Q 12-10-068JEE MainJEE Main 2024 (9 Apr, Shift 1)Medium
In a Young's double slit experiment, the intensity at a point is $\left(\dfrac14\right)^{\text{th}}$ of the maximum intensity. The minimum distance of the point from the central maximum is ______ $\mu\text{m}$. (Given: $\lambda = 600\ \text{nm}$, $d = 1.0\ \text{mm}$, $D = 1.0\ \text{m}$)
Numerical value type. Enter your answer.
Answer: 200
$$I = I_{max}\cos^2\frac\phi2 = \frac{I_{max}}{4} \;\Rightarrow\; \frac\phi2 = \frac\pi3 \;\Rightarrow\; \phi = \frac{2\pi}{3}$$
Path difference $= \dfrac\lambda3 = \dfrac{yd}{D}$:
$$y = \frac{\lambda D}{3d} = \frac{600\times10^{-9}\times1}{3\times10^{-3}} = 2\times10^{-4}\ \text{m} = 200\ \mu\text{m}$$
Solution by Sreeraj P, M.Sc Physics