In a Young's double slit experiment, the width of the one of the slit is three times the other slit. The amplitude of the light coming from a slit is proportional to the slit-width. Find the ratio of the maximum to the minimum intensity in the interference pattern.
Solution
Given, amplitude $\propto$ width of slit<br/><br/>$\Rightarrow$ A<sub>2</sub> = 3A<sub>1</sub><br/><br/>We know that,<br/><br/>$${{{I_{\max }}} \over {{I_{\min }}}} = {{{{(\sqrt {{I_1}} + \sqrt {{I_2}} )}^2}} \over {{{(\sqrt {{I_1}} - \sqrt {{I_2}} )}^2}}}$$<br/><br/>$\because$ Intensity, I $\propto$ A<sup>2</sup><br/><br/>$\therefore$ $${{{I_{\max }}} \over {{I_{\min }}}} = {{{{({A_1} + {A_2})}^2}} \over {{{(|{A_1} - {A_2}|)}^2}}}$$<br/><br/>$$ = {\left( {{{{A_1} + 3{A_1}} \over {|{A_1} - 3{A_1}|)}}} \right)^2} = {\left( {{{4{A_1}} \over {2{A_1}}}} \right)^2} = {4 \over 1}$$<br/><br/>$\therefore$ ${I_{\max }}:{I_{\min }} = 4:1$
About this question
Subject: Physics · Chapter: Optics · Topic: Reflection and Mirrors
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