Q 12-10-100JEE MainJEE Main 2022 (27 Jun, Shift 1)Medium
In Young's double slit experiment the two slits are $0.6$ mm distance apart. Interference pattern is observed on a screen at a distance $80$ cm from the slits. The first dark fringe is observed on the screen directly opposite to one of the slits. The wavelength of light will be ______ nm.
Numerical value type. Enter your answer.
Answer: 450
The point opposite a slit is at $y = \dfrac d2 = 0.3$ mm from the centre. The first dark fringe is at $y = \dfrac{\lambda D}{2d}$.
$$\frac{\lambda D}{2d} = \frac d2 \Rightarrow \lambda = \frac{d^2}{D} = \frac{(0.6\times10^{-3})^2}{0.8} = 4.5\times10^{-7}\ \text{m}$$
$$\lambda = 450\ \text{nm}$$
Solution by Sreeraj P, M.Sc Physics