Displacement of Fringes

When a film of thickness’t’ and refractive index 'm' is introduced in the path of one of the sources, then fringe shift occurs as the optical path difference changes.

 Optical path difference at

P = S₂P - [S₁P+ µt - t] = S₂P - S₁P - (µ - 1)t = y.d/D - (µ - 1)

Illustration:

Monochromatic light of wavelength of 600 nm is used in a YDSE. One of the slits is covered by a transparent sheet of thickness 1.8 x 10^-5 m made of a material of refractive index 1.6. How many fringes will shift due to the introduction of the sheet?

Solution:

As derived earlier, the total fringe shift = w/λ (µ-1)t .

As each fringe width = w,

The number of fringes that will shift = total fringe shift/fring width

(w/λ(µ-1)t)/w = (µ-1)t/λ = (1.6-1) x 1.8 x 10^-5m / 600 x 10^-9 = 18
 

Illustration:

In the YDSE conducted with white light (4000Å-7000Å), consider two points P₁ and P₂ on the screen at y₁=0.2mm and y₂=1.6mm, respectively. Determine the wavelengths which form maxima at these points.

Solution:

The optical path difference at P₁ is

p₁ =dy₁/d = ( 10/4000) (0.2) = 5 x 10^-4mm = 5000Å

In the visible range 4000 - 7000Å

n₁ = 5000/4000 = 1.25  and n₂ = 5000/7000 = 0.714

The only integer between 0.714 and 1.25 is 1

∴ The wavelength which forms maxima at P is    l = 5000Å

For the point P₂,

p₂ = dy₂/D = ( 10/4000) (1.6) = 4 x 10^-3mm = 40000Å

Here n₁ = 40000/4000 = 10 and n₂ = 40000/7000 = 5.71

The integers between 5.71 and 10 are 6, 7, 8, 9 and 10

∴The wavelengths which form maxima at P₂ are

_λ1 = 4000 Å                for   n = 10

λ₂ = 4444 Å                for   n = 9

λ₃ = 5000 Å                for   n = 8

λ₄ = 5714 Å                for   n = 7

λ₅ = 6666 Å                for   n = 6