YDSE is conducted using light of wavelength 6000 A ∘ to observe an interference pattern. When a film of some material 3 . 0 × 10 - 3 cm thick was placed over one of the slits, the fringe pattern shifted by a distance equal to 10 fringe widths. What is the refractive index of the material of the film?

Step-by-step solution

Fringe width, β = λ D d . . . i where, D: distance between screen and slit d: distance between two slits when a film of thickness t and refractive index μ is placed over one of the slit, the fringe pattern is shift by distance S and is given by S = ( μ - 1 ) tD d . . . i i Given: S = 10 β . . . i i i From equations (i), (ii) and (iii), we get ( μ - 1 ) + D d = 10 λ D d ⇒ μ - 1 = 10 λ t = 10 × 6000 × 10 - 9 cm 3 × 10 - 3 cm ⇒ μ - 1 = 0 . 2 ⇒ μ = 1 . 2