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(D) At point $A$,the incident intensity is $I$. Since $25\%$ is reflected,the intensity of ray $AB$ is $I_1 = 0.25I = I/4$. The intensity of the refra

Vedclass · Accepted Answer

$49$

Vedclass · Answer

(D) At point $A$,the incident intensity is $I$. Since $25\%$ is reflected,the intensity of ray $AB$ is $I_1 = 0.25I = I/4$. The intensity of the refracted ray is $0.75I = 3I/4$.At point $C$,the ray is reflected. Again,$25\%$ of the incident energy $(3I/4)$ is reflected. So,the intensity of the ray moving towards $A'$ is $0.25 	imes (3I/4) = 3I/16$.At point $A'$,this ray is refracted. The intensity of the refracted ray $A'B'$ is $0.75 	imes (3I/16) = 9I/64$. Thus,$I_2 = 9I/64$.The ratio of intensities is $I_2/I_1 = (9I/64) / (I/4) = 9/16$.The ratio of maximum to minimum intensity is given by:$\frac{I_{\max}}{I_{\min}} = \left( \frac{\sqrt{I_1} + \sqrt{I_2}}{\sqrt{I_1} - \sqrt{I_2}} ight)^2 = \left( \frac{1 + \sqrt{I_2/I_1}}{1 - \sqrt{I_2/I_1}} ight)^2$Substituting $I_2/I_1 = 9/16$:$\frac{I_{\max}}{I_{\min}} = \left( \frac{1 + \sqrt{9/16}}{1 - \sqrt{9/16}} ight)^2 = \left( \frac{1 + 3/4}{1 - 3/4} ight)^2 = \left( \frac{7/4}{1/4} ight)^2 = (7)^2 = 49/1$.

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