Question

In a double slit experiment the angular width of a fringe is found to be $0.2^{\circ}$ on a screen placed $1 \,m$ away. The wavelength of light used is $600\, nm$. 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 $\lambda'$ and the angular fringe width be $\theta'$, then
$d = \frac{\lambda'}{\theta'} $ or $ \frac{\lambda}{\theta} = \frac{\lambda'}{\theta'}$ or
$\theta' = \frac{\lambda'}{\lambda} $,
$ \theta =\frac{ \lambda /\mu}{\lambda}.\theta \left[\because \lambda' = \frac{\lambda}{\mu}\right]$
$ \therefore \theta' = \frac{0.2^{\circ}}{4/3}$
$ = 0.15^{\circ}$