Thin Film Interference- Transmission

AI Thread Summary
The discussion focuses on calculating the wavelength of light that will not be transmitted through a 113 nm thick layer of water (n=1.33) surrounded by air. The key points include the relationship between wavelength in air and water, with the wavelength in water being shorter due to the refractive index. The condition for destructive interference is highlighted, indicating that while transmittance is minimized at 601 nm, it is not zero. Participants clarify that there will always be some transmission due to the nature of thin film interference, and misunderstandings about total internal reflection are addressed. The minimum transmittance for the water layer is approximately 92%, emphasizing that destructive interference does not equate to complete transmission loss.
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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 wavelenght 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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