Let OQRT be any ray, draw QM parallel to SH; then if i, r be the angles of incidence and refraction

Since a very nearly, and a is small, this may be approximately written

SH=2Ta (1-μ1).

Having obtained the value of SH, the calculation proceeds as before.

21. A fourth method, which was also employed by Fresnel', consists of three mirrors, L, M, N, placed so that L and N intersect at a point 0 on M.

B

S

A

The light proceeds from a source S, and is reflected at the first mirror L, and is then reflected from the third mirror N. After reflection from N, the light will appear to diverge from a focus A such that

where is the inclination of SO to L; and w, w' are the angles which L and N make with M.

The light reflected at M appears to diverge from a focus B, such that

DISPLACEMENT OF FRINGES.

23

Hence if w' is small, the distance AB will be small, and the two pencils proceeding from A, B will be in a condition to interfere.

22. When interference fringes are viewed through a prism, or through a plate of glass held obliquely to the screen, the fringes will be displaced, and we shall now calculate the displacement.

P

M

R

H

Let T be the thickness of the plate, μ its index of refraction, Bits inclination to the screen; also let SQRP be any ray, and draw RM parallel to CP. Let R be the retardation, and let CP=x, OS= c; then

R1 = SQ + μQR + RP.

In calculating R1, we shall consider B, and i the angle of incidence, to be small quantities, and we shall neglect cubes and higher powers.

Now d2 + (x − c + RM)2 = Sp2 = (SQ+QM+RP)2,

whence

-

or

We must now find i in terms of x. We have

whence

approximately, accordingly

R1 =d+

1

2μ3d {μ (x − c) + T (μ − 1) ß}3 + ¿T (μ − 1) (2 − 2).

The value of R, the retardation of a ray proceeding from H to P, is obtained by changing c into - c; whence

The original central band was x = 0, and the central band which is determined by 80 is now given by

=

which shows that it is shifted through a distance - T(1-μ1) B.

23. When interference fringes are examined through a prism, the displacement of the central band is different from the theoretical result given by (6). This difficulty was first explained by Airy', who drew attention to the fact that when no prism is used, the central band is the locus of the points for which all colours of the light composing the two pencils have travelled over equal paths. Now from (6) it appears, that the displacement of the points which formerly constituted the central band, depends upon μ the index of refraction of the prism; and this quantity is different for different colours, being greatest for violet and least for red light. Since the original central band consists of a mixture of light of every colour, it follows from (6) that the displacement of the red portion of the band will be less than that of the violet, and consequently the portion of the central band which is nearest C will be red, whilst the farthest portion will be violet. This band

1 Phil. Mag. 1833, p. 161.

LLOYD'S EXPERIMENT.

25

can therefore no longer be considered the central or achromatic band.

The actual achromatic band is determined from the consideration, that if the bands of all colours coincide at any particular part of the spectrum, they will coincide at no other part; hence if v be the displacement, measured in the direction CP, of the original central band, the distance an of the nth band after displacement will be

=

Xnvxλd/2c.

The achromatic band occurs when an is as nearly as possible independent of λ, that is when dan/dλ = 0, in which case n must be the integer nearest to

2c dv ́d dr・

Since the width h of a band is equal to Ad/2c, this may be written

so that the apparent displacement of the achromatic band is'

24. The following method of producing interference fringes was devised by Lloyd'.

A luminous point A is reflected from a plane mirror CD at nearly grazing incidence. The reflected rays accordingly emerge

from a virtual focus B, and the arrangement is therefore equivalent to two small sources of light very close together. Let BC,

1 See also Cornu, Jour. de Phys. vol. 1. p. 293 (1882). Lord Rayleigh, "On achromatic interference bands," Phil. Mag. (5), vol. xxvi. pp. 77 and 189. 2 Trans. Roy. Ir. Acad. vol. xvII.

BD meet the screen in Q, q; then since interference is due to the mixture of the two streams of light, the bands will only exist between the points Q, q. Moreover since the difference of path is never zero, there can be no achromatic band.

The achromatic band may however be rendered visible by placing a thin plate of glass in the path of the direct pencil. Putting AB 2c the retardation at x is

=

and consequently the position of the achromatic band, which is determined by dan/dλ = 0, will be given by n, where n is the integer nearest to

One peculiarity must be noticed, and that is that the band, which corresponds to a zero difference of path, is not white but black. Now when we consider the dynamical theory of reflection and refraction, it will be found that at grazing incidence, the amplitude of the reflected light is very nearly equal to that of the incident light, but is negative. From (2) we see that when A' A, the intensity of the mixture is proportional to 442 sin πe, which vanishes when e = 0. The adjoining bright band is given by eλ, or

x=T(μ-1) d/2c+λd/4c.

EXAMPLES.

1. A small pencil of light is reflected at three mirrors, so that the images form a small triangle ABC, of which C is a right angle. Prove that the intensity at any point (x, y) on a parallel screen at a distance d, is proportional to

where AC-a, BC= c; and the projection of C on the screen is

the origin, and CA is the axis of x.