Question

In Young’s double slit experiment, find the ratio of intensities at two points on a screen when waves emanating from two slits reaching these points have path differences (i) \(\frac{\lambda}{6}\) and (ii) \(\frac{\lambda}{12}\).

Hint:

In Young's double slit experiment, slight changes in the path difference lead to significant changes in intensity due to the sensitive dependence on the cosine squared relationship.

Answer 1

Step 1: The phase difference \( \phi_1 \) for path difference \( \frac{\lambda}{6} \) is calculated as: \[ \phi_1 = \frac{2 \pi}{\lambda} \times \frac{\lambda}{6} = \frac{\pi}{3} \] The intensity \( I_1 \) for this path difference is given by: \[ I_1 = 4 I_0 \cos^2 \frac{\phi_1}{2} \] \[ I_1 = 4 I_0 \cos^2 \left( \frac{\pi}{6} \right) \] Since \( \cos \left( \frac{\pi}{6} \right) = \frac{\sqrt{3}}{2} \), we get: \[ I_1 = 3 I_0 \] Step 2: For the second path difference \( \frac{\lambda}{12} \), the phase difference \( \phi_2 \) is calculated as: \[ \phi_2 = \frac{2 \pi}{\lambda} \times \frac{\lambda}{12} = \frac{\pi}{6} \] The intensity \( I_2 \) for this path difference is: \[ I_2 = 4 I_0 \cos^2 \frac{\phi_2}{2} \] \[ I_2 = 4 I_0 \cos^2 \left( \frac{\pi}{12} \right) \] Using \( \cos \left( \frac{\pi}{12} \right) \approx 0.9659 \), we get: \[ I_2 = 4 I_0 \times (0.9659)^2 = 4 I_0 \times 0.933 \] Thus, \( I_2 = 4 I_0 \times 0.933 \). Step 3: The ratio of intensities is: \[ \frac{I_1}{I_2} = \frac{3 I_0}{4 I_0 \cos^2 15^\circ} \] \[ \frac{I_1}{I_2} = \frac{3}{4 \cos^2 15^\circ} \] Thus, the ratio of intensities is \( \frac{3}{4 \cos^2 15^\circ} \).