Question

In a double slit experiment, when light of wavelength 400 nm was used, the angular width of the first minima formed on a screen placed I m away, was found to be 0.2$^\circ$. What will be the angular width of the first minima, if the entire experimental apparatus is immersed in water ? ($\mu _{water}=4/3$)

• A. 0.05°
• B. $0.1^\circ$
• C. $0.266^\circ$
• D. $0.15^\circ$

Answer 1

Answer: (D)

In air angular fringe width $\theta_0 = \frac{\beta}{D}$
Angular fringe width in water
$\theta_{w} = \frac{\beta}{\mu D} = \frac{\theta_{0}}{\mu} $
$ = \frac{0.2^{\circ}}{\left(\frac{4}{3}\right)} = 0.15^{\circ} $