Question

In young's double slit arrangement, a thin
transparent sheet of thickness t and
refractive index u is introduced in the path of
one of the beams. Obtain the expression for
the distance through which each of the
maxima gets shifted.

Answer 1

Let S1 and S2 be two slits separated by a distance d. Consider a point P on XY plane such that CP = x. The nature of interference between two waves reaching point P depends on the path difference S₂P-S₁P.

We using figure:

$$S_1P^2=D^2+(x-\frac{d}{2})^2 (1)$$

$$S_2P^2=D^2+(x+\frac{d}{2})^2 (2)$$

We get using (1) and (2):

$$S_2P^2 - S_1P^2=2xd (3)$$

$$(S_2P - S_1P)= \frac{2xd}{ S_2P + S_1P } (4)$$

for x, d<<< D , S₁P+S₂P =2D with negligible error included , path difference would be

$$(S_2P - S_1P)= \frac{xd}{D} (5)$$

Phase difference between wave for constructive interference is equal to

$$n \lambda$$

In this case, we can write

$$n \lambda=\frac{xd}{D} (6)$$

where n=1, 2, 3, …

Let a thin transparent sheet of thickness t and refractive index μ be introduced in the path of wave from one slit S₁. It is seen from the figure that light reaching the point P from source S₁ has to traverse a distance t in the sheet and a distance (S₁P−t) in the air. If c and v are velocities of light in air and in transparent sheet respectively, then the time taken by light to reach from S₁ to P is given by

$$\frac{(S_1P-t) } {c } + \frac{t} {v}=\frac{(S_1P-t) } {c } + \frac{\mu t} {c} (7)$$

The effective path difference at any point P on the screen

$$Δ= S_2P - (S_1P+(\mu-1)t) (8)$$

Using (5) and (8) we can write for constructive interference

$$n \lambda=\frac{xd}{D} - (\mu-1)t (9)$$

where n=1, 2, 3, …