Total Solar Eclipse 1999 European Track
Solar Eclipse Exposure Guide
This Total Solar Eclipse Photography Exposure Guide was downloaded off the WWW from part of the NASA / Goddard Space Flight Center, Eclipse Home Page. It contains most of the information needed to obtain a photograph of the eclipse. It is a useful guide for getting an exposure of various parts of the event which you can be proud of. Please remember it is only a guide and the photographer should have tested all the equipment and run off test exposures with the same type of film well before the day to check that the camera and all lenses are working correctly. Nothing will be as disappointing has not getting a shot through faulty equipment.
Exposure Formula: t = f2 / [ I x 2Q ]
where: t = exposure time (seconds)
f = f number or focal ratio
I = ISO film speed
Q = brightness exponent
ND = Neutral Density Filter
Rs = Solar Radii
1 Exposures for partial phases are also good for annular eclipses.
2 Baily's Beads are extremely bright and change rapidly.
3 This exposure also recommended for the "Diamond Ring" effect.
4 Use a coloured highlighter pen to mark the line that contains the film speed you are using and then draw down from the aperture to the shutter speed list for the various subjects. This will help to give a guide to the correct exposures.
Solar Eclipse Photography
Edited by Ivor Clarke, from the WWW
Solar eclipses can be easily photographed provided that basic eye safety precautions are followed. Almost any kind of camera with manual controls can be used to capture excellent views of this rare event. While a fixed lens compact camera will provide a result of some sort, the image of the sun will be very small and the automatic exposure system in such cameras will struggle with the extreme contrast of the scene. If this is the only camera you have, then try to rest the camera on a firm support during the total phase as the camera's automatic exposure system will give an exposure longer than it is possible to hold the camera still. This applies to all cameras during totality, use a tripod if possible for all types of camera as this will leave your hands free and guarantee sharper pictures.
An SLR (Single Lens Reflex) 35mm camera is the best choice for capturing this type of event as manufactures provide a full range of focal length lenses for their cameras and many types of filters. However, a lens with a fairly long focal length is recommended to produce as large an image of the Sun as possible. A standard 50mm lens yields a minuscule 0.5mm image, while a 200mm telephoto or zoom produces a 1.9mm image. A better choice would be one of the small, compact catadioptric or mirror lenses that have become widely available in the past ten years. The focal length of 500mm is most common among such mirror lenses and yields a solar image of 4.6mm. With one solar radius of corona on either side, an eclipse view during totality will cover 9.2mm. Adding a 2x tele-converter will produce a 1000mm focal length, which doubles the Sun's size to 9.2mm.
Focal lengths in excess of 1000mm usually fall within the realm of amateur telescopes. If full disk photography of partial phases on 35mm format is planned, the focal length of the optics must not exceed 2600mm. However, since most cameras don't show the full extent of the image in their viewfinders, a more practical limit is about 2000mm.
Longer focal lengths permit photography of only a magnified portion of the Sun's disk. In order to photograph the Sun's corona during totality, the focal length should be no longer than 1500mm to 1800mm (for 35mm equipment). However, a focal length of 1000mm requires less critical framing and can capture some of the longer coronal streamers. For any particular focal length, the diameter of the Sun's image is approximately equal to the focal length divided by 109.
A mylar or glass solar filter must be used on the lens throughout the partial phases for both photography and safe viewing. Such filters are most easily obtained through manufacturers and dealers listed in astronomy magazines. An abbreviated list of solar filter manufacturers is also available online. These filters typically attenuate the Sun's visible and infrared energy by a factor of 100,000. However, the actual filter factor and choice of ISO film speed will play critical roles in determining the correct photographic exposure. A low to medium speed film is recommended (ISO 50 to 100) since the Sun gives off abundant light.
The easiest method for determining the correct exposure is accomplished by running a calibration test on the uneclipsed Sun. Shoot a roll of film of the mid-day Sun at a fixed aperture (f/8 to f/16) using every shutter speed between 1/1000 and 1/4 second. After the film is developed, note the best exposures and use them to photograph all the partial phases. The Sun's surface brightness remains constant throughout the eclipse, so no exposure compensation is necessary except for the narrow crescent phases which may require two more stops due to solar limb darkening. Bracketing by several stops may also be necessary if haze or clouds interferes on eclipse day.
