Question

In a Young’s double-slit experiment, the intensity at the central maximum in the interference pattern on the screen is \( I_0 \). Find the intensity at a point on the screen where the path difference between the interfering waves is \( \frac{\lambda}{6} \).

Hint:

In Young's double-slit experiment, when a path difference \( \Delta x \) is given, convert it to phase difference using \( \delta = \frac{2\pi}{\lambda} \Delta x \). Use the cosine squared formula to compute intensity.

Answer 1

The intensity at a point in Young's double-slit experiment is given by: \[ I = I_0 \cos^2\left(\frac{\delta}{2}\right) \] where \( \delta \) represents the phase difference between the two waves. Given a path difference of \( \frac{\lambda}{6} \), the corresponding phase difference is: \[ \delta = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{6} = \frac{2\pi}{6} = \frac{\pi}{3} \] The intensity can now be calculated as: \[ I = I_0 \cos^2\left(\frac{\pi}{6}\right) = I_0 \left(\cos\frac{\pi}{6}\right)^2 = I_0 \left(\frac{\sqrt{3}}{2}\right)^2 = I_0 \cdot \frac{3}{4} \] Therefore, the required intensity is: \[ \boxed{I = \frac{3}{4} I_0} \]