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The Brightness of an Image in a Simple Camera (or Eye)

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 dvs, 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)

Simple camera image formation

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)    file:///media/disk/home/slawrence/website/physics_insights/physics/formulas/eqe_temp_image_nOGosR.png

Consequently the area of our tiny piece of image, "dve", which will be proportional to the square of the magnification, will be

(2)    file:///media/disk/home/slawrence/website/physics_insights/physics/formulas/eqe_temp_image_s2qdx3.png

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, dvs, emitted a little bit of light, dEs, 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 dvs 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:eqn-4).  If we write dEe for the amount of light falling on area dve, then we'll have,

(3)    file:///media/disk/home/slawrence/website/physics_insights/physics/formulas/eqe_temp_image_7LQoI5.png

Image brightness is just the ratio of these, or

(4)    file:///media/disk/home/slawrence/website/physics_insights/physics/formulas/eqe_temp_image_QPFvc9.png

But for small angles, we have

(5)    file:///media/disk/home/slawrence/website/physics_insights/physics/formulas/eqe_temp_image_5kfL3R.png

so, as long as the distance to the scene is much larger than the radius of the lens, we can write (4) as

(6)    file:///media/disk/home/slawrence/website/physics_insights/physics/formulas/eqe_temp_image_IBG7RC.png

(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)    file:///media/disk/home/slawrence/website/physics_insights/physics/formulas/eqe_temp_image_tTLAjk.png

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 small-angle 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)    file:///media/disk/home/slawrence/website/physics_insights/physics/formulas/eqe_temp_image_Q8fXYk.png

and for the image brightness we have

(9)    file:///media/disk/home/slawrence/website/physics_insights/physics/formulas/eqe_temp_image_6M92jv.png

Now if we set the image size equal to the scene size (1:1 magnification), then l1= le, 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 Sun-Burn 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

(S-1)     file:///media/disk/home/slawrence/website/physics_insights/physics/formulas/eqe_temp_image_5APCm0.png

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

(S-2)    file:///media/disk/home/slawrence/website/physics_insights/physics/formulas/eqe_temp_image_ltyKn4.png

Consequently, the radiation intensity of sunlight at the Earth will be about ρ12 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/ρ1)2 times the area of the lens.  The radius of the spot projected by the lens will consequently need to be about 1/ρ1 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:

(S-3)    file:///media/disk/home/slawrence/website/physics_insights/physics/formulas/eqe_temp_image_ZoNVhg.png

where we've used θs in place of the more exact tan(θs), since it's a very small angle.

We want the radius, RI, to be 1/ρ1 times the size of the lens.  Let RL be the radius of the lens.  Then, plugging in the desired value for RI along with the value of ρ1 from equation (S-2), and canceling the fraction Rs/Ro which appears on both sides, we see

(S-4)    file:///media/disk/home/slawrence/website/physics_insights/physics/formulas/eqe_temp_image_mExxJL.png

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 "Sun-Burn 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 front-surface 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.