We're told that a real image is formed when light rays actually converge to a point. That's all good. But what happens if a screen isn't there to take the image on? Is it still there?
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The rays will continue to move as if there wasn't any obstruction. – Yashas Feb 24 '17 at 16:17
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Whether the screen is there or not the image is there but the problem is focussing your eye on a region of air where the image is formed without a screen.
Try the following set up:
Illuminate a 35 mm slide with a light bulb and adjust the lens so that a sharp real inverted image is formed on a screen with the eye to the left of the screen.
Now replace the screen with a sheet of tissue paper (something which allows light through and will at the same time produce a visible image) and form a sharp image on the tissue paper.
Now observe the image from the other side of the tissue paper (to the right of it as in the diagram).
Your eye will have to be at least 25 cm from the screen if you have normal eyesight.
Keep focussing on the image and slowly move the tissue paper slowly to one side so that some of the image is on the tissue paper and some in "mid air".
With a little practice you should be able see the image of the 35 mm slide without the tissue paper being there at all.
The tissue paper was used to enable you to focus on the correct area of space to view the sharp image.
Update as a result of some comments
Set up with a $3.5\,\rm cm$ focal length hand magnifier as the converging lens.
The object is a pin (white) illuminated by a torch which is switched off for the photograph to be taken without contrast problems.
The other pin (red) will be used to located the image of the white pin.
A white screen was placed next to the image pin to show the real inverted image formed by the lens.
Viewing point now from the top of the first picture ie on the other side of the lens from the position of the object pin.
Image is distorted due to a variety of cheap lens defects.
The position of the real image can be conformed by moving ones eye up and down and seeing that the tips of the image and the tip of the image pin do not move relative to one another - a position of no parallax.
In the end if you know what you are looking for and approximately where to look just looking through the lens on the side remote from an object will enable you to see the real image in mid air.
This is more difficult if the image is highly magnified.
Further update
Note that in the third photograph the image is in focus so the camera "knew" where the image was.
The image pin did help with location but by telling the camera to focus at a certain distance away and no image location pin I would still have been able to get a sharp image on the photograph.
Consider the diagram below which shows the formation of an image of an object $ABC$ on the retina of your eye.
A sharp image is formed if all the light which leaves point $B$ on the object arrives at the same point on the retina $B'$.
So all the rays in the cone of light with apex $B'$ shown in the diagram arrive at the same point on the retina $B'$.
The same being true of all points on the object eg $A$ and $C$ which will arrive ar $A'$ and $C'$.
The light from object $ABC$ originates either from the object itself or as a result of light which has been reflected off it.
So you "see" object "ABC" with your eye.
Now how is that different from the arrangement below?
What was the object in the first diagram is now an image of the object $A''B''C''$ formed by the converging lens.
That intermediate image $ABC$ forms an image $A'B'C'$ on the retina which is no different to that in the first diagram.
You "see" intermediate image $ABC$.
Because there are no reference points around it (just air) it is difficult to decide exactly where that image is and you do indeed see it as though it is "in the lens".
By using an image pin (or your finger) you can easily show by the method of no parallax that you are actually looking at an image in mid air.
When you use an optical instrument you are looking at an image in the air but you have the advantage of being allowed to move the eyepiece to form an image of that image in the air as well as possibly having some cross hairs which are in the image plane.
Farcher
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I am sceptical about this explanation. You seem to be saying that the in-focus image exists at a plane in space without any screen, and that you are training your eye to focus on that plane and ignore all other planes, at which out-of-focus images are formed. I suspect that what you are seeing is an after-image of a brightly lit object. – sammy gerbil Feb 24 '17 at 23:48
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@sammygerbil Try it and see. I can assure you that what you observe is the real inverted image and not any after-image. – Farcher Feb 25 '17 at 00:37
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@KunalPawar You can see an image "in the air" and you can locate it by using a pin and the method of no parallax as described above. – Farcher Feb 25 '17 at 11:11
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@Farcher I've performed this experiment in school. But no one ever explained why that's happening. That's pretty much why I asked this question. – Kunal Pawar Feb 25 '17 at 15:44
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Because in the image plane all the rays from one point on an object arrive at one point in the image plane. Then when the rays continue onwards they appear to come from a point in the image plane i.e. Act like an object. So the rays collected by the eye appear to come from a region in the image plane which is now acting like an object. – Farcher Feb 25 '17 at 15:52
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@KunalPawar You think that it is the convex lens but it is not as I showed in my experiment. – Farcher Feb 25 '17 at 15:54
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Re, "in air" vs. "in the lens." The rays that make up the image must come through the lens, and they all follow straight lines (that's what "ray" means). When you do @Farcher's experiment you won't be able to see the whole real image, you will only be able to see those parts of it that lie on a straight line between some part of the aperture of your eye (a.k.a., your "pupil") and some part of the aperture of the lens. If the lens is tiny, then you will only be able to see a tiny part of the real image. If the lens is bigger, you'll be able to see a bigger part of it.... – Solomon Slow Nov 07 '19 at 14:26
