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.

The single-lens telescope

In my first post I mentioned that Roger Bacon has written, in his Opus Majus, of  the use of lenses to magnify distant objects.  He does not go into any detail but it is indeed possible that he did observe that  a single convex lens of sufficiently long focal length before his eye would make remote objects appear nearer.  In the figure below we represent  the observation of the image of a distant object formed by a convex lens at its focus.

Fig.1 

Write $\alpha$ for the angle subtended by the image at the eye, and $\beta$ for the angle subtended by the object at the lens.  Since rays through the optical centre of a thin lens are undeviated, $\beta$ is also the angle subtended by the image at the lens. Furthermore, if the object is taken to be effectively at infinity $\beta$ is the angle subtended by the object at the eye.  We now readily see that the angular magnification is
\[ m=\frac{\alpha}{\beta}= \frac{f}{d} \]
where f is the focal length  IL  of the lens, and EI to the the least distance of distinct vision d . 
Thus taking d as the conventional value 10″, a focal length of 20″ would give us a magnification of two.

If it had been recognised that the above shows the true state of affairs, that the lens produces a real image, hanging in space, as it were, then it would have been a small step to take a second lens of small focal length to examine and magnify that image and the telescope might have been born much earlier than it was.  That state of affairs was unrecognised until after Kepler, but that is another story.


In 1585, in a memoir on optics written at the request of Lord Burghley, William Bourne described glasses for improving both near and distant vision,   Henry C King, in his  The History of the Telescope (1955)  comments that Bourne's eyes were undoubtedly hypermetropic (for which, see the previous two posts) since as he moved his eye further and further back from the focus of the lens, distant objects became "of marvellous bignesse".  With his eye at the focus nothing could be seen, while nearer the lens he could see the distant object 'reversed and turned the other way'.  

'Reversed and turned the other way' is what Bourne saw in the situation as shown in figure 1 above, with his eye accommodating to the image - how well, is not mentioned.  Why he was  also able to see images "of marvellous bignesse", by implication the right way round and the right way up, is depicted in figure 2.  

Fig.2

In the absence of the eye an image is formed by the lens at its focus, its focal length being IL=$f$. When the eye is interposed the image becomes a virtual object at a distance IE=$d$ behind the eye.  As we have seen in previous posts, since rays through the optical centre of a lens are undeviated, such a ray through any point on the object passes through the corresponding point (the  conjugate point) on the image.  Hence object and image subtend the same angle at the centre of the lens. It will be seen that if $\alpha$ is the angle subtended by the image at the eye, and $\beta$ the angle subtended by the object at the lens, then the angular magnification is 
\[ m=\frac{\alpha}{\beta}= \frac{f}{d} \]

the same expression as before.  We do not know what the values of $f$ and $d$ were in Bourne's case.

The magnifying glass

The picture on the left shows three forms of magnifying glass, the two on the left being intended to be held close to the eye, and the one the right the familiar Reading Glass or “spyglass”.  The latter is sometimes called a burning glass, as by its means one can concentrate the rays of the sun to a small spot so as to heat and possibly set on fire a suitably combustible material such as paper.  All three magnifiers are convex lenses and when inserted into a parallel beam of light, such a lens bends the individual rays to meet at a focus; when it is inserted perpendicular to the beam the rays are brought to meet at the principal focus. The figure below represents a burning glass concentrating the rays of the sun on to a piece of card  held perpendicular  to the optical axis, the line, shown in green, running through the centre C and the principal focus F.  In red we show two rays of light from the centre of the disc of the sun.  Since the sun is so faraway it is effectively at infinity and the rays are effectively parallel and so are brought to the principal focus. In blue we show  rays from a spot on the limb (the edge of the disc) of the Sun, brought to a focus on the card a short distance from F.  The card is said to lie in the focal plane of the lens since all parallel rays at any particular angle to the optical axis are focused to a point on this plane.  In this way we obtain an image of the Sun on the card.  To obtain the size of this image we note that any ray going through the optical centre is sensibly undeviated if the lens is sufficiently thin – just at the centre opposite sides of the lens are parallel and so any ray passes through without change of direction.

