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Let the amount of energy in one pulse of (laser) light be $E$, and the wavelength be $\lambda$.

This pulse goes straight to the mirror, and it is reflected by the mirror.

Let the reflectivity of this mirror be $r_{\lambda}$.

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Question. How can I get the formula of $r_{\lambda}$?

Any answers (or hints) will be appreciated, thank you.

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What do You mean with "show the light emergy absorption"? You set r, the rest to 100 % is absorbed, so what is Your question? –  Georg Jul 22 '11 at 8:59
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1) this lacks necessary detail to solve the question, 2) this sounds somewhat like homework... –  Tim Jul 22 '11 at 13:11
    
I agree with Tim that this lacks necessary detail to solve the question. You haven't specified what data you would like to use to find $r_\lambda$ - did you want to compute it from the properties of the reflecting material, as in whoplisp's answer, or by taking measurements of the laser? –  David Z Jul 22 '11 at 18:13
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Unfortunately this question is really unanswerable in it's current form. I'm sorry you don't see this, but the fact is that you have not given enough information to answer the question. You simply describe a laser beam reflecting off a mirror, and then ask us to tell you the reflectivity of the mirror. How could we know that? You must tell us something about the mirror, or about how you plan to measure the reflectivity, or something that tells us what you are really asking. We simply cannot guess at the reflectivity of a mirror sitting in your lab someplace. –  Colin K Jul 22 '11 at 18:25
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Personally I suspect this is all one big trolling attempt. –  Colin K Jul 22 '11 at 18:25

1 Answer 1

You can obtain the reflection coefficient using the Fresnel equations: http://en.wikipedia.org/wiki/Fresnel_equations

For a metal mirror and perpendicular incidence the (intensity) reflectivity is given by:

$r_\lambda=|R_\perp|^2=\frac{(1-n(\lambda))^2+\kappa(\lambda)^2}{(1+n(\lambda))^2+\kappa(\lambda)^2}$

Here $n$ is the real part of the refractive index (e.g. n(530 nm)=0.95 for Aluminium at green (530 nm) light) and $\kappa$ is the imaginary part (e.g. $\kappa(530 nm)=6.4$ for Aluminium). So for Aluminium you obtain $r_\textrm{green}=91\%$.

You can find values for other metals and wavelengths here: http://refractiveindex.info/?group=METALS&material=Aluminium .

In general $r$ increases with the frequency of light. The electrons can't follow the field for high frequencies. One can measure $n$ and $\kappa$ of a material by Ellipsometry http://en.wikipedia.org/wiki/Ellipsometry . $\kappa$ is related to the electrical conductivity $\sigma(\omega)$ of the metal.

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Can you write more of your application? If it's a pulsed laser it is quite hard to predict how it will destroy the surface. –  whoplisp Jul 22 '11 at 16:52
    
In your current question you asked for r. The energy that is deposited into the mirror is just (1-r)E. So for green light 9% of the energy would heat up the mirror. –  whoplisp Jul 22 '11 at 17:31
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He asked for a formular! :=) –  Georg Jul 22 '11 at 17:34
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@kso830: A explanation more "simple" and "complete" than this would require whoplisp to write a substantial portion of a textbook. My copy of Landau and Lifshitz book on the subject is about 400 pages, and they are known for their terseness. –  dmckee Jul 23 '11 at 1:50

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