
nλ
=
=
1.55 − 1
0.55
(µ − 1)
–8
t = 1071.45 × 10 cm.
3. In double slit arrangement, a strong green light of wavelength 5460 Å is used. The slits are
0.01 cm apart and the screen is placed 20 cm away. What is the angular position of the first
minima?
Solution: Given Data
For first minimum n = 0 and
distance between slits (d) = 0.01 cm.
wavelength of light (l) = 5460 × 10–8 cm.
\t=
1

The condition for minima is d sin q =  n +  . λ
2

or

sin q =  n +

1. λ

2 d
1  5460 × 10

= 0 + 
2
0.01

−8
= 0.0027
Since when q is very small, sin q ≅ q
\ q = 0.0027 radians
= 0.16
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4. A parallel beam of light (l = 5890 × 10–8cm) is incident on plate (m = 1.5) such that the
angle of refraction into the plate is 60°. Calculate the smallest thickness of the glass plate,
which will appear dark by reflection.
Given that
m = 1.5,
r = 60°,
cos 60° = 0.5
n = 1,
l = 5890 × 10–8 cm
Applying 2µt cos r = nl
nλ
1 × 5890 × 10 −8
=
2µ cos r
2 × 1.5 × 0.5
\ The minimum thickness of the film
We get t =
t = 4.207 × 10–5 cm
5. A soap film 4 × 10–5 cm thick is viewed at an angle of 35° to the normal. Find the wavelengths
of light in the visible spectrum which will be absent from the reflected light (m = 1.33).
Let i be angle of incidence and r the angle of refraction
Give that
i = 35°
and µ = 1.33,
r=?
sin i
sin 35°
or
1.33 =
sin r
sin r
We get
r = 25.55° and cos r = 0.90
Apply the relation
2 µt cos r = nl
and taking t = 4 × 10–5 cm
(i) For the first order, n = 1
\
l = 2 × 1.33 × 4 × 10–5 × 0.90
= 9058 × 10–5 cm
which lies in the infra-red (invisible) region.
(ii) For the second order, n = 2
2l2 = 2 × 1.33 × 4 × 10–5 × 0.90
l2 = 4.79 × 10–5 cm
which lies in visible region.
(iii) Similarly, taking n = 3
l3 = 3.19 × 10–5 cm
which also lies in the ultraviolet range.
Hence, absent wavelength in the reflected light is 4.79 × 10–5 cm
6. A parallel beam of light (l = 5890 Å) is incident on a thin glass plate (m = 1.5) such that the
angle of refraction is 60°. Calculate the smallest thickness of the plate which will appear
dark by reflection.
Given that µ = 1.5 and r = 60°; cos 60° = 0.5, l = 5890 Å or l = 5890 × 10–10 m
For minimum thickness n = 1
Applying 2µt cos r = nl
Applying µ =
We have
nλ
t = 2µ cosr
Wave Optics
17
1 × 5890 × 10 −10
t =
2 × 1.5 × 0.5
t = 3926 × 10–10 m
t = 3.926 × 10–4 mm
7. A soap film of refractive index 1.33 is illuminated with light of different wavelengths at an
angle of 45°. There is complete destructive interference for l = 5890 Å. Find the thickness
of the film.
Given that
m = 1.33
r = 45°
cos 45° = 0.707
l = 5890 Å = 5890 × 10–10 m
n =1;t= ?
Applying 2µt cos r = nl
Thickness of the film t =
nλ
2µ cosr
1 × 5890 × 10 −10
2 × 1.33 × 0.707
t = 3.132 × 10–7m
t = 3.132 × 10–4 mm
8. A thin film of soap solution is illuminated by white light at an angle of incidence, i =
 4
sin–1   . In reflected light, two dark consecutive overlapping fringes are observed
 5
4
corresponding to wavelengths 5.1 × 10–7 m and 5.0 × 10–7 m. m for the soap solution is .
3
Calculate the thickness of the film.
Here
nl 1 = (n + 1)l2
n(5.1 × 10–7) = (n + 1) × 5 × 10–7
n = 50
t =
sin i =
µ =
sin r =
4
5
4
sin i
=
3
sin r
4/5
sin i
=
= 0.6
4/3
µ
cos r = [1 − sin 2 r ]1 / 2 = 0.8
Engineering Physics
18
Apply 2µt cos r = nl1
We have t =
nλ1
50 × 5.1× 10−7
=
2µ cos r
2 × (4 / 3)0.8
