Question

In a Young's double slit experiment, the intensities at two points, for the path difference \( \frac{\lambda}{4} \) and \( \frac{\lambda}{3} \) (\( \lambda \) being the wavelength of light used) are \( I_1 \) and \( I_2 \) respectively. If \( I_0 \) denotes the intensity produced by each one of the individual slits, then \( \frac{I_1 + I_2}{I_0} \) is equal to:

• A. \( 3 \)
• B. \( 5 \)
• C. \( 7 \)
• D. \( 10 \)

Hint:

In Young's double slit experiment, intensity is given by \( I = 4I_0 \cos^2\left(\frac{\Delta \phi}{2}\right) \). Use the relation between path difference and phase difference to simplify calculations.

Answer 1

The Correct Option is A

Resultant intensity in Young's double slit experiment \[ I = 4I_0 \cos^2 \left( \frac{\Delta \phi}{2} \right) \] For path difference \( \lambda/4 \), phase difference: \[ \Delta \phi = \frac{2\pi}{\lambda} \times \frac{\lambda}{4} = \frac{\pi}{4} \] \[ \therefore I_1 = 4I_0 \cos^2 \left( \frac{\pi}{4} \right) = 2I_0 \] For path difference \( \lambda/3 \): \[ I_2 = 4I_0 \cos^2 \left( \frac{2\pi}{\lambda} \times \frac{\lambda}{3} \right) = I_0 \] \[ \therefore \frac{I_1 + I_2}{I_0} = 3 \]