# Can't seem to reconcile geometric optics and wave optics

I was looking at a physics situation involving light, and I can make the correct derivation assuming light is a ray of a given intensity (geometric optics), energy conservation checks out, everything. But when I try to move to a wave model of light to take into account interference, nothing works out anymore - I get completely different results. And I don't know why. It's driving me insane.

Here is the situation. Find the total reflected and transmitted light intensity from/across this layer.

External medium has refractive index $n_0$, layer $n_1$ and internal medium $n_2$. Using Fresnel equations, call the reflectance of interface $n_i \to n_j$, $\text{R}(n_i \to n_j)$ and call the transmittance $\text{T}(n_i \to n_j)$.

The Fresnel properties apply, $\text{R}(n_i \to n_j) = \text{R}(n_j \to n_i)$, $\text{T}(n_i \to n_j) = \text{T}(n_j \to n_i)$, and:

$$\text{R}(n_i \to n_j) + \text{T}(n_i \to n_j) = 1$$

I'll just do the transmitted light derivation since its enough to show my problem. The transmitted light is:

$$\text{T} = \text{T}(n_0 \to n_1) \cdot \text{T}(n_1 \to n_2) \cdot \sum \left (\text{R}(n_1 \to n_0) \cdot \text{R}(n_1 \to n_2) \right )^k$$

$$\text{T} = \frac{\text{T}(n_0 \to n_1) \cdot \text{T}(n_1 \to n_2)}{1 - \text{R}(n_1 \to n_0) \cdot \text{R}(n_1 \to n_2)}$$

A similar derivation gets the reflected intensity, and they correctly add up to 1, as expected.

Now assume light is a wave, the layer has thickness $\delta$ and the incident wave (in red) has wavelength $\lambda$.

Now use the Fresnel equations for amplitudes, call them $\text{r}(n_i \to n_j)$ and $\text{t}(n_i \to n_j)$.

The phase shift for the $k$th transmitted wave is (assume no phase shift due to reflection):

$$\frac{2 k \pi}{\lambda} \left ( 2 n_1 \delta \cos{\theta} \right ) = k \varphi$$

Where $\theta$ is the angle made by the light wave inside the layer (depends on incident angle).

Then, the transmitted intensity is:

$$\text{T} = \left | \text{t}(n_0 \to n_1) \cdot \text{t}(n_1 \to n_2) \cdot \sum \left (\text{r}(n_1 \to n_0) \cdot \text{r}(n_1 \to n_2) \cdot e^{i \varphi} \right )^k \right |^2$$

$$\text{T} = \frac{| \text{t}(n_0 \to n_1) |^2 \cdot | \text{t}(n_1 \to n_2) |^2}{| \text{r}(n_1 \to n_0) |^2 \cdot | \text{r}(n_1 \to n_2) |^2 - 2 \cdot \text{r}(n_1 \to n_0) \cdot \text{r}(n_1 \to n_2) \cdot \cos(\varphi) + 1}$$

Which isn't even close to the right result! This is completely meaningless.

I understand this isn't an elegant way to do electromagnetics and it makes a lot of assumptions (I am not a physics student, this is for computer graphics) but I don't understand why the two models don't give the same answer? I know I'm making a mistake in the second one but I can't find it. I thought this is how you add waves and I feel my sum is correct but the results still don't agree with the geometric optics result. Note I removed the $\omega t$ part from the wave since it cancels out in the end.

I guess an equivalent question would be, could someone show me the correct wave optics derivation (which should average out to the geometric optics solution over all possible phase shifts, as noted in comments)? Because I've spent days on this problem and I'm really at my wit's end at this point.

EDIT: after putting Mathematica to work on this, it turns out the wave optics average is equal to the geometric optics solution, with one small issue: it's telling me that $|\text{t}|^2$ is equal to the transmittance $\text{T}$, which is apparently wrong according to the Wikipedia Fresnel article, where:

$$\text{T}(n_i \to n_j) = \frac{n_j \cos{\theta_j}}{n_i \cos{\theta_i}} |\text{t}(n_i \to n_j)|^2$$

But at least it looks as if I wasn't completely all wrong. So now I just need help on understanding why the transmittance coefficient doesn't work out as it should..