Fig.1 The burning glass

 Thus the blue ray going through C remains at the same angle α to the optical axis and at the focal plane its perpendicular distance from the optical axis is αf, where  f is the focal length of the lens.The Sun subtends an angle of approximately 0.5 ° , i.e, about 0.09 radians, so that the image in the focal plane will be 0.09f.  For a lens of  focal length 25cm (4 inches) we thus find the image is about 2.25 mm  across.  Quite a small spot of light.

If a small object is placed on the optical axis of a convex lens at a shorter distance from the optical centre C than 

the focal length and the eye placed on the other side of the lens then an image of  the object  is seen.  A detailed ray diagram for the production of this virtual  image is shown in the previous post.  As shown in the figure below, where the object is at O, the image will appear to be behind the  lens at point I.  The image being virtual, the rays from any point on the image do not actually pass through it, the point being located where the refracted rays meet when produced back, as shown by the dotted lines.   Rays from the focus all emerge from the lens parallel to the optical axis and accordingly as the object O is moved towards the focus the virtual rays become closer to being parallel and the image I recedes.  Thus if the object is exactly at the focus the image is at infinity, all rays from one point on the object being exactly parallel.  This is the configuration most comfortable to the normal eye.

Examination of the rays shown going through the optical centre shows that at the lens the image subtends the same angle as the object so that if the eye is close to the lens the final image on the retina will be of the same size  as the image of the object in the absence of the lens.  However, if the object is closer than the near point, then by

definition of that point it will appear blurred: it is too close to the eye for a sharp image to be formed with the naked eye.  Given that the distance of distinct vision (the distance of the near point from the eye) is d then a small object of height h will subtend an angle of  ε=h/d.  If, however, it is placed at the focus of our lens a sharp image at infinity can be seen with the normal eye, with an angular size of α=h/f  (see fig.3 below).  The ratio of angular sizes, and hence of sizes of final sharp images on the retina, obtainable with aided and unaided  eye is α/ε =d/f.    The value of d is conventionally taken as 10 inches so that a lens of  focal length f=1 inch will be quoted as having a magnification m=10.  Actually, many people, especially when young, can focus down to 4″  or less.  If we took d=4″ then the magnifier in question might be said only to have m=4.  In this sense the younger person benefits less from using the magnifier.  It is a leveller – for both young and old  the angular size of the image is the same.

Figure 3. Reading glass with image at infinity, the object being at the focus.

The  two magnifiers on the left in the photograph have short focal lengths (1 inch) and are intended to be held close to the eye,  In contrast the reading glass has a focal length of 8′.  According to our prescription above the magnification is then m=1.2.  Hardly worth the bother, one might say.   The answer to this is twofold.  Firstly the reading glass is likely to be used by long sighted people, where d, the distance of distinct vision, may be 2 ft or more, so that a figure m=3 would be appropriate.   Secondly, it might well not be convenient to bring the object near to the eye, or vice versa, for example when the object is a book.  One can see from figure 3 that when the image is at infinity the distance between eye and lens does not affect the angular magnification.   As this distance becomes greater the field of view does become less, but the longer the focal length the larger the lens can be made without significant distortion of the image.
 The expression m=d/f  we have given for the magnification is true only if the image is at infinity. Referring to Fig.4, we compare this situation with that where eye and lens are positioned to yield a virtual image at finite distance.

Fig. 4 Magnifiers

Write the distance PN of the image from the lens as d, the focal length NF as f  and the heights of object and image as h and h′ respectively.  The  angle subtended at the lens is α = h′/d  so that  from the figure we can see that
\[   \alpha=h\frac{f+d}{df}  \]
 We compare that with the angle β = h/d_n subtended by the object at the naked eye when it is placed at the near point d_n..   We find a value for the magnification of 
\[  m=\alpha/\beta= \frac{d_n}{f}+\frac{d_n}{d} \]
The maximum value of this is obtained by setting the distance of the image to that of the near point, since it cannot be seen clearly nearer than that. We obtain  
\[m=\frac{d_n}{f}+1 \]
(c.f. the value  d_n,/ f for image at infinity). This will be true for both normal and myopic eyes. The myopic eye cannot, of course clearly see an object at infinity, but when relaxed sees an image clearly when it is at the far point, distance d_f..   The magnification is then
\[ m=\frac{d_n}{f}+\frac{d_n}{d_f} \]