Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

An optical lens focuses the light according to a rule (due to its special shape) and we know that rule. Now, instead of making bigger and bigger lenses, can we use an array of optical sensors and the above mentioned rule to focus the light without actually using a lens?

share|cite|improve this question
up vote 1 down vote accepted

Lenses work by changing the relative phase of the light rays as they pass through the lens, and anything that changes the phase in the same way as a lens will cause the light to focus. An example would be a Fresnel lens. Another vaguely related area is metamaterials - these are behind the invisibility cloaks that are currently fashionable.

In principle you could replace the lens by an array of light detectors on one side and an array of light emitters on the other side, but only if you could control the phase change between the detector and emitter. You couldn't just use an array of photodiodes on one side and LEDs on the other because the light from the LEDs would preserve the original phase.

share|cite|improve this answer

Well you are not correctly representing just what a lens does.

Let's start by doing away with the lens, and just having a hole. So imagine a dark room with a small hole in the middle of an outside wall. Make it a one inch hole and assume the wall is much thinner than that.

Light coming from several regions outside the wall; some street lights perhaps, will arrive at the hole from a different direction for each object, and proceed in a straight line towards the opposite wall of the room. So we will get spots of light, about one inch diameter, in places on the wall, that match their apparent two dimensional arrangement outside. So the spots on the wall are a crude "image" of the scene outside; a fuzzy one, so our "lens hole" not only determines how much light gets in from each object, but it also directs them to a geometrical likeness of the outside scene. If we make the hole smaller, say a tenth of an inch. The image spots are now smaller too, so we have a sharper "image". Unfortunately, it is only 1% as bright as our first case because the hole area is 1/100 of the first hole. So now we put a lens in the hole, which we can now make bigger again, to gather more light. Now we need the lens do do something else, in addition to putting the light spots at the correct image points. We want the lens to steer any ray, that strikes anywhere on the lens aperture, to the exact same point as the one going through the center of the lens. So we have to tilt each point on the lens aperture, in a different direction and amount, like a miniature prism, to steer each ray to the same point. A spherical surface can do this quite well. For simplicity, we keep one surface flat.

So now we have collected a quite large beam of light from EACH object direction, and concentrated it to a single point at each corresponding image location.

So now we ask the question; can we get rid of the lens and put an array of sensors somewhere, and get the same result ? Well where would we place these sensors; and what now determines which rays of light fall on which sensor ? Well I think you can see that there really is nothing sorting out the rays the way the lens does. So the short answer, is that we can't get the same result by the method you propose.

share|cite|improve this answer
Well, I was thinking of a computer program to analyze the light reaches the sensors and then construct a magnified image! – richard Jun 24 '13 at 14:32

What kind of sensors are you talking about? CCDs? You could use a number of CCDs to capture the image and then add the images together in software and map the points to a magnified image which you then project.

Of course, depending on your application, a lens would be cheaper!

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.