Answer to Question #90078 in Optics for Shivam Nishad

Question #90078
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.
1
Expert's answer
2019-05-29T09:37:53-0400

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:


S1P2=D2+(xd2)2(1)S_1P^2=D^2+(x-\frac{d}{2})^2 (1)

S2P2=D2+(x+d2)2(2)S_2P^2=D^2+(x+\frac{d}{2})^2 (2)

We get using (1) and (2):


S2P2S1P2=2xd(3)S_2P^2 - S_1P^2=2xd (3)


(S2PS1P)=2xdS2P+S1P(4)(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


(S2PS1P)=xdD(5)(S_2P - S_1P)= \frac{xd}{D} (5)

Phase difference between wave for constructive interference is equal to

nλn \lambda

In this case, we can write



nλ=xdD(6)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



(S1Pt)c+tv=(S1Pt)c+μtc(7)\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=S2P(S1P+(μ1)t)(8)Δ= S_2P - (S_1P+(\mu-1)t) (8)

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


nλ=xdD(μ1)t(9)n \lambda=\frac{xd}{D} - (\mu-1)t (9)

where n=1, 2, 3, …




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