In Young's double slit experiment, the fringe width is 12 mm. If the entire arrangement is placed in water of refractive index 4/3, then the fringe width becomes (in mm):

Solution

Physics sir · 3 steps

In water the light itself gets shorter in wavelength, so the fringes squeeze closer by a factor of μ.

1. Fringe width. The fringe width is β = (λ D)/d. Putting the setup in water does not change D or d; only the wavelength of the light changes.
2. Wavelength in water. The frequency stays the same, but light slows down by a factor of μ. So the wavelength becomes λ' = λ/μ, and the fringe width falls by the same factor.

   β' = β/μ = 12/(4/3) = 12 × 3/4 = 9 mm

3. So. The new fringe width is 9 mm, which is option (b).

The idea to keep. Any denser medium filling the whole YDSE setup shrinks the wavelength to λ/μ, so every length that depends on λ, including the fringe width, gets divided by μ.

Answer from the NTA final answer key. Solution by PrepWiser.