Index of refraction of different wavelengths

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SUMMARY

The discussion centers on calculating the minimum thickness of a thin plastic film that causes maximum reflection for monochromatic light at wavelengths 482.9 nm and 676.0 nm. The established solution indicates that the minimum thickness is 525 nm, derived from the interference of light reflected from both surfaces of the film. The phase shift introduced by the first reflection and the path difference due to the film's refractive index, estimated at n=1.61, are critical to the calculations. The equations used include 2nt = (m + 1/2)λ, which is essential for determining the thickness based on the wavelengths provided.

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  • Understanding of optical interference principles
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  • Knowledge of wavelength and refractive index concepts
  • Basic algebra for solving equations
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Homework Statement



Monochromatic light of variable wavelength is incident normally on a thin sheet of plastic film in air. The reflected light is a maximum only for 482.9 and 676.0 in the visible spectrum. What is the minimum thickness of the film ?

Homework Equations


2nt=m*lambda 2nt=(m+1/2)*lambda


The Attempt at a Solution


Ive attemped using both wave lengths in the equation doesn't work.

Answer is 525nm.
 
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This effect is caused by interference between light reflected from the front surface of the film (the air/film surface) and the back surface of the film (film/air surface)
The first reflection introduces a phase shift of pi.
The light passing into the film introduces a path difference.
Hope this shows you how to proceed
 
The order of interference (m) for both wavelengths differ by 1. But you can not get the thickness without the refractive index. Is it n=1.61?

The reflected light is maximum if

2tn=(m+1/2)λ,and this is true for two wavelengths =>
(m1+1/2)=(2tn)/λ1
(m2+1/2)=(2tn)/λ2

Subtracting the equations we get that 2tn(1/λ1-1/λ2)=1. Solve for t.

ehild
 
Last edited:

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