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

The intensity in Young’s double slit experiment is given by:

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

Where:

- \( I_0 \) is the intensity of the central maximum,

- \( \Delta x \) is the path difference between the two waves,

- \( \lambda \) is the wavelength of the light.

Given: \( \Delta x = \frac{\lambda}{3} \), we substitute into the equation:

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

Since \( \cos\left( \frac{\pi}{3} \right) = \frac{1}{2} \), we get:

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

Final Answer:
The intensity at the point where the path difference is \( \frac{\lambda}{3} \) is: \[ I = \frac{I_0}{4} \]