A parallel beam of light of intensity I_0 is | vedclass

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Question

(A) Given that $25 \%$ of the total intensity of incident light is reflected from the upper surface. This implies that if the intensity of incident li

Vedclass · Accepted Answer

$\left[\frac{\frac{1}{2}+\sqrt{\frac{3}{8}}}{\frac{1}{2}-\sqrt{\frac{3}{8}}}\right]^2$

Vedclass · Answer

(A) Given that $25 \%$ of the total intensity of incident light is reflected from the upper surface. This implies that if the intensity of incident light is $I_0$,the intensity of the first reflected ray is $I_1 = 0.25 I_0 = \frac{I_0}{4}$.The intensity of light reaching the lower surface of the plate is $I_0 - 0.25 I_0 = 0.75 I_0 = \frac{3}{4} I_0$.Since $50 \%$ of this intensity is reflected from the lower surface,the intensity of the second reflected ray is $I_2 = 0.50 	imes \frac{3}{4} I_0 = \frac{3}{8} I_0$.The ratio of maximum to minimum intensity is given by the formula:$\frac{I_{\max }}{I_{\min }}=\frac{\left(\sqrt{I_1}+\sqrt{I_2}ight)^2}{\left(\sqrt{I_1}-\sqrt{I_2}ight)^2}$Substituting the values of $I_1$ and $I_2$:$\frac{I_{\max }}{I_{\min }}=\frac{\left(\sqrt{\frac{I_0}{4}}+\sqrt{\frac{3 I_0}{8}}ight)^2}{\left(\sqrt{\frac{I_0}{4}}-\sqrt{\frac{3 I_0}{8}}ight)^2} = \left(\frac{\frac{1}{2} \sqrt{I_0} + \sqrt{\frac{3}{8}} \sqrt{I_0}}{\frac{1}{2} \sqrt{I_0} - \sqrt{\frac{3}{8}} \sqrt{I_0}}ight)^2 = \left(\frac{\frac{1}{2}+\sqrt{\frac{3}{8}}}{\frac{1}{2}-\sqrt{\frac{3}{8}}}ight)^2$

$A$ parallel beam of light of intensity $I_0$ is incident on a glass plate. $25 \%$ of light is reflected by the upper surface and $50 \%$ of light is reflected from the lower surface. The ratio of maximum to minimum intensity in the interference region of the reflected rays is

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