Q. In a Young’s double slit experiment, $I_{o}$ is the intensity at the central maximum and $\beta$ is the fringe width. The intensity at a point $P$ distant $x$ from the centre will be
A) $I_{o}\cos\frac{\pi x}{\beta}$
B) $4I_{o}\cos^{2}\frac{\pi x}{\beta}$
C) $I_{o}\cos^{2}\frac{\pi x}{\beta}$
D) $\frac{I_{o}}{4}\cos^{2}\frac{\pi x}{\beta}$
Explanation
(c) Path difference at point $P = \frac{xd}{D}$
Phase difference at point $P = \frac{2\pi}{\lambda}\frac{xd}{D} = \frac{2\pi x}{\beta}$
$I_{0} = 4I_{1}$, intensity at point $P$
$I = I_{1} + I_{1} + 2I_{1}\cos\frac{2\pi x}{\beta} = 2I_{1}\left\lbrack 1 + \cos\frac{2\pi x}{\beta} \right\rbrack$
$= I_{0}\cos^{2}\frac{\pi x}{\beta}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.