Question

In Young's double-slit experiment using monochromatic light of wavelength \( \lambda \), the intensity of light at a point on the screen, where path difference is \( \lambda \), is \( K \) units. Find the intensity of light at a point on the screen where the path difference is \( \lambda / 6 \).

Hint:

When the path difference is not an integer multiple of \( \lambda \), use the formula for intensity and calculate using the cosine squared function for accurate results.

Answer 1

We know that the intensity at a point in Young's double-slit experiment is given by: \[ I = I_0 \cos^2 \left( \frac{\pi \Delta}{\lambda} \right), \] where \( I_0 \) is the maximum intensity and \( \Delta \) is the path difference. For path difference \( \Delta = \lambda \), the intensity is \( K \). Thus, we have: \[ K = I_0 \cos^2 \left( \pi \right) = I_0. \] Now, for the path difference \( \Delta = \lambda / 6 \), we calculate the intensity: \[ I = I_0 \cos^2 \left( \frac{\pi}{6} \right) = I_0 \times \left( \frac{\sqrt{3}}{2} \right)^2 = \frac{3}{4} I_0. \] Since \( K = I_0 \), the intensity at a point where the path difference is \( \lambda / 6 \) is: \[ I = \frac{3}{4} K. \]