Question

A Young's double slit apparatus has slits separated by 0⋅28 mm and a screen 48 cm away from the slits. The whole apparatus is immersed in water and the slits are illuminated by red light \[\left( \lambda = 700\text{ nm in vacuum} \right).\] Find the fringe-width of the pattern formed on the screen.

Answer 1

Given:-

Separation between two slits,

\[d = 0 . 28  mm = 0 . 28 \times  {10}^{- 3}   m\]

Distance between screen and slit (D) = 48 cm = 0.48 m

Wavelength of the red light,

\[\lambda_a  = 700\text{ nm in vaccum} = 700   \times  {10}^{- 9}   m\]

Let the wavelength of red light in water = \[\lambda_\omega\]

We known that refractive index of water (μ_w =4/3),

\[μ_w = \frac{\text{Speed of light in vacuum}}{\text{Speed of light in the water}}\]

So, \[ \mu_w  = \frac{v_a}{v_\omega} = \frac{\lambda_a}{\lambda_\omega}\] 

\[ \Rightarrow \frac{4}{3} = \frac{\lambda_a}{\lambda_\omega}\] 

\[ \Rightarrow  \lambda_\omega  =   \frac{3 \lambda_a}{4} = \frac{3 \times 700}{4} = 525  nm\]

So, the fringe width of the pattern is given by

\[\beta = \frac{\lambda_\omega D}{d}\]

\[       = \frac{525 \times {10}^{- 9} \times \left( 0 . 48 \right)}{\left( 0 . 28 \right) \times {10}^{- 3}}\]

\[       = 9 \times  {10}^{- 4}  = 0 . 90  mm\]

Hence, fringe-width of the pattern formed on the screen is 0.90 mm.