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Standard XII

Physics

Intensity Amplitude Relationship

A ray of ligh...

Question

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 $A B$ and $A^{'} B^{'}$ undergo interference. The ratio of $I_{m a x}$ and $I_{m i n}$ is

49 : 1

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7 : 1

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4 : 1

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8 : 1

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Solution

The correct option is B $49 : 1$
From figure $I_{1} = \frac{I}{4}$ and $I_{2} = \frac{9 I}{64} \Rightarrow \frac{I_{2}}{I_{1}} = \frac{9}{16}$
By using $\frac{I_{m a x}}{I_{m i n}} = {⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ \frac{\sqrt{\frac{I_{2}}{I_{1}}} + 1}{\sqrt{\frac{I_{2}}{I_{1}}} - 1} ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠}^{2} = {⎛ ⎜ ⎜ ⎜ ⎜ ⎝ \frac{\sqrt{\frac{9}{16}} + 1}{\sqrt{\frac{9}{16}} - 1} ⎞ ⎟ ⎟ ⎟ ⎟ ⎠}^{2} = \frac{49}{1}$ .

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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

The correct option is B 49:1From figure I1=I4 and I2=9I64⇒I2I1=916By using ImaxImin=⎛⎜ ⎜ ⎜ ⎜ ⎜ ⎜⎝√I2I1+1√I2I1−1⎞⎟ ⎟ ⎟ ⎟ ⎟ ⎟⎠2=⎛⎜ ⎜ ⎜ ⎜⎝√916+1√916−1⎞⎟ ⎟ ⎟ ⎟⎠2=491.

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 $A B$ and $A^{'} B^{'}$ undergo interference. The ratio of $I_{m a x}$ and $I_{m i n}$ is