The minimum thickness of the film t = 1.2 × 10–5 m
9. In Newton’s rings experiment, what will be the order of the dark ring which will have
double the diameter of that of 20th dark ring. The wavelength of incident light is 5890 Å.
(OU, 1999)
Solution: Given Data
order of dark ring n2 = 20
wavelength (l) = 5890 × 10–8 cm
We know that the radius of the ring is given by ‘r’.
r=
λ×R×m
or Diameter (d) = 2r = 2
λRm
or
= 4 lRm
2
\ D1 = 4lR × n1
D2
and D22 = 4lR × n2 = 4lR × 20
= 80 lR
Given that :
D1 = 2D2
(or) \ D22 = 4D22
...(1)
...(2). On substituting we have
4 l Rn1 = 4 × 80 × lR
\ n1 = 80
10. In Newton ring’s experiment, the diameter of the 5 th and 15 th rings respectively was
0.336 cm and 0.590 cm. If the wavelength of light is 5890 Å. Find the radius of curvature
of lens surface in contact with plane glass plate.
(OU, 2000)
Solution: Given data
Dm = D15 = 0.590 cm
Dn = D5 = 0.336 cm
l = 5890 × 10–8 cm
m = 15 n = 5
We know that in the Newton’s rings experiment (R)
R=
=
2
D15
Dm2 − Dn2
− D52
=
4λ(m − n)
4λ(10)
(0.590) 2 − (0.336) 2
4 × 5890 × 10 −8 × 10
= 99.8 cm
Wave Optics
19
11. Newton’s ring arrangement is used with a source emitting two wavelengths l1 = 6000 Å and
l2 = 4500 Å and it is found that the nth dark ring due to l1 coincides with (n + 1) th dark ring
for l2. Find the diameter of nth dark ring of l1 if the radius of curvature of the lens
R = 90 cm.
(OU, 2002)
Solution: Given data
Wavelength l1 = 6000 × 10–8 cm
l2 = 4500 × 10–8 cm
radius of curvature R = 90 cm
Let dn be the diameter of nth ring corresponding to wavelength l1, then diameter of
(n + 1)th dark ring corresponding to wavelength l2 will also be dn
dn2 = 4nλ1 R
...(1)
d n2 = 4(n + 1)λ 2 R
and
...(2)
Dividing eq. (2) by (1)
λ1
n +1
6000 × 10−8
=
=
λ2
n
4500 × 10−8
1+
60
1
=
n
45
15
1
1
60
−1 =
=
=
n
45
3
45
\ n = 3. Putting the value of n in eq. (1)
d n2 = 4 × 3 × 6000 × 10–8 × 90
\ dn = 0.2545 cm
12. Newton’s rings are observed in reflected light of wavelength 5900 Å. The diameter of 10th
dark ring is 0.50 cm. Find the radius of curvature of the lens and the thickness of the air
film.
(OU 2001, 2003)
Solution: Given data
Wavelength of light (l) = 5900 × 10–8 cm
Diameter of the mth ring (D10) = 0.50 cm
Say, Diameter of the nth ring (Do) = 0
2
Dm2 − Dn2
Dm
=
Therefore radius of curvature (R), R =
4λ ( m − n)
4λ m
R=
0.50 × 0.50
4 × 5900 × 10
−8
× 10
=
0.25
4 × 59 × 10−5
25
= 105.96 cm.
4 × 59 × 10 − 3
\ R = 105.93 cms.
=
Engineering Physics
20
13. Newton’s rings are formed with reflected light of wavelength 5.895 × 10–5 cm. with a liquid
between the plane and the curved surface. The diameter of the 5th dark ring is 0.3 cm and
the radius of curvature of the curved surface is 1 metre. Calculate the refractive index of the
liquid.
Solution: Wavelength of light (l) = 5.895 × 10–5 cm
Diameter of 5th dark ring D5 = 0.3 cm
Radius of curvature R = 1m = 100 cm
For the ring system to be dark, we have
d n2 µ
= nl
4R
4 Rnλ
µ=
d n2
=
4 × 100 × 5 × 5.895 × 10 −5
0.3 × 0.3
= 1310 ×10–5+2
= 1.31
\ Refractive index of liquid is 1.31
14. In Newton’s rings experiment the diameter of 10th ring changes from 1.40 cm to 1.20 cm
when a liquid is introduced between the lens and the plate. Calculate the refractive index of
the liquid.
For liquid medium D12 =
4nλR
µ
...(i)
For air medium D22 = 4nlR
...(ii)
Divide (ii) by (i)
 D2 

