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JEE Mains · Physics · STD 12 - Wave optics
  • JEE Mains 2024
  • Hard
  • MCQ

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 _______.

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Answer & Solution

Correct Answer

(A) \(9: 1\)

Step-by-step Solution

Detailed explanation

Since, Intensity \(\propto\) width of slit \((\omega)\)

\(\text { so, } I_1=I, I_2=4 \mathrm{I}\)

\(I_{\min }=\left(\sqrt{I_1}-\sqrt{I_2}\right)^2=I\)

\(I_{\max }=\left(\sqrt{I_1}+\sqrt{I_2}\right)^2=9 \mathrm{I}\)

\(\frac{I_{\max }}{I_{\min }}=\frac{9 \mathrm{I}}{\mathrm{I}}=\frac{9}{1}\)

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