## Abstract

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### Figures (5)

Fig. 1

Optical arrangements of Koenig-Martens photometer. In the upper half of Fig. 1 the paths of the rays from two apertures, A, B, are shown; in the lower half the paths of the rays when only one aperture, A, is employed.

Fig. 2

Photograph of the Koenig-Martens photometer as modified in the instrument-shop of the Geophysical Laboratory. In this figure the following attachments are shown: (1) reflecting prism arrangement B; (2) device used in the measurement of the amount of light reflected at approximately vertical incidence from a plane surface, (D and E); (3) device used in the measurement of the light transmission of an optical instrument, C.

Fig. 3

Graphical solution of equation (5). In this chart the sloping lines represent the angles α1; the abscissae the angle α2; and the ordinates the percentage transmissions.

Fig. 4

Graphical solution of equation (7). In this diagram the abscissae represent the thickness of the plate, the ordinates the absorption in per cent. per centimeter, and the sloping lines the percentage transmission of the plate.

Fig. 5

Optical arrangement of polarization photometer employing a calcite rhomb in place of the Wollaston cube of Fig. 1. In this figure A, B are entrance pupils; C1, C2, lenses situated at focal distance from the entrance pupil; R, the cleavage rhomb of calcite, F1, F2, two biprisms; N, analyzer; D, a low power eyepiece; E, exit pupil.

### Tables (1)

Table I In this table are listed the values of a computed from equation (1) for different intensity ratios, 11/12, of the two fields from 0.01 to 1.00.

### Equations (13)

$1 1 1 2 = tan 2 α$
$I = β l ( 1 - ρ ) 2 ( 1 + β 2 ρ 2 + β 4 ρ 4 + - - - ) = β l ( 1 - ρ ) 2 1 - β 2 l ρ 2$
$I · ρ 2 · β 2 l + ( 1 - ρ ) 2 · β l - I = 0$
$( n - 1 ) 4 · I · β 2 l + 16 n 2 β l - I · ( n + 1 ) 4 = 0$
$β l 1 · ( 1 + I 1 · β l 1 ) · ρ 2 - 2 β l 1 · ρ + β l 1 - I 1 = 0 β l 2 · ( 1 + I 2 · β l 2 ) · ρ 2 - 2 β l 2 · ρ + β l 2 - I 2 = 0$
$I 1 I 2 = β l 1 β l 2 or β = ( I 1 I 2 ) 1 l 1 - l 2$
$I = tan 2 α 2 · cot 2 α 1 , ( α 2 < α 1 )$
$I = tan 2 α 4 · cot 2 α 3 , ( α 4 < α 3 )$
$log β = 1 l 2 - l 1 · log I 1 I 2 = 2 l 2 - l 1 ( log tan α 2 + log cot α 1 )$
$I 1 I 0 = tan 2 α 0 · cot 2 α 2 · n 2 + 1 2 n$
$I 1 I 0 = tan 2 α 4 · cot 2 α 0 · n 2 + 1 2 n$
$I 1 I 0 = tan α 4 · cot α 2 · n 2 + 1 2 n ( c 4 < a 2 )$
$I = I 2 I 1 = tan 2 α 2 cot α 1 , ( α 2 < α 1 )$