polarized to find the intensity of the light after reflection from B, and the form of the coloured rings.

which is perpendicular to the plane of first polarization,

parallel to that plane. The latter of these, by (136), has its phase increased by 90°, and therefore on coming out of the rhomb the vibrations may be represented by

parallel to that plane. Resolving these in directions perpendicular and parallel to the principal plane of the crystal, we find;

Vibration which produces Ordinary ray

On emerging from the crystal, the Ordinary vibration being represented by the same expression, the Extraordinary vibration must be represented by

The resolved parts of these perpendicular to the plane of analyzation (which are the only parts that reach the eye)

and the intensity of the light, or the sum of the squares, is

161. Since a does not enter into this expression, the appearance will not be altered on turning B round its spindle. When sin 240, that is when y = 0, or 90°, or 180°, or 270o, the intensity is 2: this shews that there is a cross with light of mean intensity interrupting the rings. When

is > 0 < 90o, or > 180° < 270°,

the expression is maximum when

When is 90° 180° or > 270° 360°, the expression is maximum when

Thus it appears that of the four quadrants into which the cross divides the image, each opposite pair is similar, but each adjacent pair is dissimilar: the bright rings in one quadrant having the same radii as the dark rings in the next quadrant. And on comparing these expressions with those în (157), it will be seen that the effect of placing Fresnel's

rhomb has been to push the rings outward by of an order in two opposite quadrants, and to pull them in by of an order in the other two opposite quadrants. At the same time the cross which was perfectly black has now some light. The most important difference of character however which the use of Fresnel's rhomb produces is the unchangeability of appearances as B is turned round.

If we compared the rings produced with the same position of Fresnel's rhomb by two crystals, in one of which c2 was > a2 and in the other of which c2 was <a2, then for a given I order of rings, that is for those in which the magnitude of

without respect to its sign, is the same, sin would be

λ

positive for the first and negative for the second, or vice versâ. Consequently the bright rings of one crystal would correspond to the dark ones of the other. But we have seen that the bright rings of one quadrant correspond to the dark rings of the neighbouring quadrant. Consequently the rings presented by one of the crystals would be the same as those presented by the other, supposing the latter rings turned round 90°. This affords a convenient method of determining whether the double refraction of a uniaxal crystal is of the same kind as that of a standard crystal (for instance Iceland spar) or of the opposite kind.

PROP. 34. A plate of a biaxal crystal whose optic axes make a small angle with each other (as nitre or arragonite) is bounded by planes perpendicular to the plane passing through the axes and nearly perpendicular to each axis; light is incident at a small angle of incidence: to find the difference of retardation of the two rays.

162. The accurate solution of this problem leads to some rather complicated expressions: and we shall therefore content ourselves with a very approximate solution analogous to that found in (153). We have found there that the difference of retardation was nearly

where the difference of the squares of velocities of the two waves was (c-a) sini: or that the difference of retardation

=

Tv 2as

was nearly × difference of squares of velocities of the two waves. As the difference of retardation arises solely from the difference of velocities, we shall suppose the same proportion to be true here. Now by (125) neither of the rays undergoes Ordinary refraction, or has a constant velocity. Still, even in extreme cases, the velocity of one is so nearly a constant = a, that in a calculation depending almost wholly on the difference there will be no sensible error in considering one as constant and = a. And by (123), putting v' for the velocity of the other,

where m' and n' front with the two small).

are the angles made by the normal to its optic axes of the crystal (C being always

Hence the difference of retardations

=

TCav
2

sin m'. sin n'.

Now let us consider the system of rays in air which on entering the crystal will pass in the directions that we have described. Let m and n be the angles made by the same ray in air with the rays which on entering the crystal will pass in the directions of the optic axes. As all the refracted rays (represented by the normals to the fronts of the waves) are in the same planes perpendicular to the refracting surface as the incident rays, and as all the angles of refraction are very nearly in the same proportion to the angles of incidence, it follows that all the other small angles depending on them, and their sines, are nearly in the same ratio.