I was trying to prove that during reflection from a concave mirror, all rays parallel to the principal axis will meet the principal axis at a point (focus) that is equidistant from the centre of curvature and the pole.

I thought of proving this using coordinate geometry (and some trigonometry). I thought of proving it by proving that for a ray at an arbitrary height from the principal axis, parallel to it, the reflected ray will intersect the principal axis at a distance of $\dfrac{R}{2}$ from the centre of curvature, where $R$ is the radius of curvature of the concave mirror.

So, I plotted a concave mirror on a Cartesian Plane using the equation $x^2+y^2=R^2$ and also added a little constraint to make it look more like a concave mirror. The constraint was $x > 0$, $-R+1 \leq y \leq R-1$, but that's not the point here. Here's what the mirror looked like :

Concave mirror is plotted

Now, I assumed that the incident ray is parallel to the principal axis at a height of $h$ from it. So, it's equation becomes $y=h$. Let's take values of $h$ such that $0 \leq h \leq R -1.5$ for the sake of simplicity here. Now, to find the point of incidence of the ray on the mirror, I found out the point of intersection of the curves of the two equations obtained above. As the point of incidence is a part of the incident ray, so, it's $y$ coordinate is $h$. Let it's $x$ coordinate be $a$. So, $a^2+h^2=R^2 \implies a = \sqrt{R^2-h^2}$. Hence, the point of incidence becomes : $(\sqrt{R^2-h^2},h)$.
Here's what the graph looked then :

Concave mirror with incident ray and point of incidence

$\Big($Note : I have added a little constraint to $y=h$ to limit it till the mirror only, the constraint is $x \leq \sqrt{R^2-h^2}\Big )$

Now, I made the tangent at the point of incidence and the normal too. The tangent was made using the equation $y=\dfrac{-ax+a^2+b^2}{b}$ where $(a,b)$ is the point where the tangent touches the circle. In this case, $a = \sqrt{R^2-h^2}$, $b = h$, so $y = \dfrac{-x\sqrt{R^2-h^2}+R^2}{h}$ becomes the equation for the tangent. For the normal, I just joined $(0,0)$ and $(\sqrt{R^2-h^2},h)$.
Here's what I obtained :

Tangent and normal added

Now, I thought that if I could find another point on the reflected ray, then I would have two points, that point that I will obtain and the point of incidence. Using these two points, I can figure out the equation of the reflected ray. Now, to find the other point I took a point $(0,h)$ on the incident ray and then evaluated what its coordinates on the reflected ray will be. I observed the behavior of the point when it will be rotated with the point of incidence as it's centre and the angle of rotation equal to $2\alpha$, where $\alpha$ is the angle of incidence, that is, the angle between the incident ray and the normal.
The following diagram would help to explain it better :

Labelled (0,h), α, (m,n)

Now, evaluating this was lengthy. I made another coordinate system, where the origin was $(\sqrt{R^2-h^2},h)$. Let's say that $f_2$ is a function that converts the coordinates of a point from the first Cartesian plane to the second. So, $f_2(x,y) = (x-\sqrt{R^2-h^2}),y-h)$. This converts the point $(0,h)$ to $(0-\sqrt{R^2-h^2},h-h) = (-\sqrt{R^2-h^2},0)$
The new Cartesian plane is shown in the diagram below. I have also drawn a dotted circle with radius equal to $\sqrt{R^2-h^2}$.

New Cartesian Plane

Now, in the Cartesian Plane, the coordinates of $(m,n)$ are $(r\cos(\pi+2\alpha),r\sin(\pi+2\alpha))$ $(\pi = \pi^c = \pi \text{ rads} = 180^o)$, and $r$ is the radius of the circle, which means that $r = \sqrt{R^2-h^2}$
Using $\sin(\pi+\theta) = -\sin\theta$ and $\cos(\pi+\theta) = -\cos\theta$, we get that the coordinates of $(m,n)$ in the second plane are $(-\sqrt{R^2-h^2}\cos (2\alpha), -\sqrt{R^2-h^2}\sin (2\alpha))$.

Now, let $f_2^{-1}$ be the function that converts the coordinates of a point from the second plane to the first. So, $f_2^{-1}(x,y) = (x+\sqrt{R^2-h^2},y+h)$.
So, $(m,n) = (\sqrt{R^2-h^2}-\sqrt{R^2-h^2}\cos(2\alpha), h - \sqrt{R^2-h^2}\sin(2\alpha))$.

Now, if we know the values of $\sin\alpha$ and $\cos\alpha$, we can use $\sin(2\theta) = 2\sin\theta\cos\theta$ and $\cos(2\theta) = \cos^2\theta - \sin^2\theta$, we can find the values of $\sin(2\alpha)$ and $\cos(2\alpha)$ in terms of $R$ and $h$.

Now, in Figure 4, $\alpha$ is the angle made by the origin, the point of incidence and $(0,h)$.
Using $\sin\theta = \dfrac{\text{Perpendicular side}}{\text{Hypotenuse}}$, we obtain : $\sin\alpha = \dfrac{h}{R}$. And as $\cos\theta = \sqrt{1-\sin^2\theta}$, we get : $\cos\alpha = \dfrac{\sqrt{R^2-h^2}}{R}$.

So, $\sin(2\alpha) = \dfrac{2h\sqrt{R^2-h^2}}{R^2}$, $\cos(2\alpha) = \dfrac{R^2-2h^2}{R^2}$ (with some calculations that I omitted because the question is long enough already).

