Answer in Optics for VIKASH KUMAR #87068

1 Obtain the conditions for observing maxima and minima in a Young two-slit interference pattern.

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:

“S1P^2=D^2+(x-frac{d}{2})^2 (1)”“S2P^2=D^2+(x+frac{d}{2})^2 (2)”

We get using (1) and (2):

“S2P^2 – S1P^2=2xd (3)”

“(S2P – S1P)= frac{2xd}{ S2P + S1P } (4)”

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

“(S2P – S1P)= 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=0, pm1, pm2, pm3, …”

Phase difference between wave for destructive interference is equal to

“frac{(2n+1)lambda }{2}”

Similarly, for destructive interference,

“frac{(2n+1)lambda }{2} =frac{xd}{D} (7)”

2 Show that these conditions change when a thin transparent sheet of thickness t and refractive index μ is introduced in the path of one of the superposing beams.

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{(S1P-t) } {c } + frac{t} {v}=frac{(S1P-t) } {c } + frac{mu t} {c} (8)”

The effective path difference at any point P on the screen

“u0394= S2P – (S1P+(mu-1)t) (9)”

Using (5) and (9) we can write for constructive interference

“n lambda=frac{xd}{D} – (mu-1)t (10)”

for destructive interference

“frac{(2n+1)lambda }{2}=frac{xd}{D} – (mu-1)t (11)”