On this page we treat the case of a camera viewing a real image
produced by a positive lens, with the scene on the other side of the
lens from the camera. We show that the brightness (per unit area)
of the image formed in the camera is unchanged by the lens.
On this page, we will be
assuming that the image brightness in a camera, when viewing a scene directly, is
independent of the distance from the camera to the scene. That was shown when we discussed
camera image brightness.
We'll
now present a geometric argument that the image brightness is not
affected by a simple positive lens producing a real image. The
argument, though purely intuitive, is somewhat labored, which
detracts substantially from its appeal (which is sad, considering
how long it took to do the illustrations for it). Consequently,
we'll also provide a simple algebraic proof down at the
end of the page.
In
figure 1 we show the camera, lens, and scene (which is a magenta arrow). The lens, labeled "
Lens 1" (vertical blue line near the right side) focuses the light from the image, forming a "real image", which we'll call the
lens image. That is in turn focused by the camera's lens to form the image in the camera, the "
camera image". The distance from the scene to the lens is
l_{1}, the distance from Lens 1 to the "lens image" is
l_{2}, and the distance from the lens image to the camera is
l_{3}. The magenta lines show the path of the light which enters the camera from an infinitesimal part of the scene,
δv, located at the base of the arrow.
We've traced the path of the light from the region
δv from the scene, through Lens 1, and on to the camera.
Cone 3 is the
cone of light actually entering the camera from the projected "lens image" of
δv. (The lenses are, of course,
disks  they're not really just line segments, as they're drawn! Hence the bundle of light from
δv forms a
cone, not a triangle.)
Cone 2
is the path the light followed going from Lens 1 to the lens image.
Cone 2 was drawn by simply extending the lines from cone 3 all
the way back to Lens 1.
Cone 1 is the path which was followed by that light as it went from
δv in the scene to Lens 1. It was formed by observing that all the light entering
cone 2 must have come from
δv, so we can just draw lines from the base of cone 2 to the scene. Light from
outside cone 1, conversely, won't enter the camera, as it won't enter cone 2, and hence can't go through the projected image of
δv and into cone 3 on a path leading to the camera's lens.
Figure 1: Scene, Lens image, Camera image equidistant

In
figure 1, we've arranged the scene, lens, and camera such that the scene is the same distance from the
lens as the projected image, and the camera and lens are the same
distance from the projected image. So, we have:
(1)
By
construction, the apex angle of Cone 2 is equal to the apex angle of
Cone 1 (which is φ). Since Cone 2 and Cone 3 are the same
height, and have the same angle, they are identical. Since Cone 1
and Cone 2 have the same base size, and they are also the same height,
they're also identical. So, Cone 1, Cone 2, and Cone 3 are all
identical in figure 1.
Consequently, in figure 1, we also have
(2)
The amount of light passing through Lens 1 and going on to the camera is the light from
δv
which enters Cone 1. Since Cone 1 is identical to Cone 3, that's
identical to the amount of light that would enter the camera if the
scene itself were at the location of the lens image. Since the
scene and the lens image are the same size in figure 1, we conclude
that the camera image brightness is the same as it would be if the
camera were "viewing" the scene directly.
I've used a lot of
words here, which may have obscured the basic simplicity of the
situation in figure 1. We've contrived, by keeping
l_{1},
l_{2}, and
l_{3}
the same length, to make it completely clear that the lens is making no
difference to the brightness in this case. If this doesn't seem
clear, just forget the text and look at figure 1, and think about how
the light goes.
In
figure 2, we've allowed the distance from the lens image to the camera to vary. (It's shown as smaller than
l_{1} and
l_{2} in the figure, but it could just as well have been larger; the point is it's no longer fixed.)
As long as we keep
l_{1} and
l_{2}
equal, the size of the lens image and the scene will be the same.
Furthermore, since Cone 1 and Cone 2 are identical in this case,
and they both must have the same apex angle as cone 3, the light passing through the lens image of
δv and going on to the camera doesn't depend on the value of
l_{1} and
l_{2}. So, since the amount of light received from
δv doesn't depend on
l_{1} and
l_{2},
and the size of the image doesn't depend on them, the image brightness
can't depend on them either. So, the brightness will be unchanged
if we set
l_{1} =
l_{2} =
l_{3}, and that brings us back to the case shown in
figure 1.
In conclusion, varying
l_{3} alone doesn't change the camera image brightness.
Figure 2: Varying length l_{3}

Finally, in
figure 3 we allow
l_{1}
to vary as well. (Relative placement of Lens 1 between the scene
and camera is now unconstrained.) Looking at the figure, it's
simplest to think of what we're doing here as dragging the scene to the
right or the left, while
varying the height of the scene in
order to keep the lens image and camera images unchanged. Does
the brightness of the camera image change as we do that?
The fraction of the
light gathered into the camera from
δv varies inversely with the square of
l_{1}. But the
magnification also varies inversely with
l_{1}. Specifically, the magnification of the lens image is
l_{2}/l_{1}, so the area of the image of
δv will vary as the square of
l_{2}/l_{1}.
Since the brightness varies as the ratio of the total light
gathered to the area over which it's spread, the net brightness won't
be affected.
Figure 3: Varying length l_{3} and length l_{1}

A Symbolic Proof
That
completes the "geometric" proof. If you got lost in the English
explanations accompanying the figures, or found my explanations to be
hopelessly unclear, you may have found the proof less than
satisfactory. So we'll now redo it purely algebraically.
The total light emitted from
δv will be
δE. We'll assume that the light is radiated evenly over a full hemisphere. This is the same assumption we made on the
camera image brightness
page. With that assumption, and with Lens 1 in place, we're going
to compute the brightness of the camera image and see if it's the same
as the image brightness for the camera alone.
Light gathered by the camera will be the
solid angle
subtended by Cone 1 divided by the solid angle of a full hemisphere,
times the light emitted by the image. We'll call that amount
δE_{e}. We're going to assume that the scene is
distant relative to the lens diameter, so we can use smallangle approximations. Then we have:
(
3)
The
area of the camera image of
δv will be the square of the magnification due to Lens 1 times the magnification due to the camera's lens, or
(
4)
From our page on ideal lenses,
case (1), we can write the magnifications as
(
5)
Looking at figure 3, we can see that we can write p
_{1} as
(
6)
Now we divide (
3) by (
4), and plug in the values from (
5) and (
6), to obtain,
(7)
Almost everything cancels; after taking the limit in small
δv, we're left with:
(8)
which is the same formula we found in the simple camera brightness page, equation (
SC6), which was to be shown.
Page added on 09/25/2007