Understanding Reflection of Light on a Concave Mirror

[SHASHWAT PRATAP SING]

I just did it. At a distance, you see yourself a long way off, upside down. You go in close and you see your eyeball the right way up. In between, there's the explosion and you see nothing because the image of your eye is behind you at first, then too near your eye to focus and then, there it is. You can tell it's the right way up when you blink.
Probably a teaspoon would be too small but I used a tablespoon and a soupspoon (spherical and better). Better still, a shaving mirror - or visit a funfair hall of mirrors.

I think you may be having a problem with confidence about understanding that video. Whatever you manage to make of it, the theory is supported by what you can actually see for yourself so you do not have to work too hard to reconcile what the guy says in the video. He is just peripheral to the real issue.
Quite frankly the video is very bad. If he wanted to show you what he was actually seeing then he should have carried the camera with him. As it is, he's actually lying to you about what he and the camera see! He's not too sharp, actually. He started off mixing up convex and concave names and then he fell over his floor marker - duh! Not worth worrying about it. Just repeat the experiment with the spoon and see your eye invert after the exploding region. Admittedly you don't see your eye in focus, up close (unless you have very strong accommodation muscles) but when you can see an image, it HAS to be in front of you.

Don't forget. You are not obliged to believe what you are shown in any video when it clearly doesn't make sense.

A big thankyou sophiecentaur by explaining this you have given me great satisfaction and relief. You mean a lot to me, really thankyou...

Now I clearly understand your post-(#21) ...

I understand your problem entirely. It was the same for me at first. Only when I used a mirror could I actually appreciate what's happening.

I think your problem is that you think the image is real and behind you when you can see it. When you get that condition, you are standing near the focus of the mirror and, on either side of that position, you will see an image that's either virtual and behind the mirror (when you're close) or real and in front of you (you're further away). Around that critical point, the image 'explodes' and you don't actually see it - because it's formed behind you. It's only when you go further away that the image position is actually in front of you (near and highly magnified).
Around that position your poor brain has a serious problem and can't actually make sense of what you see.
Note: This effect is most obvious with large mirrors (clearly the radius needs to be large) and the region around the exploding point is quite large. Looking in the bowl of a shiny spoon, you need to put your eye in very close and you can see your eye explode as it goes from inverted to non inverted.

sophiecentaur Please tell me now am I correct-
In real practice when we stand between the centre of curvature and focus of a large concave mirror we will see a large blurry image of ourself as our brain can't actually make sense of what we see, the actual real and inverted image is formed behind us ...

sophiecentaur Please correct me if you think...

 

[jbriggs444]

IMO, that's a red herring. The same thing happens with all images and all rays entering the eye. Light coming in from above the horizontal always hits the lower half of the retina.

It is an important point if one is under the mistaken impression that an inverted image on the retina means "seeing things upside down".

 

[SecularSanity]

I just did it. At a distance, you see yourself a long way off, upside down. You go in close and you see your eyeball the right way up. In between, there's the explosion and you see nothing because the image of your eye is behind you at first, then too near your eye to focus and then, there it is. You can tell it's the right way up when you blink.
Probably a teaspoon would be too small but I used a tablespoon and a soupspoon (spherical and better). Better still, a shaving mirror - or visit a funfair hall of mirrors.

I think you may be having a problem with confidence about understanding that video. Whatever you manage to make of it, the theory is supported by what you can actually see for yourself so you do not have to work too hard to reconcile what the guy says in the video. He is just peripheral to the real issue.
Quite frankly the video is very bad. If he wanted to show you what he was actually seeing then he should have carried the camera with him. As it is, he's actually lying to you about what he and the camera see! He's not too sharp, actually. He started off mixing up convex and concave names and then he fell over his floor marker - duh! Not worth worrying about it. Just repeat the experiment with the spoon and see your eye invert after the exploding region. Admittedly you don't see your eye in focus, up close (unless you have very strong accommodation muscles) but when you can see an image, it HAS to be in front of you.

Don't forget. You are not obliged to believe what you are shown in any video when it clearly doesn't make sense.

