Relation between spectral intensity and spectral energy density

[IcedCoffee]

TL;DR Summary

How is the following relation between spectral intensity and spectral energy density derived?

In Principles of Lasers by Svelto, while deriving the Planck radiation formula, equation 2.2.3 says $$I_{\nu} = \frac {c_0} {4n} \rho_\nu$$
where $I_\nu$ is the spectral intensity at some hole in the cavity wall (energy per time per area per frequency),
$c_0$ is the speed of light in vacuum,
$n$ is the refractive index of the medium inside the cavity,
and $\rho_\nu$ is the spectral energy density inside the cavity (energy per volume per frequency).

I understand that in the case of monochromatic wave propagating in one direction, $I = \frac {c_0} {n} \rho$
since multiplying both sides by $dt$ would give the amount of energy passing through a given area (perpendicular to the direction of propagation),
which must originate from the volume filled with the energy density $\rho$.

However, in this case, where does the factor of $\frac 1 4$ come from?

 

[IcedCoffee]

Here's a picture:

Since the entire cavity is in a steady-state, the energy leaving through the circular hole of radius $\delta r$ per unit time must be equal to the energy crossing the hemisphere $S_2$ per unit time. (I guess we are also saying that the medium does not exist between $S_2$ and the hole?) Now, in time $dt$ such energy must come from the shaded area. Since radiation inside the cavity is not directional like plane wave, only a fraction of energy inside the shaded area will cross the surface $S_2$.

In such integration, how do I justify the factor 1/4 ?

 

[IcedCoffee]

Found it: https://physicsisbeautiful.com/reso...15/solutions/4.15.pdf/fs8GhQ3hnKFu2t8QkKQPEM/