how to model the exponential growth of luminosity when you flick a light on My original thought was:
$$ L = L_0 \left( 1 - e^{-t/\tau} \right) $$
where $L$ = luminosity and $L_0$ = initial luminosity.   But when I tested this I found that the time constant $\tau$ was higher for smaller values of Voltage, so I have no idea and all I can find is the luminosity of planets etc. Any help would be good.
 A: I was just about to post this, but John Rennie beat me to it, but I'd like to add that the model that he.and I use does appear to explain your observation that the equalization time $\tau$ increases as the voltage is decreased.
The problem can be crudely modeled as
$$T'(t)=-\alpha T(t)^4+\alpha T_0^4+\gamma \frac{V^2}{R(T(t))}$$
where $\alpha$ gives the rate of blackbody equalization, $T_0$ is the background temperature, $V$ is the applied voltage, $\gamma$ is the heat capacity of the filament, and $R(T(t))$ is the filament resistance as a function of temperature.
According to the Hypertextbook facts on tungsten, the resistivity of tungsten increases roughly linearly with temperature, so to a fair approximation we have $R(T)=\beta T$. 
Numerically solving the resulting system with the initial condition $T_0=1,\alpha=1,\gamma/\beta=1,V=1$ yields:

Repeating this, but with the initial condition $T_0=1,\alpha=1,\gamma/\beta=1,V=5$ yields:

As is clearly visible, as the voltage is lowered, the equalization time $\tau$ becomes longer.
Here is Mathematica code to solve the equation:
eqn = T'[t] == \[Alpha] (T0^4 - T[t]^4) + \[Gamma] V^2/(\[Beta] T[t]^2) && T[0] == T0;
rule = {\[Alpha] -> 1, \[Beta] -> 1, \[Gamma] -> 1, V -> 1, T0 -> 1};
t1 = 1;
sol = NDSolve[eqn /. rule, T, {t, 0, t1}][[1]];
Plot[T[t] /. sol, {t, 0, t1}, Frame -> True, PlotRange -> All]

Naturally the numbers I used above were all set to 1 to see how the system behaved, but it may be difficult to estimate what you think $\alpha$ and $\gamma$ ought to be for a light bulb filament ($\beta$ can be determined from the Hypertextbook link on tungsten).
A: This seems a fun experiment, but I very much doubt the simple equation you've provided will give you a good description of the luminosity as the system is really quite complicated.
When you apply a current $I$ to the lamp the heat dissipated in the filament will be:
$$ W = \frac{V^2}{R} $$
but the filament will also lose heat by radiation at a rate given by the Stefan Boltzmann law:
$$ W = A T^4 $$
where $A$ is some constant that you'll have to measure as it's a complicated function of the filament design - you could estimate it from the rate of cooling when you turn the power off (this assumes cooling is dominated by radiation). The filament will start heating up at a rate given by:
$$ \frac{dT}{dt} = C \left(\frac{V^2}{R} - AT^4\right)$$
where $C$ is the specific heat of the filament. Solve this equation and you will have the temperature as a function of time. Then you can calculate the radiated power as a function of time using the Stefan Boltzmann equation.
But note that the resistance $R$ changes with temperature, so the previous equation should really be written as:
$$ \frac{dT}{dt} = C \left(\frac{V^2}{R(T)} - AT^4\right)$$
The equation for the resistance as a function of temperature is even more complicated, though you could probably use a simpler approximation to the Bloch–Grüneisen formula.
Finally note that the colour of the light depends on temperature, and the sensitivity of your light meter almost certainly changes with the colour of the light, and the (peak) colour of the light is gien by Planck's law.
The point of all this is that there is some really interesting physics here, but I think your chances of calculating the behaviour from scratch are close to zero. Instead I would use the equation you started with and maybe look to see how the time constant varies with voltage and relate that back to some of the behaviour described above. You could also look at the cooling curves, and maybe measure the filament resistance as a function of voltage to get some idea of how it changes with temperature.
