Thin Film Interference- Transmission

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Homework Help Overview

The problem involves thin film interference, specifically regarding a thin layer of water surrounded by air and determining the wavelength of light that will not be transmitted through the water at a specific thickness of 113 nm.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the conditions for constructive and destructive interference, with some questioning the interpretation of the problem regarding transmission and reflection. There are attempts to clarify the relationship between wavelength in air and water, as well as the implications of the refractive index on transmission.

Discussion Status

The discussion is ongoing, with various interpretations being explored. Some participants have provided insights into the nature of transmission and reflection in thin films, while others are seeking clarification on the setup of the problem and the relevant formulas.

Contextual Notes

There is a noted confusion regarding the nature of transmission at destructive interference and the assumptions made about total internal reflection. The refractive indices of the materials involved are also a point of discussion.

jayjay713
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Homework Statement



A thin layer of water (n=1.33) is surrounded by air what wavelength of light will not be transmitted through the water at the point where the water is 113 nm thick?

The Attempt at a Solution



- In phase
- Path difference of 2T.
- Constructive Interference

I just don't know how to set up the formula to find the wavelength. I know to find the wavelength in water I divide wavelength in air by the index of refraction of water.

Any help appreciated, cheers.
 
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Are you certain the question asks for light not TRANSMITTED. ?
 
yes! The answer is 601 nm.

If you take the wavelength and times it by the index of refraction of water (1.33), you get 150.29. Multiply that by 4 and get ~601 nm. Don't know how that fits into the formula of 2T = wavelength / something (in example its over 4 or 2, don't know how to get that value)
 
there is some confusion here.
The wavelength of light in water is shorter than in air =λair/1.33
 
For the transmittance of a thin layer surrounded by air, the condition for destructive interference is 2Tn=(m+1/2)lambda. The transmittance is minimum at λ=601 nm but it is not zero. The reflectance can be zero because of destructive interference, the transmittance cannot.

ehild
 
It is not a weird total internal reflection question... is it?
I agree with ehild ...there will always be some transmission... mostly transmission.
It is possible to get no reflection as a result of interference but this requires some reflection from the air/water and water/air boundaries. This is never 100% (unless it involves total internal reflection)
Look forward to hearing more posts on this one.
 
technician said:
It is not a weird total internal reflection question... is it?
I agree with ehild ...there will always be some transmission...

It is a weird thin layer interference question. The maker of the problem might think that all destructive interference is totally destructive.
At constructive interference,the maximum transmittance of a stand-alone thin layer is 100% (ignoring the slight absorption), but the minimum transmittance at destructive interference depends on the refractive index both of the layer and of the surrounding medium. If the refractive index of the ambient is n0 and that of the layer is n1, the minimum transmittance is 4n02n12/(n02+n12)2, about 92% for the water layer.


ehild
 

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