Question

What must be the thicknessof a thin film which,when kept nearone of the slits shifts the central fringe by 5 mm for incident light of wavelength 5400 Å in Young’s double slit interference experiment? The refractive index of the material of the film is 1.1 and the distance between the slits is 0.5 mm​

Answer 1

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Solution: Given λ = 5400 Å, the refractive

index of the material of the film = 1.1 and

the shift of the central bright fringe = 5 mm.

Let t be the thickness of the film and P

be the point on the screen where the central

fringe has shifted. Due to the film kept in

front of slit S1 say, the optical path travelled

by the light passing through it increases

by t (1.1-1)= 0.1t. Thus, the optical paths

between the two beams passing through the

two slits are not equal at the midpoint of

the screen but are equal at the point P, 5

mm away from the centre. At this point thedistance travelled by light from the other slit

S2 to the screen is larger than that from S1 by

0.1t. The difference in distances

S2P – S1P = y λ /d,

where y is the distance along the screen

= 5 mm = 0.005 m and d is given to be 0.5

mm = 0.0005 m.

This has to be equal to the difference in

optical paths introduced by the film.

Thus, 0.1t = 0.005 x 5400 x 10-10/0.0005.

t = 5.4 x 10 -5m = 0.054 mm

Answer 2

Step-by-step explanation:

what must be the thickness of a thin film which when kept near one of the slits shifts the centralfringe by 5mm