Q. In Young’s double slit experiment intensity at a point is (1/4) of the maximum intensity. Angular position of this point is
A) $\sin^{- 1}{(\lambda/d)}$
B) $\sin^{- 1}{(\lambda/2d)}$
C) $\sin^{- 1}{(\lambda/3d)}$
D) $\sin^{- 1}{(\lambda/4d)}$
Explanation
(c) $I = 4I_{0}\cos^{2}{(\phi/2) \Rightarrow \phi = 2\pi/3}$
$\Rightarrow \Delta x \times (2\pi/\lambda) \Rightarrow 2\pi/3 = \lambda/3$
$\sin\theta = \Delta x/d \Rightarrow \sin\theta = \lambda/3d$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.