Question

In Young’s double-slit experiment, a glass slide of refractive index n_g and thickness b is placed in front of one of the slits. What happens to the interference pattern and fringe width ? Derive an expression for the positions of the bright fringes in the interference pattern.

Answer 1

Suppose, in Young’s double-slit experiment, a glass slide of refractive index n_g and thickness b is placed in front of the slit S₁.

Young's experiment

If the experiment is set up in free space, this introduces an additional optical path length (n_g – 1)b in the path S₁P

Then, the optical path difference to the point P from S₁ and S₂ is,

∆l = S₂P – [S₁ P + (n_g – 1)b] 

= (S₂P – S₁P) – (n_g – 1)b 

= y d/D – (n_g – 1)b … (1)

where y = PO’, d is the distance between S₁ and S₂ , and D is the distance of the screen from S₁ and S₂ . Thus, point P will be bright (maximum intensity) if ∆l = nλ, 

where n = 0, 1,

2, … . The central bright fringe, corresponding to n = 0, will be obtained at a distance y₀ from O’ such that,

∆l = y₀ d/D– (n_g – 1)b = 0

∴ y₀ d/D = (n_g – 1)b 

∴ y₀ = d/D (ng-l) b …(2)

Therefore, the central bright fringe and the interference pattern will shift up (towards P) by a distance y₀ given by EQ. (2).

The distance of the nth bright fringe from O’ towards P is,

The distance of the (n + l)th bright fringe from O’ towards P is

Therefore, the fringe width,

w = y_{n + 1} – y_n = \(\cfrac{λD}d\)….(5)

Thus, the fringe width remains unchanged.