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Let's say I place a tennis ball 1 m in front of a plane mirror. The mirror will form a virtual image of the tennis ball, and if I look in the mirror, it appears to me that there is a tennis ball sitting 1 m behind the mirror. If you asked me what the apparent location of the image was as I looked in the mirror, I could give you a reasonable estimate of the image's distance from me.

Now let's say I place a tennis ball some distance behind a converging lens and look through the lens. Let's assume that the ball is further from the lens than the len's focal length. We know that the lens will form a real image of the ball on the near side of the lens (the side from which I'm viewing). My eye will focus the rays coming from the lens onto my retina, and I'll be able to see the image of the tennis ball.

My question is: where does it appear to me that the image of the tennis ball I am seeing is located when I look through the lens? In other words, what would I estimate is the distance of the ball from me?

One thought I have is that, if I am viewing the image from a distance further than the image distance, the image will appear to be at the image location. My gut feeling is that this is probably not the right answer. Certainly if I view the image from a point closer to the lens than the image distance, it won't look like the image is at the image distance, because the image distance is behind me.

Another thought is that the apparent image location is just based on the apparent size of the image. If I calculate the angular size of the image, I can predict the apparent image location based on the distance at which a tennis ball would have that same angular size. I think this is probably closer to the right answer, but I'm not confident in this.

(I am not sure this question is 100% on topic, because ultimately it's as much about how our brain infers the position of objects as it is about optics, but I am interested in to what extent this question can be answered using optics.)

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  • $\begingroup$ A "real image" is not something you see by looking through an optical component. If you see an image at all when you do that, then you are looking at a virtual image. If you want to see a "real image," you have to hold a white card in the image plane, and look at the card. Of course, you could complicate the situation, (e.g., by peering through a telescope at a card, you could see a virtual image of a real image of some subject. $\endgroup$ Commented May 19 at 21:35
  • $\begingroup$ Object and Image Space $\endgroup$
    – mmesser314
    Commented May 19 at 21:40
  • $\begingroup$ @SolomonSlow I don't think what you say is true. You can look through a converging lens that is producing a real image and see the image. Wikipedia has an example: en.wikipedia.org/wiki/Real_image#/media/… (Obviously there is a camera involved here, but we would also see the image with our naked eye.) $\endgroup$
    – d_b
    Commented May 19 at 21:46
  • $\begingroup$ I am unconvinced that the image seen by the camera in that example is the real image that is formed by the lens.* Unfortunately, I don't have time right now to draw a ray diagram that would answer the question one way or the other. Maybe somebody else will weigh in. Au Revoir! [* For one thing, the real image usually is larger than the aperture of the lens that forms it, but the image in the Wikimedia picture is exactly bounded by the aperture of the lens.] $\endgroup$ Commented May 19 at 22:40
  • $\begingroup$ @SolomonSlow Even if the image is larger, an observer looking at the image can only see the parts of it that overlay the lens. Or in other words, for a screen, all points of the image receive at least some rays from part of the lens and therefore the entire image shows up. For an eye, the same ray that goes from the eye to a point on the image has to land on the lens for the eye to see that part of the image. The eye may have to move around to reveal the whole image. $\endgroup$
    – HTNW
    Commented May 20 at 2:39

3 Answers 3

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enter image description here

The figure describes the situation, but in words, the real image will be formed on the near side of the lens (same side as observer). This was pointed out in the question. It will appear to the viewer that the object is actually physically located at that position - somewhat "floating" in front of the lens, although inverted and resized (amount dependent on the distance that the actual object is from the lens).

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    $\begingroup$ The OP appears to have some concerns about what happens when the real image is viewed from closer to the lens than the image. What happens in this case is that the images has properties different from any real object: most noticeably, its parallax is in the wrong direction. In other words, the image truly has a negative distance (optically) to the viewer if the viewer is close to the lens, even though the viewer can see it. $\endgroup$
    – HTNW
    Commented May 20 at 1:24
  • $\begingroup$ @HTNW Thanks, yes, that was really the source of my confusion. I probably should have condensed the question down to just asking about that case. If you care to post your comment as an answer I would accept it. It is quite surprising to me! $\endgroup$
    – d_b
    Commented May 20 at 1:45
  • $\begingroup$ @d_b: if you try to observe the image from a point between the lens and the image, you won't see it as all the light rays from the image are going away from you. $\endgroup$ Commented May 20 at 4:57
  • $\begingroup$ @RossMillikan Could you elaborate on this? In particular, I don't understand what "all the light rays from the image are going away from you" means. $\endgroup$
    – d_b
    Commented May 20 at 6:18
  • $\begingroup$ To see the image from where you say your eye is between the lens and the image. We ignore that your head blocks the rays that would form the image, but after they form the image they keep going away from you. No light enters your eye from the image. $\endgroup$ Commented May 20 at 13:11
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Real images look like real objects. Think about it: your eyes cannot distinguish, having access only to the light rays entering it, a real object from a real image. In both cases, your eyes receive light rays that appear to be diverging from a given point. That is all the information available to you in the case of a real object, and anything that replicates that situation (i.e. an image, virtual or real) will appear like an object, too.

If you view a real image from a point behind (w.r.t the lens) the place where the image forms, you simply see what looks like a (scaled) copy of the object, in front of the lens. The accommodation depth cue (that is, your sense of what focal length you have to set your eyes to in order to focus the image) will be as if for a real object at the distance of the image. Similarly, the binocular depth cue due to parallax between the two views of the object in either eye will be as for a real object. But there will be one strange thing: you will not be able to see parts of the image that lie (in your field of vision) outside of the outline of the lens. That is, it is just like looking through a window, where objects behind the window are cut off by the outline of the window. So there is a conflicting depth cue in that the image appears to be occluded by the lens. This situation, where the accommodation and binocular cues say the image is front of the lens but the occlusion cue says the image is behind the lens, is not possible for a system of ordinary objects. In my experience the first two cues are stronger and the image does "feel" like it is in front of the lens.

Optically, the "distance" of an image (whether a real object or real image or virtual image) from an observation point is defined according to parallax, i.e. according to what the accommodation and binocular cues of the eye say.

If you view a real image from closer to the lens than where the image forms, things get weird. Your eyes now need to relax more than they need to relax to view something infinitely far away. The view of the image in your left eye is further to the left than the view in your right eye, where for ordinary objects the situation should be reversed. These are indicative of viewing an image that is at negative distance in front of your eye. Mathematically, it makes sense to consider distances in optics as lying on the extended real line $\mathbb{R}\cup\{\infty\},$ interpreted as a circle. The negative numbers connect to the positive numbers through zero, but then the positive numbers connect back to the negative numbers through infinity. If you see an image that is at negative distance in front of you, it just appears to be really far away, behind even the sky (which is at infinite distance to you).

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If you are viewing real image with your eye further away from the real image than the least distance of distinct vision (ie further than the near point) then the answer is give here, Real images and their formatiom.

However if your eye is closer than the least distance of distinct vision (ie closer than the near point) then you eye will be unable to focus the real image on your retina.

If your eye is closer to the lens than the real image then you now must consider the lens as adding to the optical system of your eye and making the optical system more powerful as you are adding a biconvex lens to it.

What happens all depends on the condition of your eye and whether it has sufficient accommodation to adjust so as to produce a sharp real image on the retina.
With a relatively weak, +0.2D (f = 5 m) lens I saw a relatively sharp image with my unaided eye but with a +10 D (f = 10 cm) lens I saw a fuzzy image of the object I was looking at.

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