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
\[ m=\frac{d_n}{f}+\frac{d_n}{d_f} \]