Wave Optics

JEE Main level, single correct, medium

91 of 191

Question 91 of 191

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.

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Answer: (D) 4:14:1

⇒A2=3A1\Rightarrow {A}_{2}=3{A}_{1}
Imax⁡Imin⁡=(I1+I2∣I1−I2∣)2\dfrac{{I}_{\max }}{{I}_{\min }}={\left(\dfrac{\sqrt{{I}_{1}}+\sqrt{{I}_{2}}}{\left|\sqrt{{I}_{1}}-\sqrt{{I}_{2}}\right|}\right)}^{2}
∵∵ Intensity I∝A2I\propto {A}^{2}
⇒Imax⁡Imin=(A1+A2∣A1−A2∣)2\Rightarrow \dfrac{{I}_{\max }}{{I}_{min}}={\left(\dfrac{{A}_{1}+{A}_{2}}{\left|{A}_{1}-{A}_{2}\right|}\right)}^{2}
=(A1+3A1∣A1−3A1∣)2={\left(\dfrac{{A}_{1}+3{A}_{1}}{\left|{A}_{1}-3{A}_{1}\right|}\right)}^{2}
=(4A12A1)2=4:1={\left(\dfrac{4{A}_{1}}{2{A}_{1}}\right)}^{2}=4:1

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