Recently I was studying about optical instruments and in my book I came across a point which stated that

When non axial parallel rays are incident on an ideal spherical mirror at a small angle or a parabolic mirror, the rays meet at a point on the focal plane rather than the focus point.

Also in the section about refracting telescopes, diagrams drawn show that the non axial parallel rays meet at a point on the focal plane.

Please note that I'm talking about concave mirrors and convex lenses and aberration and other non idealities are ignored.

I looked at some of the websites like this and this to understand why it is the case but an explanation is lacking everywhere.

I mean, can we show that non axial parallel rays will always converge at a point on the focal plane?

I tried to proceed by attempting to find the equation for the focal plane but I don't think that will work at all.

Could anybody please tell me a way to do this?

How do we guarantee that a group of parallel (non axial) rays incident on a spherical mirror or a lens will always converge at the focal plane (assuming ideality)?

I also saw this SE post and tried to follow the "Physics Teacher" link mentioned in comments but couldn't reach it.

Any kind of approach, either geometric or mathematical will help. Thanks for any suggestions.

  • $\begingroup$ They will not. That sentence is better changed to, "When non axial parallel rays are incident on an ideal spherical mirror at a small angle or a parabolic mirror, the rays meet approximately at a point on the focal plane rather than the focus point." You may take a look of en.wikipedia.org/wiki/Coma_(optics) $\endgroup$
    – verdelite
    Sep 14, 2019 at 18:45
  • $\begingroup$ @verdelite Yes, I saw that aberrations can complicate things hence I decided to ignore them for a while. $\endgroup$ Sep 15, 2019 at 1:06

1 Answer 1


This sketch may give you an idea for the case of the ideal thin lens: The ray AC will continue in a straight line (going through the center of the lens), while BD, which passes through focal point F, will refract to become parallel after the lens. Simple construction shows that the point where these lines intersect is the focal plane.

enter image description here

A spherical mirror M can be thought of as a part of a sphere - and when you are incident “at an angle to your mirror” you could equally say you are hitting “another part of the sphere”, mirror M’. As such, the rays would go through the “original” focal point F; if we consider the “actual” focal point F’ for the segment of the mirror at an angle, we see that F and F’ lie approximately (but not exactly) in the focal plane of M’ (cyan dashed line). I indicate the error (aberration) in my diagram with $\epsilon$. enter image description here

  • $\begingroup$ $+1$, Thanks for the answer. I have some follow-up questions. Could you please address them as well? $1)$ In the lens image an incident ray passing through the center of curvature won't hit the lens, but if the lens was a bit thinner and longer how would that light ray refract and meet the other two at the point below $F'$? $2)$ In the mirror case, when a parabolic mirror is considered, can it be shown that all non axial parallel ray will focus on the latus rectum? Thanks. $\endgroup$ Sep 15, 2019 at 2:34
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    $\begingroup$ @user8718165 I am not sure why you bring in a "center of curvature" for the lens - that is a concept that is only interesting for a mirror (for a lens it only makes sense if you are also considering its refractive index; an "ideal" lens should have infinite refractive index so the curvature becomes infinitesimal - "looks flat but behaves curved"). But after that the general rules of constructing rays apply: rays that pass through focus on one side go parallel to axis on the other side, and rays through the optical center go straight. $\endgroup$
    – Floris
    Sep 17, 2019 at 19:59
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    $\begingroup$ @user8718165 as for your second question - please see this earlier answer $\endgroup$
    – Floris
    Sep 17, 2019 at 20:01
  • $\begingroup$ Thank you so much for the answer :-) I actually meant the optical center instead of the center of curvature. Sorry for that mistake. $\endgroup$ Sep 18, 2019 at 2:44

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