# Are reflectance $R$ and transmittance $T$ independent of the direction in which the light travels?

If we have two different optical mediums $$1$$ and $$2$$, and we know the reflectance and transmittance when the light gets to the interface from the first one to the second, $$R_{1 \rightarrow 2}$$ and $$T_{1 \rightarrow 2}$$, would they have the same value when the light gets to the interface on the contrary sense, from $$2$$ to $$1$$?

$$R_{1 \rightarrow 2}\stackrel{?}{=} R_{2 \rightarrow 1}$$ $$T_{1 \rightarrow 2}\stackrel{?}{=} T_{2 \rightarrow 1}$$

For example, for the particular case of glass prism: if a light beam enters and exits it, will T and R take the same values when the light enters (air to glass) and when it leaves (glass to air)?

• I think the term you are wanting to learn about is "non-reciprocal medium", where you get different behavior one way compared to the other. Commented Feb 8, 2020 at 19:44
• Could be, but I think this is simpler: if a light beam enters and exits a glass prism, will $T$ and $R$ take the same values when the light enters (air to glass) and when it leaves (glass to air)? Commented Feb 8, 2020 at 22:33
• I see, you should add that to the question then. I thought you were considering the total transmission through an object (i.e. transmission through the front and back interfaces). Commented Feb 8, 2020 at 23:41

This question can be answered by looking at the Fresnel equations for reflectance and transmission, which are shown below in the case of normal incidence and ignoring polarization.

$$r = \frac{n_1-n_2}{n_1+n_2} \\ t = \frac{2n_1}{n_1+n_2}$$

and the total reflectance and transmission are $$R=\vert r \vert^2$$ and $$T=\frac{n_2}{n_1}\vert t \vert^2$$.

We can see that the complex reflectance coefficient $$r$$ changes its sign when you swap $$n_1$$ and $$n_2$$, but the overall reflectance $$R$$ is unchanged. This tells you that get the phase shifts reflecting from $$1\rightarrow2$$ and $$2\rightarrow1$$ have opposite signs. Thus $$r_{12}=-r_{21}$$ but $$R_{12}=R_{21}$$. This makes sense since waves reflecting from a "denser" medium get a phase shift, but when going to a "thinner" one you don't.

For transmission, again we see that swapping $$n_1$$ and $$n_2$$ does change the complex transmission coefficient, but not the total transmission. So $$t_{12} \neq t_{21}$$ but $$T_{12} = T_{21}$$

This is because:

$$T=\frac{n_2}{n_1}\vert \frac{2n_1}{n_1+n_2} \vert^2=\frac{4n_1 n_2}{(n_1+n_2)^2}$$

So swapping $$1\rightarrow2$$ does nothing to the overall $$T$$.

Edit: Per request I will discuss the case of non-normal incidence, now $$1\rightarrow2$$ and $$2\rightarrow1$$ are inequivalent in general. In the case of total internal reflection we have the extreme case of light being able to enter the higher-index material, but not able to leave.

Let's ignore polarization again, an only consider reflection (we can always get $$T=1-R$$ anyways).

Now the expression for $$r$$ is:

$$r=\frac{n_1 \mathrm{cos}(\theta_i)-n_2 \mathrm{cos}(\theta_t)}{n_1 \mathrm{cos}(\theta_i)+n_2 \mathrm{cos}(\theta_t)}$$

Where $$\theta_i$$ is the incident angle (normal incidence is $$\theta_i=0$$), and $$\theta_t$$ is the transmitted angle given by Snell's law $$n_1 \mathrm{cos}(\theta_i) = n_2 \mathrm{sin}(\theta_t)$$.

Now we can rewrite this expression for $$r$$ purely in terms of the incident angle $$\theta_i$$.

$$r=\frac{n_1 \mathrm{cos}(\theta_i)-n_2 \sqrt{1-\left(\frac{n_1}{n_2} \mathrm{sin}(\theta_i)\right)}}{n_1 \mathrm{cos}(\theta_i)+n_2 \sqrt{1-\left(\frac{n_1}{n_2} \mathrm{sin}(\theta_i)\right)}}$$

Now we can see that swapping $$1\rightarrow2$$ does change things so that $$r_{12} \neq r_{21}$$ and $$R_{12} \neq R_{21}$$.

As a concrete example, take $$\theta_i=\frac{\pi}{4}$$, or 45 degrees, and $$n_1=1$$ and $$n_2=2$$. Then $$r_{12}\approx -0.45$$ or \$R_{12}\approx 0.20, meaning we get incomplete reflection going from the low-index to higher index.

Now compare this to $$n_1=2$$ and $$n_2=1$$, we get $$r_{12} = e^{-1.23i}$$, or $$R_{21}=1$$, so total internal reflection from the higher index to lower index!

• Perhaps you could discuss the non-normal incidence case? Commented Feb 9, 2020 at 9:24
• @Rob Jeffries, does it change in that case? Commented Feb 9, 2020 at 10:23
• I don't think so, but it is certainly less obviously true Commented Feb 9, 2020 at 11:13
• For example, consider total internal reflection. Commented Feb 9, 2020 at 11:24
• @RobJeffries, yes I should do that. In the case of non-normal incidence you do not have this symmetry. Commented Feb 9, 2020 at 22:20