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
Concept. In water the wavelength becomes $\dfrac{\lambda}{\mu}$, and $\beta \propto \lambda$.
Step 1. New width. $\beta_w = \dfrac{\beta}{4/3} = \dfrac34\beta$.
Answer. Becomes $\dfrac34$ of its value in air (Option B).
Solution by Sreeraj P, M.Sc Physics