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;

\({\theta _0} = \frac{\beta }{D}\)

\(\theta_o\) is the angular fringe, \(\beta\) is the fringe width and D is the distance between the slits and screen.

CALCULATION:

Given: \(\theta_o\) = 0.2^o

and \(\mu_W=\frac{4}{3}\) 

In air angular fringe width \({\theta _0} = \frac{\beta }{D}\)

⇒ \(\beta= \theta_oD\) ----(1)

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

\({\theta _W} = \frac{\beta }{{\mu D}}\)

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

 \( \theta_W= \frac{\theta_oD}{\mu D }\)

⇒ \(\theta_W= \frac{{{\theta _0}}}{\mu }\)

Here, \(\theta_W\) is the angular fringe with water, \(\mu\) is the refractive index.

Now, putting all the given values we get;

\(\theta_W = \frac{{0.2^\circ }}{{\left( {\frac{4}{3}} \right)}} \)

⇒\(\theta_W= 0.15^\circ\)