Verified Solution • 2025–26 Session moderate

QJEEM24Ph0f20: JEE Main 2024 Physics - Class 12 JEEM24Ph0f20 Solution: The width of one of the two slits in a Young's double slit…

The width of one of the two slits in a Young's double slit experiment is 4 times that of the other slit. The ratio of the maximum of the minimum intensity in the interference pattern is:

Multiple Choice Options

  • A 1:11: 1
  • B 4:14: 1
  • C 16:116: 1
  • D 9:19: 1

Step-by-Step Verified Solution

amp;Imax⁡=(4I0+I0)2=(3I0)2=9I0amp;Imin⁡=(4I0−I0)2=I0amp;Imax⁡Imin⁡=9:1\begin{aligned}& I_{\max }=\left(\sqrt{4 I_0}+\sqrt{I_0}\right)^2=\left(3 \sqrt{I_0}\right)^2=9 I_0 \\& I_{\min }=\left(\sqrt{4 I_0}-\sqrt{I_0}\right)^2=I_0 \\& \frac{I_{\max }}{I_{\min }}=9: 1\end{aligned}

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