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

Answer 1

Answer: (C)

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