In the given figure,light of intensity I is i | vedclass

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(D) Let $I$ be the intensity of the incident light.At point $A$,the first reflected ray $AB$ has intensity $I_1 = 25\% 	ext{ of } I = \frac{I}{4}$.Th

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$49$

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(D) Let $I$ be the intensity of the incident light.At point $A$,the first reflected ray $AB$ has intensity $I_1 = 25\% 	ext{ of } I = \frac{I}{4}$.The transmitted ray enters the slab,reflects at point $C$ (bottom surface),and then reaches point $A'$ (top surface).At point $C$,$25\%$ of the incident light is reflected. The intensity of light reaching $C$ is $I - I_1 = \frac{3I}{4}$.Reflected intensity at $C = 25\% 	ext{ of } \frac{3I}{4} = \frac{1}{4} 	imes \frac{3I}{4} = \frac{3I}{16}$.At point $A'$,this light is partially transmitted to form ray $A'B'$. Since $25\%$ is reflected at $A'$,$75\%$ is transmitted.Intensity $I_2 = 75\% 	ext{ of } \frac{3I}{16} = \frac{3}{4} 	imes \frac{3I}{16} = \frac{9I}{64}$.Now,the ratio of maximum to minimum intensity is given by:$\frac{I_{max}}{I_{min}} = \frac{(\sqrt{I_1} + \sqrt{I_2})^2}{(\sqrt{I_1} - \sqrt{I_2})^2} = \frac{(\sqrt{\frac{I}{4}} + \sqrt{\frac{9I}{64}})^2}{(\sqrt{\frac{I}{4}} - \sqrt{\frac{9I}{64}})^2} = \frac{(\frac{1}{2} + \frac{3}{8})^2}{(\frac{1}{2} - \frac{3}{8})^2} = \frac{(\frac{4+3}{8})^2}{(\frac{4-3}{8})^2} = \frac{(\frac{7}{8})^2}{(\frac{1}{8})^2} = \frac{49}{1}$.

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