Q. In a double slit experiment, the screen is placed at a distance of 1.25 m from the slits. When the apparatus is immersed in water $\ (\mu_{w} = 4/3),$ the angular width of a fringe is found to be$\ 0.2{^\circ}$. When the experiment is performed in air with same set up, the angular width of the fringe is
A) $0.4{^\circ}$
B) $0.27{^\circ}$
C) $0.35{^\circ}$
D) $0.15{^\circ}$
Explanation
(d) When the approatus is immersed is water the angular width of a fringe $\theta = \frac{\lambda}{d}$ and $\theta = 2{^\circ}$ and the angular width of a fringe in air
$\theta^{'} = \frac{\lambda^{'}}{d}$
$\frac{1}{\mu_{\omega}} = \frac{\lambda^{'}}{\lambda}$
$\frac{\lambda^{'}}{\alpha} = \frac{3}{4}$
Now, $\frac{\theta^{'}}{\theta} = \frac{\lambda^{'}}{\alpha}$
$\theta^{'} = \frac{\lambda^{'}}{\alpha} \times \theta$
$\theta^{'} = \frac{3}{4} \times 0.2{^\circ}$
$\theta^{'} = 0.15{^\circ}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.