On this page we determine the brightness, per unit area, of an
image formed in a simple camera, or in a human eye. We'll need
this for reference, in order to show that an optical system consisting
of simple lenses can't produce a
brighter image in your eye 
it can only produce a larger (or smaller) image of, at most, the same
brightness as you could see with the naked eye.
In
figure 1, we show an image in a simple camera; see (
lens images:case 1) for a description of the image formed by a simple positive lens.
Light
from the scene (which is a magenta arrow in our picture), on the right,
passes through the camera's lens (shown as a blue line at the front of
the camera) and forms an inverted image at the back of the camera, on
the left. The thin pale blue lines are shown for reference, and
trace two rays from head of the arrow in the scene, one passing
straight through the center of the lens, the other passing through the
front focus of the lens and then exiting parallel to the lens axis.
The rays intersect at the head of the arrow in the image.
The
amber lines in figure 1 are the ones we're primarily interested in
here. Imagine a tiny portion of the scene, of total area
dv_{s}, right at the base of the arrow. The amber lines leading from the scene to the camera show the
cone of light from that piece of the scene which enters the camera, and the amber lines within the camera show the
cone
formed by that light as it passes from the lens to the image, where it
forms an image of a tiny part of the base of the arrow. We now
just need to determine how much light arrives at that piece of the
image, and how big that piece of the image is; from that, we can
determine brightness per unit area in the image.
Figure 1: An image formed in a simple camera (or eye)

Again referring to (
lens images:case 1), or just looking the triangles formed by the pale blue line through the center of the lens in
figure 1, we have the magnification of the scene given by
(1)
Consequently the
area of our tiny piece of image, "
dv_{e}", which will be proportional to the
square of the magnification, will be
(2)
To
determine how much light falls on that part of the image, we need to
know how much light was emitted by that part of the scene, and where it
went. We'll assume that the piece of scene,
dv_{s}, emitted a little bit of light,
dE_{s},
and that the light was distributed evenly over the hemisphere on the
camera's side of the scene. In other words, we're assuming the
scene is
flat and all light from it goes out to the front.
This assumption isn't actually realistic, as a given area in a
flat scene doesn't radiate evenly throughout a hemisphere, but it's
sufficient for our purposes here.
The amount of light from
dv_{s} which actually enters the camera will be the amount which passes
through the lens of the camera  in other words, it's the light which
travels down the cone outlined by amber lines in figure 1. The
fraction of the light from the scene which passes down that cone will
be the
solid angle of the cone,
divided by the solid angle of the complete hemisphere in front of the
scene. We'll assume that the size of the lens is small compared
to the distance to the scene, so we can use the small angle
approximation for the solid angle subtended by a cone, given in (
solid angles:eqn4). If we write
dE_{e} for the amount of light falling on area
dv_{e}, then we'll have,
(
3)
Image brightness is just the ratio of these, or
(4)
But for small angles, we have
(5)
so, as long as the distance to the scene is much larger than the radius of the lens, we can write (4) as
(6) 
(
N.B.
 Don't take that factor of "1/2" too seriously; it depends on the
(unrealistic) assumption that a flat scene radiates evenly throughout a
full hemisphere. A more exact solution would have a different 
but still constant  factor there.)
Note that the brightness is
a function of the original scene brightness (obviously!) and of the
ratio of the lens radius to the distance from the lens to the image.
The lens focal length does
not enter into it, nor does the distance from the scene to the lens.
The ratio of the focal length to the lens diameter is the
f number of the lens, or if the lens is stopped down, then the
f stop is the lens focal length divided by the diameter of the diaphragm. If we write "
F" for the F stop, then we have
and
so, for a distant scene, where the focal length is approximately the
distance from the lens to the film plane, we can write (
6) as
(7)
It's
interesting that at some finite, fixed F number, the brightness of the
image on film will be equal to the surface brightness of the scene, and
for larger lenses, it will actually
exceed the surface
brightness of the scene. Unfortunately, without a more exact
model of the scene brightness we can't say what that critical F number
is. Our value of "1/8" in (7) is surely too small, however,
as we assumed there is no falloff in scene brightness at positions far
off the axis, which is not true; consequently, we can take (7) as a
lower bound on image brightness, and we can conclude that an F/0.35
lens would produce an image at least as bright as the surface
brightness of the original scene. (For whatever it's worth, the
"fastest" camera lens I've ever seen was F/1, and the largest I've
heard of was around F/0.9. An F/0.35 lens would be a bit of an
absurdity.)
I should mention that there's no contradiction here,
even though one's intuition may be offended. To achieve image
brightness equal to scene brightness, it seems like we'd need to
capture
all the light from the scene. If I focus the Sun
through an F/0.35 lens, will the image brightness really equal the
brightness of the surface of the sun?? Apparently  but keep in
mind that the camera is very far from the scene, and the image is very
much reduced from the size of the Sun. Nearly all the Sun's light
is going other places than the camera, and the image we've produced is
tiny relative to the original! If, on the other hand, we ask what
it would take to produce a
large image, one as large as
the original, which is just as bright as the original, then we need to
move the camera very close to the scene  and we need to abandon the
smallangle approximation we made in equation (
3).
If we rewrite the results using the exact value for the angle
subtended by the pupil as viewed from the location of the scene, then,
for the light falling on the image, we have
(8)
and for the image brightness we have
(9)
Now if we set the image size equal to the scene size (1:1 magnification), then
l_{1}=
l_{e},
and in order to achieve image brightness equal to scene brightness we
need to have cos(θ) = 0, which implies θ = π/2. But that
implies our lens covers the
entire hemisphere as viewed from the location of the scene! That in turn implies the lens is infinitely large, which is as we would expect.
