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JEE MAIN 2021
01-09-21 S2
Question
Proportional to the slit-width, the ratio of minimum to maximum intensity in the interference pattern is x : 4 where x is The width of one of the two slits in a Young's double slit experiment is three times the other slit. If the amplitude of the light coming from a slit is pr
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Solution
Given amplitude $\propto$ slit width Also intensity $\propto(\text { Amplitude })^2 \propto(\text { Slit width })^2$ $$ \begin{aligned} & \frac{I_1}{I_2}=\left(\frac{3}{1}\right)^2=9 \Rightarrow I_1=9 I_1 \\ & \frac{I_{\max }}{I_{\max }}-\left(\frac{\sqrt{I_1}-\sqrt{I_2}}{\sqrt{I_1}+\sqrt{I_2}}\right)^2-\left(\frac{3-1}{3+1}\right)^2-\frac{1}{4}-\frac{x}{4} \\ & \Rightarrow x=1.00 \end{aligned} $$
Question Tags
JEE Main
Physics
Easy
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