No, that's not correct. The image is not formed behind you at all. All of the graphs are just depicting the magnification relative to the distance.

 

[sophiecentaur]

No, that's not correct. The image is not formed behind you at all. All of the graphs are just depicting the magnification relative to the distance.

The formula
1/f = 1/u+1/v
always applies (using the correct signs, of course)
Forget magnification and just use an ordinary object / image system. Don't forget, when he's at f, the image is at infinity behind him. He even implies that when he describes what f is.
I think you are confused because you are still trying to go along with some very pretty but misleading graphics. Moving the object and the observer at the same time is a needless complication for getting to grips with the very basic system. You are trying to make two steps at once when you should really do them one at a time. That video actually does only one step because the camera doesn't move.

The fact is that there are positions in which you do not see an image as you move in and out. If you still don't believe this then that video is useless because it doesn't present the right experiment. For some reason you don't seem to want to get anything out of my spoon suggestion. In a spoon you CAN see a virtual image behind the reflecting surface and you CAN see an image in front of the surface. Not ideal but it shows the effect. In between those situations you cannot see an image. It flips behind you and you can only see it again when you are in close enough. But there's no point in going that far with the thing if you don't appreciate the 1/f = 1/u + 1/v equation.

 

[SHASHWAT PRATAP SING]

Sophiecentuar could you please reply on my post-#91 I am eagerly waiting for your reply. ...

 

[SecularSanity]

The fact is that there are positions in which you do not see an image as you move in and out. If you still don't believe this then that video is useless because it doesn't present the right experiment. For some reason you don't seem to want to get anything out of my spoon suggestion. In a spoon you CAN see a virtual image behind the reflecting surface and you CAN see an image in front of the surface. Not ideal but it shows the effect. In between those situations you cannot see an image. It flips behind you and you can only see it again when you are in close enough. But there's no point in going that far with the thing if you don't appreciate the 1/f = 1/u + 1/v equation.

I think that someone else has already pointed out that when the object is located at the focal point no image is formed because the light rays neither converge or diverge. They run parallel.
Image Characteristics for Concave Mirrors

 

[jbriggs444]

I think that someone else has already pointed out that when the object is located at the focal point no image is formed because the light rays neither converge or diverge. They run parallel.
Image Characteristics for Concave Mirrors

True, but not relevant for whether the person can see themselves in the mirror when positioned at the focal point. They can. The image that is formed is "at infinity". Images at infinity can be seen just fine.

 

[A.T.]

... when the object is located at the focal point no image is formed because the light rays neither converge or diverge. They run parallel.

True, but not relevant for whether the person can see themselves in the mirror when positioned at the focal point. They can. The image that is formed is "at infinity". Images at infinity can be seen just fine.

Much of the confusion seems to come from failing to make the distinction between "image formed by the mirror" vs "image formed by the eye/camera lens".

 

[SecularSanity]

The image isn't formed behind you.

I could be wrong, but it seems to me that if that were the case, and I was beside you, I’d be able to see your real image behind you. It’s not, it’s magnified. Look at the basic ray diagrams for a magnifying glass. When you look through it, does the object appear further away? Nope.

 

[sophiecentaur]

sophiecentaur Please tell me now am I correct-
In real practice when we stand between the centre of curvature and focus of a large concave mirror we will see a large blurry image of ourself as our brain can't actually make sense of what we see, the actual real and inverted image is formed behind us ...

Light from every part of you (over a limited range, of course) will go out and return along the same path. AS you say, that will not make any sense to your brain because the image will all be spread out, appearing to be in all directions at once.
Once you are slightly closer, there will be a normal, enlarged virtual image just like the limit where the mirror has an infinite curvature (plane). Image distance increases.

