A ray of light of intensity I is incident on a parallel glass slab and refraction. At each reflection, 25% of incident energy is reflected. The rays AB and A′B′ undergo interference. The ratio of Imax and Imin is
A
49:1
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B
7:1
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C
4:1
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D
8:1
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Solution
The correct option is B49:1 From figure I1=I4 and I2=9I64⇒I2I1=916 By using ImaxImin=⎛⎜
⎜
⎜
⎜
⎜
⎜⎝√I2I1+1√I2I1−1⎞⎟
⎟
⎟
⎟
⎟
⎟⎠2=⎛⎜
⎜
⎜
⎜⎝√916+1√916−1⎞⎟
⎟
⎟
⎟⎠2=491.