Question

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

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

Answer 1

Answer: (C)

$I=I_{max}co{s}^2\left(\frac{\varphi }{2}\right)$
$\therefore \frac{I_{max}}{4}=I_{max}co{s}^2\frac{\varphi }{2}$
$cos\frac{\varphi }{2}=\frac{1}{2}$
or $\frac{\varphi }{2}=\frac{\pi }{3}$
$\therefore \varphi =\frac{2\pi }{3}=\left(\frac{2\pi }{\lambda }\right)\Delta x$ ...(i)
where $\Delta x=dsin\theta$
Substituting in Eq. (i), we get
$sin\theta =\frac{\lambda }{3d}$
or $\theta =si{n}^{-1}\left(\frac{\lambda }{3d}\right)$
$\therefore$ Correct answer is (c).