µ = 
 D1 
Here D1 = 1.20 cm, D2 = 1.40 cm
2
2
 1.40 
 = 1.361
m= 
 1.20 
\
15. In a Newton’s rings arrangement, if a drop of water (µ = 4/3) is placed in between the lens
and the plate, diameter of the 10th ring is found to be 0.5 cm. Obtain the radius of curvature
of the face of the lens in contact with the plate. The wavelength of light used is 6000 Å.
Dn2 =
4
, Dn = 0.5 cm
3
n = 10, l = 6000 Å = 6 × 10–5 cm
R =?
µ =
Here
4nλR
µDn2
or R =
µ
4nλ
Wave Optics
21
4 × (0.5) 2
R =
3 × 4 × 10 × 6 × 10 −5
= 139 cm
16. Newton’s rings are formed by reflected light of wavelength 5895 Å with a liquid between
the plane and curved surface. It the diameter of the 6th bright ring is 3 mm and the radius of
curvature of the curved surface is 100 cm, calculate the reflective index of the liquid.
Here, for the nth bright ring,
Given that n = 6, l = 5895 × 10–8 cm, R = 100 cm, r =
3
mm = 0.15 cm
2
To find µ = ?
Applying µ =
µ =
(2n − 1)λR
2r 2
, we get
(2 × 6 − 1) × 5895× 10 −8 × 100
2(0.15) 2
µ = 1.441
17. In a Newton’s rings experiment the diameter of the 15th ring was found to be 0.590 cm and
that of the 5th ring was 0.336 cm. If the radius of the plano–convex lens is 100 cm, calculate
the wavelength of light used.
Here
D 5 = 0.336 cm = 33.6 × 10–3 m
D 15 = 0.590 cm = 5.90 × 10–3 m
R = 100 cm = 1m, l = ?
l =
( Dn + m ) 2 − Dn2
D2 − D52
= 15
4mR
4 × 10 × R
(5.9 × 10 −3 ) 2 − (3.36 × 10 −3 ) 2
4 × 10 × R
= 5.880 × 10–7m
l = 5880 Å
18. In a Newton’s rings experiment the diameter of the 12th ring changes from 1.50 cm to
1.30 cm when a liquid is introduced between the lens and the plate. Calculate the refractive
index of the liquid.
Given that
D 1 = 1.50 cm
D 2 = 1.30 cm
For air medium
l =
D12 = 4nlR
...(i)
Engineering Physics
22
For liquid medium
D22 =
4nλR
µ
...(ii)
Dividing (i) by (ii)
 D1 

