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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 three times the other slit. If the amplitude of the light coming from a slit is proportional to the slit-width, the ratio of minimum to maximum intensity in the interference pattern is \(x: 4\) where \(x\) is ..... .

  1. A \(1\)
  2. B \(1.4\)
  3. C \(14\)
  4. D \(10\)
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Answer & Solution

Correct Answer

(A) \(1\)

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Detailed explanation

Given amplitude \(\propto\) slit width Also intensity \(\propto(\text { Amplitude })^{2} \propto\) (Slit width \()^{2}\) \(\frac{{I}_{1}}{{I}_{2}}=\left(\frac{3}{1}\right)^{2}=9 \Rightarrow {I}_{1}=9 {I}_{2}\)…
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