Certainly the most spectacular and awe inspiring phase of the eclipse is totality. For a few brief minutes or seconds, the Sun's pearly white corona, red prominences and chromosphere are visible. The great challenge is to obtain a set of photographs which captures some aspect of these fleeting phenomena. The most important point to remember is that during the total phase, all solar filters must be removed!
The corona has a surface brightness a million times fainter than the photosphere, so photographs of the corona are made without a filter. Furthermore, it is completely safe to view the totally eclipsed Sun directly with the naked eye. No filters are needed and they will only hinder your view. The average brightness of the corona varies inversely with the distance from the Sun's limb. The inner corona is far brighter than the outer corona. Thus, no one exposure can capture its the full dynamic range. The best strategy is to choose one aperture or f/number and bracket the exposures over a range of shutter speeds (i.e. - 1/1000 down to 1 second). Rehearsing this sequence is highly recommended since great excitement accompanies totality and there is little time to think.
Exposure times for various combinations of film speeds (ISO), apertures (f/number) and solar features (chromosphere, prominences, inner, middle and outer corona) are summarised in table above. The table was developed from eclipse photographs made by Espenak as well as from photographs published in Sky & Telescope.
To use the table, first select the ISO film speed in the upper left column. Next, move to the right to the desired aperture or f/number for the chosen ISO. The shutter speeds in that column may be used as starting points for photographing various features and phenomena tabulated in the 'Subject' column at the far left. For example, to photograph prominences using ISO 100 at f/11, the table recommends an exposure of 1/500. Alternatively, you can calculate the recommended shutter speed using the 'Q' factors tabulated along with the exposure formula at the bottom of the table. Do keep in mind that these exposures are based on a clear sky and a corona of average brightness. You should bracket your exposures one or more stops to take into account the actual sky conditions and the variable nature of these phenomena.
Another interesting way to photograph the eclipse is to record its various phases all on one frame. This is accomplished by using a stationary camera capable of making multiple exposures (check the camera instruction manual). Since the Sun moves through the sky at the rate of 15° degrees per hour, it slowly drifts through the field of view of any camera equipped with a normal focal length lens (ie. 35 to 50mm). If the camera is oriented so that the Sun drifts along the frame's diagonal, it will take over three hours for the Sun to cross the field of a 50mm lens. The proper camera orientation can be determined through trial and error several days before the eclipse. This will also insure that no trees or buildings obscure the camera's view during the eclipse. The Sun should be positioned along the eastern (left in the northern hemisphere) edge or corner of the viewfinder shortly before the eclipse begins. Exposures are then made throughout the eclipse at, say, five minute intervals. The camera must remain perfectly rigid during this period and may be clamped to a wall or fence post since tripods are easily bumped. If you're in the path of totality, you'll want to remove the solar filter during the total phase and take a long exposure (~1 second or more) in order to record the corona in your sequence. Don't forget to re-mount the filter if you want to capture the remaining portion of the eclipse as the moon exits the sun. The final photograph will consist of a string of Suns, each showing a different phase of the eclipse.
Finally, an eclipse effect that is easily captured with point-and-shoot or automatic cameras should not be overlooked. Use a leafy tree to see the small spots of sunlight cast upon the ground and allow a spot or two to fall on a piece of white card-board placed several feet away. The holes between the leaves act like pinhole cameras and each one projects its own image of the Sun. The effect can also be duplicated by forming a small aperture with one's hands and watching the ground below. The pinhole camera effect becomes more prominent with increasing eclipse magnitude. Virtually any camera can be used to photograph the phenomenon, but automatic cameras must have their flashes turned off since this would otherwise obliterate the pinhole images.
Finally the choice of film. Buy fresh film, well in date, whatever is your favourite make, as this event will not be repeated take as many shots has possible until you feel you have the shot you want. Film type, this is down to the individual, if you would like a set of prints or transparencies. I would only use ISO 100 speed film in colour negative format and ISO 50 or 100 for transparency use. I will properly use Fuji Velva tranny film as I like its colour rendering. Don't forget that on the whole, colour neg film is softer than tranny and may be better at showing the corona, but it will be expensive to get a set of transparencies off neg film. From transparency film prints can be made easily and by selecting just the best exposures, print your best shots to show around. Many large prints of the eclipse will no doubt be gracing many walls this autumn.