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....In a real sense, you are looking through a hole (the lens) to see the image, but it will feel and look strange because your eye has to focus on a plane that is in front of the lens/hole as seen from your point of view. – Solomon Slow Nov 07 '19 at 14:28
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@Farcher what I didn't understand is when I see any object directly my eyes(or brain) know how to focus those rays that are coming from it on the retina but when i see its real image in air why my eyes don't know how to focus it? – Arun Bhardwaj May 14 '22 at 12:39
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@ArunBhardwaj It is because the image in the air is “transparent” and does not block other rays from reaching the eye. This is not so if the object is opaque so the eye has only one set of rays to cope with. – Farcher May 14 '22 at 15:27
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@Farcher So, as I understand what u tryna say is that the light rays from other sources say a bulb is getting mixed with the rays of real image at those points that's why that area is looking no different from other areas of space?? Am I understanding correctly? – Arun Bhardwaj May 14 '22 at 15:40
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but if that's the case then how screen is blocking those rays? I mean light rays from that bulb which is illuminating the room would still strike on the screen and get mixed at those areas also where the real image is forming? – Arun Bhardwaj May 14 '22 at 15:42
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@ArunBhardwaj True but if the intensity of the light from the image is large enough you will see it having focussed on the screen (and the image). – Farcher May 14 '22 at 21:13
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@ArunBhardwaj Perhaps the only way that you can be convinced is for you to perform the experiment? Is the eyepiece with crosshairs not a good example of seeing a image in “mid-air” (with the help of the crossfires)? – Farcher May 15 '22 at 04:37
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@Farcher thank you so much, I think now I understand.....please can u help me with one more question I posted here https://physics.stackexchange.com/questions/708571/why-image-of-vertical-object-is-always-a-straight-line-and-perpendicular-to-prin/708600?noredirect=1#comment1585077_708600 most textbooks internally assumes certain things without giving any reason while deriving the mirror formula ,,please help me with this – Arun Bhardwaj May 15 '22 at 05:49
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@Farcher while deriving the equation for a vertical object they assumes that paraxial rays even from the topmost point would converge at a single point.... they don't give any mathematical proof of it ,,,proof for only the point lying on the principal axis is given – Arun Bhardwaj May 15 '22 at 05:55
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The answer given by @OfekGillon is a good one which says that if such an assumption is made the theory and experiment agree to a good approximation in certain limited cases. – Farcher May 15 '22 at 05:58
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Please note that my answer is totally in the framework of geometrical optics.
The point where the real image is formed is where rays converge. After this point, rays will naturally diverge, as if they were coming from a point source of spherical waves. You can see an example of this phenomenon here.
JackI
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@Yashas Samaga Will I be able to see the image if a screen isn't there? – Kunal Pawar Feb 24 '17 at 17:03
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@KunalPawar Yes, you will see an image. Rays coming from the image will go to your eyes and then be focused again by your pupil on your retina. You will see it as if they were originating from a point source. The problem for positions different from the image point is that if you put there a screen you won't see it. – JackI Feb 24 '17 at 17:13
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So you mean to say that if I have a concave mirror and a light source to illuminate an object in front it, I can distinctly see the real image of the object (in air?) ? Given that the object is suitably placed for the formation of a real image. – Kunal Pawar Feb 25 '17 at 03:04
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@KunalPawar I have never tried this experiment, so I don't know whether there are other effects that make it not possible. If you are in a geometrical optics framework, you can theoretically predict this possibility. I invite you to take a look at this link http://www.physicsclassroom.com/class/refln/Lesson-3/Ray-Diagrams-Concave-Mirrors. – JackI Feb 25 '17 at 06:58
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When you see an image, what happens all the light that your eye intercepts ( reflected or refracted ), Is focused on retina and then the brain assigns suitable position to it, so if your eye intercepts some light after reflection or refraction, then again an image is formed on your retina and you must see it, but in air it would look like blurred light patterns I think... – Sarthak Sharma Feb 25 '17 at 14:28
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@SarthakSharma You are right, is your eye that focus again what is just blurred rays – JackI Feb 25 '17 at 14:34
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@KunalPawar You can see the image formed in air as long as you're within the cone of light that is originating from the object and is reflected by the mirror. You can't be at any random place, let's say behind the mirror and expect to see the image floating in air like a hologram. The purpose of the screen is light rays from the mirror will convege on the screen and then be reflected in all directions by the screen, so that it can be observed from any location not limited to the light cone – Rupesh Routray May 20 '23 at 07:09