We saw in the last post that convex lenses are used to correct hyperopic eyes. We see in the figure above we show the image of a distant tree a convex spectacle lens would produce if the eye were not interposed. This may be said to be a virtual object for the lens of the eye and we can analyse the system in terms of ray diagrams by splitting it into two parts. First, the spectacle lens gives rise to a virtual object for the convex lens of the eye and second the action of the eye lens on this object gives rise to the final image on the retina.
In the figure below we represent a converging lens upon which falls a system of rays which, in the absence of the lens, would form an image, at O on the optical axis, and at which we take the virtual object in the form of this image. We draw a ray, passing through the front focus F$_1$ and falling on the lens at P. If it were undeviated it would meet the virtual object at S but in actuality emerges from
the lens parallel to the optical axis. We draw a second ray QR which again would pass through S in the absence of the lens but in fact passes through the second focus F$_2$. We take $f$ as the focal length of the lens, $x$ as the distance of the object from the first focus and $x'$ the distance of the image from the second. Referring to the diagram we note that $\triangle$F$_1$CP and $\triangle$F$_1$SO are similar and hence $h/h'=x/f$. Furthermore, $\triangle$F$_2$CR and $\triangle$F$_2$IJ are similar and hence $h/h'=f/x'$. We arrive at Newton's formula
\[ f^2=x'x\]
The lens formula is readily obtained from this, in a manner similar to that detailed in the post on virtual images. In fact, the two situations are the same geometrically, the directions of rays being reversed while object and image are interchanged. This is also the case for the diverging lens with virtual object. The ray diagram is shown below. We shall not discuss it further as Newton's formula is obtained in exactly the same way as before.
Virtual objects were included in the summary of cases of the lens formula in the post for the 21$^\mathrm{st}$ of November.
The above shows (very) schematic eyes. When relaxed, the crystalline lens of the normal eye focusses light from an object at infinity on the retina and can expand to bring a nearer object into focus. Some eyes, however, have difficulty in accommodating to near vision, although distant objects may still be clearly seen. The defect increases with age and is termed presbyopia (Gr. presbys, old, ops, opis, the eye). It is further possible that even when relaxed the eye's focus is beyond the retina, a condition termed hyperopia or hypermetropia ( Gr. hyper, over, metron, measure, ops, eye). The opposite can occur, that the eye focusses light from infinity before it reaches the retina, and objects can be clearly seen only when close at hand. This condition is termed myopia (Gr. myops. short-sighted - myein, to shut, ops, the eye ). "Short-sight" is of course the common English term, while "long-sight" (or "far-sight" in American English) is the corresponding term for presbyopia and hyperopia. With reference to the derivation of "myopia" from the Greek, people with short-sight often squint to improve their vision in the absence of glasses. It is well known to photographers that reduction of aperture increases depth of field, that is, the range of distances for which the camera can be said to be in focus - a pinhole camera has infinite depth of field. For the same reason squinting can help long-sighted people too, as can bright illumination, where the pupil is contracted to be very small.
In Europe convex lenses were used to correct long-sightedness from the twelfth century and concave lenses to correct short-sightedness somewhat two centuries later. It would appear that this was done almost purely empirically, with no knowledge of how the improvement to vision came about. A partial explanation of the action of spectacles was provided later still by Franceso Mauricolo, in a work published posthumously in 1611 (Lindberg 1976). He recognised that myopia occurs when the convex crystalline humour (lens) is excessively curved, so that the light passing through converges too much or too soon, while in hyperopia the lens of the eye is insufficiently curved so that convergence is delayed. The function of spectacles is therefore to hasten or delay the convergence, as the case may be.
The diagrams above illustrate the application of the lens formula
\begin{equation} \frac{1}{u}+ \frac{1}{v}=\frac{1}{f} \end{equation}
to the various possibilities of real and virtual objects and images and the corresponding conventions for the signs of $u$, object distance, $v$, image distance, and $f$, focal length. The conventions are
- Focal lengths of converging lenses are positive
- Focal lengths of diverging lenses are negative
- Distances of real objects and images from the optical centre are positive
- Distances of imaginary objects and images from the optical centre are negative.
