Question on light intensity at the focus of a lens.

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 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
.

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