Path:  physics insights > basics > simple optics > image brightness > Brightness of a Virtual Image, formed from 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 negative lens; 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 (5).  If the rather telegraphic description given here is not clear, 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.

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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 instead forms an inverted virtual image, labeled "Lens Image" in figure 1.  (Again, this is discussed at greater length in Lens Images case (5).)

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 in the "Source Cone" in figure 1.

We've traced the lines back from the camera lens to the lens image of δv.  From the camera lens to Lens 1, the magenta lines show the path actually followed by the light;  from Lens 1 to the lens image, the dotted lines show the path traced back to the virtual image.  The path of the incoming light in the source cone is shown as magenta lines.  The tan lines show the path the light would have followed from Lens 1 to the projected scene, had Lens 1 not been there.

Figure 1: Camera viewing virtual image formed by lens from image projected through lens:

A Simple Geometric Argument

If l₁ = l₂, then the light received from δv by the camera is equal to the light received if the scene were simply pushed back to the location of the lens image.  Since h₁ = h₂ in that case, the scene size is also the same as it would be if it were simply pushed back; consequently the effect of the lens is identical to the effect we'd get just by either moving the projected scene to the right, or sliding the camera to the left by  l₁ + l₂ units.  We know that moving the camera forward or back doesn't change the camera image brightness, so the camera image brightness is unaffected by the lens in that case.

Now allow  l₁ to vary.  The light received as l₁ varies will scale as (1 / l₁)².  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 brief; we'll now essay something a little more detailed.

A Symbolic Proof

Begin by assuming the lens has infinite focal length (it's a plane sheet of glass, or null lens).  In that case, the camera images the projected scene directly.  We still have a source cone of light coming in from the right, passing through Lens 1 (which doesn't bend the light), passing through δv, and proceeding on to the camera's lens.  The radius of the "pupil" in Lens 1 through which the light passes is still p₁.  However, for this case only, we would have

(1)    

The magnification of the camera image, for this case only, would be

(2)    

If the intensity of the light from the scene is I = d²E/(dθ 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:

(3)    

Note that we used the small-angle approximations there.  Continuing to use small angle approximations, and plugging in the magnification from equation (2), the brightness of the camera image in the case of a null lens would be

(4)    


Now let's put back a "real" Lens 1, so the lens image appears as it's shown in figure 1.  We now have a different value for p₁ and a different value for the magnification.  We break out the magnification due to Lens 1 and the magnification due to the camera's lens separately:

(5)    

Plugging the values for (5) into (3), we obtain:

(6)    

Almost everything cancels, and we're left with

(7)    

which is the same as (4).  Consequently, the image brightness is unchanged by Lens 1, which was to be shown.



Page created on 09/26/2007