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)?
 A: 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!
