Question

In Young's double-slit experiment intensity at a point is (3/4)^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. $\sin ^{-1}\left(\frac{\lambda}{6 \mathrm{~d}}\right)$
• D. $\sin ^{-1}\left(\frac{\lambda}{4 d}\right)$

Answer 1

Answer: (C)

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