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.