Question

In Young's double slit experiment intensity at a point is $(1/4)$ of the maximum intensity. Angular position of this point is (separation between slits is $d$)

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

Answer 1

Answer: (C)

If $a$ is the amplitude of the wave then $\frac{I_{\max }}{4}=a^2+a^2+2as\cos \varphi$
or $\cos \varphi =-\frac{1}{2}$
or $\varphi =\frac{2\pi }{3}$
Corresponding path difference,
$\Delta x=\frac{\varphi \times \lambda }{2\pi }=\frac{(2\pi /3)\times \lambda }{2\pi }=\frac{\lambda }{3}$
So $d\sin \theta =\frac{\lambda }{3}$
or $\theta ={\sin }^{-1}\left(\frac{\lambda }{3d}\right)$