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MHT CET · Physics · Wave Optics

In Young's double slit experiment, the intensity of light at a point on the screen where the path difference is \(\lambda\) is ' \(I\) '. The intensity at a point where the path difference is \(\lambda / 6\) is \(\left[\cos \frac{\pi}{6}=\frac{\sqrt{3}}{2}\right]\) \([\lambda=\) wavelength of light \(][\cos \pi=-1]\)

  1. A I
  2. B \(\frac{3 \mathrm{I}}{4}\)
  3. C \(\frac{\mathrm{I}}{2}\)
  4. D \(\frac{\mathrm{I}}{4}\)
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Answer & Solution

Correct Answer

(B) \(\frac{3 \mathrm{I}}{4}\)

Step-by-step Solution

Detailed explanation

\(I_{max} = I\) \(I' = I_{max} \cos^2\left(\frac{1}{2} \cdot \frac{2\pi}{\lambda} \cdot \frac{\lambda}{6}\right) = I \cos^2\left(\frac{\pi}{6}\right)\)
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