On the Formation of Rings around Planets
by Amara Graps
(Max Planck Institute of Nuclear Physics, Heidelberg, Germany)
Edited by Paritosh Maulik
Not only Saturn, but also Jupiter, Uranus and Neptune are known to have
rings around them. Here is a brief discussion on some of the current
thinking on the formation of a ring system around a planet.
Basically there are two theories about the ring formation:
#1 Ring formed around the same time as the planet and the
moon(s). This theory was developed in the 1970s and was called the
Cosmogonic Ring Hypothesis. According to this theory, the ring is
coplanar with planet; it can explain the dynamics and kinetics of the system,
and also the ring growth by accretion. This theory, however, fails
to explain properly the fact that over the life period of the ring, the
ring has not dissipated by impact from meteor or similar objects.
#2 Another theory was developed around 1980 and 1990, which
suggested that rings formed when a moon was destroyed by a comet or meteoroid
and the ring is around 108 years or younger and thus can survive the dissipative
The structure of the ring is rather complicated and a single theory
fails to model all the features. Satellites can guide the ring to a well
defined orbit and may provide materials for the maintaining the ring(s).
Gaps, ringlets and broad ring plateau may be caused by the embedded moonlets
within the ring and the collision induced damage of the moonlets may provide
materials for the continuation of the ring(s).
It is now being asked if all the rings of a planet are of same age, or developed over a period. Rings are thin compared to their width. The appearance of the ring depends on the size range of the particles in the ring, for example, the rings of Saturn are highly reflective, while those for the other planets are dark. From time to time in the Saturn and Jupiter system, satellites may enter into and out of the ring, while Uranus ring have satellites adjacent to the ring.At the beginning of the ring formation, it is likely that a cloud of particles were moving about the central mass of the planet and then interacted with each other to form the ring as Theory 1 and according to Theory 2, the remains or fragments of the satellite(s), which entered the Roche* limit of the planet, were destroyed either by tidal forces or meteoric impacts. Some of the particles are electrically charged and these interact with electromagnetic radiation and magnetic fields; this interaction, in turn aids the particles to form 'streams. With time, the micron size charged particles would be dragged around the planet and other particles would go through processes like accretion, breaking up of micrometeoroids by impacting the rings, erosion of nearby moons and other dissipative and mass transport phenomenon. Images from the Galileo mission show that the small inner moons of Jupiter are heavily createred from meteoroid impact. These images also suggest that the dust coming off from these moons feeds the rings of Jupiter. The dust source of the transparent Gossamer ring appears to be Amalthea and Thebe and for the main ring the sources are Adrastea and Metis. The source of the particles is due to erosion of the moonlets and other impact processes. This is also responsible for filling the up the gaps, while the continuation of the rings are primarily due to the gravitational influence of the shepherding moons. If the breaking up of the large particles leads to finer particles, then in about 108 to 109 years, the rings may become very sparse.
Watch this space!
* Roche limit: pronounced as Rosh, a French astronomer, pointed out that when a satellite approaches a planet, under the influence of gravity of the planet, it can become distorted like a rugby ball and eventually breaks up to form separate satellites in an obit of about 2.5x the radius of the planet. The exact distance is however depends on the density and radius of the planet.
by Mike Frost
'Whenever I cover the sky with clouds, and the rainbow appears, I will
remember my promise to you and all the animals, that a flood will never
again destroy all living beings ...'
(Genesis, Chapter 9, v14-15)
'sunshine on a rainy day, makes my soul trip, trip, trip, trip away...'