The results are for a point object O on the optical axis. The angle $D$ is the deviation of a ray meeting the lens at a distance $h$ from the optical axis. Taking the first diagram, we can see that we have taken the angles $\alpha$ and $\beta$ as $h/u$ and $h/v$ respectively, whereas strictly speaking $\tan\alpha=h/u$ and $\tan\beta=h/v$. In practice however, we are restricted to small angles, as otherwise optical aberrations become significant. We are then justified in approximating the tangents of angles by the angles themselves, in which case the deviation may be taken as $D=h/f$, irrespective of the angle at which the ray meets the lens. .
The telescope that Hans Lippershey attempted to patent in 1608 (see post for 18th September)) had a converging objective and diverging (concave) eyelens. We term such an instrument a Galilean telescope because Galileo made a number of them
and used them to such good effect in observing the Moon, Jupiter and Venus. The objective takes a parallel beam of light from an object (effectively) at infinity to the left and brings it to a focus, as shown in the first figure below. When correctly positioned between the objective and its
focus(we say,when the telescope is focussed) the eyepiece will then give back a parallel beam of light, as shown in the second figure. The focus of the converging lens is shown at F.
We say this is
also the (principal) focus of the diverging lens. Any ray which would pass through this focus if it were not deviated by the lens is rendered parallel to the optical axis by its passage through the lens. Conversely, because of the reversibility of light, a ray parallel to the optical axis is deviated so that appears to pass through the focus. We accordingly analyse the optical properties of the diverging lens making the following assumptions.
- A beam of light parallel to the optical axis diverges when incident on the face of a lens and all rays of the beam appear to originate from a single point, the focus (the principal focus). There are two principal foci, since the beam may be incident on either face.
- The distance of the focus from the optical centre, termed the focal length, is the same on either side of the lens.
- The principle of the reversibility of light is true. Thus if a ray
of light travels from A to B, light from B to A can travel on the same
path.
- All light passing through the lens from a particular point P of an
object are brought to a focus J. That is, all rays passing through the
lens from P meet at a point J, the image of P. These two points are said
to be conjugate.
- The lens is thin, by which we mean that the value of its thickness is small compared with the other dimensions in the problem.
In the figure above we represent a lens perpendicular to the plane of the figure, with optical centre at C. The object is represented by the arrow OP. In accordance with the discussion above a ray of light from P directed to the far focus F2 is rendered parallel to the optical axis when it strikes the lens at R. A second ray from P, parallel to the optical axis, is deviated at Q so that it appears to have passed through the the focus F1. The rays QT and RS meet at J, when produced back, so that it is the image point of P. Triangles F1IJ and F1CQ are similar, from which we find $h'/h=x'/f$. Furthermore, triangles F2CR and F2 are similar, so that $h'/h=f/x$. Comparing the two equations we see that
\begin{equation} f^2=xx' \end{equation}
This is Newton's equation for the thin concave lens.
Taking the distances of object and image form the lens as $u$ and $v$ respectively, we readily find from Newton's equation that \[1/u-1/v=-1/f\]. With the convention that the distances are measured from left to right, and the focal length of a divergent lens is negative we change the signs of $v$ and $f$ to bring the equation in the standard form
\begin{equation} \frac{1}{u}+ \frac{1}{v}=\frac{1}{f} \end{equation}
(c.f. our previous equation for the converging lens).
Finally we should note that, as with the converging lens, rays through the optical centre are undeviated.
We saw in the previous post how by a ray diagram we can find the position and size of the image of an object placed behind a converging lens at a distance greater than the focal length of the lens. If the object is placed at a distance from the lens less than the focal length then an image is still seen, but it can no longer be projected onto a screen. We must therefore now distinguish between types of image. If the light, passes through the image it will be said to be real. In contrast, one for which the light by which it is viewed does not pass through, will be said to be virtual. The questions now arise as to the nature of this image and how its properties may be found.
We may answer these questions by applying what we have already learnt to the lens system in the figure below.
