Question

In Young's double-slit experiment intensity at a point is $\left(\frac{3}{4}\right)^{th}$ of the maximum intensity. The possible angular position of this point is

• A. $\sin ^{-1}\left(\frac{\lambda}{3 d}\right)$
• B. $\sin ^{-1}\left(\frac{\lambda}{2 d}\right)$
• C. $\sin ^{-1}\left(\frac{\lambda}{6 d}\right)$
• D. $\sin ^{-1}\left(\frac{\lambda}{4 d}\right)$

Answer 1

Answer: (C)

$I=I_{\max } \cos ^{2}\left(\frac{\Phi}{2}\right)$
$\therefore \frac{3 I_{max}}{4}=I_{max}\left(cos\right)^{2}\left(\frac{\phi}{2}\right)$
$cos\left(\frac{\phi}{2}\right)=\frac{\sqrt{3}}{2}$
or $\frac{\phi}{2}=\frac{\pi }{6}$
$\therefore \phi=\frac{\pi }{3}=\left(\frac{2 \pi }{\lambda }\right)\Delta x...\left(i\right)$
where, $\Delta x=dsin\theta $
Substituting in the equation $\left(i\right)$ , we get,
$sin\theta =\frac{\lambda }{6 d}$
or $\theta=\sin ^{-1}\left(\frac{\lambda}{6 d}\right)$