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

Answer 1

Answer: (C)

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