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MHT CET · Physics · Wave Optics

In Young's double slit experiment, the light of wavelength ' \(\lambda\) ' is used. The intensity at a point on the screen is 'I' where the path difference is \(\frac{\lambda}{4}\). If ' \(\mathrm{I}_{\mathrm{o}}\) ' denotes the maximum intensity then the ratio of ' \(\mathrm{I}_{\mathrm{o}}\) ' to ' I ' is \(\left(\cos 45^{\circ}=1 / \sqrt{2}\right)\)

  1. A \(2:1\)
  2. B \(4:1\)
  3. C \(8:1\)
  4. D \(12:1\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(2:1\)

Step-by-step Solution

Detailed explanation

\( \phi = \frac{2\pi}{\lambda} \Delta x = \frac{2\pi}{\lambda} \left( \frac{\lambda}{4} \right) = \frac{\pi}{2} \) \( I = I_o \cos^2 \left( \frac{\phi}{2} \right) = I_o \cos^2 \left( \frac{\pi/2}{2} \right) = I_o \cos^2 \left( \frac{\pi}{4} \right) \)