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

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

$I=I_{max}{\left(cos\right)}^2\left(\frac{\delta }{2}\right)$
Given $I=\frac{I_{max}}{2}.$ Hence
${\left(cos\right)}^2\left(\frac{\delta }{2}\right)=\frac{1}{2},$ 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
$\frac{yd}{\lambda d}=\frac{1}{4}$
Or $\frac{y}{D}=\frac{\lambda }{4d}$ .....(iii)
Using Equation. (i) in Equation. (iii), we have
$\sin \theta =\frac{\lambda }{4d}$ or $\theta ={\sin }^{-1}\left(\frac{\lambda }{4d}\right)$