Does Kirchhoff's law of thermal radiation fail for optically thin sources?

Question

A box made of any material with a small hole in it will give blackbody radiation. According to my textbook by ARNAB RAI CHOUDHURI, If you place an optically thick source of same temperature as the box just inside the hole, we would still get black body radiation from the hole of the box. Which leads to the Kirchhoff's law for thermal radiation.
$I_v=S_v$ for optically thick materials.
Since the intensity coming out of the box is $B_v(T)$(blackbody radiation intensity) , we get
$S_v=B_v$
Which is the kirchhoffs law for thermal radiation.Here $S_v$ is the source function of the material.

The assumption that our box with material in it will give blackbody radiation comes from the following argument. We take another box at temperature T(same as our initial box) with no material behind the hole and connect it with our box and place a filter that allows only radiation of frequency v to pass between them. We know this new box will give black body radiation $I_v$.Now if our box with the material behind the hole doesn't give radiation of this same intensity there will be a net energy transfer between the boxes. Since the two boxes are maintained at the same temperature the net energy transfer will be a violation of second law of thermodynamics.
This logic works for any material placed behind the hole,even an optically thin one. Using the same procudure I used above can I say.
$I_{v}=j_v L$ for an optically thin body.L is the length of the ray path through the material
$\alpha_v L=B_v$. Can I write this, although L here does not seem right.

Answer 1

For an optically thin material in thermodynamic equilibrium, $S_\nu = B_\nu$. For simple radiative transfer $$\frac{dI_\nu}{dl} = j_\nu - \alpha_\nu I_\nu\ . $$

If you put an optically thin slab of material in front of the hole, and the slab is in thermodynamic equilibrium such that $j_\nu = \alpha_\nu B_\nu$ and the radiation incident on the slab from the cavity is $I_\nu = B_\nu$, then you can see that $dI_\nu/d\nu =0$ and $I_\nu$ is not changed by the slab at all. Just as much radiation is emitted as absorbed.

Your expression for $I_\nu$ is incomplete, since you have not included radiation from the cavity that passes through the optically thin material.