Is it dark inside the Sun? This might sound like a strange question, but something got me thinking about it recently.

The opacity of plasma in stellar interiors can get quite high, making for shorter free-paths for photons. In these conditions I guess that the light you could theoretically gather, supposing you have a pair of indestructuble eyes submerged in the solar interior, would be the one emitted by the plasma in your immediate surroundings, right? So if the opacity is high enough I can imagine places inside a star like the Sun where there is the same ambient illumination as a typical moonless night here on Earth.

My questions are:


Is this line of reasoning correct?
Are these conditions actually possible inside a star?
Where exactly inside a star are these conditions possible?
No, it's not. The radiation field in the interior of the Sun is very close to a blackbody spectrum.
If you look in any particular direction the brightness (power per unit area) you see is $\sigma T^4$, where $\sigma$ is Stefan's constant. Even at any particular wavelength it is always the case that a blackbody of higher temperature is brighter than a blackbody at lower temperature.
Given that the interior temperature might be $10^7\ \mathrm K$, then the surface brightness is $5.7 \times 10^{20}\ \mathrm{W/m^2}$, compared with the $1400\ \mathrm{W/m^2}$ you would get by looking directly at the Sun (please don't do this). Note that most of this power comes out at X-ray wavelengths, but because of the properties of a blackbody, the brightness at visible wavelengths will still be plenty brighter than that of the solar photosphere (see below).
A possible source of confusion is this term &quot;opacity&quot;. When things are in thermal equilibrium, which the interior of the Sun is, then they emit the same amount of radiation as they absorb. So high opacity also means high emissivity.
Details for interest:
The opacity, $\kappa$ in the solar interior ranges from 1 cm$^2$ g at the centre to about $10^5$ cm$^2$ g just below the photosphere. To estimate the mean free path of photons we need to multiply this by the density $\rho$ and take the reciprocal:
$$ \bar{l} = \frac{1}{\kappa \rho}\ .$$
The density varies from 160 g/cm$^3$ at the centre to about 0.001 g/cm$^3$ just below the photosphere. Thus the mean free path is about  6 micrometres at the center and is actually quite similar just below the photosphere (it peaks at around 2 mm about three quarters of the way out towards the surface).
Thus your &quot;view&quot; of the stellar interior is of a foggy sphere with radius of no more than a few times $\bar{l}$. The fog however is tremendously bright - as outlined above.
The brightness at particular wavelengths is proportional to the Planck function
$$B_\lambda = \frac{2hc^2}{\lambda^5} \left(\frac{1}{\exp(hc/\lambda k_B T) -1}\right).$$
Thus at $\lambda=500$ nm (visible light), the ratio of brightness for blackbodies at $10^7$ K (solar interior) to 6000 K (solar photosphere) is $4.2\times 10^{4}$. i.e. Even just considered at visible wavelengths, the interior of the Sun is about 40,000 times brighter than the photosphere.