# B Real image appears in front of the mirror?

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1. May 9, 2018

### Fizzizist

Is there a simple way to determine or prove this? Real images are always inverted, and unlike virtual images can be projected onto a screen (I'm not even sure what this means to be honest). If I look at the back end of a spoon (convex mirror), the image is always upright and therefore virtual, in back of the mirror. Makes sense, since I'm in front of the mirror. Then I turn to the concave side of the spoon and keep it less than one focal length away from me, still the image is virtual and upright, and supposedly still in back of the mirror. But after moving the mirror more than a focal length away from me the image is now inverted, and therefore real. Does this mean it's in front of the mirror? I can't tell the difference. Please clarify and correct any misstatements/misassumptions.

2. May 9, 2018

### Staff: Mentor

This video may explain it:

3. May 9, 2018

### sophiecentaur

That is not always practical when a curved mirror is concerned. Light from an object would very often have to pass through the screen in order for it to reflected onto the screen. Spoons are a great example of where it really doesn't work very well. They are handy though and you can play at the dinner table - people may well think you are loopy though. Take some time and you will make it work.

You can resolve all this if you use the Parallax Method for finding the position of an image. This YouTube movie shows how the virtual image in the mirror appears to be at the same depth as the (finder) red pole that is actually placed behind the mirror. There is only one position for the finder where the image coincides with it from all viewing angles so that's where the image is. This demo can be scaled down and a small rectangular mirror can be stood on a soft wood board and pins can be used to get the effect. In school, we did this easily and could locate the position of the image by moving our heads from side to side. The same method works for concave and convex mirrors and lenses. If you hold a spoon up with your thumb near the centre of the bowl, you can put it at a place where there is no parallax with the image of a window or lamp.

Better still. a concave 'shaving mirror' has less curvature and you can easily put your finger in the no-parallax position with the image of a distant object. If you put a thin tissue or tracing paper, covering about half the mirror, at the right spot, you can also focus the image of a window or lamp on the tissue and you can see it through the tissue.

Nowadays, kids are often given laser pointers for this sort of experiment but they actually fail to beat the Pin method when you need understanding.

4. May 9, 2018

5. May 9, 2018

### Merlin3189

How can your brain know what direction individual rays are coming from, if they are all focused to a single spot? All it can know is that spot is getting light from somewhere.
If the rods and cones could (they can't) in some way tell the directions of at least some rays, they would see converging rays, not diverging ones. It does not matter how far away the source, nor how divergent the rays from it *, the lens of the eye bends them so that they converge at the same rate, to form a single point on the retina. The only thing the brain could learn from the angle of convergence, is the distance between the retina and the lens.

I think the only information the eye can get about the divergence of the rays coming from the object, is the amount it has to stretch the lens to get the rays to focus to a point. This is a recognised cue to depth perception, but is limited in its range and IMO not a major cue. Spectacle wearers may experience a momentary sensation of movement when they put on or remove their specs, but I don't think spectacles generally impair ones depth perception, even though they significantly change the divergence of rays and the degree of accommodation of the lens. Even wearing varifocal lenses does not seem to interfere with depth perception.

What I suspect the brain has learned, is to adjust the stretching of the lens to get the greatest contrast in the centre of the image, much as digital cameras do.

It's hard to get a single point of light without any other cues to distance, to test one's distance perception using just accommodation of the lens. The nearest I can think of at the moment, is looking at a star through a telescope. I don't think I'm aware of how far away the image is, but perhaps that's because I've never thought about it. **

Other than that I'd agree that there is no property of the eye and the formation of the retinal image that distinguishes between real and virtual images. To the eye (as to another lens) an image is as real as an object. You can tell the position of an image in the same way as you can tell the position of an object and if you understand a bit about optics, your brain deduces whether it is real or virtual.

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* - All assuming the object is within the range that can be focused: otherwise they don't converge to a single spot anyway.
** - It's not because I have the telescope in normal adjustment - I'm myopic, so I need an image about 50 cm away.

6. May 9, 2018

### kuruman

Individual rays stimulate individual spots on the retina that can be assigned separate spherical angles θ and φ, if you wish. When I look up to the night sky, I can see one distant star by looking right at it and one with the corner of my right eye. Each star forms an image on my retina at different θ and φ. My conclusion is that one star is straight ahead and the other to my right.

