Question

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 ${{0.2}^{o}}$ . When the experiment is performed in air with same set up, the angular width of the fringe is

• A. ${{0.4}^{o}}$
• B. ${{0.27}^{o}}$
• C. ${{0.35}^{o}}$
• D. ${{0.15}^{o}}$

Answer 1

Answer: (D)

${{0.15}^{o}}$

Answer 2

When the apparatus is immersed in water the angular width of a fringe
$\theta =\frac{\lambda }{d}$
and $\theta ={{0.2}^{o}}$
and the angular width of a fringe in air
$\theta =\frac{\lambda }{d}$
$\frac{1}{{{\mu }_{w}}}=\frac{\lambda }{\lambda }$
$\frac{\lambda }{\lambda }=\frac{3}{4}$
Now, $\frac{\theta }{\theta }=\frac{\lambda }{\lambda }$
$\theta =\frac{\lambda }{\lambda }\times \theta$
$\theta =\frac{3}{4}\times {{0.2}^{o}}$
$\theta =0.15{}^\circ$