# B Imaging by cylindrical concave mirror

1. Sep 24, 2016

### Pete_L

Consider an upright cylindrical concave mirror with a point source located at the focal point of the mirror. By upright I mean that the mirror is plane in the vertical direction and a concavity in the horizontal plane.

Now it seems likely that you would see a virtual image of the point source looking at the mirror in the vertical plane coincident with the principle axis of the mirror. The point source radiating to the vertex of the mirror "sees" a plane reflective surface both vertically and horizontally.

So however what happens if the viewer looks to one side (laterally) of the principal axis? Off to one side, the point source as before is in the vertical plane reflected according to reflection from a plane mirror. But horizontally, the reflected light ray is parallel with the principal axis as the point source is located at the focal point of the mirror. Will the virtual image of the point source in this case disappear completely? Or, if a virtual image does occur, then where is it located?

This is an interesting problem I think that I haven't seen discussed in the standard treating of reflection by concave mirrors.

Let me have your thoughts, please-

Pete

2. Sep 24, 2016

### Simon Bridge

The image looks like a horizontal line of whatever the source is. The position of the image is at focal length away on the other side of the mirror in the horizontal plane passing through the object.

3. Sep 25, 2016

### hsdrop

you know i hate to ask but does anyone have a pic or can draw up a quick sketch of the question

4. Sep 25, 2016

### Simon Bridge

Trying to draw ... it's annoying. Did not manage to find a pic in image search :(
Lets see if I have it:

Consider the cylinder $x^2+y^2=R^2$, so the x-y plane is "horizontal" and the +z direction is "vertical".
Keep the section of the cylinder in the +x side of the y-z plane, with y between y=+w/2 and y=-w/2, with w<<2R (because par-axial approximation). Remove the rest.
This is the mirror surface and w is the width of the mirror.
A point source is placed at position $O=(1,0,0)R/2$ - being on the focal line of the mirror.

The OP asks about the nature of the image formed by reflection in the mirror surface.

I am asserting that the image is a horizontal line segment length $w$ centred on $x=3R/2$ and parallel to the y axis, which can be viewed from any line of sight that passes through the mirror surface.

If this remains unclear I'll see if I can draw a diagram later (or, by preference, someone else can).

Last edited: Sep 25, 2016
5. Sep 25, 2016

### sophiecentaur

No need. You can split the problem into two parts. If you look at the diagrams for spherical reflectors, that will tell you what happens in the xy plane. The image of a point source, in a concave mirror will be real for distant objects and virtual for objects closer than the focal length of the sphere (r/2)
with distant objects and virtual for lose-in objects. The image distance is given by the concave lens formula. This link has a nice picture to show what happens. There will also be spherical aberration and that will spread the image out; it will not be perfect.
For a cylindrical reflector, the image position is problematical because it will always be 'behind' the z axis (treating it like a plane mirror but 'in front of' the reflector.
So if you look at the image and move your head parallel to the z axis, it will move in the same direction as your head. Parallax will tell you it is behind the reflector. But if you move your head side to side, the image may move the same way (when it's close to the cylinder) but in the contrary direction if it's distant. An image will 'explode' into a LINE as the object goes through the focal point. (As suggested b y the OP) It's the only location where spherical / circular aberration won't occur. We have all played with a shaving mirror or a soup spoon and seen this effect. Anyone who hasn't, should go straight to the bathroom / cutlery drawer and do the experiment ASAP.
I agree with the OP that it is an interesting situation. But there is not a 'good answer' to answer the practical side of the question because you would actually need a true point source because the situation is a singularity. I think that one interesting point that has emerged is the simultaneous appearance of the image in front and behind the reflector for object distances beyond f. That will tickle your brain nicely as it tries to make sense of what it is seeing.

6. Sep 26, 2016

### Pete_L

Thanks for all of the replies. Today I made a drawing, but converting the drawing to a picture file did not go well.

Simon Bridge I think is correct in his saying that what you would see is a line at a distance equal to the focal length, behind the mirror. The line should follow the curvature of the concavity of the mirror.

My initial thought was that the line image would not be visible unless the viewer were looking directly at the mirror because all of the reflected rays are parallel with the principle axis of the mirror (due to the point source at the focal point). That is wrong I believe, because any ray striking the eyeball at any angle is a visible ray. However it would seem that if the viewer were to move too far to either side of the mirror so that none of the rays strike the eyeball, then the line image would no longer be visible.