The rainbow is a familiar part of our culture, so much so that we take it for granted. Ours is a world of rainbow nations, rainbow warriors, Judy Garland, Richie Blackmore, Bungle and Zippy. Yet do you remember the first time you ever saw a rainbow? Like me, perhaps, you'll recall a long forgotten sense of wonder. I'm a great fan of 'child-like wonder', which I think is an underrated emotion. We live in a society which values the novel and ignores the familiar. Some people (myself included) go to great lengths to experience new phenomena such as eclipses or auroras. Yet there is wonder still to be gained from staying at home and viewing afresh familiar sights such as the rainbow.Do you think you understand rainbows? Most people have some idea that they form from light bouncing around raindrops. Many people will have seen a double rainbow. Have you ever seen a triple bow? Stick around and I'll tell you something wondrous about the third rainbow.
Let's start with the absolute basics. What do you need for a rainbow? Sunshine and rain, of course. Well not quite. Any distant source of light will do. The Sun is by far the most common cause of rainbows, but it is certainly possible to see moonbows from time to time. And the drops of water certainly need not be raindrops; rainbows can easily be observed in sprinkler sprays (drought orders permitting) and even in dewdrops. But do we need drops of water, or will ice crystals do? And does the drop size matter? To answer the first question, ice crystals do not lead to rainbows; instead they produce a whole new set of phenomena such as sundogs, sub-suns and halo effects. So rainbows are seen much less frequently (in this country at least) in the winter. If you think back to rainbows you have observed, you'll perhaps realise that they are also far more common in spring and autumn than in summer. Partly this is because rain is less frequent in the summer, but there is another reason. Where does a rainbow appear relative to the Sun? If the sunshine is strong enough to cast a shadow, you may notice something interesting; your shadow points towards the centre of the rainbow! This isn't just co-incidence: everybody sees their own personal rainbow. If you walk towards the rainbow to try and gain a better view, it recedes from you at just the rate you walk forward. For this reason, the theory that there is a pot of gold at the end of a rainbow has never been verified, although I would be delighted to receive evidence to the contrary! Just as frustratingly, if you try and take a photograph with a conventional camera, a full rainbow is always too big for the shot, and if you try moving back to fit it in, the rainbow moves with you. Take a close look at a rainbow. Rainbows, you will see, are circular arcs. The centre of the circle is the point exactly opposite the Sun in the sky. This is why your shadow points towards the centre of the rainbow, and it's also why the centre of the circle is below the ground. From an aeroplane, and even from a mountain, it is sometimes possible to see a full 360 degree rainbow and quite a sight that must be! The angular diameter of the bow (the angle subtended in the sky from the centre of the circle to the bow) is always the same; forty degrees to the inner violet edge and fortytwo degrees to the outer red edge. This is why you can never fit the full bow in a conventional camera field. It is also explains why rainbows are less often seen in summer, and why rainbows are more common in the early morning and late afternoon. If the sun is higher than fortytwo degrees above the horizon, it cannot produce a rainbow. For a sun setting from a height of fortytwo degrees it may be possible to see a 'rising rainbow' which gradually peeks above the horizon. The order of colours in the rainbow is fixed. Red at the top, violet at the bottom, and in between the familiar spectrum of colours: red, orange, yellow, green, blue, indigo, violet. Note that the colours do not run in discrete bands, like the Central TV logo: rather, the colours run into each other in a continuum. If much of the sunlight has been scattered away, for example from a setting sun, some of the colours may be missing; giving, say, a red rainbow at sunset. Now look at the secondary, external, bow. This is also a circular arc, centred exactly on the same anti-solar point as the main rainbow. The secondary rainbow is always fainter than the primary bow (often so faint that you can't see it, although it is surprising how often you see it is present when you know where to look). Like the primary bow, the secondary rainbow always subtends the same angle; fifty one degrees from centre to edge. Look at the colours; they are in the opposite order to the primary bow. Now examine the regions around the rainbow. Perhaps the most striking thing is how bright the sky is inside the primary rainbow. On the outside of the primary rainbow it is much darker. There isn't so much of a contrast visible on either side of the secondary bow, but in fact the outside of the secondary bow is slightly brighter than the inside. The area between the two bows is called