Question

Interference fringes were produced in Young’s double slit experiment using light of wavelength 5000Å. When a film of material 2.5×10^-3cm thick was placed over one of the slits, the fringe pattern shifted by a distance equal to 20 fringe widths. The refractive index of the material of the film is:

• A. 1.25
• B. 1.33
• C. 1.4
• D. 1.5

Answer 1

Answer: (C)

1.4

Answer 2

: Fringe width $\beta = \frac{\lambda D}{d}$……. (i)

Where D is the distance between the screen and slit and d is the distance between two slits.

When a film of thickness t and refractive index $\mu$is placed over one of the slit, the fringe pattern is shifted by distance S and is given by

$S = \frac{(\mu - 1)tD}{d}$…… (ii)

Given : $S = 20\beta$…… (iii)

From equation (i), (ii) and (iii) we get,

$(\mu - 1)t = 20\lambda$

Or $(\mu - 1) = \frac{20\lambda}{t} = \frac{20 \times 5000 \times 10^{- 8}cm}{2.5 \times 10^{- 3}cm}$

$\mu - 1 = 0.4or\mu = 1.4$