Planck radiation law of a dielectric layer Suppose we have a rectangular slab of thickness $h$, width $a$ and length $b$. The upper surface of the slab is put at constant temperature $T$ while all the rest is at initial temperature $T_0$. Obviously the temperature of this slab will increase according to the heat equation:
$$\dfrac{\partial T}{\partial t}=K\left(\dfrac{\partial^2 T}{\partial x^2}+\dfrac{\partial^2 T}{\partial y^2}+\dfrac{\partial^2 T}{\partial z^2}\right)$$ where  $K$ is the thermal diffusivity of the material.
Because:
$$\dfrac{\partial^2 T}{\partial x^2}=\dfrac{\partial^2 T}{\partial y^2}=0$$
the previous equation becomes:
$$\dfrac{\partial T}{\partial t}=K\left(\dfrac{\partial^2 T}{\partial z^2}\right)$$
What is the total radiance seen from the bottom of the slab vs. time, assuming the refractive index of the material is $n$?
 A: If I understand your question, you’re asking about the flux of thermal radiation from a heated side of a dielectric slab (index $n$), through the slab, incident to the other side. Correct me if I’m wrong.
I would analyze this starting with the relation
$$A=E,$$
where $A$ is the absorptivity and $E$ is the emissivity of the dielectric material (both as a function of wavelength, etc, depending on the details of $n$). This relation is a consequence of time-reversal symmetry.
Note that if $E$ is large, then actually very little light will make it to the back of a slab with non-negligible thickness because $A$ is also large. The emitted light will be absorbed exponentially into the depth, proceeding to heat up the regions further along. Whether or not the additional heat diffusion process is significant would depend on the details, I suppose.
Note also that if the slab is totally transparent ($A=0$), well then there is no thermal emission anyway.
There are a number of directions you can take the analysis from here, and I’ll leave it as an exercise to the reader to proceed as desired.
