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## Brightness of a Real Image Cast Through a Lens

We assume there is an optical system somewhere off to the right.  It's projecting a real image of a scene through a lens, which may be either negative or positive; the projected image would fall beyond the lens's focus, if the lens were not present. The image which is formed is discussed on our Lens Images page, in case (3) (for a positive lens) and case (6) (for a negative lens).  If the rather telegraphic description given here is not sufficient, you should probably check that page.

On this page, we're going to show that the brightness of the image formed by a simple camera, when it views the scene, is unchanged by the presence of the lens.

The arguments given here follow the same general form, and use the same terminology, as those given to show that brightness is unchanged when viewing a virtual image and when viewing a real image; you should probably look at those pages first.

In figure 1,  as described above, there's an optical system somewhere off to the right.  It's projecting an image of a scene (a simple magenta arrow).  The projected scene would appear as the arrow we've labeled "Projected Scene", if Lens 1 were not present.  However, after passing through Lens 1, the light forms a real image which lies either closer to or farther from Lens 1 than the projected scene, depending on whether Lens 1 is positive or negative.  The (real) image formed by Lens 1 is labeled "Lens Image" in figure 1.  (Again, this is discussed at greater length in Lens Images, case (3) and case (6).)

We're interested in a tiny piece of the scene, δv, at the base of the arrow in the projected scene.  The light which would pass through that point in the projected scene, and which enters the camera, is the light which enters the "Source Cone" in figure 1.

We've traced the lines from the camera lens through the lens image of δv.  From the camera lens to Lens 1, the magenta lines show the path actually followed by the light.  The tan lines from Lens 1 to the projected scene show the path the light would have followed if Lens 1 had not been there.

 Figure 1:  Real image cast through lens

### A Simple Geometric Argument

If l1 = l2, then the lens has "infinite focal length" -- it's a flat sheet of glass.  In that case, the lens has no effect on the image, and image brightness is likewise unaffected.

Now allow  l1 to vary.  In short, the magnification and gathered light scale together, so the brightness doesn't change.  Here's a brief argument supporting that:

If we keep  l2 and l3 fixed, then  p1 will also remain fixed.  The source cone apex angle is proportional to p1 / l1, and so the light received scales as (1 / l1)2 as we vary l1.  The magnification of the lens image relative to the projected scene is  l2 / l1, so the linear magnification will scale as (1 / l1), and the magnification of the area δv will also scale as  (1 / l1)2.  The brightness varies as light received divided by magnification of the area; so, the factors of (1/l1)2 cancel, and the brightness doesn't change as we vary the placement of the projected scene.

That was rather sketchy; we'll now essay something a little more detailed.

### A Symbolic Proof

If the intensity of the light from the scene is I = d2E/( dv), where θ is the solid angle over which we're receiving the light and v is the scene area from which we're receiving it, then in general, for any lens used in place of Lens 1, the brightness of the camera image (per unit area) will be I, times the solid angle over which we're receiving light, divided by the overall magnification:

(1)

Note that we used the small-angle approximations here, and we will continue to do so throughout this page.

We need the radius of p1, the "effective pupil" in Lens 1, and we need the magnification of the lens image caused by Lens 1, and we need the magnification of the camera image as a result of the action of the camera lens.  From figure 1, using similar triangles, we can see

(2)

Plugging in the magnification from equation (2), the brightness of the camera image will be,

(3)

Almost everything cancels, and we're left with

(4)

This is unaffected by the choice of  l1 and l2, and in particular it is the same if we plug in a "null lens" (with infinite focal length) in place of Lens 1.  Consequently, the image brightness is unchanged by Lens 1, which was to be shown.

Page created on 09/26/2007