Q 12-10-013NEETJEE MainMedium
When the whole Young's double slit apparatus is immersed in water ($\mu = \dfrac{4}{3}$), the fringe width
Answer: (B) becomes $\dfrac{3}{4}$ of its value in air
The wavelength in water is $\dfrac{\lambda}{\mu}$, and $\beta \propto \lambda$, so $\beta_w = \dfrac{3}{4}\beta$.
Solution by Sreeraj P, M.Sc Physics