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A parallel beam of light of intensity \mathrm{\mathrm{I}} is incident on a glass plate. \mathrm{25 \%} of light is reflected in any reflection by upper surface and \mathrm{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

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


Option: 2

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


Option: 3

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


Option: 4

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


Answers (1)

best_answer

\mathrm{\begin{aligned} & \therefore \quad \frac{\mathrm{I}_1}{\mathrm{I}_2}=\frac{8}{9} \text { or } \sqrt{\frac{\mathrm{I}_1}{\mathrm{I}_2}}=\frac{2 \sqrt{2}}{3} \\\\ & \therefore \quad \frac{\mathrm{I}_{\max }}{\mathrm{I}_{\min }}=\left(\frac{\sqrt{\frac{\mathrm{I}_1}{\mathrm{I}_2}}+1}{\sqrt{\frac{\mathrm{I}_1}{\mathrm{I}_2}}-1}\right)=\left(\frac{2 \sqrt{2}+3}{2 \sqrt{2}-3}\right)=\left(\frac{1+\frac{3}{\sqrt{8}}}{1-\frac{3}{\sqrt{8}}}\right)^2 \end{aligned}}

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Rishi

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