Is brightness proportional to power? In an exam, I am given a situation where a student investigates how the current varies with potential difference for two bulbs of the same type.
For the same voltage, the current of bulb A is twice the current of bulb B.
The conclusion made in the mark scheme is that since the power dissipated by bulb A is twice the power dissipated by bulb B, the brightness of bulb A is twice the brightness of bulb B.

My question is: is power directly proportional to brightness? The question definitely seems to assume so, but where does this relationship break down and when is it a fair assumption?

This is GCSE level physics (so physics in the UK for 16 year olds).
 A: "Brightness" isn't very well defined, but it's commonly used as if proportional to the power received by the eye (or other detector). I assume the question writers didn't want to make a lengthy and confusing digression into the details of this. More rigorously, the luminous intensity is a measure of power, and what we perceive as brightness is related (though not really linearly) to this.
A: 
The conclusion made in the mark scheme is that since the power dissipated by bulb A is twice the power dissipated by bulb B, the brightness of bulb A is twice the brightness of bulb B.

Certainly bad conclusion. Here's why. At first tungsten filament bulbs emits more heat than electromagnetic radiation. So that it's effectiveness defined as :
$$ \eta = \frac {P_{light}}{P_{light}+P_{heat}} $$
is quite low. I don't remember exact numbers, but it certainly less than $50\%$. You can experience that fact by trying to grab bulb with your hand.
So biggest part of electric power will be dissipated into non-useful heat, not as visible light.
Second point is that your given explanation assumes a linear relationship between a black body temperature and emitted radiance power, which according to the Stefan–Boltzmann law is incorrect.
Actually, a relationship between total power radiated vs body (tungsten filament) temperature is a non-linear one :
$$ P_{radiant} \propto T^4 $$
And then it's a separate question about how electric current in a tungsten filament maps to it's temperature. If we assume a Steinhart–Hart relationship :
$$ {\frac {1}{T}}=A+B\ln R+C(\ln R)^{3}$$
then it's also a non-linear one. All in all,- your exam question makes too much assumptions and over-simplifications.
A: Brightness is a vague term. One common usage of this word is to denote luminance. It's similar to radiance, which is proportional to power, but to compute luminance, the spectral power distribution of radiation is integrated with a weight to account for luminous efficiency of different wavelengths of EM radiation.
If we are talking about incandescent bulbs, peak wavelength of their thermal radiation changes according to Wien's displacement law. This moves the peak from IR to visible to UV range as emission power increases, and thus makes luminance not proportional to radiance and thus to power.
Particularly, at low filament temperatures luminance is essentially zero, and its ratio to radiance is also very close to zero. At higher temperatures the filament starts to glow dim red, which makes luminance/radiance ratio nonzero. Even higher temperatures move the peak to UV range, and the visible tail doesn't grow as fast, so this again destroys proportionality. The result is the following, decidedly non-constant-value, curve for the ratio of luminance to radiance (assuming ideal blackbody filament):

A: I'm going to make some assumptions:  1. We are talking about bulbs with tungsten filaments.  2. They are designed to operate at the same temperature (somewhat below the melting point of tungsten).  3. The filaments are the same length.  4. The brightness is proportional to the area of the filament.  Then the current is: V/R = Vπ$r^2$/(ρL) = [Vπ/(ρL)]$[A/(2πL)]^2$ where A = 2πrL (the surface area of the filament).  It would appear that the current (and power) is proportional to the square of the surface area (and brightness).
