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## Brightness of a Real Image

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 l1, the distance from Lens 1 to the "lens image" is l2, and the distance from the lens image to the camera is l3.  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 l1, l2, and l3 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  l1 and l2 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  l1 and l2 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 l1 and l2.  So, since the amount of light received from δv doesn't depend on l1 and l2, 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  l1 = l2 = l3, and that brings us back to the case shown in figure 1.

In conclusion, varying l3 alone doesn't change the camera image brightness.

 Figure 2:  Varying length l3

Finally, in figure 3 we allow l1 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 l1.  But the magnification also varies inversely with l1.  Specifically, the magnification of the lens image is l2/l1, so the area of the image of δv will vary as the square of l2/l1.  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 l3 and length l1

### 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 δEe.  We're going to assume that the scene is distant relative to the lens diameter, so we can use small-angle 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 p1 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 (SC-6), which was to be shown.