# Put a sensor at the focal length, behind, or in front?

My intuition tell me that at the focal length a convex lens all the light converges to a point. Following that logic, it would make sense to me that a camera chip would either need to go slightly in front or slightly behind the focal point in order to expose the entire chip to the incoming light. This would produce a normal or inverted image, respectively.

Upon investigating this question, I read through these articles:
1) Wikipedia - Focal Length
2) Paragon Press - Focal Length
3) The Photo Forum

That all agree that the sensor MUST be placed AT the focal length, not in front or behind.

How does a chip placed at the focal length not just produce a white spot of light at the center?

When lenses are modeled (with the geometric optics approximations) what is usually shown is a single, mathematical, point source emitting equally in all directions (or in all forward directions), and the lens then transforms that point source to another point behind the lens.

When the rays are coming in parallel, as shown above (courtesy of Wikipedia), you can think of that as putting the point source before the lens essentially at a distance far enough that it appears to be at $\infty$, like a star.

But, rays don't have to come from $\infty$, most cameras have a focal range which you can change to do what is called imaging a finite conjugate shown below (courtesy of Wikipedia) for a single point source:

This is what image formation is, transforming some emitting point source out in front of the lens to another point at the focal plane. If the detector is displaced from the focal plane, you can look at the rays in those pictures and see that they don't intersect. In practice, this results in defocus blur, where a point is imaged to some larger "blob." It is essentially why out of focus objects appear blurry.

In practice a lens doesn't just image one point, as is usually shown in examples, it images an entire scene. That scene can be thought of as a three dimensional volume of many, many point sources all emitting at the same time! Then, depending on the distance to the lens from the detector, there will exist a plane for which all point sources lying in that plane will be imaged to points on the detector, and that will be the "in focus" part of the image, and all other points will be "blurred." In the picture above, the green line labeled "object" would be such a plane. Most cameras have the detector parallel to the lens, which creates focus planes essentially perpendicular to whatever direction you are pointing the camera. Tilt-Shift lenses get around this by essentially "tilting" the detector plane, and therefore the "in focus plane" in the scene.

I think this point is where your confusion lies, the examples are only showing how a lens works with a point source (like a tiny LED), but real scenes are collections of basically an infinite number of point sources.

• +1 and accept. This is the clearest example to me. I now understand perfectly! – dinkelk Jan 19 '13 at 1:44

More detailed diagrams might help.

It is only rays from "infinity" that come to a focus at a point at the "focal length" distance from the lens. These are parallel rays, or very close to parallel, coming from some point of the object far away. The direction from which they come determines where on the focal plane they converge. Some illustrations that claim to explain optics show rays like the pink ones, on the optical axis, but don't show anything like the green rays, so non-experts in optics might not grok what's going on.

A source of light not so far away emits (or reflects) rays that diverge. The lens is not far enough away that we can pretend the rays are parallel. The rays converge someplace farther away than the standard focal plane. There's an equation relating the distances:

1/D1 + 1/D2 = 1/F


Keep in mind that rays of light from any point of any object are going out in all directions; we only illustrate and talk about the ones that happen to hit the camera lens and thereby get focused (we hope) onto film or CCD.

When we photograph something far away, like a sunrise and trees, the rays are nearly parallel, so they will focus at F. The scenery is at D1=infinity. The equation simplifies to D2=F. Thus we want the film or CCD at the focal plane.

This being the internet, I must show a cat photo. The cats are close to the camera (D1 = 3 feet) and so the rays of light will converge beyond the plane at F, at D2 from the formula. This is why we "focus" the camera.

In either case, we see a full image, not a single blob of light at the center, because different points in the scene send light from different directions toward the camera (and toward everything else, but ignore those) and therefore land at different points on the film or CCD.

• What did you use to make those diagrams? I like them. – Colin K Jan 19 '13 at 5:04
• Inkscape. I did these quick & sloppy, late at night. Just lines, a few bezier curves. The pink and green light areas are semi-transparent for the nice overlap effect. Photos are ones I took a few years ago, tweaked a bit in GIMP before using here. – DarenW Jan 20 '13 at 2:20
• Best description so far! This really helped me getting to grips with these things! +1 – gablin Jul 15 '14 at 14:20

Yes - it focus a parralel beam of light from an object at infinity to a point (in an ideal world). So a truly parralel beam (from eg a star) would go to a point just like the first example in the image.

Now imagine a parralel pair of rays coming in from a slightly different angle (a different star), they would also come to a point slightly above or below the first set at F but still on the same focal plane (the vertical line).

This is what makes an image

• +1 A star is an excellent example since the angular divergence of a star is so small (i.e. the light so nearly parallel) that it does indeed look like a point. However if you take a pair of stars, then as a composite object they span a measurable angle and the light from the pair isn't parallel, so the image of the pair isn't a spot. – John Rennie Jan 18 '13 at 7:59
• Why "imagine ..."? Show it. The illustration is the sort the OP has already seen. – DarenW Jan 19 '13 at 2:46