Wave Optics

JEE Main level, single correct, medium

62 of 191

Question 62 of 191

In a Young’s double slit experiment, the ratio of amplitude of light coming from slits is 2:12:1. The ratio of the maximum to minimum intensity in the interference pattern is

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Answer: (D) 9:19:1

The ratio of the maximum to minimum intensity is given by
Imax⁡Imin⁡=(I1+I2)2(I1−I2)2\dfrac{{I}_{\max }}{{I}_{\min }}=\dfrac{{\left(\sqrt{{I}_{1}}+\sqrt{{I}_{2}}\right)}^{2}}{{\left(\sqrt{{I}_{1}}-\sqrt{{I}_{2}}\right)}^{2}}
It is given that A1A2=21\dfrac{{A}_{1}}{{A}_{2}}=\dfrac{2}{1}
ImaxImin=(A1+A2A1−A2)2=A22(2+1)2A22(2−1)2=(2+12−1)2=91\dfrac{{I}_{max}}{{I}_{min}}={\left(\dfrac{{A}_{1}+{A}_{2}}{{A}_{1}-{A}_{2}}\right)}^{2}=\dfrac{{{A}^{2}}_{2}{\left(2+1\right)}^{2}}{{{A}^{2}}_{2}(2-1{)}^{2}}={\left(\dfrac{2+1}{2-1}\right)}^{2}=\dfrac{9}{1}

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