7. May 9, 2018

### Merlin3189

Sorry, I thought you were talking about all the rays from a single point (or single star) diverging to your pupil and then being converged to a single point on your retina. I shall have to reread your post.
I was assuming, for objects such as your arrow, each point on the object sent out millions of rays which were then focused (mapped?) to a single point on the retina.
I suppose, if it were dim enough to emit a single photon at a time, or if your pupil were small enough (& ignoring diffraction), then maybe we could say there was only a single ray from each point on the object to each point on the image.

I can understand you can tell two stars apart, as even if they each emit countless rays, those rays may end up at two distinct spots on the retina.

But it's not worth discussing further, as it relates to visual perception rather than real and virtual images that op asked about.

As far as that goes, I don't think where the image is nor what it looks like is important. Simply whether the light passes through the image, or would do so in the absence of other obstructions (including such as a second mirror or lens.) An objective criterion rather than a subjective one.

8. May 9, 2018

### ZapperZ

Staff Emeritus
Concave mirror gives you a real image at the focal point (for object at infinity). When you are looking at the concave side of the spoon, you are way beyond the focal point. It is why you saw an image with your eye. Inside the focal point, you'll never be able to see a focused image with your eye.

Draw a proper ray diagram to convince yourself of this, or do a Google search.

Zz.

9. May 9, 2018

### kuruman

I think it is dangerous to put too much faith in the ray model in that limit. I prefer to model it as many rays distributed over an area element dA of the pupil much like electric field lines are distributed over an area when one considers electric flux. To me a single ray is as meaningful as a single electric field line.
I agree.

10. May 9, 2018

### Fizzizist

Great info everyone, thanks. So the eye can't tell the difference. A movie is a real image, while a mirage is virtual. A plane mirror is always upright and therefore virtual: the source of light is in front of the mirror but it appears in back. Does the curvature reflect the light more than once in the concave mirror, changing the source so it's in back? Somewhat like the reason letters appear backwards in a mirror?

11. May 9, 2018

### Mister T

Try to form an image on a screen with a convex mirror, you never can. Try with a concave mirror and you sometimes can.

Make the ray-tracing diagrams. When the rays "really" converge to form an image, that's a real image. When they don't, it isn't.

12. May 9, 2018

### Mister T

The use of "therefore" doesn't seem right to me. The image is always upright and the image is always virtual, but I don't see how either one of those is the cause of the other.

I don't follow. The image is behind the mirror and you are in front of the mirror. But I don't understand what connection between the two leads you to conclude that that makes sense. It does make sense, but I wonder why it makes sense to you. The reason it makes sense to me is that when I draw the ray diagram I see that the rays appear to have originated from an image, and the place where that happens is behind the mirror.

No supposition involved. Draw the ray diagram and see for yourself that the rays appear to come from an image, that that image is indeed behind the mirror and that that image is upright.

Being inverted is not the cause of it being real. What makes it real is that the rays "really" do converge to form an image. And when you draw the ray diagram you do indeed see that the image is inverted.

13. May 9, 2018

### Staff: Mentor

This is easier with a large-diameter mirror and a small-diameter screen. When I was teaching, as part of my lecture-demonstration equipment, I had a concave mirror about 40-50cm in diameter. I could hold up a card about 5x5cm in size, at the image location, and get a nice image on it.

Last edited: May 10, 2018
14. May 10, 2018

### Fizzizist

Is there any way to hold the spoon near a wall so it does? Or is some other method needed?

I don't either, it's just something to accept without proof and hope everything reconciles later on. A common but suboptimal learning technique :O

How do I draw the ray diagram? Measure the object and image distances from the mirror, and also that of the focal length? Is it even possible to measure the image distance without a sensor/computer?

15. May 10, 2018

### Mister T

I just typed "drawing a ray diagram" into the google search engine. The first hit is a lesson in how to do it.

The task itself is not particularly hard, although it does take a lot of practice, but it is hard to explain it in a post. The time spent learning this is time well spent because it helps you understand what the equations are telling you.

16. May 10, 2018

### sophiecentaur

Are you doing a course on optics or is this a 'one-off' enquiry? There is far too much to learn about optics for you to progress by a Q and A method, using questions that happen to strike you as important.
As @Mister T says, you can find loads of sources that will show you what to do and give you examples to solve - which will show you how well you have understood what they tell you.

17. May 10, 2018

### Staff: Mentor

18. May 16, 2018 at 8:25 PM

### Fizzizist

Googling is nice, but I like direct answers to questions (which you all have been kindly providing!) This isn't homework, just me trying to understand the concepts better.