What would happen if the entire inner surface of a Dyson sphere (not necessarily a solid shell, maybe a very dense swarm) would be coated into a highly reflective material?
Let's say it can reflect 99.9% at all wavelengths.
Would it cook the star in its own light?
Building on L.Dutch's answer:
Light emitted from the star would travel until the mirrors and then would be reflected back, bouncing back and forth. Due to enormous scale of the Dyson sphere, you can neglect cavity effects and related wavelenght selection.
This would build up energy into the sphere, which can only dissipate through the mirrors.
Basically such a configuration would act as a cosmic scale black body (mind I say black body, not black hole).
If your mirrors can withstand the energy contained inside the sphere, the ensemble will behave like a new star with the size of the Dyson sphere, emitting like a black body at the temperature of the mirror.
If your mirrors are not so sturdy they will simply evaporate and join the stellar wind.
And now, to do away with pointless talk and actually do some science.
Theoretical body which absorbs all the light is called black body, theoretical body which doesn't absorb all the light but absorption efficiency doesn't depend on wavelength is called grey body. Object for which absorption depends on wavelength is called coloured body.
Curious property of grey bodies is that they not only absorb light less effectively than black bodies, they also emit light less effectively, assuming same temperature.
Energy emission per unit of surface for grey body is: $$ j = \epsilon * \sigma * T^4[\frac{W}{m^2}] $$ where $\epsilon$ is absorptivity/emissivity, $\sigma$ is Stefan-Boltzmann constant and $T$ is temperature in Kelvins. Square brackets contain dimensions.
Your theoretical sphere will emit: $$ P = 2*4\pi R^2\epsilon\sigma T^4 [W] $$ Where $R$ is radius of sphere. Notice factor of 2 at the start. That's because it will emit to the outside and to the inside (I deliberately wrote it as $2*4$ instead of just $8$).
Meanwhile, star emits: $$ P_s = 4\pi r^2\sigma t^4 [W] $$
Considering that entire shell is reflective, we can assume that reflected star light does not fall on other parts of the shell and instead returns to star to be fully absorbed (stars are with good approximation black bodies). However, internal emission of the shell with be into half-full spatial angle, thus we can't make such assumption.
Star will absorb back all the emitted light which falls on it, but shell will again absorb only $\epsilon$. Since from each infinitesimal part of the shell star obscures only part of the full angle we can see that star will absorb $\frac{\pi r^2}{2\pi R^2}=\frac{r^2}{2R^2}$ of total internal emission. This means that $\epsilon(1-\frac{r^2}{2R^2})$ will be absorbed by shell while $(1-\epsilon)(1-\frac{r^2}{2R^2})$ bounce again, thus star will again absorb $\frac{r^2}{2R^2}(1-\epsilon)(1-\frac{r^2}{2R^2})$. This looks like a geometric sequence with first term of $a=\epsilon(1-\frac{r^2}{2R^2})$ and multiplicative factor of $q=(1-\epsilon)(1-\frac{r^2}{2R^2})$. Since obviously $q<1$ sum of the sequence converges. Summing from 0 to infinity we get: $$ A_{shell}=\frac{\epsilon(1-\frac{r^2}{2R^2})}{1-(1-\epsilon)(1-\frac{r^2}{2R^2})} $$ Now we need to calculate same series for absorption by star and we will be able to calculate total fraction of internal emission absorbed by star to internal emission absorbed back by shell. This time we get $a=\frac{r^2}{2R^2}$ and $q=(1-\epsilon)(1-\frac{r^2}{2R^2})$, thus sum is: $$ A_{Star}=\frac{\frac{r^2}{2R^2}}{1-(1-\epsilon)(1-\frac{r^2}{2R^2})} $$
Since obviously all the internal emission has to be absorbed over course of infinite bounces, $A_{Shell}+A_{Star}=1$ has to be true. And indeed it is, verifying that no mistakes were made.