µ = 
 D2 
2
2
 1.50 
 ,
µ = 
 1.30 
m = 1.331
19. Newton’s rings are observed in reflected light of l = 5.9 × 10–5 cm. The diameter of the
10th dark ring is 0.52 cm. Find the radius of curvature of the lens and the thickness of the air
film.
Given that l = 5.9 × 10–5 cm = 5.9 × 10–7 m
n = 10
Radius of the ring r = 5.2 × 10–3 m
Apply r2 = nlR
\ We get
R =
(5.2 × 10 −3 ) 2
10 × 5.9 × 10 − 7
(ii) Thickness of the air film = t
2t = nl
nλ
t =
2
10 × 5.9 × 10 −7
2
t = 2.95 × 10–6 m
=
EXERCISE PROBLEMS
1. In Newton’s ring experiment, diameter of the 5th ring is 0.336 cm and that of the 15th ring
is 0.590 cm. Find the radius of curvature of the plano-convex lens if the wavelength of light
used is 5890 Å.
2. Light of wavelength 6 × 10–5 cm falls on a screen at a distance of 100 cm from a narrow slit.
Find the width of the slit if the first minimum lie 1mm on either side of the central maxima.
3. The diffraction maxima due to single slit diffraction is at q = 30° for a light of wavelength
5000 Å. Find the width of the slit.
Wave Optics
23
4. Monochromatic light of wavelength 6.56 × 10–5 cm falls normally on a grating 2 cm wide;
the first order spectrum is produced at an angle of 18° 14' from the normal. What is the total
number of lines on the grating.
5. Newton’s rings are observed in reflected light of wavelength 5900 Å. The diameter of 10th
dark ring is 0.50 cm. Find the radius of curvature of the lens and the thickness of the air
film.
6. Green light of wavelength 5100 Å from narrow slit is incident on a double slit if the overall
separation of 10 fringes on a screen 200 cm away is 2 cm. Find the slit separation.
7. A biprism is placed at a distance of 5 cm in front of a narrow slit and is illuminated by
sodium light of wavelength 5890 Å. The distance between the virtual sources is found to be
0.05 cm. Find the width of the fringes observed in a light eyepiece placed at 75 cm from the
biprism.
8. Newton’s rings are observed in a reflected light of wavelength 6000 Å. The diameter of 15th
dark ring is 7 mm. Find the radius of curvature of the lens.
9. Find the ratio of intensity at the center of a bright fringe to the intensity at a point 1/4th of
the distance two fringes from the center.
10. In Newton’s rings experiment, the diameter of 10th ring changes from 1.5 to 1.3 cm when a
liquid is introduced between the lens and the plane. Calculate the refractive index of the
liquid.
11. In Young’s double slit experiment two parallel slits 1 mm apart are illuminated by
monochromatic light. If the width is 0.50 mm and the screen is held at a distance of 80 cm
from the slits, what is the wavelength of light?
12. In Newton’s rings experiment the diameter of 5th and 15th rings are 0.336 and 0.590 cm
respectively. If the wavelength of light is 5896 Å, find the radius of curvature.
13. A thin film 4 × 10–5 thick is illuminated by white light normal to its surface. Its refractive
index is 1.5. What wavelength within the visible spectrum will be intensified in the reflected
beam?
14. White light falls normally on a film of soapy water whose thickness is 5 × 10–5 cm and
refractive index is 1.33. Which wavelength in the visible region will be reflected more
strongly?
15. Newton's rings are formed with reflected light of wavelength 5895 Å with a liquid between
the plane and the curved surface. The diameter of the 5th dark ring is 0.3 cm and the radius
of curvature of the curved surface is 100 cm. Calculate the refractive index of the liquid.
16. A thin parallel liquid film of refractive index 1.28 and thickness 0.48 µm is illustrated by
visible light and observed at an angle of 35°. Determine the wavelengths(s) which are absent
in reflected light.
17. In Newton’s rings experiment, the wavelength of light used is 600 nm and radius of curvature
of lens is 1 m. Determine the diameter of the tenth dark ring.
18. In Newton’s rings pattern, the diameter of a certain ring is 0.38 cm. A liquid of refractive
index 1.38 is introduced between the glass plate and the lens in the same set up. Determine
the diameter of the same ring.
Engineering Physics
24
19. In Newton’s rings pattern, the diameter of the fifth ring is 0.3 cm and that of the tenth ring
is 0.5 cm. Wavelength of light used is 589 nm. Determine the radius of curvature of the lens.
20. Newton’s rings are observed in reflected light of wavelength 589 nm with the liquid film
formed between plane glass plate and a plano-convex lens. The diameter of 9th bright ring is
105 cm. Determine the refractive index of the liquid.
QUESTIONS
1. Explain constructive and destructive interference. Explain in detail Young's double slit
experiment.
2. Discuss the necessary theory of interference in thin films.
3. Discuss the formation of Newton’s rings and calculate their diameters. How do you determine
the wavelength of monochromatic light using Newton’s rings ?
4. What is interference of light ? Deduce the conditions for maxima and minima of interference
fringes formed by thin films.