The width of one of the two slits in a Young's double slit experiment is three times the other slit. If…

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 proportional to the slit-width, the ratio of minimum to maximum intensity in the interference pattern is x: 4 where x is _______.

Solution

Given amplitude $\propto$ slit width
Also intensity $\propto(\text { Amplitude })^2 \propto(\text { Slit width })^2$
$\frac{I_1}{I_2}=\left(\frac{3}{1}\right)^2=9 \Rightarrow I_1=9 I_2$
$\frac{I_{\min }}{I_{\max }}=\left(\frac{\sqrt{I_1}-\sqrt{I_2}}{\sqrt{I_1}+\sqrt{I_2}}\right)^2=\left(\frac{3-1}{3+1}\right)^2=\frac{1}{4}=\frac{x}{4}$
$\Rightarrow x=1$

Asked in: JEE Main 2021 (01 Sep Shift 2)

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