Thus, over infinite reflections of internal emission, shell will absorb back: $$ A_{Shell}*P=\frac{\epsilon(1-\frac{r^2}{2R^2})}{1-(1-\epsilon)(1-\frac{r^2}{2R^2})} 4\pi R^2\epsilon\sigma T^4 $$ While star will absorb: $$ A_{Star}*P=\frac{\frac{r^2}{2R^2}}{1-(1-\epsilon)(1-\frac{r^2}{2R^2})} 4\pi R^2\epsilon\sigma T^4 $$
Thus total power absorbed by shell will be: $$ A_{Shell}P+\epsilon P_s=\frac{\epsilon(1-\frac{r^2}{2R^2})}{1-(1-\epsilon)(1-\frac{r^2}{2R^2})} 4\pi R^2\epsilon\sigma T^4 + \epsilon 4\pi r^2\sigma t^4 $$ Which for equilibrium has to be equal to total emitted power: $$ P = 2*4\pi R^2\epsilon\sigma T^4 $$ Combining those equations we get T as a function of t,r,R and $\epsilon$: $$ T=t\sqrt{\frac{r}{R}}\sqrt[4]{\frac{1}{2-\frac{\epsilon(1-\frac{r^2}{2R^2})}{1-(1-\epsilon)(1-\frac{r^2}{2R^2})}}}=t\sqrt{\frac{r}{R}}\sqrt[4]{\frac{\epsilon (1-\frac{r^2}{2R^2})+\frac{r^2}{2R^2}}{\epsilon (1-\frac{r^2}{2R^2})+2\frac{r^2}{2R^2}}} $$
Unfortunately, for star it's more complicated. Simplified equilibrium requires that temperature raises enough so that total emission is equal to original star emission plus reflected starlight plus absorbed internal shell emission. In practice, it will increase temperature, increasing rate of fusion, which increases internal power generation, increasing temperature even further. I can not at this point make predictions on this. So I will continue with grossly oversimplified equilibrium conditions. Thus, in grossly oversimplified conditions, star temperature has to raise so that following are true:
$$ P'_s=P_s+P_s(1-\epsilon)+P_s(1-\epsilon)^2+...+A_{star}P=\frac{P_s}{\epsilon}+A_{star}P $$ Term $\frac{P_s}{\epsilon}$ represents infinite series of starlight bouncing from shell, being absorbed by star, emitted again, bounced, absorbed and so on.
Which after using expressions, using expression for T(t) and simplifying a bit: $$ t'^4=\frac{t^4}{\epsilon}+\frac{\frac{r^2}{2R^2}}{1-(1-\epsilon)(1-\frac{r^2}{2R^2})}\epsilon \frac{\epsilon (1-\frac{r^2}{2R^2})+\frac{r^2}{2R^2}}{\epsilon (1-\frac{r^2}{2R^2})+2\frac{r^2}{2R^2}} t^4 =\frac{t^4}{\epsilon}+ \frac{\frac{r^2}{2R^2}}{\epsilon (1-\frac{r^2}{2R^2})+2\frac{r^2}{2R^2}}\epsilon t^4=t^4(\frac{1}{\epsilon}+\frac{\frac{r^2}{2R^2}}{\epsilon (1-\frac{r^2}{2R^2})+2\frac{r^2}{2R^2}}) $$ Which means that simplified equilibrium temperature of star will be: $$ t'=t\sqrt[4]{\frac{1}{\epsilon}+\frac{\frac{r^2}{2R^2}}{\epsilon (1-\frac{r^2}{2R^2})+2\frac{r^2}{2R^2}}} $$ And final temperature of shell will be T'=T(t'): $$ T'=t'\sqrt{\frac{r}{R}}\sqrt[4]{\frac{\epsilon (1-\frac{r^2}{2R^2})+\frac{r^2}{2R^2}}{\epsilon (1-\frac{r^2}{2R^2})+2\frac{r^2}{2R^2}}}=t\sqrt[4]{\frac{1}{\epsilon}+\frac{\frac{r^2}{2R^2}}{\epsilon (1-\frac{r^2}{2R^2})+2\frac{r^2}{2R^2}}} \sqrt{\frac{r}{R}}\sqrt[4]{\frac{\epsilon (1-\frac{r^2}{2R^2})+\frac{r^2}{2R^2}}{\epsilon (1-\frac{r^2}{2R^2})+2\frac{r^2}{2R^2}}} $$
Now it's just a trivial matter of calculating unimportant details. Feel free to put in whatever values you want.
Obviously, you can calculate external emission of the shell to know how much power will that pseudo-star output. Simply use $P=4\pi\sigma\epsilon T'^4$.
EDIT:
Disclaimer: expression $\frac{r^2}{2R^2}$ comes from assumption that shell is significantly larger than star. If you want shell to be merely slightly larger, replace it with $\frac{r^2}{R^2}$
It is also conceivable that the mirrors are not absorbing the remaining 0.1% of the radiation, but are letting it through. In that case, the intensity of the radiation outside the sphere won't change, while the intensity inside the sphere will increase 1000-fold (assuming the mirrors are able to withstand it).
Hopefully not. I do not like to think about what would happen to my tyrannical plans should the sun be overcooked. I mean, the Dyson sphere would reflect the 99.9% energy to the solar energy batteries set up in space to collect the energy, not back at the sun correct? Isn't that the concept of the Dyson sphere?
We wouldn't be reflecting the energy back towards the sun at any rate. And in all probability we would not be able to reflect 99.9% of all the energy coming towards our solar mirrors. There would most likely be at least a little bit of energy escaping or refracting off because of space debris damaging our mirrors AND collectors. Now with that model the internal temperature of the sun would not be as high as our highly intelligent friend calculated for us. We MUST think of our Dyson Sphere in a constant state of disrepair, because that would be the most likely outcome.
