How can I determine transmission/reflection coefficients for light?

When light rays reflect off a boundary between two materials with different indices of refraction, a lot of the sources I've seen (recently) don't discuss the relation between the amplitude (or equivalently, intensity) of the transmitted/reflected rays and the original ray. Mostly they just discuss the phase difference induced by the reflection, for instance to calculate thin film interference effects.

Is it possible to calculate the transmission coefficient $T$ and reflection coefficient $R$ based on other optical properties of the materials, such as the index of refraction? Or do they need to be looked up from a reference table?

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Let me see if my thinking is right here. This doesn't account for angle, but it's still interesting. Couldn't you treat the barrier as a square potential step with height equal to the difference in the kinetic energy of the light in each medium (which should be easy to calculate from the refractive indeces)? And then solve Schroedinger's equation on each side? –  ZachMcDargh Nov 8 '10 at 22:45

In addition to Fresnel equations, and in response to your question regarding the "... relation between the amplitude of the transmitted/reflected rays and the original ray":

$$T_{\parallel}=\frac{2n_{1}\cos\theta_{i}}{n_{2}\cos\theta_{i}+n_{1}\cos\theta_{t}}A_{\parallel}$$

$$T_{\perp}=\frac{2n_{1}\cos\theta_{i}}{n_{1}\cos\theta_{i}+n_{2}\cos\theta_{t}}A_{\perp}$$

$$R_{\parallel}=\frac{n_{2}\cos\theta_{i}-n_{1}\cos\theta_{t}}{n_{2}\cos\theta_{i}+n_{1}\cos\theta_{t}}A_{\parallel}$$

$$R_{\perp}=\frac{n_{1}\cos\theta_{i}-n_{2}\cos\theta_{t}}{n_{1}\cos\theta_{i}+n_{2}\cos\theta_{t}}A_{\perp}$$

where $A_{\parallel}$ and $A_{\perp}$ is the parallel and perpendicular component of the amplitude of the electric field for the incident wave, respectively. Accordingly for the $T$ (transmitted wave) and $R$ (reflected wave). I think the notation is straightforward to understand. This set of equations are also called Fresnel equations (there are three or four representations).

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Thanks, good answer. (+1) I wasn't really particular about whether I got the coefficients for amplitude or for intensity, since it's easy to go from one to the other, but I can see how that may not have been clear from the question. I edited it a bit to perhaps clarify that. –  David Z Nov 8 '10 at 5:38
+1, but I think these formulae are only valid for $\mu=1$ (which is usually the case in optics) –  Tobias Kienzler Nov 8 '10 at 8:37
True, Wikipedia does mention that assumption. I'm fine with that. (Although if there are more general equations out there for $\mu \neq 1$, I would be happy to have them posted as another answer) –  David Z Nov 8 '10 at 16:53
Note that the amplitude coefficients will give you the phase shift in reflection or transmission, and the intensity coefficients won't. So you can go easily from amplitude to intensity, but you can only go from intensity to the modulus of the amplitude. –  ptomato Apr 11 '12 at 7:26

This was intended as a comment, but for the sake of clarity, I'd better use an answer.

Regarding to the case $\mu \neq 1$, we can start using the following set of equations, which are derived from the Maxwell equations and after applying boundary conditions that demand that across the boundary the tangential components of $E$ and $H$ should be continuous.

$$\cos\theta_{i}(A_{\parallel}-R_{\parallel})=\cos\theta_{t}T_{\parallel}$$ $$A_{\perp}+R_{\perp}=T_{\perp}$$ $$\sqrt{\frac{\epsilon_{1}}{\mu_{1}}}\cos\theta_{i}(A_{\perp}-R_{\perp})=\sqrt{\frac{\epsilon_{2}}{\mu_{2}}}\cos\theta_{t}T_{\perp}$$ $$\sqrt{\frac{\epsilon_{1}}{\mu_{1}}}(A_{\parallel}+R_{\parallel}=\sqrt{\frac{\epsilon_{2}}{\mu_{2}}}T_{\parallel}$$

Then, adding together the first and fourth equation, you obtain

$$T_{\parallel}=\frac{2\cos\theta_{i}\sqrt{\epsilon_{1}\mu_{2}}}{\cos\theta_{t}\sqrt{\mu_{2}\epsilon_{1}}+\sqrt{\epsilon_{2}\mu_{1}}\cos\theta_{i}}A_{\parallel}$$

Adding the second and third equation, you have

$$T_{\perp}=\frac{2 \sqrt{\mu_{2}\epsilon_{1}}\cos\theta_{i}}{\sqrt{\epsilon_{2}\mu_{1}}\cos\theta_{t}+\cos\theta_{i}\sqrt{\mu_{2}\epsilon_{1}}}A_{\perp}$$

Accordingly for $R_{\parallel}$ and $R_{\perp}$ (in which we have to substitute the value we already found for $T_{\parallel}$ and $T_{\perp}$)

$$R_{\parallel}=\frac{\sqrt{\epsilon_{2}\mu_{1}}\cos\theta_{i}-\sqrt{\epsilon_{1}\mu_{2}}\cos\theta_{t}}{\sqrt{\epsilon_{2}\mu_{1}}\cos\theta_{i}+\sqrt{\epsilon_{1}\mu_{2}}\cos\theta_{t}}A_{\parallel}$$

$$R_{\perp}=\frac{\sqrt{\epsilon_{1}\mu_{2}}\cos\theta_{i}-\sqrt{\epsilon_{2}\mu_{1}}\cos\theta_{t}}{\sqrt{\epsilon_{1}\mu_{2}}\cos\theta_{i}+\sqrt{\epsilon_{2}\mu_{1}}\cos\theta_{t}}A_{\perp}$$

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+1 for the derivation. :) –  Noldorin Nov 8 '10 at 21:47
(And yes, please don't try and fit this into comments!) –  Noldorin Nov 8 '10 at 21:48
@Noldorin: Haha, nice comment :-) –  Robert Smith Nov 9 '10 at 5:09