When you are more distant than the radius, I cannot help thinking it's easier to start by thinking of yourself at infinity. You are at infinity so your image will appear at f (inverted, small and definitely in front of you so you can see it). Come in a bit closer than that and your image will move out from f, towards you and get bigger. Now consider the reciprocal path; if you are at a point just outside f, you will have an image that's way behind your head.
When you are at 2f, your image will coincide with your face. (1/f = 1/2f +1/2f). Closer than 2f your image will be behind you. But, as has been pointed out and I only just really got this, your eye can still make sense of what it sees, which is a (can be highly) magnified image which is very unstable and you can only view it with one eye. Much further in and you see less and less of a proper image until you go inside the r distance and the regular shaving mirror image kicks in.
PS I just actually did this with a low quality makeup mirror and can confirm that the image is very confusing as you approach nearer to 2f. You can just see that it's inverted all the way to r then it starts to behave itself with a virtual image

 

[hutchphd]

Much of the confusion seems to come from failing to make the distinction between "image formed by the mirror" vs "image formed by the eye/camera lens".

I would take this point one step farther.
At every point along the axis (except perhaps at the focus exactly), I could design an eyeball/camera which would, in fact, project a real image of itself onto the retina (or CCD) therein. As it happens a standard camera or eyeball is not designed to do this for the contested region between between focus and cc.

 

[sophiecentaur]

The image isn't formed behind you.

I could be wrong, but it seems to me that if that were the case, and I was beside you, I’d be able to see your real image behind you. It’s not, it’s magnified. Look at the basic ray diagrams for a magnifying glass. When you look through it, does the object appear further away? Nope.

I think you must be wrong for the reason above. We are more familiar with convex lenses but the same thing happens. A close up object produces a distance image - a projector - and a distant object produces a close up image - a camera (see the formula and what I wrote above). Judging where an image actually is is not too easy. You really need to do the parallax operation as they do in school with pins on an optical bench.
This 'looking at yourself' exercise adds a lot of confusion.

 

[sophiecentaur]

I would take this point one step farther.
At every point along the axis (except perhaps at the focus exactly), I could design an eyeball/camera which would, in fact, project a real image of itself onto the retina (or CCD) therein. As it happens a standard camera or eyeball is not designed to do this for the contested region between between focus and cc.

I agree. I made the point way up the thread that a correcting lens would do the job by actually changing the focal length of the system. Problem with experimenting is that you need a very good spherical mirror to see anything much at all when closer than 2f. We all know about this stuff when applied to lenses - it's just the mirror context and the reciprocal path that makes it hard.

 

[SecularSanity]

I think you must be wrong for the reason above. We are more familiar with convex lenses but the same thing happens. A close up object produces a distance image - a projector - and a distant object produces a close up image - a camera (see the formula and what I wrote above). Judging where an image actually is is not too easy. You really need to do the parallax operation as they do in school with pins on an optical bench.
This 'looking at yourself' exercise adds a lot of confusion.

Ah, dang it! I would be able to see your real image behind you. You were right and I was wrong. I hate when that happens.

Well done!

 

[SHASHWAT PRATAP SING]

Light from every part of you (over a limited range, of course) will go out and return along the same path. AS you say, that will not make any sense to your brain because the image will all be spread out, appearing to be in all directions at once.
Once you are slightly closer, there will be a normal, enlarged virtual image just like the limit where the mirror has an infinite curvature (plane). Image distance increases.

When you are more distant than the radius, I cannot help thinking it's easier to start by thinking of yourself at infinity. You are at infinity so your image will appear at f (inverted, small and definitely in front of you so you can see it). Come in a bit closer than that and your image will move out from f, towards you and get bigger. Now consider the reciprocal path; if you are at a point just outside f, you will have an image that's way behind your head.
When you are at 2f, your image will coincide with your face. (1/f = 1/2f +1/2f). Closer than 2f your image will be behind you. But, as has been pointed out and I only just really got this, your eye can still make sense of what it sees, which is a (can be highly) magnified image which is very unstable and you can only view it with one eye. Much further in and you see less and less of a proper image until you go inside the r distance and the regular shaving mirror image kicks in.
PS I just actually did this with a low quality makeup mirror and can confirm that the image is very confusing as you approach nearer to 2f. You can just see that it's inverted all the way to r then it starts to behave itself with a virtual image

Thankyou sophiecentuar for your help and all you did for me... really thankyou

sophiecentuar I just wanted to confirm one thing by saying "Exploding Point " you mean the focus of the concave mirror, Right ?