-
The thickness of the layer surely will matter, due to interference, via the phase shift $\varphi$, and this is a thing you can only see in the wave optics. Geometric optics can't see such thing; it is only applicable in the limit when the wavelength goes to zero i.e. the layer is much thicker than the wavelength and all the phase shifts are effectively $\varphi\to\infty$. If one does things right, a geometric-optics result should be calculable as the averaging of the full wave-optics result over $\varphi$. Otherwise your calculations look impressive for a non-physicist. – Luboš Motl Feb 15 '13 at 20:58
@LubošMotl Yes, that is what I would anticipate. But averaging the result I got over $0 \leq \varphi < 2 \pi$, I do not get the result I got for the geometry optics approach, so I am thinking I did not "do things right". And thank you. – Thomas Feb 15 '13 at 21:06

Both results are correct. As Luboš Motl pointed out in his comment, we can get the geometrical optics approach answer from the wave method answer by averaging it over 1 full period.

Perhaps you made a mistake somewhere when averaging.

If we calculate the average carefully:

$\bar{T}=\frac{1}{2\pi}\int _{-\pi}^{\pi}T d\varphi$

$\bar{T}=\frac{1}{2\pi} \times T_{01}T_{12} \int_{-\pi}^{\pi}\frac{1}{(r_{10}r_{12})^2-2r_{10}r_{12}cos\varphi+1}d\varphi$

we'll get this

$\bar{T}=\frac{1}{2\pi} \times T_{01}T_{12}\left|\frac{2tan^{-1}\left(\frac{1+r_{10}r_{12}}{1-r_{10}r_{12}}tan{\frac{\varphi}{2}}\right)}{1-(r_{10}r_{12})^2}\right|_{-\pi}^{\pi}$

you can check here

now it's not hard to see that

$\bar{T}=\frac{1}{2\pi} \times T_{01}T_{12}\frac{(2\pi)}{1-R_{10}R_{12}}=\frac{T_{01}T_{12}}{1-R_{10}R_{12}}$

same as the one obtained from geometric optics method

EDIT(to respond to the EDIT part of the question):

First of all, let's make everything clear so that the equation that you quoted from Wikipedia make sense. The amplitude here means the magnitude of electric field. The $t$ simply means the ratio of transmitted and incoming electric field (We were using different definition of $t$ in the calculations above, the $t$ that we use above includes other factors beside the ratio of electric fields). And the transmittance $T$ here represents the fraction of power transmitted to the medium 2. Here the power $P$ is proportional to

$P\propto IA$

where the intensity is proportional to $I\propto nE^2$. And the beam area is proportional to $A\propto cos\theta$, it's because the beam cross section gets smaller as it bends toward the boundary plane (see this image). Putting them together we have

$P\propto nE^2cos\theta$

so from the new definition of transmittance $T$ we get

$T=\frac{n_2 cos\theta_t}{n_1 cos\theta_i}t^2$

and since the incoming and reflected rays have the same $cos\theta$ and $n$, the reflectance is simple

$R=r^2$

we have calculated that

$\bar{t^2}=\frac{t_{01}^2t_{12}^2 }{1-R_{10}R_{12}}$

multiply both sides with $\frac{n_2 cos\theta_t}{n_0 cos\theta_i}$

$\bar{T}=\frac{t_{01}^2t_{12}^2 }{1-R_{10}R_{12}}\frac{n_2 cos\theta_t}{n_0 cos\theta_i}$

We can check

$T_{01}T_{12}=\left(\frac{n_1 cos\theta_m}{n_0 cos\theta_i}t_{01}^2\right)\left(\frac{n_2 cos\theta_t}{n_1 cos\theta_m}t_{12}^2\right)=\frac{n_2 cos\theta_t}{n_0cos\theta_i}t_{01}^2t_{12}^2$

Thus again we have

$\bar{T}=\frac{T_{01}T_{12}}{1-R_{10}R_{12}}$

-
Thanks, that makes sense. So the Fresnel transmittance is equal to the amplitude coefficient for transmission squared, i.e. $|\text{t}|^2$? What happens to the $\frac{n_j \cos{\theta_j}}{n_i \cos{\theta_i}}$ term which is supposed to be in front of it (for conservation of energy)? – Thomas Feb 17 '13 at 23:40
I do not know how much you want to calculate with waves and exponential functions but it is excellent explained in Classical Electrodynamics Third Edition: John David Jackson Chapter 7.3. Reflection and refraction of Electromagnetic Waves at a Plane Interface Between Dielectrics. Unfortunately, it requires a little bit of mathematics but as you are asking that I assume you know the required basics. Jackson calculates the electro-magnetic wave composed of electric and magnetic field.