Q 12-10-057JEE MainJEE Main 2024 (5 Apr, Shift 1)Easy
In Young's double slit experiment carried out with light of wavelength $5000\ \text{\AA}$, the distance between the slits is $0.3\ \text{mm}$ and the screen is at $200\ \text{cm}$ from the slits. The central maximum is at $x = 0\ \text{cm}$. The value of $x$ for the third maximum is ______ mm.
Numerical value type. Enter your answer.
Answer: 10
$$x_3 = \frac{3\lambda D}{d} = \frac{3\times5\times10^{-7}\times2}{3\times10^{-4}} = 10^{-2}\ \text{m} = 10\ \text{mm}$$
Solution by Sreeraj P, M.Sc Physics