Question

In a double slit experiment the angular width of a fringe is found to be $0.2^{\circ }$ on a screen placed $1m$ away. The wavelength of light used is $600nm$. The angular width of the fringe if entire experimental apparatus is immersed in water is (Take ${\mu }_{water}=\frac{4}{3})$

• A. $0.15^{\circ }$
• B. $1^{\circ }$
• C. $2^{\circ }$
• D. $0.3^{\circ }$

Answer 1

Answer: (A)

Since a fringe of width $\beta$ is formed on the screen at distance $D$ from the slits, so the angular fringe width
$\theta =\frac{\beta }{D}=\frac{D\lambda /d}{D}$
$=\frac{\lambda }{d}\left[\because \beta =\frac{D\lambda }{d}\right]$
$\Rightarrow d=\frac{\lambda }{\theta }$
If the wavelength in water be ${\lambda }^{\prime }$ and the angular fringe width be ${\theta }^{\prime }$, then
$d=\frac{{\lambda }^{\prime }}{{\theta }^{\prime }}$ or $\frac{\lambda }{\theta }=\frac{{\lambda }^{\prime }}{{\theta }^{\prime }}$ or
${\theta }^{\prime }=\frac{{\lambda }^{\prime }}{\lambda }$,
$\theta =\frac{\lambda /\mu }{\lambda }.\theta \left[\because {\lambda }^{\prime }=\frac{\lambda }{\mu }\right]$
$\therefore {\theta }^{\prime }=\frac{0.2^{\circ }}{4/3}$
$=0.15^{\circ }$