Alexander's dark band (and has nothing to do with Alexander's ragtime band). They aren't particularly visible on my shots, but occasionally, if you examine the sky just within the primary rainbow, you see peculiar interior arcs of brightness and darkness. These are called 'supernumary' bows. Clearly there are some odd effects going on at the edge of the rainbow. Very occasionally, when observing rainbows close to reflecting surfaces such as lakes, it's possible to see unexpected effects due to reflections. Do rainbows have reflections? Yes and no. As we'll see, rainbows are virtual images, so there's nothing physically present to reflect. However reflected light can also form a rainbow, and this looks just like the reflection of rainbow, so for practical purposes rainbows can have reflections. From time to time a rainbow and it's reflection can cross, leading to some very odd pictures. I asked one unanswered question a while back. Does raindrop size matter? The answer to this one is 'yes' but not much. Very large droplets may be distorted by the pull of gravity, whilst small droplets, which are kept spherical by surface tension, will scatter light through diffraction effects if they are too small. In fog it is sometimes possible to see a fogbow or white rainbow. However, the relative independence of rainbow formation on droplet size means that, usually, the colours in rainbows are easily distinguished. How lucky we are! This property independence from droplet size was a key factor in the first attempts to understand how rainbows formed. Theodoric of Freiburg, in 1304, was the first to carry out experiments, but the real breakthrough was made independently by Rene Descartes in 1637. He reasoned that it might be possible to see rainbow effects in very large flask sized droplets. So he built a spherical glass ball, filled it with water, and shone a light on it. What do you think he saw? Before you go and annoy the goldfish, let me take you through the optics involved, which are really very simple, and consequently open to study by mathematical and computer techniques. Let's list carefully the ingredients necessary for a rainbow. These consist firstly of a distant source of multi-coloured light, such as the Sun. It's a good enough approximation that rays of light from the Sun arrive parallel to each other. Then we have a large number of spherical water drops of indeterminate sizes. So the obvious first step is to look at what happens when a parallel set of rays hits a spherical water drop. What happens to a ray of light when it hits the surface of a water drop? Well, in principle we should carry out a full quantum mechanical analysis of the interaction of photon and water surface. This has been done but mercifully we can take a much simpler view. If a ray of light hits the surface of a water drop it is either reflected back or transmitted on into the water. Fortunately, in both cases, the laws of physics are very simple. For a reflected ray, the angle of reflection equals the angle of incidence. For transmission, the text books tell us that the angle of incidence and angle of refraction are related by Snell's law: sin i / sin r = const
This law wasn't known in Theodoric's time, but Descartes, crucially, was aware of it. The constant is different for each different wavelength of light, varying from 1.331 for red light to 1.343 for violet light. If you've studied physics at school these equations may be very familiar to you. If you haven't, don't worry, there's no more maths unless you want it! So how do we get to rainbows? Well, let's simplify again to the case of parallel light of one single colour (red, say) falling on a collection of raindrops. What sort of interactions can take place? Here is where a computer can come in handy. Using a computer, it is fairly straightforward to compute the ray paths for a large number of rays of light falling on a raindrop and undergoing a set series of interactions. I used the Matlab (Mathematical Laboratory) package to analyse the following cases. (You can see the mathematics I used in a separate table, if you're interested). Case 0:
Light misses the raindrops altogether. Not very interesting but don't forget that this is what happens to much of the light from the Sun is usually visible from the far side of a rain cloud. Case 1:
Light hits a raindrop and is reflected off. Note that light is scattered in all directions. Case 2:
Light hits the first surface of the drop, is refracted into the water drop, and then refracted out of the far side. Note that the raindrop acts as a rather inaccurate spherical lens. Immediately behind the drop all the rays congregate in a small area, but are then scattered in widely different directions.
So far things are hardly exciting. Case 0 is dull in the extreme, and all cases 1 and 2 do is to redistribute light around the sky. Take heart! Things liven up considerably for... Case 3:
Light is refracted into the droplet, reflects off the internal surface of the droplet, and is then refracted back out of the droplet surface. For clarity, I have only drawn rays to one side of the spherical surface.
