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JEE Mains · Physics · STD 12 - 10. Wave optics

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 _______.

  1. A \(9: 1\)
  2. B \(16: 1\)
  3. C  \(1: 1\)
  4. D \(4: 1\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(9: 1\)

Step-by-step Solution

Detailed explanation

Since, Intensity \(\propto\) width of slit \((\omega)\) \(\text { so, } I_1=I, I_2=4 \mathrm{I}\) \(I_{\min }=\left(\sqrt{I_1}-\sqrt{I_2}\right)^2=I\) \(I_{\max }=\left(\sqrt{I_1}+\sqrt{I_2}\right)^2=9 \mathrm{I}\)…
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