Question

In Young’s double slit experiment, intensity at a point is $\left(\frac{1}{4}\right)$ of the maximum intensity. Angular position of this point is

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

Answer 1

Answer: (C)

In Young’s double slit experiment, intensity at any point $P$ is given by
$I=I_{max}co{s}^2\left(\frac{\varphi }{2}\right)$
where $\varphi$ is the phase difference
According to question, $I=\frac{I_{max}}{4}$
$\frac{I_{max}}{4}=I_{max}co{s}^2\left(\frac{\varphi }{2}\right)$ or $cos\frac{\varphi }{2}=\frac{1}{2}=cos\frac{\pi }{3}$
$\therefore \varphi =\frac{2\pi }{3}$
If $\Delta x$ is the path difference between two waves, then
$\frac{\varphi }{2\pi }=\frac{\Delta x}{\lambda }$
or $\Delta x=\frac{\lambda }{2\pi }\varphi =\frac{\lambda }{2\pi }\times \frac{2\pi }{3}=\frac{\lambda }{3}$
Phase difference, $\Delta x=dsin\theta$
or $dsin\theta =\frac{\lambda }{3}$ or $\theta =si{n}^{-1}\left(\frac{\lambda }{3d}\right)$