**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 tâ€™ 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 â€˜tâ€™ and refractive index â€˜Î¼â€™

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

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

Optical path difference = Î”

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

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

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

But MG=t

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

Similarly

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

Now, In Î”BHD

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

But

Therefore, In Î”BGM

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

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 Ï€ is

introduced. Correspondingly, path difference Î»/2 is introduced, therefore

The effective path difference = âˆ† = 2Î¼ t cos r â€“ Î»/2

**Condition for Brightness and Darkness:**

1. When path difference âˆ† = n Î», where n = 0,1,2,………

Constructive interference takes place and film appears bright.

Therefore 2Î¼ t cos r â€“ Î»/2 = n Î» (for maximum intensity)

2Î¼ t cos r = (2n +1) Î»/2

2. When path difference âˆ† = (2n+1) Î»/2, where n = 0, 1, 2,

Destructive interference takes place and film appears dark.

Therefore 2Î¼ t cos r â€“ Î»/2 = (2n+1) Î»/2

2Î¼ t cos r = (n+1) Î»

Where n is integer, therefore (n+1) can be written as n

Therefore,

2Î¼t cos r = n Î» (for minimum intensity)

The number n is called the order of interference.

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,

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

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

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