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Determine transmission of light through silver.

  • Thread starter Bhumble
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  • #1
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Homework Statement


The complex index of refraction of silver at 532 nm is [tex]\tilde{n} = 0.13 + 3.19i[/tex]. What is the transmission for this light through a layer of silver that is one wavelength (in the medium) thick? In this problem, don't worry about reflections from the boundaries, just consider the absorption loss in the silver itself.

Homework Equations



The Attempt at a Solution


So I'm assuming that I can neglect the real part altogether since the problem states to ignore the reflections from boundaries.
I also know that [tex]\tilde{n} = a + \frac{\alpha}{2k_0}[/tex].
I know the fresnel coefficients for reflection but they were to annoying to put in. So can I just calculate reflection then get T = 1 - R?
If I'm on the right path just let me know cause I feel lost with this problem.
 

Answers and Replies

  • #2
gneill
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Hmm. The real component of the index of refraction is 0.13, so that would make the wavelength of the 532 nm light about 4092 nm in the silver. The ratio of the wavelengths will be 1/Re(n), or about 7.7.

That puts the attenuation exponent at -4π*Im(n)/Re(n) for a one wavelength passage through the silver, or about -308. That's a lot of attenuation! (I/Io = e-308 via Beer-Lambert law).

Of course, I could be off base here; there may be some reason that I'm not aware of that straight-forward Beer-Lambert won't work for a silver coating... Free advice: you get what you pay for!
 
  • #3
ehild
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The light intensity in an absorbing medium decreases exponentially with the thickness d according to I=I0exp(-4πkd/λ), where λ is the wavelength in vacuum and k is the imaginary part of the refractive index.
With the data given, the exponent is big enough to make the transmittance very law.
The problem said to ignore reflectance at the boundaries, but the reflectance is quite hight to make the transmitted intensity considerably lower.
The reflectance of an air-solid interface at normal incidence is

R=[(n-1)2+k2]/[(n+1)2+k2].

Taking reflectance into account, the transmitted intensity for such highly absorbing medium is T=(1-R)exp(-4πkd/λ) .

ehild
 
  • #4
gneill
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Nifty. So in this case, if [tex]\tilde{n} = 0.13 + 3.19i[/tex] is the complex index of refraction,

[tex] R = \left| \frac{\tilde{n} - 1}{\tilde{n} + 1}\right|^2 = 0.955 [/tex]

and only fraction 0.045 of the incident light even makes it "under the skin", only to be further attenuated to the tune of about 134 orders of magnitude.
 
  • #5
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Thank you both very much. I really appreciate the help.
So if I were to consider the boundary conditions then I would have the transmitted as calculated above and then I would determine the attenuation due to absorption according to I= I_0 exp(-4πkd/λ).
 
  • #6
gneill
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Thank you both very much. I really appreciate the help.
So if I were to consider the boundary conditions then I would have the transmitted as calculated above and then I would determine the attenuation due to absorption according to I= I_0 exp(-4πkd/λ).
Yes, that's the way I see it.
 

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