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Consider a spherical (say concave) mirror. Let an object be placed beyond its focus so that a real image is formed on reflection. The location of the image is found by considering a point on the object as a point source, drawing a couple of light rays from it, and finding their point of intersection on reflection.

My question is how do we know that all the other infinite light rays that can be drawn from the point source will also intersect at that same point on reflection?

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  • $\begingroup$ Do you mean spherical, hemispherical or just curved? $\endgroup$ – JMLCarter Sep 7 '17 at 3:29
  • $\begingroup$ Where the mirror is a small part of a sphere. $\endgroup$ – User Sep 7 '17 at 3:36
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The answer is that you are dealing with approximations which are reasonable in some circumstances but not in others.

If you look at a derivation of the mirror formula $\frac 1 u + \frac 1 v = \frac 1 f = \frac 2 R$ you will always find an approximation made as shown in the diagram below.

enter image description here

However there is another flaw in the diagram in that the assumption which has been made is that all the rays parallel to the principal axis meet at the focal point $F$ where the focal length is half the radius of curvature of the mirror.

If you have ever looked at tea, coffee or water in a mug you will know that this meeting at a point for all parallel rays does not happen.
A caustic curve is formed.

enter image description here

This type of defect of a spherical mirror produced when the rays are not mear to the principal axis is called spherical aberration.

Correction is possible but not for all rays.
The most common type of correction is for use in reflecting telescopes where either a parabolic reflector is used or a aspherical reflector is used with a Schmidt corrector plate in front of it.

Even that does not completely eliminate spherical aberration as is shown below with parallel rays not parallel to the principal axis.

enter image description here

So the title to your question "Why do light rays intersect (or appear to intersect) at a specific point on reflection from spherical mirror?" is a good one in that it has in it the word "appear" which you can interpret as being related to the degree of accuracy of reproduction of the object as an image.

Have you ever looked at yourself in a magnifying shaving/make-up mirror and notices the distortion?

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You don't know that, and it isn't really important to build an image.

To build an image what you need is each point on the image to look like (to be illuminated by) a corresponding point on the object.

to say another way:

It's not that you need all the light from one point on the object to reach the image...

What you need is some light from each point on the object to be directed to a corresponding point on the image.

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With a bit of geometry, you can show that if the concave mirror has a spherical profile, all of the light rays emanating from your source will converge at a single point - as long as the point of reflection is near the center of the mirror. (This is a bit like the small angle approximation you use when you're talking about a pendulum.)

However, rays which reflect from points near the edges of the mirror will indeed fail to intersect at a single consistent location, causing the image to become distorted. This effect is called spherical aberration.

Interestingly, for rays which are nearly parallel (such as those coming from a faraway star or galaxy), the image can be improved by using a parabolic mirror rather than a spherical mirror, which is why such mirrors are commonly used in telescopes.

Lenses have the same problem, but it's made worse by the fact that the index of refraction of a transparent material depends slightly on the wavelength of the light. This means that, even with zero geometrical aberration, every color of light will focus through a lens at a slightly different location, causing distortion of the image. This is called chromatic aberration.

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  • $\begingroup$ I know about spherical aberration and the parabola approximation, the 'bit of geometry' is where I'm stuck. :-P $\endgroup$ – User Sep 8 '17 at 7:33

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