Verified Solution • 2025–26 Session moderate

QJEEM21Ph5b41: JEE Main 2021 Physics - Class 12 JEEM21Ph5b41 Solution: In a Young's double slit experiment, the width of the one of…

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.

Multiple Choice Options

  • A 4:01
  • B 2:01
  • C 1:04
  • D 3:01

Step-by-Step Verified Solution

Given, amplitude ∝\propto width of slit

⇒\Rightarrow A2 = 3A1

We know that,

Imax⁡Imin⁡=(I1+I2)2(I1−I2)2{{{I_{\max }}} \over {{I_{\min }}}} = {{{{(\sqrt {{I_1}} + \sqrt {{I_2}} )}^2}} \over {{{(\sqrt {{I_1}} - \sqrt {{I_2}} )}^2}}}

∵\because Intensity, I ∝\propto A2

∴\therefore Imax⁡Imin⁡=(A1+A2)2(∣A1−A2∣)2{{{I_{\max }}} \over {{I_{\min }}}} = {{{{({A_1} + {A_2})}^2}} \over {{{(|{A_1} - {A_2}|)}^2}}}

=(A1+3A1∣A1−3A1∣))2=(4A12A1)2=41= {\left( {{{{A_1} + 3{A_1}} \over {|{A_1} - 3{A_1}|)}}} \right)^2} = {\left( {{{4{A_1}} \over {2{A_1}}}} \right)^2} = {4 \over 1}

∴\therefore Imax⁡:Imin⁡=4:1{I_{\max }}:{I_{\min }} = 4:1

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