We show two identical lenses placed parallel to one another, their focal length $f$ apart. An object (a miniature tree) is shown at a distance $u$ from the first lens, where $u$ is less than $f$. A real inverted image of the tree may be seen or projected upon a screen and its position and size may readily be found by drawing just two rays. We draw one ray from P at the top of tree parallel to the optical axis to meet the lens at Q, where it is deviated to pass through the focus F. Since this point is the optical centre of the second lens it passes through undeviated. The second ray is drawn through the optical centre of the first lens, where it is undeviated, to meet the second lens at R. Here since C is at the focus of the second lens, the ray is deviated to be parallel to the optical axis. The image of P is at J, where the two rays we have constructed meet, as shown. To find its position we note that the triangles FCQ and JRF are similar and FR is equal to $h_1$. Hence $h/h_1=f/w$. We can also see that triangles COP and CFR are similar, so that $h/h_1=u/f$. Equating the two expressions for $h/h_1$ we find
\begin{equation} f^2=uw \end{equation}.
We now observe that in the absence of the first lens the same image at I could be obtained by placing a suitable object at M, its position being found by producing the rays PR and QF back to meet at K, as shown. Through the second lens this object would look precisely similar to the image, of the original tree at O, produced by the first lens. We therefore locate this latter image at M. It is a virtual image, since no rays of light actually pass through or originate from it. We may find its location either using equation (1), or more simply from the figure below.
We wish to find the value of $v$, giving the location of the image. We shall see the reason for the minus sign presently. Triangles KLM and QCF are similar , so that $-(h_2-h)/h=-v/f$. Triangles KMC and POC are similar and so $-v/u=h_2/h$. Comparing the two equations we arrive at the result
\begin{equation}\frac{1}{u}+\frac{1}{v}=\frac{1}{f} \end{equation}
and we now see the reason for introducing the minus sign as this equation is exactly the form of the lens equation obtained for real images in the last post. We accordingly adopt the convention that the distance of a virtual image from the lens is negative.
In the previous post we saw that the telescope was invented by what must have been a happy accident, as there was very little theoretical foundation to guide design. Once the telescope did appear, Kepler began the laying of such a foundation but before discussing his and other work it will be helpful to give a short account of the foundations of Geometrical Optics as it stands today.
The figure above the passage of light from a small part of an object (an apple on a miniature tree) through a converging lens, the lens being at a distance from the object greater than its focal length. Kepler would have understood that light from the object would form an image (Kepler called it a pictura) on the back of an eye (shown schematically on the right) and would also have been aware that an inverted image of the tree would be formed upon a screen suitably placed between the lens and the eye. However, he did not consider the significance of this intermediate image for determining what is directly seen or, for example, the effect of placing a second lens between the image and the eye. Since all rays of light going through both lenses from a particular apple go through the image then if the position of the image has been found we can take it to be the object of the second lens in determining the paths of the rays through it. Let us consider, then, how we may find a relation between the distances of image and object. We shall proceed semi-empirically, not seeking to prove that the lens does produce an image, or how it does so, but simply assume that it does. We define the optical axis as a line perpendicular to the (circular) lens and passing through its centre, termed in this context the optical centre. We shall assume
- A beam of light parallel to the optical axis and incident on the face of a lens is brought to a focus (the principal focus). There are two principal foci, since the beam may be incident on either face.
- The distance of the focus from the optical centre, termed the focal length, is the same on either side of the lens.
- The principle of the reversibility of light is true. Thus if a ray of light travels from A to B, light from B to A can travel on the same path.
- All light passing through the lens from a particular point P of an object are brought to a focus J. That is, all rays passing through the lens from P meet at a point J, the image of P. These two points are said to be conjugate.
- The lens is thin, by which we mean that the value of its thickness is small compared with the other dimensions in the problem.
In the figure above we represent a lens perpendicular to the plane of the figure. The line OI is the optical axis and C the optical centre. The object is represented by the arrow from O to P at a distance x from the focus F; The ray PQ is parallel to the optical axis and hence passes through F2, as shown. Since we are assuming the lens is thin, we neglect the change in the lateral position of the ray as it passes through the lens. A second ray PR passes through the focus F1 and, invoking the principle of the reversibility of light, travels parallel to OI after deviation by the lens. The point J where the two rays meet is, under our assumptions, the image of P. Now triangles POF1 and RCF1 are similar so that $h/h'=x/f$, where $h$=OP, $h'$=CR, and $x$=OF1. Furthermore, triangles QCF2 and JIF2are similar, where I is such that IJ is perpendicular to the optical axis. Then since CQ=$h$ and IJ=$h'$, we find $h/h'=f/x'$. Equating the two expressions we have found for $h/h'$, we see that $x/f=f'/x'$ or
\begin{equation} f^2=xx' \end{equation}.
