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

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(B) Let the incident intensity be $I$. At point $A$,$25\%$ of $I$ is reflected as ray $AB$. So,$I_1 = 0.25I = I/4$. The intensity of the refracted ray

Vedclass · Accepted Answer

$49$

Vedclass · Answer

(B) Let the incident intensity be $I$. At point $A$,$25\%$ of $I$ is reflected as ray $AB$. So,$I_1 = 0.25I = I/4$. The intensity of the refracted ray is $0.75I$. At point $C$,this ray is reflected. $25\%$ of the incident energy $(0.75I)$ is reflected. So,the intensity of the ray reaching $A^1$ is $0.25 	imes 0.75I = 0.1875I = (3/16)I$. At point $A^1$,this ray is refracted out as $A^1B^1$. Since $25\%$ is reflected at $A^1$,$75\%$ of the incident energy is transmitted. So,$I_2 = 0.75 	imes (3/16)I = (3/4) 	imes (3/16)I = (9/64)I$. The amplitudes are $a_1 = \sqrt{I_1} = \sqrt{I/4} = (1/2)\sqrt{I}$ and $a_2 = \sqrt{I_2} = \sqrt{9I/64} = (3/8)\sqrt{I}$. The ratio of amplitudes is $a_1/a_2 = (1/2) / (3/8) = 4/3$. $I_{	ext{max}} / I_{	ext{min}} = (a_1 + a_2)^2 / (a_1 - a_2)^2 = ((4+3)/ (4-3))^2 = (7/1)^2 = 49/1$.

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