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D = z OTS = supplement of the angle of deviation of any ray (this angle is frequently itself called, incorrectly, the deviation),

q. and D, the values of p and D corresponding to the minimum deviation, or the maximum of its supplement D.

Let O be the centre of the sphere and origin of polar co-ordinates, OP=p, z POT=9; let QOAT be the ray incident perpendicularly on the sphere, q PSpT any other ray emergent at S, OS = radius = r, also z OST= Z of incidence, = p, by property of a refracting sphere; let also z OPT = y = tr—(D+6), then Op = p. sin y = r. sin p;

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Now when the point P is the intersection of consecutive rays, p. and 6 remain constant, whilst p and y vary;

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This (3) with the equations (1) and (2) would suffice to eliminate p and D, and give p in terms of 6 and constants, if the transcendental relations of p, q, and 9 did not prevent it: that expression would be the polar equation of the caustic for the primary bow: we may however trace this curve from equation (3).

Taking a = #. as is usually done for red light, we find two values of p which give p-r,

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which, as seen in elementary books, is the angle of minimum deviation, the ray at this angle being an asymptote to the caustic.

The deviation diminishing from p = 0 to p = q, the intersections of

consecutive rays are behind the sphere, but from p = p, to p =;

the deviation increases, and the intersections are in front. So that the two branches of the caustic are as at Aa, Bb in the Fig. 2. being perpendicular to the sphere at A, and tangential to it at B, and the

line ec being the ray which has the minimum deviation and an asymptote to the two branches.

The section of the luminiferous surface at any position will be

Fig. 2.

similar to ege', the branch eg being an involute of the caustic Bb, and eg of the caustic Aa, or more accurately the part of eg as far as the asymptote is the involute of the virtual branch Aa, and the remainder of the other branch.

For want of knowing the equation of the curve ege', I have used this approximation of considering the two branches near g for the small angular distances which we employ as coinciding with their osculating circles, at the given points.

Then finding p and p, which correspond to this angular distance, and taking rectangular co-ordinates parallel and perpendicular to ce, we

easily get the values of the co-ordinates (a, b) (a, b) of the centres of these osculating circles. Returning thence to polar co-ordinates for the branches eg and e.g. to the same pole O, we find the space (pp.) between the branches at the required point. The caustic in any observation of the rainbow may be considered at the eye as coinciding with its

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asymptote, as a spherule of water of % inch diameter subtends no sensible angle at 1000 yards distance.

Let A be centre of circle osculating at p,

a, b rectangular co-ordinates of A,

a', " ................................... 4',

Op = radius vector = u,
Op'-.................. = u'.
Also za Op = a.

Now if (a a )" + (y –b) = R be equation of circle whose centre is A,

(a'-a') + (y–b) = R* ................................................. 4',

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To establish the condition that the two branches are in contact at the cusp g, and remembering at the same time that, in an observation, the rain-drop is at a great distance, and that therefore R and R are large quantities compared with a, b, a', b', we have

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Or, since A" lies on the side of the negative a ; therefore a' is negative: also b > b , and they are also both negative, in our problem; therefore

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To put this into a form for use in calculation, let
(1 = (11, a' = - a, b = -b, b' = b,

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The quantities a, a, b, b, are to be calculated from the values found for p and 9 corresponding to any particular value of a, which I have taken, for example, as the angular distance from the red to the purple;

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