Question

In a young's double slit experiment $I_0$ is the intensity at the central maximum and $\beta$ is the fringe width. The intensity at a point P distant $x$ from the centre point of the screen will be.

• A. $I_0\cos \frac{\pi x}{\beta }$
• B. $4I_0{\cos }^2\frac{\pi x}{\beta }$
• C. $I_0{\cos }^2\frac{\pi x}{\beta }$
• D. $\frac{I_0}{4}{\cos }^2\frac{\pi x}{\beta }$

Answer 1

Answer: (C)

$I_0{\cos }^2\frac{\pi x}{\beta }$

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 $P$
$I=I_1+I_1+2I_1\cos \frac{2\pi x}{\beta }=2I_1[1+\cos \frac{2\pi x}{\beta }]$
$=I_0{\cos }^2\frac{\pi x}{\beta }$