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Is there formula that gives reflectance of very thin film of given metal (tens of nanometers) to the visible light of given wavelength(808nm) ? Which properties of metals are needed for the formula ?

I would like to draw a plot of reflectance that is a function of titanium film thickness. Thanks

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You need to know the index of refrection of the metal AND the substrate. (You didn't mention the substrate, but I assume your 10nm-thick film is not floating in space!!)

This is a three-layer structure: Air, thin film, substrate. You need to know wavelength, the incident angle, the refractive index of all three layers, and the thickness of the film. With those parameters in hand, you can do the calculation.

The calculation is conventionally set up using the transfer matrix method. You can find many programs online. (This page alone has 2 programs, plus links to 9 others on various different websites.) But with only three layers, the formula is sufficiently simple that you can probably do the calculation more quickly by hand.

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Have a look at my answer to Make a semi transparent mirror with copper. To a good approximation the transmission falls exponentially with thickness. Just work out what tranmission you need, e.g. if you want 80% reflectance choose 20% transmission, and work out the film thickness accordingly.

You can find the optical constants for titanium at http://refractiveindex.info/?group=METALS&material=Titanium

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  • $\begingroup$ Thanks sir for your prompt reply. I am concerned of is what you suggested me to do, I actually did this to find absorption because I thought absorption would be equal to 1-transmittance. What gave me confidence to pursue my calculations are the following: Transmittance, T= Output intensity/initial intensity. And, absorption, A=(initial intensity-output intensity)/(initial intensity). After simplifying the equation, one should get A= 1-T. This is what is confusing me a lot. It would be much better if I could find an equation of transmittance that takes into account the absorption. $\endgroup$ Commented Jul 28, 2012 at 20:57
  • $\begingroup$ FYI. I actually used e^(-αx) for transmittance. $\endgroup$ Commented Jul 28, 2012 at 21:04

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