Question

In Young's double slit experiment, the wavelength of the monochromatic light used is \( \lambda \), the distance between the slits is \( 5\lambda \) and the distance of the screen from the plane of the slits is 100 cm. If the maximum intensity on the screen is \( I_0 \), then the intensity at a point on the screen which is at 5 cm from the central maximum is:

• A. \( \frac{I_0}{2} \)
• B. \( \frac{3I_0}{4} \)
• C. \( \frac{I_0}{4} \)
• D. \( I_0 \)

Hint:

In Young's double slit experiment, the intensity at any point is based on the cosine squared of the angle \( \theta \), which depends on the distance and wavelength. If the distance is small enough, it affects the intensity proportionally.

Answer 1

The Correct Option is A

We know that the intensity at a point in Young’s double slit experiment is given by: \[ I = I_0 \cos^2 \left( \frac{\pi d \sin \theta}{\lambda} \right) \] Where \( d \) is the distance between the slits, \( \lambda \) is the wavelength, and \( \theta \) is the angle of diffraction. The angle \( \theta \) for a distance \( x = 5 \, \text{cm} \) from the central maximum is given by: \[ \theta = \frac{x}{L} = \frac{5}{100} = 0.05 \] Substitute the values for \( d = 5\lambda \) and the given value for \( \theta \): \[ I = I_0 \cos^2 \left( \frac{\pi \cdot 5\lambda \cdot 0.05}{\lambda} \right) \] \[ I = I_0 \cos^2 \left( \frac{\pi \cdot 0.25}{1} \right) = I_0 \cos^2 \left( \frac{\pi}{4} \right) \] Since \( \cos^2 \left( \frac{\pi}{4} \right) = \frac{1}{2} \), the intensity is: \[ I = \frac{I_0}{2} \]