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 \( \phi_1 \) corresponding to a path difference of \( \frac{\lambda}{6} \) is: \[\phi_1 = \frac{2 \pi}{\lambda} \times \frac{\lambda}{6} = \frac{\pi}{3}\]The intensity \( I_1 \) for this path difference is: \[I_1 = 4 I_0 \cos^2 \frac{\phi_1}{2}\]Substituting the value of \( \phi_1 \): \[I_1 = 4 I_0 \cos^2 \left( \frac{\pi}{6} \right)\]Given that \( \cos \left( \frac{\pi}{6} \right) = \frac{\sqrt{3}}{2} \), the intensity becomes: \[I_1 = 3 I_0\]Step 2: For the second path difference of \( \frac{\lambda}{12} \), the phase difference \( \phi_2 \) is: \[\phi_2 = \frac{2 \pi}{\lambda} \times \frac{\lambda}{12} = \frac{\pi}{6}\]The intensity \( I_2 \) for this path difference is: \[I_2 = 4 I_0 \cos^2 \frac{\phi_2}{2}\]Substituting the value of \( \phi_2 \): \[I_2 = 4 I_0 \cos^2 \left( \frac{\pi}{12} \right)\]Using the approximation \( \cos \left( \frac{\pi}{12} \right) \approx 0.9659 \): \[I_2 = 4 I_0 \times (0.9659)^2 = 4 I_0 \times 0.933\]Therefore, \( I_2 = 4 I_0 \times 0.933 \).Step 3: The ratio of the intensities \( I_1 \) to \( I_2 \) is calculated as: \[\frac{I_1}{I_2} = \frac{3 I_0}{4 I_0 \cos^2 15^\circ}\]Simplifying the expression: \[\frac{I_1}{I_2} = \frac{3}{4 \cos^2 15^\circ}\]Thus, the ratio of intensities is \( \frac{3}{4 \cos^2 15^\circ} \).