This is Newton's equation relating the distance of object and image from the foci.
It will be noted that for a given value of $x$ the position of I (given by the value of $x'$) is independent of of $h$, so that if the object is perpendicular to the optical axis, then so will be the image. Furthermore, $h/h'$ is independent of $h$, so that there is no lateral distortion of the image.
If $u=f+x$, the perpendicular distance of the object from the lens, and $v=f+x'$, the perpendicular distance of the image, then is readily shown from Newton's equation that
\begin{equation}\frac{1}{u}+\frac{1}{v}=\frac{1}{f} \end{equation},
which is known as the lens equation. It is also readily shown that
\begin{equation} \frac{h}{h'}=\frac{u}{v} \end{equation}.
It will be useful to note that it can now be seen that a ray (as shown dotted in the figure) from any point on the object to its conjugate passes through the optical centre.
Telescopes of course need lenses. Lenticular Bronze Age objects of rock crystal have been excavated which from their shape were very probably used as magnifiers but unequivocal written record of such objects used as optical aids appears to begin with Roger Bacon's Opus Majus (1268) in which he noted that a crystal or glass lens formed as a small portion of a sphere will magnify small objects beneath it when viewed from the convex side and that such an instrument is thus very useful for reading by anyone with weak eyes . In 1289 in a
manuscript entitled Traite de con uite de la famille,
di Popozo wrote: "I am so debilitated by age that
without the glasses known as spectacles, I would no
longer be able to read or write. These have recently
been invented for the benefit of poor old people
whose sight has become weak". It thus appears that the first spectacles were made between 1268 and 1289, though it is very possible that monocles or "perspective glasses" were made before that time.
To correct di Popozo's presbyopia converging (convex ) lenses would be used. It was nearly two hundred years after his time before diverging (concave) lenses began to be used to correct myopia (short-sightedness). Nicholas of Cusa is widely accredited with this innovation, in 1451, but on very slender grounds. It is certain that by 1466 spectacles giving "distant vision for the young" were being made in quantity in Florence but, as with spectacles using convex lenses, we do not know who designed or made the first pair.
In a flight of fancy, Roger Bacon had predicted that by refraction distant things would be made to appear near, but 340 years were to pass before the telescope was invented. When they eventually appeared, the telescope and (compound) microscope were invented within twenty years of one another, both in the Dutch town of Middelburg . We intend to discuss the invention of the microscope in a separate post. As far as that of the telescope is concerned, we note that in early October, 1608, Hans Lippershey of Middelburg in Zeeland, with a view of obtaining a patent, submitted for inspection of the States General in the Hague details of a small telescope consisting of a tube with a convex lens at one end and a concave at the other, magnifying three times. Three weeks later, a submission was made by Jacob Metius, of Alkmaar, a hundred miles or so to the north of Middelburg, that he could make a telescope quite equal to Lippershey's and of cheaper materials. Neither patent was granted, on the grounds that the ideas had already become known, but Lippershey was asked by the States General if he could make a device though which one could look with both eyes. He obliged with the first binocular, and made several, for which he was well paid. Interest in binoculars soon waned, probably because of the difficulty in making two identical telescopes and holding them in exact alignment.
It is perhaps surprising lenses had been used singly for several
centuries before they were combined to form the compound microscope and
the telescope. Two possible reasons for this are first, the lack of technical means to
make lenses of high sufficiently high quality and second, the lack of a
useful theory of light - as it was, progress appears to have been made largely by trial and error.
Be that as it may, once the cat was out of the bag, detailed news of the invention of the telescope spread rapidly throughout Europe and Thomas Harriot made observations of the moon in August 1609 and of sunspots in December of that year, while in January 1610 Galileo viewed the four largest moons of Jupiter.