Is it correct to examine separately how a ray from the source is reflected in horizontal and vertical planes, and then to attribute those two angles to the reflected ray? (There can be only one reflected ray for each ray from the light source)

If you find this behavior of cylindrical mirrors interesting, here is the link to short article about it. The authors consider what happens only for the case where the source is at a greater distance from the vertex than the focal length.
https://www.redlands.edu/contentassets/e8193043b14545aca28afd326b500d3d/cylindrical.pdf
Regards,
Pete

7. Sep 26, 2016

### Pete_L

Here is my drawing of the concave cylindrical mirror. This isn't strictly accurate in the horizontal ("top view") as the cylindrical curvature in reality produces aberration for reflections occurring at the outside edges of the mirror, that is, when the aperture is significant relative to the radius of the cylinder.
S = point source of light (for example, 5 mm LED)
F = focal point
V = vertex
I = image

#### Attached Files:

• ###### Cyl-Reflect.png
File size:
2.6 KB
Views:
88
8. Sep 27, 2016

### sophiecentaur

Taking the same construction as is used in spherical mirrors, the mirror equation tells you that an object in the focal plane will form an image at infinity. (I have posted a link about this already and it must be valid for a cylinder too; please read it and elsewhere) Ignoring spherical aberration, the image will cover the whole of the mirror and will be at infinity until the object is moved slightly away from that point. There is no defined image position for a cylinder because this is an extreme case of astigmatism when your eye lens has different curvature on two axes.
The xy component of this image will be a line (parallel to xy plane) but the cylindrical aberration will mean that its apparent position will not be at one depth. The image will, as I have pointed out previously, be in an undetermined place. Some of the image will be real and formed behind the observer and some of it will be virtual and in front. As soon as the object moves away (in or out) from the the focal point, the position of the image will be better resolved and your eye will 'grab hold of' the information that there is an extended spot with no defined place in the xy plane. Moving your head up and down will produce the parallax that you get in a plane mirror and your brain will tell you that the image you are seeing is distance f behind the mirror. But that is only one version of where the image is because the rays coming from the curved surface are coming from all directions; the image appears to be 'everywhere' (infinitely magnified) and, from the spherical case, it is at around infinity.
I repeat, however, that, due to the aberration, there is no fixed image position in an xy plane.
I have a Newtonian (reflecting) telescope and I can vouch for the fact that what I say is true. The exit pupil of the eyepiece is a small omni source (point), viewed via the secondary (but irrelevant) mirror. I can see that the main mirror appears to be uniformly illuminated by light entering the eyepiece; the image is spread out. The reflector is a paraboloid so the aberration is much less and I am looking at an image that appears at infinity (using the parallax method of moving my head from side to side).
PS the Mirror equation gives you different image positions, according to which plane you choose to cut through the cylinder with (any arbitrary angle) as you can get a not perfect but approximate variation of radius from r to infinity.

9. Sep 27, 2016

### Pete_L

Responding to your above post (sophiecentaur)-

Assuming that there is no aberration of the cylindrical mirror, then do you see the virtual line image at a fixed position behind the mirror? That is, assume that the aperture of the mirror is very small compared to the radius of the concavity.

In theory maybe light rays that are parallel both vertically and horizontally form an image at infinity. But practically there is no image. With regards to the cylindrical mirror of my example, two rays from the source at an equal lateral angle, one rising and one falling from the principal axis, diverge vertically upon reflection. Extensions of the reflected rays meet at a point behind the mirror (as in my drawing). Thus a virtual image is formed, regardless of the fact that horizontally the reflected rays are in parallel and thus neither the reflected or extension of the reflected rays converge in the horizontal plane.

See the article linked to in my post # 6 for an explanation of image formation by the cylindrical mirror. The authors of that article only consider what happens for a point source at a greater distance than the focal length, but how they analyze the mirror can I think be applied to what I'm asking about.

-Pete

10. Sep 28, 2016

### sophiecentaur

It's not as easy as that. As I said before, the position of the line in the z axis direction will be at f behind the mirror surface and this will be perceived by the parallax as you move your head parallel to the z axis. (Or with binocular vision when the eyes are parallel to the z axis).
However, the 'position' on the xy plane is totally indeterminate. It is 'everywhere' along a curved line. You can only say there is an image position if that image is in a single place. With a perfect point and cylinder, you will get a point at infinity behind you and then at infinity in front of you - as you move in and out from the focus.
You still seem to be asking for a definite answer to this one, for a situation that is indeterminate. Not all questions in Science have defined answers.

11. Sep 28, 2016

### Pete_L

Okay, I understand, the line image is a distortion and because it is a line and not a point-image there is indeterminacy.

-Pete

12. Sep 29, 2016

### sophiecentaur

But I think there is more to it than just that. When the point object is not at f, in a spherical mirror, the image will be a point but its position will never be the same as in a plane mirror. As the distance approaches zero, the two images will approach zero - which is not surprising because you are effectively dealing with no significant curvature in any plane - i.e plane in every axis.
With a spherical / plane mirror (i.e. a cylindrical reflector) as the object distance increases , the image will split, with the 'plane mirror' version (virtual) retreating behind the mirror at the same rate as the object distance in front and the cylindrical version (real) following the mirror equation. At f, the image is a line and at r (i.e. 2f), the image will be r in front and r behind.
To my mind this is a much more interesting case because it is no longer an indeterminate situation. It is a nonsense as far as your visual system is concerned, of course, and two different people could see different results, depending upon the rotation angle they happen to be looking from.
You asked this, in the OP:
I think you need to consider what I have been banging on about before you can get a satisfactory answer. Nothing will "disappear"; you will 'see' light coming from various directions and you will / can interpret what you see in various ways.
I remember, a long while ago, discussions about the formation of a rainbow and, I think we concluded (or at least accepted) that the bow is, in fact, an image of the Sun - just suffering from extreme aberration.

