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  • A parallel beam of light of intensity I is incident on a glass plate. 25 % of light is reflected in any reflection by upper surface and 50 % of light is reflected by any reflection from l

A parallel beam of light of intensity I is incident on a glass plate. 25 % of light is reflected in any reflection by upper surface and 50 % of light is reflected by any reflection from lower surface. Rest is refracted. The ratio of maximum to minimum intensity in interference region of reflected rays is:

 

Option: 1

\left(\frac{\frac{1}{2}+\sqrt{\frac{3}{8}}}{\frac{1}{2}-\sqrt{\frac{3}{8}}}\right)^2


Option: 2

\left(\frac{\frac{1}{4}+\sqrt{\frac{3}{8}}}{\frac{1}{2}-\sqrt{\frac{3}{8}}}\right)^2


Option: 3

\frac{5}{8}


Option: 4

\mathrm{\text { None of these }}


Answers (1)

best_answer

\mathrm{\begin{aligned} & I_1=\frac{I}{4} \text { and } I_2=\frac{9 I}{32} \\ & \therefore \frac{I_1}{I_2}=\frac{8}{9} \\ & \text { or } \sqrt{\frac{I_1}{I_2}}=\frac{2 \sqrt{2}}{3} \\ & \therefore \frac{I_{\max }}{I_{\min }}=\left(\frac{\sqrt{I_1} / I_2+1}{\sqrt{I_1} / I_2-1}\right)^2=\left(\frac{2 \sqrt{2}+3}{2 \sqrt{2}-3}\right)^2 \end{aligned}}

Posted by

Suraj Bhandari

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