A ray of light of intensity I is incident on | vedclass

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(A) The intensity of the first reflected ray $AB$ is $I_1 = 25\% 	ext{ of } I = \frac{I}{4}$.The intensity of the second ray $A'B'$ is determined by

Vedclass · Accepted Answer

$49$

Vedclass · Answer

(A) The intensity of the first reflected ray $AB$ is $I_1 = 25\% 	ext{ of } I = \frac{I}{4}$.The intensity of the second ray $A'B'$ is determined by the transmission and reflection processes:$1$. At point $A$, $75\%$ of $I$ is transmitted into the slab: $I_{trans} = \frac{3}{4}I$.$2$. At point $C$, $25\%$ of $I_{trans}$ is reflected: $I_{refl} = \frac{1}{4} 	imes \frac{3}{4}I = \frac{3}{16}I$.$3$. At point $A'$, $75\%$ of $I_{refl}$ is transmitted out of the slab: $I_2 = \frac{3}{4} 	imes \frac{3}{16}I = \frac{9}{64}I$.The ratio of maximum to minimum intensity is given by:$\frac{I_{max}}{I_{min}} = \left( \frac{\sqrt{I_1} + \sqrt{I_2}}{\sqrt{I_1} - \sqrt{I_2}} ight)^2$Substituting the values:$\sqrt{I_1} = \sqrt{\frac{I}{4}} = \frac{1}{2}\sqrt{I}$$\sqrt{I_2} = \sqrt{\frac{9}{64}I} = \frac{3}{8}\sqrt{I}$$\frac{I_{max}}{I_{min}} = \left( \frac{\frac{1}{2} + \frac{3}{8}}{\frac{1}{2} - \frac{3}{8}} ight)^2 = \left( \frac{\frac{4+3}{8}}{\frac{4-3}{8}} ight)^2 = \left( \frac{7}{1} ight)^2 = \frac{49}{1}$

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