 

[sophiecentaur]

Thankyou sophiecentuar for your help and all you did for me... really thankyou

sophiecentuar I just wanted to confirm one thing by saying "Exploding Point " you mean the focus of the concave mirror, Right ?

I would say that when you are at r and the image spreads out everywhere would be the exploding point. Thereafter, nothing is easy to see until you are at 2f. Then you get a clear image right in front of you.
But you have to do it yourself to be really convinced. A spoon really does work once you identify what you are seeing.

 

[SHASHWAT PRATAP SING]

I would say that when you are at r and the image spreads out everywhere would be the exploding point. Thereafter, nothing is easy to see until you are at 2f. Then you get a clear image right in front of you.
But you have to do it yourself to be really convinced. A spoon really does work once you identify what you are seeing.

Thanks Sophiecentuar,
Here r is the radius of curvature,right? Just to confirm

This means that on Radius of curvature of a concave mirror,Until we reach the centre of curvature from focus, everywhere would be the exploding point...
thanks for help

 

[swampwiz]

An easy to think about it is that any time the rays appear to be emanating from a single point - even if they got reflected or refracted from some other point (which could be the actual point on the object or a point from which they appear to be emanating due to other previous reflection/refraction) - if the observer is in front of that emanation point, then he will see the light as if it were being originally the object from which the light is emanating. The original point on the object from which the light emanates is termed the "object", and the point from which the rays appear to be emanating (i.e., which is due to the optical system) is termed the "image" - and when considering a compound optical system, the image point from one optical system is considered to be the object for the following one.

In the 3rd case for the figure posted, the image is behind the observer, and thus there is no point of emanation in front of him. Now of course, there is light emanating to him, but it is not from a point, and thus, it will be a blur - and is the same blur that someone gets when trying to focus on something too close to his eye. Someone with hyperopia (i.e., farsightedness) has the condition in which he can focus light coming from a hyper-infinite point - i.e., light that appears to be emanating from a point behind him (i.e., the light is coming at him, but it appears to emanate from a point behind him), so he would be able to focus in that 3rd case (i.e., up to some limit where his hyperopia is not strong enough to focus in at a "closer" hyper-infinite point.

The reason that a standard flat mirror always shows an image that is twice as deep as the distance to the mirror is that the image "point of emanation" is always the same distance (but behind) the mirror as the object is from the mirror, so the image appears to be deep into the mirror at the same distance that the object is in front of the mirror.

 

[SHASHWAT PRATAP SING]

I would say that when you are at r and the image spreads out everywhere would be the exploding point. Thereafter, nothing is easy to see until you are at 2f. Then you get a clear image right in front of you.
But you have to do it yourself to be really convinced. A spoon really does work once you identify what you are seeing.

sophiecentaur here r is the radius of curvature of the concave mirror, right ? Just to confirm...
Please reply whenever you are free...

 

[sophiecentaur]

sophiecentaur here r is the radius of curvature of the concave mirror, right ? Just to confirm...
Please reply whenever you are free...

You really shouldn't need me to say right or wrong. Just put yourself at the centre of curvature and think where an image could be. Rays are coming from everywhere. If the word 'exploding' doesn't apply then what other word would you use?
Another point is that our brain does its very best to make sense of what is going into the eye. It will try hard to deal with any image - even when it is defocussed or distorted so there will be some vague idea about what you see in other places.

 

[SHASHWAT PRATAP SING]

You really shouldn't need me to say right or wrong. Just put yourself at the centre of curvature and think where an image could be. Rays are coming from everywhere. If the word 'exploding' doesn't apply then what other word would you use?
Another point is that our brain does its very best to make sense of what is going into the eye. It will try hard to deal with any image - even when it is defocussed or distorted so there will be some vague idea about what you see in other places.

Thank you sophiecentuar,
The word Exploding fits here...

 

[A.T.]

As it happens a standard camera or eyeball is not designed to do this for the contested region between between focus and cc.

