A parallel beam of light of intensity I_0 is | vedclass

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(A) Let $I_0$ be the intensity of the incident light.Intensity reflected from the upper surface, $I_1 = 25 \% 	ext{ of } I_0 = \frac{I_0}{4}$.Intensi

Vedclass · Accepted Answer

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

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(A) Let $I_0$ be the intensity of the incident light.Intensity reflected from the upper surface, $I_1 = 25 \% 	ext{ of } I_0 = \frac{I_0}{4}$.Intensity transmitted into the glass plate = $I_0 - \frac{I_0}{4} = \frac{3I_0}{4}$.Intensity reflected from the lower surface, $I_2 = 50 \% 	ext{ of } \frac{3I_0}{4} = \frac{1}{2} 	imes \frac{3I_0}{4} = \frac{3I_0}{8}$.The ratio of maximum to minimum intensity is given by the formula:$\frac{I_{\max}}{I_{\min}} = \frac{(\sqrt{I_1} + \sqrt{I_2})^2}{(\sqrt{I_1} - \sqrt{I_2})^2} = \left( \frac{\sqrt{\frac{I_0}{4}} + \sqrt{\frac{3I_0}{8}}}{\sqrt{\frac{I_0}{4}} - \sqrt{\frac{3I_0}{8}}} ight)^2 = \left( \frac{\frac{1}{2} + \sqrt{\frac{3}{8}}}{\frac{1}{2} - \sqrt{\frac{3}{8}}} ight)^2$.

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