Question

In Young's double slit experiment, carried out with light of wavelength 5000Å, the distance between the slits is 0.3 mm and the screen is at 200 cm from the slits. The central maximum is at x = 0 cm. The value of x for third maxima is ............. mm.

Answer 1

Correct Answer: 10

The fringe width, \(\beta\), is calculated as \(\beta = \frac{\lambda D}{d} = \frac{5 \times 10^{-7} \times 2}{3 \times 10^{-4}} = 10 \times 10^{-3} \, \text{m}\).

For the 3rd maxima:

The position of the 3rd maxima, \(y_3\), is \(y_3 = 3\beta = 10 \times 10^{-3} \, \text{m} = 10 \, \text{mm}\).