Look what happens here! For all rays, light ends up travelling back in the general direction from which it came. For light which hits the raindrop 'dead centre' this is exactly true. As we move away from the centre, the deflection from a 180 degree rebound becomes greater and greater until a critical point, when light rays start being deflected back towards a 180 degree rebound. What is the maximum deviation from the rebound angle? Fortytwo degrees, of course. So that is what Rene Descartes discovered when he shone red light on a flask. At all angles of viewing there is some light visible (through Cases 1 and 2) but there is a concentration at one particular angle; of light which has entered the drop, bounced off the inside, and then exited the drop again. What happens for different coloured light? Qualitatively, the effect of case 2 is the same there is a critical angle at which light is concentrated. But because the constant varies in Snell's Law, the critical angle is different for each wavelength of light. Finally, now, we can see what causes a rainbow. As we gaze at the sky fortytwo degrees in any direction from the anti-solar point, we see the concentration of light which has bounced around raindrops along our line of sight, and has been redirected towards us. At slightly different angles, slightly different colours are at the critical angle. As you move in the direction of the rainbow, you see light which has been deflected towards you from different raindrops, which are now at the critical angle. So the rainbow is a virtual image, consisting solely of light which has been redirected towards the observer by intervening drops of rain. Note that it isn't the case that all light of a particular colour is deflected at the critical angle. Looking inside the bow, light of all colours, which emerges at less than the critical angle, will be directed towards you. So inside the bow the sky appears bright, because all this multi-coloured light mixes together. However, the critical angle is a maximum; no light is deflected towards you from outside the bow. This is why the sky outside the rainbow appears much darker. And very close to the bow, light can arrive in your eyes through two different paths of almost the same length, leading to interference effects which give the 'supernumary' bows I described earlier. See, I told you Case 3 would be illuminating. To my mind it's fascinating how very simple laws of physics can, almost out of the blue, explain a very complicated phenomenon such as a rainbow. But let's move on. How do secondary rainbows occur? As you might guess, we are driven to consider Case 4:
Light enters the raindrop by refraction, is internally reflected, internally reflected a second time, and finally exits by refraction.
So the cause of the secondary bow is very similar to the primary bow. There's a critical angle of fiftyone degrees or so at which light is concentrated. But look at the way in which the light has bounced round, to emerge on the same side as it entered. So now the light which doesn't emerge at the critical angle is outside the rainbow, and the order of colours in the bow is reversed. So by this simple model all of the noted features of the rainbows can be explained. I should also say that the complete explanation of interference effects, and a treatment of such matter of the polarization of rainbow light, requires a much more sophisticated approach. But it is gratifying that most of the easily observed features of rainbows yield to a simple physical model. But what about the third rainbow? And the fourth, fifth and sixth...? Be patient! Let me tell you some more history. Descartes had explained the primary and secondary rainbows by the laborious calculation of angles for ray tracing. The calculations for a primary bow are tedious without a calculator, and those for the secondary bow mind numbingly so. Descartes therefore never carried out calculations for higher order bows. Almost certainly he would have agreed with the later assertion of Jean Bernoulli that the third rainbow was likely to be exceedingly faint and visible only to "eagles or lynxes". The mathematical ability to be able to analyse third and higher order rainbows had to wait for Isaac Newton and his invention of the calculus. In the mathematical table I state the formula which Newton derived in his book 'Opticks' to calculate the critical angle for any order bow. There is no evidence, however, that Isaac Newton ever explicitly carried out the calculation to see exactly what the critical angle was for the third rainbow. This calculation was first done by his friend and collaborator Edmund Halley (of comet fame). Edmund Halley must have had quite a surprise when he carried out the calculation.
Here is the ray diagram for Case 5, where light is reflected three times internally. The critical angle is 140 degrees. What does this mean? Simply, Descartes, Newton, and anyone else who ever wondered about the third rainbow, have looked for it in completely the wrong direction. The third rainbow would be a ring around the Sun! Isn't that amazing! It is unlikely that the third rainbow can be seen in nature, because the light of the Sun will drown it out, but far be it from me to stop you looking (but as with any observations close to the Sun, please take care and never look at the Sun directly!). The third rainbow would have an angular radius of forty degrees, almost the same size as a primary bow. Don't confuse it with the ice-crystal related halo, which is only twenty two degrees from the Sun. The deflection of light for the third bow has been observed in the laboratory, in experiments which involve shining monochromatic laser light on a droplet suspended from spider thread. Deflections relevant to the third, fourth, fifth... all the way up to the thirteenth rainbows, have all been noted. The fourth rainbow would also be around the sun, and it's not until the fifth rainbow that the intuitive direction re-occurs; the fifth rainbow is also likely to be far too faint to observe. Perhaps you'll take a closer look at the next rainbow you come across! They are certainly worth the study. And remember your sense of wonder; could we ever have expected (do we really deserve) to see such a beautiful sight?
