Temperature increase from focusing light using a lens. (Equations required) Could someone please quote and explain the Equations required that I would need to calculate the temperature increase of an object with a lens focussing the light from a celestial body. Therefore whether combustion could occur or to work out the lens required to achieve that combustion from a certain celestial body.
I understand this has been touched lightly by the fire by moonlight debate, but I haven't yet seen the maths behind it.
 And very briefly in the black body radiation focus post in the answer by MartinJ.H
 A: Suppose the lens has an aperture radius of r and focal length f. The image of an object at infinity will then appear at a distance f from the lens. The angular separation between two points of the object is then the same as the angle between the image of the two points in the focal point seen from a distance of $f$. The area of the image of the object is thus given by $\pi\alpha^2 f^2$ where $\alpha$ is the angle between the center of the object (assumed to be spherical) and the edge, if we  assume that this angle is small.
If the object radiates as a black body, has a radius of R, a temperature of $T$ and is a distance d away, then the flux of radiation reaching the lens is:
$$F = \sigma T^4 \left(\frac{R}{d}\right)^2 = \sigma T^4 \alpha^2$$
where $\sigma$ is the Stefan–Boltzmann constant. The total power of the radiation entering the lens $P$ is the area of the lens opening times the flux:
$$P = \pi r^2 F =  \pi \sigma T^4\alpha^2r^2$$
This power ends up heating the area of the image in the focal plane. The flux of radiation there is:
$$F_{\text{im}} = \frac{P}{\pi\alpha^2f^2} = \sigma T^4\frac{r^2}{f^2}$$
Suppose then that you put a black body in the image plane, then the temperature there would be $T_{\text{im}}$ where $\sigma T_{\text{im}}^4 = F_{\text{im}}$, therefore:
$$T_{\text{im}} = \sqrt{\frac{r}{f}}T$$
The ratio of the focal length f and the lens diameter is called the F-number and this is always larger than 1. So, the factor multiplying $T$ in the above equation will always be smaller than 1, therefore you can never reach a higher temperature than the temperature of the object in this way. 
A: What you're looking for can be estimated by means of the Stephan-Boltzmann law. This law states that the total energy radiated by a black body per unit surface area across all wavelengths per unit time is given by
$$ j(T) = \left( \frac{2 \pi^5 k_B^4}{15c^2 h^3} \right) T^4 \, ,$$
where $T$ is the black body's absolute temperature. I will not derive this here, but this equation can be obtained by integrating Planck's Law with respect to the frequencies; a derivation of Planck's Law can be found here, too. Now, the factor in from of $j(T)$ has a numerical value of 
$$  \frac{2 \pi^5 k_B^4}{15c^2 h^3} \approx 5.67 \times 10^{-8} \frac{\text{watts}}{\text{metres}^2\, \text{kelvin}^4} \, .$$
As the Sun's effective photosphere temperature is $5.777$ K we have $j_\text{Sun} \approx 5.32 \times 10^7 \text{watts}/ \text{metres}^2$. Multiplying that by the area of the lens $S$,
$$P_\text{Sun} \approx S \times 6.3 \times 10^7 \text{watts}\, .$$
Since a lens concentrates all light rays in its focus, we can assume all potency that hits the lens area is transmitted to a point. Testing some lens sizes with the above equation shows that solar potency hits around $10^5$ watts per unit time in a single spot. That's surely more than enough to burn yourself.
A: In this wiki the solar energy is about $1 kW/m^2 %. So, multiply the len's area, you can get the power to heat up your object.
