Question on light intensity at the focus of a lens.

RGClark

From using a magnifying lens under the Sun, I gather it can focus all or most of the light impinging on its surface area to a small spot and that is how it is able to create a greater intensity light at its focus (disregarding absorption in the lens.) Correct? What portion of the total light falling on the lens would be delivered to the focal spot ignoring absorption?
This would be true no matter how far away the focal point. So the normal dimunition of intensity by the square of distance would not apply. So for example if you had a lens of focal distance 1 AU and put this lens right next to the Sun and directed the lens to shine toward the Earth, the full intensity of the Sun at its surface could be delivered to the Earth. True?
In a more realistic scenario if you put the lens some ten to hundreds of thousands of kilometers away from the Sun's surface so it could survive the heating then the intensity delivered to the Earth would still be many times the Sun's normal intensity at the Earth. So for example taking 1 AU as about 150,000,000 km, if we made the focus of the lens be 1 AU and put it 150,000 km from the Sun. Then the intensity of the light at the surface of the lens would be 1000^2 = 1,000,000 (one million) times greater than that normally at the Earth.
The lens would deliver all or a large portion of the light falling on it to the focal spot on the Earth. If the area of this spot was 1/1000th that of the area of the lens, then the intensity at the focal spot would then be 1,000,000,000 (one billion) times the intensity normally at the Earth.
The intensification of the light at the focus is familiar with a convergent lens, such as with a magnifying lens. But if you had a diverging lens so the focal spot was larger than the lens then the intensity would be less than at the surface of the lens. But the total amount of light delivered to the focal spot would still be all or a large portion of that falling on the surface of the lens. And this would still be true no matter how far is the focal length. Correct?


Bob Clark

russ_watters

All of it that is roughly perpendicular to the lens. Simply put, whatever the area of the lens times the solar flux at that distance is the how much you get at the focal point. The solar flux is somewhere on the order of 100w/sq ft (not sure exactly). Yes. Yes. Didn't check the math, but the concept is correct. Possibly, depending on the angles and distance. Remember, at some point the lens will be far enough away from earth that it won't capture the full sun and that complicates things quite a bit.

Claude Bile

Roughly 96% for a silica lens at visible wavelengths. You lose 4% due to reflection at the surface of the lens.
I would be cautious in making these assumptions. The longer the focal length of a lens, the larger its focal spot will be, so while the inverse square law does not apply per se, there is nontheless a reduction in intensity with distance.
No. Close to the Earth's surface you can assume that the rays from the sun are pretty much parallel. Close to the sun this assumption fails. Once this assumption fails, the amount of light you can capture depends on the numerical aperture of the lens (NA for short) and the irradiance of the sun at its surface.

You can do a sort of sanity check type of experiment by testing this with an ordinary light bulb as a light source. There is no question that, by moving the lens closer to the source, you can capture more light, BUT you can't focus it as tightly with a longer focal length lens. These are the types of interplays you can test.

Finally I will remark that acheiving intesities 1 billion times greater than normal sunlight at the Earths surface is actually not that significant when you consider that high powered lasers can acheive peak intensities at and above this level.

Nonetheless an interesting thought experiment.

Claude.