Question

In a Young’s double-slit experiment, two light waves, each of intensity \( I_0 \), interfere at a point, having a path difference \( \frac{\lambda}{8} \) on the screen. Find the intensity at this point.

Hint:

In Young’s double-slit experiment, the phase difference \( \delta \) is related to the path difference by \( \delta = \frac{2\pi}{\lambda} \times \text{path difference} \).

Answer 1

The resulting intensity \( I \) of an interference pattern is expressed as: \[ I = I_1 + I_2 + 2 \sqrt{I_1 I_2} \cos \delta \] Given that \( I_1 = I_2 = I_0 \), the formula simplifies to: \[ I = 2 I_0 (1 + \cos \delta) \] The phase difference \( \delta \) is linked to the path difference by: \[ \delta = \frac{2\pi}{\lambda} \times \frac{\lambda}{8} = \frac{\pi}{4} \] Substituting this value of \( \delta \) into the intensity formula yields: \[ I = 2 I_0 \left(1 + \cos \frac{\pi}{4} \right) \] As \( \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}} \), the final intensity is: \[ I = I_0 \left( 1 + \frac{1}{\sqrt{2}} \right) \] Therefore, the intensity at this point is \( I_0 \left( 1 + \cos \frac{\pi}{4} \right) \).