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JEE Mains · Physics · STD 12 - Wave optics
  • JEE Mains 2021
  • Hard
  • MCQ

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 & Solution

Correct Answer

(C) \(4: 1\)

Step-by-step Solution

Detailed explanation

Amplitude \(\propto\) Width of slit \(\Rightarrow A _{2}=3 A _{1}\)

\(\frac{ I _{\max }}{ I _{\min }}=\left(\frac{\sqrt{ I _{1}}+\sqrt{ I _{2}}}{\left|\sqrt{ I _{1}}-\sqrt{ I _{2}}\right|}\right)^{2}\)

\(\because\) Intensity \(I \propto A ^{2}\)

\(\Rightarrow \frac{I_{\max }}{I_{\min }}=\left(\frac{A_{1}+A_{2}}{\left|A_{1}-A_{2}\right|}\right)^{2}\)

\(=\left(\frac{ A _{1}+3 A _{1}}{\left| A _{1}-3 A _{1}\right|}\right)^{2}\)

\(=\left(\frac{4 A _{1}}{2 A _{1}}\right)^{2}=4: 1\)

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