Question

In Young's double-slit experiment, the intensity at a point $P$ on the screen is half the maximum intensity in the interference pattern. If the wavelength of light used is $\lambda $ and $d$ is the distance between the slits, the angular separation between point $P$ and the center of the screen is

• A. $\sin ^{-1}\left(\frac{\lambda}{d}\right)$
• B. $\sin ^{-1}\left(\frac{\lambda}{2 d}\right)$
• C. $\sin ^{-1}\left(\frac{\lambda}{3 d}\right)$
• D. $\sin ^{-1}\left(\frac{\lambda}{4 d}\right)$

Answer 1

Answer: (D)

If $\delta$ is the phase difference between the interfering waves at point $P,$ then the intensity at point $P$ is given by
$I=I_{m a x}\left(cos\right)^{2} \left(\frac{\delta}{2}\right)$
Given $I=\frac{I_{m a x}}{2}.$ Hence
$\cos ^2\left(\frac{\delta}{2}\right)=\frac{1}{2}$, which gives $\left(\frac{\delta}{2}\right)=\frac{\pi}{4}$
Or $\delta=\frac{\pi }{2}$
The angular separation $\theta $ between points $P$ and $O$ is given by $tan \theta =\frac{y}{D}$ . Since $\theta \sim eqsin \theta .$ Hence
$sin \theta =\frac{y}{D}$ ...... (i)
If $\beta $ is the fringe width, then
$\frac{y}{\beta }=\frac{\pi / 2}{2 \pi }=\frac{1}{4}$ ...... (ii)
This is because the phase difference $\delta$ between two consecutive maxima is $2\pi $ . Now $\beta =\frac{\lambda D}{d}.$ using this in Equation. (ii), we get
$\frac{y d}{\lambda d}=\frac{1}{4}$
Or $\frac{y}{D}=\frac{\lambda }{4 d}$ .....(iii)
Using Equation. (i) in Equation. (iii), we have
$\sin \theta=\frac{\lambda}{4 d}$ or $\theta=\sin ^{-1}\left(\frac{\lambda}{4 d}\right)$