POLARIZATION BY A PILE OF PLATES.

215

208. Let P be the fraction of the light, which is reflected at the first surface of a plate; then 1-p is the fraction of the light which is transmitted.

Since the light reflected by a plate is made up of that which is reflected at the first surface, and that which has suffered an odd number of internal reflections, it follows that if the intensity of the incident light be taken as unity, the intensity of these various portions will be

p, (1 - p) pg2, (1 - p) pg', etc.

Hence if R be the intensity of the reflected light,

Similarly if T be the intensity of the transmitted light,

.(58).

.(59).

209. The value of p depends upon the particular theory of light which we adopt, but in any case it may be supposed to be a known function of the angle of incidence and μ the index of refraction; the value of g depends upon i and μ, and also upon q, which may be supposed to have been determined by experiment. To complete the solution, we have therefore to solve the following problem :-There are m parallel plates, each of which reflects and transmits given fractions R and T of the light incident upon it; light of intensity unity being incident upon the system, it is required to find the intensities of the reflected and refracted light.

Let these be denoted by (m), (m); and consider a system of m÷n plates, and imagine them grouped into two systems of m and n plates respectively. Since the incident light is represented by unity, (m) will be the intensity of the light reflected from the first group, whilst (m) will be transmitted. A fraction (n) of the latter will be reflected by the second group, whilst a portion (n) will be transmitted; and the fraction (m) of the light reflected by the second group will be reflected by the first group, whilst the fraction (m) will be transmitted, and so on. It therefore follows, that the intensity of the light reflected by the whole system will be

$ (m)+(&m)2 $ (n) + (¥m)2 $ (m) $ (n)2 + ...,

and the intensity of the light transmitted will be

↓ (m) † (n) ¦ ¥ (m) $ (n) $ (m) & (n) + ¥ (m) ($n)* (μm)2 ¥ (n) + .....

The first of these expressions is equal to $ (m + n), whilst the second is equal to (m + n); whence summing the two geometrical series, we obtain

$(m + n) = $(m) +

(m)
1-4 (m) (n)

(n)

.(60),

↓ (m) & (n)

↓ (m + n) =

.(61).

1

− 4 (m) $ (n)

In the special problem under consideration, m and n are positive integers; but we shall now show how to obtain the solution of these two functional equations, when m and n have any values whatever. From (60) we obtain

$ (m + n) {1 − $ (m) $ (n)} = $ (m) + $ (n) {(&m)2 − ($m)3}.

Since the left-hand side of this equation is symmetrical with respect to m and n, we obtain by interchanging these letters, and equating the results

1+(¢m) – (lm)_1+(n) – (n)

Since this equation is true for all values of m and n, each side must be equal to a constant; whence denoting the constant by 2 cos a, we obtain

(†m)2 = 1 − 24 (m) cos a + ($m)2 .............

Squaring (61), and eliminating the function

(62), we obtain

{1-(m) $ (n)} [1 − 24 (m + n) cos a + {$ (m + n)}°]

....(62). by means of

= {1-24 (m) cos a + (pm)} {1-24 (n) cos a + (pn)) ...(63).

In order that (60) and (61) may hold good for a zero value of one of the variables, say n, we must have (0) = 0, † (0) = 1. If however we put n=0 in (63), the equation reduces to an identity; we must therefore differentiate (63) with respect to n, and then put n=0. Accordingly we find

p' (0) $ (m) {1 − 24 (m) cos 2 + (μm)2) + p′ (m) cos a − þ (m) p′ (m) = {1-24 (m) cos a + (pm)2) ′ (0) cos a.

Dividing out by (m) – cos a, since the solution (m) = cos a (m) = C, we obtain

would lead to

p' (m) = p′ (0) {1 − 24 (m) cos a + ($m)2}

..(64).

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217

Integrating this equation, and determining the arbitrary constant from the condition that (0)=0, and writing B for '(0) sin a, we obtain

$(m)=

sin mẞ
sin (a + mẞ)

(65).

(66).

Equations (65) and (66) may be written in the form

sin mẞ sin a sin (a + mẞ)

.(67).

When m=1, (m) = R, ↓ (m) = T, where the values of R and T are given by (58) and (59); and therefore to determine the arbitrary constants, we have the equations

210. Equations (67) and (68) give the following quasigeometrical construction for solving the problem :-Construct a triangle, in which the sides represent in magnitude the intensities of the incident, reflected and transmitted light in the case of a single plate; then leaving the first side and the angle opposite to the third unchanged, multiply the angle opposite the second, by the number of plates; then the sides of the new triangle will represent the corresponding intensities in the case of a system of plates. This construction cannot however be actually effected, inasmuch as the first side of the triangle is greater than the sum of the two others, and the angles are therefore imaginary.

To adapt the formula to numerical calculation, it will be convenient to get rid of the imaginary quantities. Putting

{(1 + R + T) (1 + R − T) (1 + T − R) (1 − R − T)}'1 = ▲ ...(69), we have by ordinary Trigonometry

211.

From this equation we see, that the intensity of the light reflected from an infinite number of plates is a1; and since a is changed into a1, by changing the sign of a or ▲, we have

a ̄1 = (1 + R2 + T2 - ▲)/2R...............(73), which is equal to unity in the case of perfect transparency. Accordingly substances, such as snow and colourless compounds thrown down as chemical precipitates, which are finely divided so as to present numerous reflecting surfaces, and which are transparent in mass, are brilliantly white by reflected light.

212. The following tables, taken from Stokes' paper, give the intensity of the light reflected from, or transmitted through, a pile of m plates for the values 1, 2, 4 and of m for three degrees of transparency, and for certain selected angles of incidence. The refractive index is taken to be equal to 1.52; 81-T is the loss by absorption in a single transit through a plate at perpendicular incidence, so that 80 corresponds to perfect transparency; also the value of p is supposed to be calculated from Fresnel's formulæ, so that

=

(74),

according as the light is polarized in or perpendicularly to the plane of incidence. The angle is the polarizing angle tan1μ; and denote the intensities of the reflected and transmitted light, the intensity of the incident light being taken as 1000. For oblique incidences, it is necessary to distinguish between

DISCUSSION OF THE RESULTS.

219

light polarized in and perpendicularly to the plane of incidence, and the suffixes 1 and 2 refer to these two kinds respectively.

213. In discussing these tables Sir G. Stokes says:-"The intensity of the light reflected from a pile consisting of an infinite number of similar plates, falls off rapidly with the transparency of the material of which the plates are composed, especially at small incidence. Thus at a perpendicular incidence, we see from the above table that the reflected light is reduced to little more than one half, when 2 per cent. is absorbed in a single transit; and to less than a quarter, when 10 per cent. is absorbed.

"With imperfectly transparent plates, little is gained by multiplying the plates beyond a very limited number, if the object be to obtain light, as bright as may be, polarized by reflection. Thus the table shows, that 4 plates of the less