Now, let the equation of the reflected ray be $y=px+c$. We know that $(m,n)$ and $(\sqrt{R^2-h^2},h)$ lie on the line. So, we can form two equations that are : $$h = p\sqrt{R^2-h^2}+c$$ $$h-\sqrt{R^2-h^2}\sin(2\alpha) = p\sqrt{R^2-h^2}-p\sqrt{R^2-h^2}\cos(2\alpha) + c$$ On subtracting the second equation from the first one, we obtain : $$\sqrt{R^2-h^2}\sin(2\alpha) = p\sqrt{R^2-h^2}\cos(2\alpha) \implies p = \tan(2\alpha)$$ $$c = h-p\sqrt{R^2-h^2} \implies c = h-\sqrt{R^2-h^2}\tan(2\alpha)$$ $$\tan(2\alpha) = \dfrac{\sin(2\alpha)}{\cos(2\alpha)} \implies \tan(2\alpha) = \dfrac{2h\sqrt{R^2-h^2}}{R^2-2h^2}$$ Finally, with some simplification, we obtain the equation of the reflected ray as : $$y = \dfrac{2hx\sqrt{R^2-h^2}-R^2h}{R^2-2h^2}$$

Finally, the equation is plotted on the graph

Now, everything seemed fine till I started changing the value of $h$ to see where the reflected ray will intersect the principal axis. The point was not the same every time. In fact, only when $h \rightarrow 0$, the point approached what's supposed to be the focus. In the above diagrams, I have taken $R = 10$, so, the reflected ray should have intersected the principal axis at $(5,0)$. But, there is clearly some deviation.

I thought of visualizing the deviation by comparing my reflected ray with the ray joining the point of incidence and the focus. Here's what it looked like :

Deviation from the actual reflected ray

I observed that in my case, the angle of incidence and reflection actually appear to be equal while that is not the case when the reflected ray passes through the focus. I also used the protractor in Desmos to check this.

Now, where did I go wrong? I have recalculated this thrice and I did not find a single error. The only possible error, in my opinion, is that I have assumed that the reflection from a concave mirror happens with the line joining the point of incidence and the centre of curvature as the normal at the point of incidence. On the contrary, I am pretty sure that this assumption is correct.

I would like to know why my reflected ray is not passing through the focus.

PS : Thank You for taking the time to read this question. It took me a long time to write it and so, I was unable to recheck it for errors. So, there might be a few of them and I would be more than happy if you correct them by editing this question or inform me about them, so that I can correct them.
PPS : Here's my graph in Desmos

  • 6
    $\begingroup$ You really did take the long path to find out spherical aberration exists. Indeed, you did nothing wrong, it's a fact that for mirrors that span an angle bigger than (roughly) 10° of arc length parallel rays won't focus on a single point but will rather form a surface of intersecting rays. Amazingly, you can see this figure by positioning a half filled cup of coffee around a lamp $\endgroup$
    – ErickShock
    Jul 1, 2020 at 4:12
  • 1
    $\begingroup$ Please see the diagram of "Reflection from a spherical mirror" here $\endgroup$
    – PM 2Ring
    Jul 1, 2020 at 4:17
  • 5
    $\begingroup$ "Check my work" questions are off-topic here. But you have clearly put a lot of effort & perseverance into this. So I congratulate you for re-discovering spherical abberation. FWIW, try it with a parabolic reflector. It does focus properly, and the mathematics is easier. $\endgroup$
    – PM 2Ring
    Jul 1, 2020 at 4:24
  • 1
    $\begingroup$ Ok. It's very easy to find the slope of a tangent to a parabola. I guess you haven't learned any calculus yet. The simplest parabola equation is $y=x^2$. On that parabola, the slope of the tangent through the point $(X, Y)$ has slope $2x$. $\endgroup$
    – PM 2Ring
    Jul 1, 2020 at 5:14
  • 1
    $\begingroup$ Oops. I meant to say, the slope of the tangent through $(X,X^2)$ is $2X$. Sorry about that. But it looks like you figured out what I meant. :) Good luck! $\endgroup$
    – PM 2Ring
    Jul 1, 2020 at 5:22

1 Answer 1


This is not working out because the when your textbook says 'Rays parallel to the principal axis converge (or appear to converge) at the focus', they are referring to paraxial rays. This assumes that the the rays are very close to the principal axis. See paraxial approximation.

You have instead used marginal rays in your graphs which are quite far from the principal axis and will thus never converge properly at one fixed point. This can be easily fixed by using parabolic mirrors which will by design obey the above rule.

  • 1
    $\begingroup$ Thanks! I wish this assumption was mentioned in the book, would have saved a lot of time. $\endgroup$ Jul 1, 2020 at 4:09
  • $\begingroup$ It probably was, you would have just missed it :) $\endgroup$
    – Sam
    Jul 1, 2020 at 4:09
  • $\begingroup$ Unfortunately, no. I read it from two textbooks and none of them mentioned this. By the way, was the equation that I obtained in the end correct? $\endgroup$ Jul 1, 2020 at 4:11
  • 2
    $\begingroup$ They would converge for a parabolic mirror. $\endgroup$
    – mmesser314
    Jul 1, 2020 at 4:12
  • 1
    $\begingroup$ It looks correct to me. $\endgroup$
    – Sam
    Jul 1, 2020 at 4:19

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