Question

In Young's double-slit experiment intensity at a point is $\left(\frac{1}{4}\right)^{t h}$ 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. $\sin ^{-1}\left(\frac{\lambda}{2 d}\right)$
• C. $\sin ^{-1}\left(\frac{\lambda}{3 d}\right)$
• D. $\sin ^{-1}\left(\frac{\lambda}{4 d}\right)$

Answer 1

Answer: (C)

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