Ray diagram for a concave spherical mirror of small aperture.

In discussing optical instruments in the previous posts we have proceeded empirically.   To examine the details of image formation we must have recourse to the laws of optics and it will be helpful if before dealing with lenses in this way we take the simpler case of reflection by mirrors.   To begin with we shall proceed empirically as before, not invoking the law of reflection except that to assume that a ray falling normally upon a mirror is reflected back along its original path.

The diagram above represents a concave mirror, in the form of a spherical cap, seen in cross section, the mirror lying perpendicular to a horizontal line CI (the optical axis ) passing through its centre N and its centre of curvature C. We shall assume

1. Any ray of light passing through the centre of curvature (and so meeting the mirror normally) is reflected back to pass through the centre of curvature.
   


2.  Every ray falling on the mirror from a point source is reflected to meet at a single point, the image of the source. A beam of light parallel to the optical axis may be regarded as emanating from a point source at infinity and the point through which all the rays pass after reflection is termed the focus. This point must be on the optical axis, by the first assumption.


3. If a ray passes in succession through points A, B and C then any ray passing through C and B will pass through A (the principle of the reversibility of light).


4. The aperture of the mirror is small compared with its radius of curvature.

In the figure the object is an arrow OO$'$ and we consider a ray O$'$P$'$ from the tip of the arrow and parallel to the optical axis. This ray will be reflected to pass through the focus. We next consider second ray O$'$P  passing through the focus.  This ray will be reflected to be parallel to the optical axis, by the principle of  the reversibility of light.  The two reflected rays will meet an a point I$'$ which is accordingly the image point of O$'$.

Referring to assumption(iii),  we shall simplify matters by taking P and P$'$ to lie on the plane tangential to the mirror at N, as in the diagram. Let $h$ be the height of the arrow so that the distance P$'$N is also equal to $h$. Let FN=$f$, the focal length of the mirror FO=$x$ and FI=$x'$. Triangles $\triangle$FPN and $\triangle$FI$'$I are similar so that

$\frac{h'}{x'}=\frac{h}{f}$

while $\triangle$ FP$'$N and $\triangle$FO$'$O are similar so that

$\frac{h'}{f}=\frac{h}{x}$

Combining these two equations we have Newton's equation

$f^2=xx'$

As in the diagram, we take the perpendicular distance of the object from the mirror to be $u$ and that of the image to be $v$.  Evidently, $u=f-x $ and $v=f+x'$.  Inserting these expressions into Newton's equation we obtain it in the form

$\frac{1}{u}+\frac{1}{v}=\frac{1}{f}$

If $u$=$R$, the radius of curvature of the mirror, then O is at C and by assumption(i) I must also be at C.  That is, $u=v$ and hence $f=R/2$.

Newton's equation, essentially in the form $R^2/4=xx'$, is stated in his Opticks as true when the rays of light fall perpendicularly or almost perpendicularly on the surface of the mirror.  We now equivalently say the the formula is true within the paraxial approximation, to be discussed later.