Question

In Young's double-slit experiment intensity at a point is ${\left(\frac{1}{4}\right)}^{th}$ of the maximum intensity. The angular position of this point is (separation between slits is $d$ and also two slits are similar)

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

Answer 1

Answer: (C)

If a is the amplitude of the wave then
$\frac{I_{max}}{4}=a^2=a^2+a^2+2aacos\varphi$
or $cos\varphi =-\frac{1}{2}$
or $\varphi =\frac{2\pi }{3}$
Corresponding path difference,
$\Delta x=\frac{\varphi \times \lambda }{2\pi }=\frac{(\frac{2\pi }{3})\times \lambda }{2\pi }=\frac{\lambda }{3}$
So $dsin\theta =\frac{\lambda }{3}$
Or $\theta ={\sin }^{-1}\left(\frac{\lambda }{3d}\right)$.