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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

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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.

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  • $\begingroup$ A real body in the image plane has a front surface and a back surface. Is there a factor of 2 for that? $\endgroup$ – Whit3rd Aug 31 '16 at 1:28
  • $\begingroup$ @Whit3rd Yes, but for a thick object you'll reach approximate dynamic equilibrium between the absorbed radiation and emission on one side. But the whole material slowly heats up and after a sufficiently long period of time the entire object will radiate on both sides. But note that in practice the focus area will not be the entire are of the object on one side, so you 'll then also have to take into account that the object's surface area is larger on each side... $\endgroup$ – Count Iblis Aug 31 '16 at 1:56
  • $\begingroup$ @count Ok that explains quite a lot. Would this work for the moon? I assume the moon isn't a black body? Or any non stellar celestial body. Sorry what is sigma in sigmaT? $\endgroup$ – Space Otter Aug 31 '16 at 8:37
  • $\begingroup$ @SpaceOtter Yes, but as you note, you then have to make some adjustments. The light we receive from the Moon is the light it reflects from the Sun. The Moon has a very low albedo, so it only reflects a small part fo the light it receives from the Sun. The sigma in the equations is the Stefan–Boltzmann constant $\endgroup$ – Count Iblis Aug 31 '16 at 23:41
  • $\begingroup$ Alright thanks @CountIbis I'm not sure why the fire by moonlight debate continues. I guess it's just a case of "Bad science" $\endgroup$ – Space Otter Sep 1 '16 at 8:20
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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.

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  • $\begingroup$ Thank you. What is kb in the Boltzmann Eqn? But surely it is impossible for all the potency to hit a point? It can simply focus it to a smaller area? I may have misunderstood that part. The light ray from one side of the sun would be much closer but not on the same point as a light ray from the other edge of the sun. $\endgroup$ – Space Otter Aug 30 '16 at 20:20
  • $\begingroup$ I'm assuming the lens will focus all rays that hit it. Surely that doesn't include the light ray that went the opposite direction and hit an alien in Andromeda ;) And you are right. Saying all potency will be focused at a point is not rigorous. It will actually be focused in a very small area, in a way that the agglomeration of rays starts to potentially burn something. $k_B$ is Boltzmann's constant. $\endgroup$ – QuantumBrick Aug 30 '16 at 20:27
  • $\begingroup$ Noo! XD I meant opposite edges of the sun as in the image of the sun (the rays heading towards the lens). Alright that clarifies things thanks. How would we work out the maximum output we could get from our lens death ray? (without the thermodynamics approach, I've read that too many times so I understand the second law) $\endgroup$ – Space Otter Aug 30 '16 at 20:37
  • $\begingroup$ All the lens does is to focus all light rays in (approximately) a point. $\endgroup$ – QuantumBrick Aug 30 '16 at 20:38
  • $\begingroup$ see above edit on comment $\endgroup$ – Space Otter Aug 30 '16 at 20:42
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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.

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  • $\begingroup$ That power is assuming one-way radiative heat transfer. If your object gets to be hot, transfer goes two ways. $\endgroup$ – Whit3rd Aug 31 '16 at 1:15

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