Question

In Young’s double slit experimental set-up, the intensity of the central maximum is \( I_0 \). Calculate the intensity at a point where the path difference between two interfering waves is \( \frac{\lambda}{3} \).

Answer 1

Young's double-slit experiment intensity formula:

\[ I = I_0 \cos^2\left( \frac{\pi \Delta x}{\lambda} \right) \]

Definitions:

- \( I_0 \): Intensity of the central maximum.

- \( \Delta x \): Path difference.

- \( \lambda \): Wavelength.

Given: \( \Delta x = \frac{\lambda}{3} \). Substitution yields:

\[ I = I_0 \cos^2\left( \frac{\pi}{3} \right) \]

With \( \cos\left( \frac{\pi}{3} \right) = \frac{1}{2} \), the intensity is:

\[ I = I_0 \left( \frac{1}{2} \right)^2 = \frac{I_0}{4} \]

Result:
For a path difference of \( \frac{\lambda}{3} \), the intensity is: \[ I = \frac{I_0}{4} \]