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 1 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$)
$
  {\text{A}}{\text{. 0}}{\text{.266}}^\circ \\
  {\text{B}}{\text{. 0}}{\text{.15}}^\circ \\
  {\text{C}}{\text{. 0}}{\text{.05}}^\circ \\
  {\text{D}}{\text{. 0}}{\text{.1}}^\circ \\
 $

Answer 1

Hint: The angular fringe width for a double slit experiment is equal to fringe width divided by the distance between the slits and the screen. When the apparatus is placed in water then the angular fringe width will shift and new value will be given by dividing the initial angular fringe width by the refractive index. Using this formula, we can obtain the correct answer.

Complete step by step answer:
We know that in a double slit experiment, the angular fringe width of the first minima will be given as
$\sin \theta = \dfrac{\beta }{D}$
Here the angle $\theta $ is very small, so we can write that $\sin \theta \approx \theta $. Therefore, we have
$\theta = \dfrac{\beta }{D}$
Here $\beta $ is known as the fringe width while D is the distance between the silts and the screen.
Now when whose apparatus is placed in a medium of refractive index $\mu $, then the fringe pattern shifts due to refraction of light. In this case, the new angular fringe width is given in terms of the refractive index of the medium by the following expression.
$\theta ' = \dfrac{\beta }{{\mu D}} = \dfrac{\theta }{\mu }$
We are given the value of angular fringe width when the apparatus is placed in air. Its value is
$\theta = 0.2^\circ $
The whole apparatus is placed in water whose refractive index is given as
${\mu _{water}} = \dfrac{4}{3}$
So, the new angular fringe width in water can be obtained in the following way.
$\theta ' = \dfrac{\theta }{\mu } = \dfrac{{0.2}}{{\left( {\dfrac{4}{3}} \right)}} = 0.15^\circ $
This is the required answer.

So, the correct answer is “Option B”.

Note:
It should be noted that when a light ray undergoes refraction from a rarer medium to denser medium then the velocity of light decreases and light ray bends towards the normal at the point of incidence. This leads to a shift in the direction of the light ray and the fringe pattern shifts in the double slit experiment.