Question

In Young's double slit experiment, the two slits are separated by 0.2 mm and they are 1m from the screen. The wavelength of the light used is 500 nm. The distance between 6th maxima and 10th minima on the screen is closest to

• A. 12 mm
• B. 10 mm
• C. 14 mm
• D. 8 mm

Hint:

In Young's double-slit experiment, the distance between maxima or minima is directly proportional to the wavelength and the distance from the slits to the screen.

Answer 1

The Correct Option is D

In Young's double-slit experiment, the distance between adjacent maxima (or minima) is given by: \[ y = \frac{\lambda L}{d} \] where \( \lambda \) is the wavelength, \( L \) is the distance from the slits to the screen, and \( d \) is the distance between the slits. The distance between the 6th maxima and the 10th minima is the distance between the 6th maxima and the 10th minima. The total distance is: \[ y = (6 - 10) \cdot \frac{\lambda L}{d} = 4 \cdot \frac{500 \times 10^{-9} \times 1}{0.2 \times 10^{-3}} = 8 \, \text{mm} \] Thus, the correct answer is (D).