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 interference patterns, the intensity is determined by the phase difference, which depends on the path difference between the waves.

Answer 1

The intensity of an interference pattern is quantified by \( I = I_0 \left( 1 + \cos \delta \right) \). The phase difference, \( \delta \), is determined by \( \delta = \frac{2\pi}{\lambda} \cdot \text{path difference} \). Given a path difference of \( \frac{\lambda}{8} \), the phase difference calculates to \( \delta = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{8} = \frac{\pi}{4} \). Consequently, the intensity is \( I = I_0 \left( 1 + \cos \frac{\pi}{4} \right) = I_0 \left( 1 + \frac{\sqrt{2}}{2} \right) \), which simplifies to \( I = I_0 \left( 1 + 0.707 \right) = 1.707 I_0 \). The intensity at this specific point is therefore \( 1.707 I_0 \).