Young's double slit inteference apparatus is immersed in a liquid of refractive index 1.44. It has slit separation of 1.5 mm . The slits are illuminated by a parallel beam of light whose wavelength in air is 690 nm . The fringe-width on a screen placed behind the plane of slits at a distance of 0.72 m , will be:

Solution

Physics sir · 3 steps

The liquid shrinks the wavelength by a factor of 1.44, so every fringe shrinks by the same factor.

1. Wavelength in liquid. Light slows down in a medium of refractive index μ. Its frequency stays the same, so its wavelength becomes λ' = λ/μ = 690/1.44 ≈ 479 nm.

   β' = (λ' D)/d = (λ D)/(μ d)

2. Put in numbers. β' = (690 × 10⁻⁹ × 0.72)/(1.44 × 1.5 × 10⁻³) = (4.968 × 10⁻⁷)/(2.16 × 10⁻³) = 2.3 × 10⁻⁴ m.
3. So. The fringe width is 0.23 mm, which is option (d).

The idea to keep. When a YDSE setup is immersed in a liquid, use λ/μ in place of λ. The fringe width becomes β/μ, so the fringes get closer together.

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