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A thin film of acetone n=1.25coats a thick glass platen=1.50White light is incident normal to the film. In the reflections, fully destructive interference occurs at 600nmand fully constructive interference at700nm. Calculate the thickness of the acetone film.

Short Answer

Expert verified

The thickness of the acetone coating is 840nm.

Step by step solution

01

Given Data

  • The refractive index of first medium, that is, air isn1=1.00.
  • The refractive index of the thin film isn2=1.25.
  • The refractive index of the third medium n3=1.50.
  • The maximum intensity occurs at wavelengthλ700=700nm.
  • The minimum intensity occurs at wavelength λ600=600nm.
02

Interference in thin films

Bright colours reflected from thin oil on water and soap bubbles are a consequence of light interference. Due to the constructive interference of light reflected from the front and back surfaces of the thin film, these bright colours can be seen. For a perpendicular incident beam, the maximum intensity of light from the thin film satisfies the condition:

2L=m+12λn2m=0,1,2,...(Maxima—bright film in the air)

where λis the wavelength of the light in air, Lis its thickness, andn2is the film’s refractive index.

Here, in this case, light is travelling in an air medium and incident on the acetone film whose refractive index n2=1.25is higher than air. And then the light gets reflected of the thick glass of refractive index n3=1.50while travelling through acetone and again the refractive index is higher than acetone. As a result, the condition for constructive and destructive interference is

2L=mλ700n2(Constructive)

2L=m+12λ600n2(Destructive)

The constructive interference occurs atλ700=700nm and destructive at λ600=600nm.

03

Determine the thickness of the thin layer

Equating the above two equations,

mλ700n2=m+12λ600n2mλ700-mλ600=λ6002m700nm-600nm=600nm2m=3

Inserting the value of in the constructive condition

L=mλ7002n2=3700nm21.25=840nm

The thickness of the acetone coating is 840nm.

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