Winter 2016 - Q.1 [Preview] - Grad Plus

Winter 2016 – Q.1 [Preview]

1. (a) Derive expression for path difference in reflected light and derive the conditions for constructive and destructive interference for a film of uniform thickness. [6M]

When a monochromatic light beam is incident on a transparent parallel thin film of uniform
thickness t’ the rays reflected from the top and bottom of the film interfere to from interference pattern.
These fringes are equally spaced alternate dark and bright bands.

The thin film interference is formed by division of amplitude of the ray which is partially
reflected from the top and partially from the bottom of the film. Considered a transparent film of
thickness ‘t’ and refractive index ‘μ’

Let ray AB is partly reflected along BC and partly refracted along BM. At M part of it is reflected
from lower surface of the film along MD and finally emerges along MK
These rays BC and DE interfere and interference fringes are produced. The intensity at any Point
depends on the path difference between the interfering rays.

Let GM be the perpendicular from M on BD and HD be perpendicular from D on BC and
Determination of path difference between the rays BM and MD. As shown in fig. 1

image 1 - Grad Plus

The geometric path difference between ray 1 and ray 2 = MF + FD – BH

Optical path difference = Δ

$latex \begin{array}{l}=\mu(BM+FD)-\mu BH\\\\=\mu\left(BM+FD\right)-BH\left(for\;air\;\mu=1\right)……….\left(1\right)\end{array}$

$latex \begin{array}{l}In\;the\;\triangle BMD,\angle BMG=\angle GMD=\angle r\\Further,\end{array}$

$latex \cos\;r=\frac{MG}{BM}\;\;\;\;\;\;;\;\;\;\;\;BM=\frac{MG}{\cos\;r}$

But MG=t

$latex BM=\frac t{\cos\;r}\;……….\left(2\right)$

Similarly

$latex MD=\frac t{\cos\;r}\;……….\left(3\right)$

Now, In ΔBHD

$latex \begin{array}{l}\sin\;i=\frac{BH}{BD}\\\\BH=BD\;\sin\;r\;\;……….\left(4\right)\\\\BD=BG+GD\;\;………\left(5\right)\end{array}$

But
Therefore, In ΔBGM

$latex \begin{array}{l}\tan\;r=\frac{BG}{MG}\\\\BG=MG\;\tan\;r\\\\BG=t\;\tan\;r\end{array}$

Similarly, GD=t tan r
Put the values of BG and GD in equation 5, we get

$latex \begin{array}{l}BD=t\;\tan\;r+t\;\tan\;r\\\\BD=2t\;\tan\;r\end{array}$

Therefore, equation 4 becomes,

$latex BH=2t\;\tan\;r\;\sin\;i\;\;……….\left(6\right)$

Put equation 2,3 and 6 in equation 1 we get

$latex \begin{array}{l}\triangle=\mu\left(\frac t{\cos\;r}+\frac t{\cos\;r}\right)-2t\;\tan\;r\;\sin\;i\\\\\triangle=\mu\left(\frac{2t}{\cos\;r}\right)-2t\frac{\sin\;r}{\cos\;r}\sin\;i\end{array}$

But

$latex \begin{array}{l}\mu=\frac{\sin\;i}{\sin\;r}\\\\\sin\;i=\;\mu\;\sin\;r\\\\\triangle=\mu\left(\frac{2t}{\cos\;r}\right)-\frac{2\mu t}{\cos\;r}\sin^2r\\\\\triangle=\frac{2\;\mu t}{\cos\;r}\left(1-\sin^2\;r\right)\\\\\triangle=\frac{2\;\mu t}{\cos\;r}\cos^2r\\\\\triangle=2\;\mu t\;\cos\;r\end{array}$

When light is reflected from the surface of an optically denser medium, a phase change π is
introduced. Correspondingly, path difference λ/2 is introduced, therefore
The effective path difference = ∆ = 2μ t cos r – λ/2

Condition for Brightness and Darkness:
1. When path difference ∆ = n λ, where n = 0,1,2,………
Constructive interference takes place and film appears bright.
Therefore 2μ t cos r – λ/2 = n λ (for maximum intensity)
2μ t cos r = (2n +1) λ/2

2. When path difference ∆ = (2n+1) λ/2, where n = 0, 1, 2,
Destructive interference takes place and film appears dark.
Therefore 2μ t cos r – λ/2 = (2n+1) λ/2
2μ t cos r = (n+1) λ

Where n is integer, therefore (n+1) can be written as n
Therefore,
2μt cos r = n λ (for minimum intensity)
The number n is called the order of interference.


(b) Explain any one application of ultrasonic waves. [3M]

Ultrasonic waves are extensively used in marine application like Echo Sounder. Ultrasonic waves
can be directed like light waves and they are enable to travel long distance in water because of their high
energy as compared to sound waves of normal frequency. The depth of the ocean is determined using an
echo sounder which uses a ultrasonic source of sound and a receiver of definite frequency. The instrument is kept at the bottom of the ship. The source sends out short pulses of ultrasonic sound and the receiver receives it. By simply measuring the time interval between sent and received pulse, the depth of the ocean can be determined using the formula,

$latex l\;=\frac{\;vt}2$


(c) The average reverberation time of a hall is 1.5 sec and the area of the interior surface is 3340 m2. If the volume of the hall is 13000 m3, find the absorption coefficient. [3M]

$latex \begin{array}{l}We\;have,\;\\T\;=\;0.161\;V/aS\;\;\\\\a\;=\;0.161\;V/T\;S\;\\\\\;\;\;=\;0.161\;x\;13000/1.5\;x\;3340\;\\\\\boxed{a\;=0.4177}\end{array}$


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