I just tested it with a cheap make-up mirror. It gets a bit blurry beyond F, but you can still clearly recognize your eye until you get really close to C. And it is not flipped between F and C. A smart phone camera works even better.

@SHASHWAT PRATAP SING If you haven't done it yourself yet, I encourage you to try it. If you don't know the focal length of the mirror, move an object back and forth in front of the mirror, until you find the location where its image flips for a distant eye/camera. That's your F and C it 2 times that. Then move the eye/camera between F and C.

 

[sophiecentaur]

its image flips for a distant eye/camera.

yes but it's at r that your own image flips. At that point the rays strike the normals from the other direction.
This is why 'that video' doesn't help at all because it is shot from a different location from the subject.

 

[A.T.]

Keep in mind that if the real image on the retina is flipped, then we see a not flipped picture.

IMO, that's a red herring.

The question was, what you see when your eye is between F and C. And to tell whether you see a flipped or a non-flipped image you need to understand how to relate the ray diagram to what you see.

 

[A.T.]

yes but it's at r that your own image flips.

You mean at the center of curvature or 2f? Then yes, as I said a few pages back:

- For a distant camera the image flip happens when the object is at the mirror's focal point.
- For a camera at the object the image flip happens when the object is at the mirror's center of curvature.

 

[hutchphd]

Between f and cc a camera/eyeball lens with a positive focal length will not create an image. If the lens can be made to have a negative focal length (like a concave lens) then the lens will produce a real image because the reflection from the mirror provides a projected virtual object behind the lens. Note that this real image will not be inverted by the camera/eyeball lens and so you see what you described. Confusing perhaps but not surprising.

 

[A.T.]

Between f and cc a camera/eyeball lens with a positive focal length will not create an image.

It will produce an image, which might not be exactly on the sensor/retina, and thus the picture you see is blurry. But you can still tell if your eyebrow is above or below the eye. Try it.

If the lens can be made to have a negative focal length (like a concave lens) ..

You don't need a negative focal length. You just need a positive focal length greater than the lens-to-sensor-distance/eyeball-diameter, because the rays are already converging, so you need less lensing than for parallel rays.

Look at the diagram below. The eye here still needs a convex lens to focus the converting rays more, so they meet on the retina, and not behind it, where the mirror forms a real image.

Note that this real image will not be inverted by the camera/eyeball lens and so you see what you described.

No. To see what I described (a non-inverted picture) the real image on the retina (or close to it) must be inverted (like in the diagram above).

 

[sophiecentaur]

You mean at the center of curvature or 2f? Then yes, as I said a few pages back:

With you and your eye at 'cc', rays are all incident back at your eye but there's no specific source. With you and your eye at 2f, there is an identifiable (inverted) image of each part of your face / eye formed at the eye. Those two cases are significantly different, although the rays converge in each case.

I still don's see why you stress the optics of the eye in this. 'Inverted' surely implies upside down compared with how you would normally appreciate a scene. Non - inverted is the converse. In my personal sketches, the inversion starts at >cc - just after the image spreads out everywhere (what I have called the explosion). Even very near cc, the upper and lower parts of the image can be appreciated, although there is no clear image. Only when you get just past 2f does your eye actually see a proper ( and inverted) image. In between those conditions, you can get clues about what you are looking at, depending on your powers of accommodation and the aperture of your pupil.

 

[A.T.]

With you and your eye at 'cc', rays are all incident back at your eye but there's no specific source. With you and your eye at 2f, there is an identifiable (inverted) image of each part of your face / eye formed at the eye.

By "cc" or "C" I mean the center of curvature, which is at 2f (2 * focal length) from the mirror. So "at cc" is the same as "at 2f". You seem to mean different things by this abbreviations?

 

[sophiecentaur]

By "cc" or "C" I mean the center of curvature, which is at 2f (2 * focal length) from the mirror. So "at cc" is the same as "at 2f". You seem to mean different things by this abbreviations?

Yes. Sloppy thinking of mine. 2f is from the equivalent lens treatment and is more familiar.