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Question

A narrow monochromatic beam of light of intensity I is incidents on a glass plate as shown, in the figure. Another identical glass plate is kept close to the first-one and parallel to it. Each glass plate reflects 25% of the light incident on it and transmits the remaining. Then, the ratio of the maximum and minimum intensities in the interference pattern formed by the two beams obtained after one reflection at each plate is
154976.JPG

A
7:1
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B
49:1
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C
4:1
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D
16:9
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Solution

The correct option is B 49:1
As given that at each reflection 2525% of incident energy is reflected, therefore at each refraction 7575% of incident energy will be refracted.
Intensity of incident light =I=I
Hence, Intensity of reflected ray ABAB
IAB=25I/100=I/4....................eq1

Intensity of refracted ray ACAC
IAC=75I/100=3I/4
IAC=75I100=3I4

Intensity of reflected ray CACA′
ICA=25IAC/100=1/4×3I/4=3I/16

Intensity of refracted ray ABA′B′
IAB=75ICA/100=3/4×3I/16=9I/64 .......................eq2

Now, rays AB and A'B' interfere, therefore
ImaxImin=IAB+IAB+2IABIABIAB+IAB2IABIAB
ImaxImin=I/4+9I/64+2I/4×9I/64I/4+9I/642I/4×9I/64=491

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