The magnifying mirror

Concave mirrors are commonly used in the activities of shaving and applying cosmetics.  The image is then virtual - behind the mirror - whereas up to now we have discussed only the formation of a real 
image.   To  discuss the formation of a virtual image, the (real) object being nearer the mirror than the focus, we refer to the diagram above. We do not need to repeat the argument of the last post in detail.   Considering, as before, rays from the top of the object O$'$ we again obtain Newton's equation,
$f^2=xx'$
Setting $u=f-x$ and $v=x'-f$, the respective perpendicular distances of object and image from the centre of the mirror,  the equation may be rewritten as
$\frac{1}{u}-\frac{1}{v}=\frac{1}{f}$
This differs from the comparable equation of the last post in sign of the the second term on the left hand side but he equations become identical if one adopts the sign convention that distances measured in the direction of the focus from the mirror are taken to be positive while those measured in the opposite direction are taken to be negative.

For a discussion of the magnifying property of the concave mirror the ratio $h'/h$ of heights of image and object is relevant.  This is obtainable from the formulae already obtained and is simply
$\frac{h'}{h}=\frac{v}{u}$,
of the same form as that obtained earlier for a thin lens.

In using the concave mirror for shaving  the relevant comparison is the angular sizes of image in the concave mirror and a plane mirror in the same position, as shown in the following figure. In a plane mirror, of course, the image is as far behind the mirror surface as the object is in front.

The angle subtended by the image (of an arrow as shown) in a plane mirror is $\alpha=h/2u$ while that subtended in the concave mirror is $\beta=h'/(u+v)$, where $h'=hv/u$. We therefore say the magnification is

$m=\frac{\beta}{\alpha}=\frac{h'}{h}\frac{2u}{u+v}=\frac{2v}{u+v}$

Substituting in for $v$ from our previous equation relating $f$, $u$, and $v$ we find

$m=\frac{2f}{2f-u}$

Which is the equation we require.   Since, as we have seen, a virtual image obtains only if $u<f$, the maximum value of the magnification is two.  Typically in shaving the face will be of the order of 1ft from the mirror.  If the focal length is 3ft, the magnification is $6/5$, a modest but significant result.  A possible second benefit from using the mirror is the movement of the image beyond the near point.  In our example, v=3/2ft  so that the image is moved back 6 inches from its position in a plane mirror, again a modest amount, but possibly significant to a person somewhat longsighted.

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.