'Rainbows, Haloes, and Glories' R.H. Greenler (C.U.P. 1980)
'Opticks' I. Newton (1730, reprinted Dover 1952)
'The Theory of the Rainbow' H.M. Nussenzveig Scientific American
236, p116 (1977)
'What is it about Mathematics' A.J. Davies Bulletin of the I.M.A.A,
Vol 31 (1995)
Matlab is produced by The MathWorks Inc (email@example.com)
The Mathematics of Rainbows
I make no claims to be doing any original maths here! This analysis follows that in an article by A.J. Davies in the Bulletin of the Institute of Mathematics and Its Applications, which in turn was inspired by an Open University lecture given by Dr John Mason (not Dr Mason of the B.A.A.); also, the diagrams in Robert Greenler's "Rainbows, Halos and Glories" inspired the very similar diagrams accompanying this article.
The diagram shows the path taken by a ray of light which enters a raindrop, undergoes two reflections within the raindrop and is refracted back out again. (Fortunately all the refractions and reflections take place in a horizontal plane of the paper). Ultimately we want to know the angle through which the ray is deflected, but as a step towards this let us calculate the angles around the circle at which the ray hits the raindrop surface. Let the original angle of incidence be θ. We can use this to define the angle around the circle for subsequent interceptions between ray and surface. θ = 0 for a ray which hits the raindrop dead centre. For the purposes of later analysis the angles are best expressed in radians rather than degrees (2 π radians = 360 degrees). The ray is refracted as it enters the raindrop; the angle of refraction ϕ can be calculated from Snell's Law.
sin (θ) / sin (ϕ) = k (1)
The problem may look complicated but in fact there are several simplifications. By symmetry, for example, each internal reflection takes place with an angle of incidence equal to ϕ. So each reflection takes place π - 2*ϕ radians further round the circle. If we label successive contacts between ray and surface as being at angles A0, A1, A2.... then
A0 = θ (2)
A1 = θ + π - 2*ϕ (3)
and so on up to ...
Am = θ + m*π - 2*m*ϕ (4)
If the ray leaves the raindrop after n reflections, its last contact with the drop is at An+1, where it undergoes one further refraction. We don't have to use Snell's law again because the angle of incidence is ϕ, from the inside, and so the external angle of refraction is θ. We have to add this to the angle round the circle to give the total deflection Dn, viz:
D1 = 2*θ + 2*π - 4*ϕ (5)
D2 = 2*θ + 3*π - 6*ϕ (6)
and so on up to n reflections:
Dn = 2*θ + (n+1)*π - 2*(n+1)*ϕ (7)
Armed with formulae (4) and (7) you can easily produce the ray diagrams for the nth order rainbows which accompany this article. But how do we calculate the critical angle? To do this it is easiest to cast the formula for deflection in terms of h, the impact parameter.
h = sin(θ) (8)
From Snell's Law, sin ϕ = h/k, and so equation (7) can be rewritten:
Dn(h) = 2*sin-1(h) + (n+1)*π - 2*(n+1)*sin-1(h/k) (9)
It is at this point that Isaac Newton's differential calculus is used, giving
and the critical angle is that for which dD/dh is zero. Simplifying
for n=1, 2, 3... it is now possible to calculate the critical impact parameter. Given the refractive index; substituting back into equation (9) gives the critical angle. For example, n=1 and k=1.331 gives a critical impact parameter for red light of
h = sqrt((4-1.3312)/3) = 0.864
D1 = 2 sin-1(0.862) + 2π - 4 sin-1(0.862/1.331)
= 2.078 + 6.283 - 2.818
= 5.543 radians or 317.6 degrees
so that the deflection for the red rainbow is 360.0 - 317.6 = 42.4 degrees as expected.