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

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.

  1. A \(1: 4\)
  2. B \(3: 1\)
  3. C \(4: 1\)
  4. D \(2: 1\)
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Answer & Solution

Correct Answer

(C) \(4: 1\)

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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 \(I \propto A ^{2}\)…
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