Question

In Young's double slit experiment, the intensity at a point where the path difference is \( \frac{\lambda}{6} \) ( \( \lambda \) being the wavelength of the light used) is \( I \). If \( I_0 \) denotes the maximum intensity, \( \frac{I}{I_0} \) is equal to

• A. \( \frac{1}{\sqrt{2}} \)
• B. \( \sqrt{\frac{3}{2}} \)
• C. \( \frac{3}{4} \)
• D. \( \frac{3}{4} \)

Hint:

In Young's double slit experiment, use the intensity formula \( I = I_0 \cos^2\left( \frac{\pi \Delta}{\lambda} \right) \) to find the intensity at any point, where \( \Delta \) is the path difference.

Answer 1

The Correct Option is D

In Young's double slit experiment, the intensity at a point is related to the path difference \( \Delta \) by the formula: \[ I = I_0 \cos^2\left( \frac{\pi \Delta}{\lambda} \right) \] Given that the path difference is \( \frac{\lambda}{6} \), we substitute this value into the formula: \[ I = I_0 \cos^2\left( \frac{\pi \times \frac{\lambda}{6}}{\lambda} \right) = I_0 \cos^2\left( \frac{\pi}{6} \right) \] Since \( \cos\left( \frac{\pi}{6} \right) = \frac{\sqrt{3}}{2} \), we get: \[ I = I_0 \left( \frac{\sqrt{3}}{2} \right)^2 = I_0 \times \frac{3}{4} \] Thus, \( \frac{I}{I_0} = \frac{3}{4} \).