Q 12-10-114JEE MainJEE Main 2021 (26 Feb, Shift 1)Easy
In a Young's double slit experiment two slits are separated by $2$ mm and the screen is placed one meter away. When a light of wavelength $500$ nm is used, the fringe separation will be:
Answer: (D) $0.25$ mm
$$\beta = \frac{\lambda D}{d} = \frac{500\times10^{-9}\times 1}{2\times10^{-3}} = 2.5\times10^{-4}\ \text{m} = 0.25\ \text{mm}$$
Solution by Sreeraj P, M.Sc Physics