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°. What will be the angular width of the first minima, if the entire experimental apparatus is immersed in water? (μ_water = 4/3)

CONCEPT:

Angular fringe width for the double-slit experiment is defined as the angular fringe which is equal to the ratio of fringe width and distance between the slits and screen. It is written as;

 is the angular fringe,  is the fringe width and D is the distance between the slits and screen.

CALCULATION:

Given:  = 0.2^o

and  

In air angular fringe width 

⇒  ----(1)

Now when the apparatus is placed in a medium with a refractive index  then we have the fringe pattern shifts due to the refraction of the light. Therefore the angular fringe width in water is written as;

Now, on putting equation (1) in the above equation, we get;

⇒ 

Here,  is the angular fringe with water,  is the refractive index.

Now, putting all the given values we get;

⇒