Index of refraction of different wavelengths

In summary, the conversation discusses the phenomenon of maximum reflection of monochromatic light from a thin sheet of plastic film, with specific wavelengths of 482.9 and 676.0 in the visible spectrum. The minimum thickness of the film can be determined using the equation 2nt=(m+1/2)*lambda, but the refractive index is needed for an accurate calculation. The order of interference for each wavelength differs by 1, and subtracting the equations for the two wavelengths can be used to solve for the thickness of the film.
  • #1
piglet
4
0

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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  • #2
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
 
  • #3
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:

1. What is the index of refraction?

The index of refraction is a measure of how much a material can bend or refract light. It is defined as the ratio of the speed of light in a vacuum to the speed of light in the material.

2. How does the index of refraction vary with different wavelengths of light?

The index of refraction can vary with different wavelengths of light due to the phenomenon of dispersion. In general, the shorter the wavelength, the higher the index of refraction.

3. Why is the index of refraction important in optics?

The index of refraction is important in optics because it determines how light will behave when passing through different materials. It is also used in the design and construction of lenses and other optical devices.

4. How can the index of refraction be measured?

The index of refraction can be measured using a variety of techniques, such as the critical angle method, the spectrometer method, or the Michelson interferometer method. Each method involves measuring the angle or wavelength of light as it passes through a material and using the known properties of the material to calculate the index of refraction.

5. What are some examples of materials with different indices of refraction for different wavelengths?

Some examples of materials with different indices of refraction for different wavelengths include glass, water, and diamond. Glass has a higher index of refraction for shorter wavelengths, while water has a higher index of refraction for longer wavelengths. Diamond has a very high index of refraction for all wavelengths, making it a valuable material for optical devices.

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