Apologies for the poor wording of the question, I'm sure I'm gravely misunderstanding something here but not sure exactly what.

Suppose we have some point light source. We can see it because the rays that enter the eye will be focused by the cornea and lens onto the retina to form an image. Crucially, if we ray trace the rays backwards, they intersect at the object, which makes sense because this is where the light comes from.

If we now introduce a converging lens such that the point source is located at some arbitrary position along its optical axis, an image of the source will be formed also along the optical axis. If the object is beyond the focal point, the image will form on the other side of the lens. I now put my eye between the image and the lens and look towards the lens. If we now ray trace the rays that enter the eye backwards as we did before, these will no longer intersect anywhere. If we ray trace 'forwards' (i.e into the eye) the rays will intersect at the image but surely this cannot be seen by the eye because it's inside it. And more importantly, the rays can't go backwards from the image into the lens so as far as the eye is concerned this image doesn't exist? So how are we able to see an image? I know an image definitely is seen because this is exactly what happens with glasses or contact lenses to correct for long-sightedness, but not sure how.

EDIT: My question hinges on a certain way of looking at compound lenses being valid which it may not be so I just want to outline it here. Essentially my dilemma relies upon being able to treat the system as the first lens (i.e. glasses/contact lenses) producing an image then when we apply a second lens (i.e. the eye) we make the OBJECT for that lens the IMAGE of the first. In this case, the image of the first lens is occurring behind the eye when we take the glasses in isolation (ignoring the presence of the eye lens for now). So if we make that the object of the eye lens, it seems to me that the rays will be leaving the eye instead of entering it, so it doesn't make sense for an image to be formed at the retina.

  • $\begingroup$ There is not only one image formed, but "infinitely" many. There is however, one convergent image whose location is determined by $\frac 1f =\frac 1u +\frac 1v$ $\endgroup$
    – joseph h
    Commented Jan 3, 2023 at 0:05
  • $\begingroup$ Please, consider adding a picture of your situations, thank you $\endgroup$
    – FGSUZ
    Commented Jan 3, 2023 at 1:38

2 Answers 2


It isn't like there's no image at all if the rays don't perfectly converge. To get a good image you need the rays to converge. If you put a piece of paper at the location of a real image, the light hits the paper and makes the paper look like a picture of the object. If you put a piece of paper between the real image and the lens, the light still hits the paper and still makes the paper look like something. But it only looks like a blurry picture of the object, because the rays from one point on the object are hitting a circle (or other shape) on the paper instead of just one point on the paper. The circles from different nearby parts of the object overlap on the paper and the colours mix together.

Same for your retina. In normal vision, the rays of what you are looking at must converge/focus at your retina. Your lens adjusts until they do. If they don't, for some reason, then your vision is blurry.

Glasses or contact lenses aren't very strong lenses. They have focal lengths somewhere between 50cm and 200cm (estimation). Since the retina is just a few cm behind the lens this isn't a huge effect. We certainly aren't expecting them to make a real image 200cm behind the eye. The point is that the small correction adds onto what your eyes' own lenses do, slightly adjusting the focal length.

  • $\begingroup$ Thanks for the reply, You say that we can't make a real image 200cm behind the eye but suppose we observe some object at infinity with a lens with a 200cm focal length. Then in this case, taking the glasses in isolation, surely we do make a real image at 200cm. My question (granted poorly worded) is how are we able to see this 200cm real image if it is occurring about 200cm behind the eye, whereas before without the lens we were observing an object infinitely far away but in front of the eye. $\endgroup$
    – David
    Commented Jan 3, 2023 at 10:29
  • 1
    $\begingroup$ We can't take the glasses in isolation and at the same time keep your paradox of not being able to see the real image formed by the glasses 2 m behind your eye. I could – annoyingly – say to you: If the glasses form the image in isolation, then just turn round and look at it! I leave it to others to talk constructively about virtual objects... $\endgroup$ Commented Jan 3, 2023 at 12:19
  • $\begingroup$ @David we aren't able to see the 200cm real image formed by the glasses, because the cornea and lens are in the way. We are able to see the 5cm real image formed by the combination of glasses, cornea and lens. If you take the glasses off, you are able to see the 4.9cm real image formed by the combination of cornea and lens, with a bit of blurring because the distance isn't quite right. $\endgroup$ Commented Jan 3, 2023 at 20:18

The lens in a normal eye is a very strong lens. That is, it has a very short focal length. It focuses rays from a distant object to a retina about 1 inch behind it. That means it has a focal length of about 1 inch.

If an object is closer than "distant", the eye needs a different focal length to focus the rays onto the same retina. Muscles in the eye squeeze the lens and changes its focal length.

Some people have a lens with an incorrect shape. It cannot focus rays from a distant object on the retina, no matter how the lens is squeezed.

Glasses help here. glasses a much weaker lenses. Very strong glasses are 3 diopters, which means 1/3 meter focal length, or about a foot. If you put such a lens very close the the eye's lens, the combination is in effect a slightly stronger or weaker compound lens. It does focus a distant object onto the retina. For a closer object, muscles can squeeze the eye's lens until the compound lens is just right.

Here is a diagram to illustrate. It was taken from https://www.amherst.edu/system/files/media/1095/Lecture32%252520slides.pdf. Keep in mind the distances are distorted to make the object fit in the illustration. It would be much farther from the eye in life.

enter image description here

Sometimes you have both a near and distant object in your field of view at the same time. You can't focus both onto the retina at the same time. How does this work?

It isn't a big problem. You focus on one and pay attention to that. Then you focus on the other and pay attention to it.

The out of focus object is blurry. Rays from it don't all hit the same point on the retina. They light up a small circle.

If you trace rays backward, you would find that they all do go back where they came from. You can do this with an imaginary point source on your retina for the in focus case. For the out of focus case, you would put the imaginary light source at the off-retina focus, they would converge to a point on the object. Or you could carefully light up a patch of the retina just right so that each ray goes the reverse of a ray that arrived. This to would go back to a point on the object.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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