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For a spherical mirror, an object at the mirror's center has an image that is also at the center. Its magnification is $-1$. For a video showing this, see here.

If you stand slightly behind the center of a large spherical mirror and hold out your right hand, as if to shake, you will see a reflection of yourself, upside down, holding out its left hand. You can "touch" palm-to-palm with the reflection, or pass through it, but your hands won't be the in the "shaking" position. Instead, the thumbs point opposite directions.

Is it possible, using any combination of mirrors and lenses in the geometric optics limit, to create an image at the same location as the source object, with magnification $+1$, so that you could appear to be shaking your own hand? As you hold out your right hand, your image would need to appear right-side-up and also hold out its right hand.

Here is a faked picture of what I'm thinking of. To make it , I photographed my own hand, then used a computer to copy my hand, rotate it 180 degrees, line it up with the original, and "ghost" it out a bit.

alt text

There is no requirement that the image hand track yours if you move around - a static handshake is enough. I don't necessarily have to be able to view the image directly from where my head is, but it should be a real image so that if I make the air dusty, someone watching my hand from any angle would see the image.

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Did you want to be able to see yourself shaking your own hand, or is it enough that there exists some position from which an observer could see the self-handshake? – David Z Jan 17 '11 at 4:19
@David Good point. I would guess that the second option implies the first, simply by taking whatever apparatus you have and rotating the entire thing around, but I'll clarify the post. – Mark Eichenlaub Jan 17 '11 at 4:54
I want to see a photo of this – Casebash Feb 25 '11 at 11:20
up vote 5 down vote accepted

Yes, it is possible.

Consider a mirror in the shape of a closed hemisphere - that is, half of a concave spherical mirror combined with a flat circular mirror passing through its center.

If you are inside this hemispherical mirror, you will see a real image of yourself rotated 180 degrees around the axis of the hemisphere. Here's one way to show this is true: The flat mirror creates a virtual image of you which is your reflection in the plane (duh), and the spherical mirror takes that as an object and creates a real image which is the inversion of that through the center of the sphere. The composition of the reflection and the inversion is a rotation.

Now, for practical reasons you don't want to be entirely enclosed by the mirror (it would be hard to get in and out, and you'd have to have your own light source...), so you could build just a part of the hemisphere. But the part you built would have to contain both some of the spherical surface and some of the flat surface for it to work.

If you have trouble understanding the orientation of everything, here is a concrete description: The flat mirror is a circle in the $z=0$ (horizontal) plane, $x^2+y^2\le1$. The spherical mirror is a hemisphere described by $z>0$, $x^2+y^2+z^2=1$. You're standing inside so your hand is at a point near the axis, for example, (0.01, 0, 0.2). What the mirrors do is this: For an object at a point $(x,y,z)$, the flat mirror creates an image at $(x,y,-z)$ (this image has opposite handedness because one coordinate is flipped). Then the spherical mirror takes that image at $(x,y,-z)$ and creates another image at $(-x,-y,z)$. This image has the same handedness as the original object. It is also "right side up" because the z coordinate is not flipped. The image of your right hand at (0.01, 0, 0.2) will appear as another right hand at (-0.01, 0, 0.2) which is in the appropriate orientation for you to shake hands.

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@keenan Could you please clarify a little? Suppose I have a spherical mirror whose center is at the origin and radius $r$. I cut off half the mirror and introduce coordinates so that the remaining mirror satisfies $x^2+y^2+z^2 = r^2, x>0$ for the sphere part, or $x=0$ for the flat part. Is that right? If so, at what coordinates do I put my hand? Which direction do I point my fingers? – Mark Eichenlaub Jan 19 '11 at 2:36
That's right. You would put your hand near the z axis, at some intermediate z coordinate (say z = r/4), and point your fingers toward the z axis. A rotated image of your right hand (which is an image of another right hand rather than a left) would appear on the other side of the z axis. – Keenan Pepper Jan 19 '11 at 3:06
I posted the same solution and then noticed this one. I believe it should work. The flat mirror has to be horizontal. – Philip Gibbs - inactive Jan 19 '11 at 7:40
@Keenan Maybe I still don't understand the setup, but I don't see how this solves the problem of the real image being upside down. It does solve the problem of the image being a left hand, though. If we call the hand itself right side up (thumb on top), then the virtual image in the flat mirror also has thumb on top. Then, just as before, the real image created by the spherical mirror has thumb on bottom. It seems to me that this setup creates an image that would go palm-to-palm and fingertip-to-fingertip with my actual hand, and be upside down. Where am I off? – Mark Eichenlaub Jan 19 '11 at 8:45
Oh, I realize I misread where you said "x > 0 for the sphere, x = 0 for the flat part". I assumed you said "z > 0 ... z = 0" because that's how I was thinking of it. – Keenan Pepper Jan 19 '11 at 17:48

If you place twoflat mirrors at right angles and look at them from the line that bisects the angle you will see yourself the right way up and with the right hand in the right place. If you stretch your hand out it will be as if you are about to shake hands with yourself. However the hands can not get to the point where they would touch before the mirror gets in the way.

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Aw, you beat me by a few seconds. +1 – ptomato Jan 17 '11 at 8:50
Right, Phil. The two mirrors just rotate the person by 180 degrees around the vertical axis. By the way, using concave mirrors etc., you could probably move this image, too. – Luboš Motl Jan 17 '11 at 8:55
@Phillip True, but the problem specifies creating a real image, not a virtual one. (I already knew about this solution.) – Mark Eichenlaub Jan 17 '11 at 8:57
I think I see what you mean Mark. Its like one of those trick mirrors where you grab for an object and find its just a reflected image. – Philip Gibbs - inactive Jan 17 '11 at 16:05
Yes, that's what I was thinking. I added a picture in the question. – Mark Eichenlaub Jan 18 '11 at 22:21

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