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MHT CET · Physics · Wave Optics

The intensity of light coming from one of the slits in Young's double slit experiment is double the intensity from the other slit. The ratio of the maximum intensity to the minimum intensity in the interference fringe pattern observed is

  1. A \(9: 1\)
  2. B \(34: 1\)
  3. C \(4: 1\)
  4. D \(16: 1\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(34: 1\)

Step-by-step Solution

Detailed explanation

Two coherent sources of intensities \(I_1\) and \(I_2\) produce,
maximum intensity, \(\mathrm{I}_{\max }=\left(\sqrt{I_1}+\sqrt{I_2}\right)^2\) and minimum intensity, \(I_{\text {min }}=\left(\sqrt{I_1}-\sqrt{I_2}\right)^2\) in an interference pattern.
\(\frac{I_{\text {max }}}{I_{\text {min }}}=\frac{\left(\sqrt{I_1}+\sqrt{I_2}\right)^2}{\left(\sqrt{I_1}-\sqrt{I_2}\right)^2}=\frac{I_1+I_2+2 \sqrt{I_1 I_2}}{I_1+I_2-2 \sqrt{I_1 I_2}}\)
Given \(\mathrm{I}_1=2 \mathrm{I}_2\)
\(\therefore \quad \frac{I_{\max }}{I_{\min }}=\frac{3 \mathrm{I}_2+2 \sqrt{2} \mathrm{I}_2}{3 \mathrm{I}_2-2 \sqrt{2} \mathrm{I}_2}=\frac{3+2 \sqrt{2}}{3-2 \sqrt{2}}=34\)
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