13. Sep 30, 2016

### Pete_L

"Nothing will 'disappear'; you will 'see' light coming from various directions and you will / can interpret what you see in various ways."

Differing from most other mirrors and arrangements that usually don't place the object at the focal point, in my example it is possible to move a relatively short distance laterally away from the principal axis and then I believe you can't see any image. This is of course because of the reflected rays in the horizontal plane are parallel to the principal axis. If this is useful in any way I don't know.

Probably I'll be pursuing constructing a cylindrical mirror just because I'm curious to actually see how a point source is reflected depending on the position of the source. This will take some time as I will have to have the film for the mirror shipped to me and I would like to give a fairly accurate curvature to the mirror.

14. Oct 1, 2016

### sophiecentaur

There are many reflector shapes that place an image at the focal point. A cylinder is nothing special and it is, as I said, only an extreme astigmatism. You seem reluctant to see the parallels with a spherical reflector. That's a shame because all the answers are available in texts about spheres. I can detect a "yes but" reaction to that but can you really believe that the calculations are different? If you are reluctant to accept and use the formulae then you have a real problem.
It will be fun to make yourself a cylindrical mirror and I'm all in favour of home experiments. You can get at least as much insight from a shiny spoon, though. Do not be misled by what your eyes tell you. Evolution has not given us the ability to resolve optical conundrums like this one so you need actual measurements (like we did in school) to help you. When real image is formed behind our head, I guess you could say it has 'disappeared' but a suitable correcting lens could bring it back in front, so it hadn't disappeared - it' just 'fuzzy'.

15. Oct 1, 2016

### Pete_L

Given that I want to construct a fairly "accurate" cylindrical mirror, that is one without too much spherical aberration, whatever "too much" is, what should the ratio of the aperture to radius of the concave curvature be? Is there an equation for this (see that, I'm not avoiding equations, I am seeking them)?

A very shallow curvature will be more difficult to accurately construct than one with a smaller radius. So I would like to use as small of a radius as possible without causing a great deal of aberration at the edges of the mirror. That is, from the point of view of ease of construction, as large as possible of a ratio of aperture to radius is desirable.

By the way, although I'm not claiming to have studied optics a great deal, I haven't run across an equation relating spherical aberration to the ratio of aperture to radius of curvature, mostly searching on the web.

16. Oct 1, 2016

### sophiecentaur

I think you are missing the point that the equation for a cylinder is identical to that for a sphere in one plane. How could it be different? That tells you where that component of the image is.
"Enough" means enough for a point to look like a point to you. If you find the point is too glubby by then put card over the edges to narrow it down until it looks ok. Then move till you get a line. Move the point object closer and you will need to view from further away in order to see the (non disappearing) point image.
How will you know where the image is? By moving your head and interpreting what the parallax is telling you. Parallax, as you move along the axis of the cylinder, will place the image behind the mirror. Moving from side to side will place the image in front of the mirror. If this makes no sense to you then google terms like " image position in lens parallax method".
I have made these points before but you don't seem to want to take them on board.

Reality is not always in terms that you want to see it. You may have to unbend a little to get this.

17. Oct 1, 2016

### Pete_L

My question is how much arc of a circle can the mirror have before significant spherical aberration occurs? My preference is for reflection over the entire width or aperture of the mirror to be constructed to be essentially free of distortion. My purpose is to confirm the theory by the images that I see in the mirror.

In the texts on the subject of spherical aberration, it is said that the aperture must be small with respect to the radius of curvature to avoid aberration. My question is, how small?

18. Oct 1, 2016

### Staff: Mentor

If I remember my book on telescope optics correctly, a spherical mirror can be used when the focal ratio is f/8 or slower (so f/8, f/9, f/10+). Anything faster than that will give you significant spherical aberration. Of course, that's for a telescope that's designed to magnify objects. If you're simply going to be observing the image in the mirror by eye, you can almost certainly get away with a much faster f-ratio.

Note that you can always step down the aperture stop to reduce spherical aberration.

You need to get a book on optics. The web is great for general information, but a book on optics will contain everything you're looking for instead of having it spread around a hundred different websites. In addition, the equations involved in optics are rarely trivial and simply changing the position of a point from on-axis to off-axis can often change the equations. I just looked in a basic book for optical engineering and the only equation if gave was related to minimizing the spherical aberration of a single lens and of a bundle of rays from an object at infinity. If I find anything more related to your situation I'll post it.

19. Oct 1, 2016

### sophiecentaur

It could help if you said what it is you actually want to achieve. Until I brought up the question of SA you weren't concerned. You need to define your requirement for SA. Otherwise it' a matter of how long is a piece of string.
Do you have an actual application or are you just doing a thought experiment in which the goal posts keep moving?
You have all the facts about simple reflector equations.

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