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I have a cylindrical lens & a surface that can be burned.( for example wood , sheet ,...) In a sunny day , I wanna burn the surface but I don't know the "Energy for burning surfaces" If the $Power$ of sun is $1400 W/m^2$ , could you tell me how can I burn them?

Michaele
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2 Answers2

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The ability of a lens to concentrate power is a function of several parameters:

  • area of the lens
  • focal length of the lens
  • absorption of power in the lens
  • aberration of the lens

For an ideal (thin, spherical, circular) lens of focal length $f$ and diameter $d$ you can estimate the diameter of the image of the sun (which is roughly 0.5° across as seen from earth) as

$$s = \frac{0.5}{180}\pi f = 9\cdot 10^{-3} f$$

And the increase in power will be in proportion with the ratio of the area of the lens and the area of the image of the sun:

$$I_{focus} = I_{lens} * \frac{A_{lens}}{A_{spot}} \propto I_{lens}\left(\frac{d}{f}\right)^2$$

Now the ratio $\frac{f}{d}$ of a lens is called the F number - and as you can see it reflects the light collection ability of a lens (smaller = collects more light; a change of $\sqrt{2}$ in F number doubles the light collection, which is why the diaphragm of the lens of a camera is marked in multiples of $\sqrt{2}$: 1.4, 2, 2.8, 4, 5.6, 8, 11, 16, ...

For a 5 cm diameter lens with a focal length of 10 cm, the increase in power would be

$$\frac{I_{focus}}{I_{lens}} \approx 3000 $$

This means that the power density will increase from about 1 kW / m$^2$ (intensity at the surface of the earth, accounting for some atmospheric absorption) to around 3 MW/m$^2$ (but over a very small area).

For a cylindrical lens, the numbers are much less impressive as you compress the image of the sun in one direction only - you get roughly the square root of the improvement, or about 60x. And that would be for an ideal cylindrical lens.

If, for example, you were using a filled plastic water bottle, there is a strong aberration - not all the rays are focused to the same line. Further, a lot of the power of the sunlight (a fraction of the IR) will be absorbed in the water, further reducing the power available.

Whether you can use this to set fire to things depends on the "things" - specifically, how combustible they are, and how conductive. Material that does not conduct heat well will get hotter; material with a low flash point is more likely to burn. If you want to set fire to a piece of paper for example,, working right at the edge will help since the heat has fewer paths to escape (it can only conduct in on direction). At any rate, you will want to keep the position of the lens steady with respect to the target so it has no time to cool.

Floris
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It depends on the thermal conductivity of the material. If it is very conductive, like a metal, a lot of the heat will be conducted away and the local temperature will not rise enough to create a burn. If it is insulating, temperature rises much faster because the heat has nowhere to go.

Color is also a factor. If the material reflects light it will stay cooler than one that absorbs it. Black painted wood burns faster than white painted.

And if it does not oxidize in air it won't burn at all no matter how much you heat it.