Again, our equation (
6) is valid only for subjects which are
distant compared to the lens diameter, which is the usual case.
An Example: The SunBurn Lens
After
writing the above text, I still felt uncomfortable about the claim that
an F/0.35 lens produces an image which is as bright as the original
scene. So, let's work through an example where we can easily
calculate the correct answer, and compare it with the above results.
The
Sun's radius is about 700,000 kilometers, and the average distance from
the Sun to the Earth is about 150 million kilometers, according to
Wikipedia. It subtends an angle of about 0.0093 radians, or about
half a degree, as viewed from the Earth. We'll define
(S1)
Then the ratio of the area of a sphere of radius equal to Earth's orbit, to the area of a sphere the size of the Sun, will be
(
S2)
Consequently, the radiation intensity of sunlight at the Earth will be about ρ
_{1}^{2}
times smaller than the intensity at the Sun's surface. Assume we
have a lens of 1 square foot area. To achieve an intensity equal
to the Sun's surface brightness in an image, the light falling on the
lens will have to be concentrated in an area (1/ρ
_{1})
^{2} times the area of the lens. The radius of the spot projected by the lens will consequently need to be about 1/ρ
_{1} times the radius of the lens.
The
image of the sun, viewed from the center of the lens, will subtend the
same angle as the Sun itself. Since the Sun is (essentially)
infinitely distant, the image will fall (essentially) on the focus of
the lens. So, the radius of the image of the Sun will be:
(S3)
where we've used θ
_{s} in place of the more exact tan(θ
_{s}), since it's a very small angle.
We want the radius, R
_{I}, to be 1/ρ
_{1} times the size of the lens. Let R
_{L} be the radius of the lens. Then, plugging in the desired value for R
_{I} along with the value of ρ
_{1} from equation (
S2), and canceling the fraction R
_{s}/R
_{o} which appears on both sides, we see
(S4)
The
diameter will be twice the focal length, and we'll need to use an F/0.5
lens. If our 1 square foot lens is round, it'll have a radius of
about 6.8 inches. If its focal length is less than 7 inches,
it'll be an F/0.5 lens.
Earlier we concluded that an F/0.35 lens would produce an image
at least
as bright as the original scene; this is consistent with that result.
In fact, in this case we see that an F/0.35 lens would produce an
image significantly brighter than the surface of the Sun (if we neglect
attenuation by the atmosphere).
Is Such a "SunBurn Lens" Feasible?
The
atmosphere attenuates sunlight quite seriously, so actually achieving a
brightness equal to that of the Sun's surface is likely to require
quite a few times as much lens area as we've been talking about.
However, if we neglect that issue (we imagine ourselves on the Moon,
for example) then the answer is yes, of course this is feasible.
A 1 foot square Fresnel lens, available from Edmund's and (no doubt)
lots of other places, is already a good part of the way toward being
this "super burner" lens all by itself. On a sunny day you can
melt copper with such a lens (I've done it). However, the focal
length of such a lens is likely to be one or two feet, so by itself it
won't produce an image as bright as the Sun's surface, even on the
Moon. Either we need to shorten the focal length or increase the
effective lens diameter.
Shortening the focal length is as
simple as stacking two or three lenses together. The aggregate
focal length of a stack of N lenses will be 1/N times the focal length
of one of them. I don't know how sharp an image such a stack will
produce, but in principle, at least, it should make a really tiny,
intolerably hot spot.
Combining four lenses "in parallel" is
also pretty simple; all you need is a frame to hold them and a few
frontsurface mirrors to reflect their light to a common focus. A
sheet of plywood with cutouts will do for the frame, and the secondary
mirrors are probably also available from Edmund's.
WARNING: If you spend much time experimenting with large lenses focusing sunlight  aka a solar furnace 
wear eye protection. A welder's mask would be appropriate. Crossed polarizers
might be reasonable. Ordinary sunglasses are
not
good enough. The spot produced by a large lens is intensely
bright, and thin plastic Fresnel lenses are unlikely to filter out much
ultraviolet. The UV is going to focus nearly as sharply as the
visible light, and it can cause eye damage which may not be apparent
until much later.
Close Focusing and Image Brightness
One last point deserves mention. The image brightness depends on the distance from the lens to the image,
not
on the focal length of the lens. Human eyes focus by changing the
shape of the lens, to adjust its focal length.
Consequently, the brightness of the image in the eye does
not
change as we look from a distant object to a very close up object.
However, the same is not true for a conventional camera.
In a camera, focusing is normally done by
moving the lens.
As we focus closer, the lens is moved farther from the film.
Since it's the distance from the lens to the film which is
critical, the result is that the image brightness can fall off
significantly in an extreme closeup. In macro photography before
the advent of through the lens metering, it was necessary to take
account of how far out the lens had been moved to calculate the
exposure. Old bellows attachments, used (at least) on Exakta
cameras in macrophotography back in the 1950's, had the focus
compensation needed engraved on the rails. You'd set up the
scene and the camera, rack the bellows out until the scene was in
focus, and then read the focus adjustment from the rail next to the
lens, and adjust the exposure as needed.
Page
created on 09/23/2007. Corrected typos and renamed some variables on 9/24/2007.