Thermodynamics

JEE Main level, numerical answer, medium

119 of 190

Question 119 of 190

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:4x:4 where xx is _______.

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Answer: 1

Given amplitude ∝\propto slit width
Also intensity ∝( Amplitude )2∝( Slit width )2\propto(\text { Amplitude })^2 \propto(\text { Slit width })^2
I1I2=(31)2=9⇒I1=9I2\frac{I_1}{I_2}=\left(\frac{3}{1}\right)^2=9 \Rightarrow I_1=9 I_2
Imin⁡Imax⁡=(I1−I2I1+I2)2=(3−13+1)2=14=x4\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}
⇒x=1\Rightarrow x=1

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