Tuesday, 19 January 2016

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.




Thursday, 14 January 2016

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/dn subtended by the object at the naked eye when it is placed at the near point dn..   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  dn,/ 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 df..   The magnification is then
\[ m=\frac{d_n}{f}+\frac{d_n}{d_f} \]