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Q.

A ray of light of intensity I is incident on a parallel glass-slab at a point A as shown in fig. It undergoes partial reflection and refraction. At each reflection 25% of incident energy is reflected. The rays AB and A'B' undergo interference. The ratio $I_{\max} / I_{\min}$ is

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a

4 : 1

b

8 : 1

c

7 : 1

d

49 : 1

answer is D.

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$F r o m f i g u r e I_{1} = \frac{I}{4} a n d I_{2} = \frac{9 I}{64} \Rightarrow \frac{I_{2}}{I_{1}} = \frac{9}{16} B y u \sin g, \frac{I_{\max}}{I_{\min}} = {(\frac{\sqrt{\frac{I_{2}}{I_{1}}} + 1}{\sqrt{\frac{I_{2}}{I_{1}}} - 1})}^{2} = {(\frac{\sqrt{\frac{9}{16}} + 1}{\sqrt{\frac{9}{16}} - 1})}^{2} = \frac{49}{1}$

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