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Mr. W. Crookes on the Measurement of the 



pare the photometric intensities of D and C. (It is necessary that neither 

 D nor C should contain any polarized light, but that the light coming from 

 them, represented on each disk by the two lines at right angles to each 

 other, forming a cross, should be entirely unpolarized.) Let H represent a 

 double refracting achromatic prism of Iceland spar ; this will resolve the 

 disk D into two disks d and d\ polarized in opposite directions ; the plane 

 of d being, we will assume, vertical, and that of d' horizontal. The prism 

 H will likewise give two images of the disk C ; the image c being polarized 

 horizontally, and c' vertically. The size of the disks D, C, and the sepa- 

 rating power of the prism H, are to be so arranged that the vertically 

 polarized image d } and the horizontally polarized image c, exactly overlap 

 each other, forming, as shown in the figure, one compound disk c d, built 

 up of half the light from D and half that from C. 



The measure of the amount of free polarization present in the disk c d 

 will give the relative photometric intensities of D and C. 



The letter I represents a diaphragm with a circular hole in the centre, 

 just large enough to allow the compound disk c d to be seen, but cutting 

 off from view the side disks c' } d'. In front of the aperture in I is placed 

 a piece of selenite, of appropriate thickness for it to give a strongly con- 

 trasting red and green image under the influence of polarized light. K is 

 a doubly refracting prism, similar in all respects to H, placed at such a 

 distance from the aperture in I that the two disks into which I appears to 

 be split up are separated from each other, as at g> r. If the disk c d con- 

 tains no polarized light, the images g, r will be white, consisting of oppo- 

 sitely polarized rays of white light ; but if there is a trace of polarized light 

 in c d, the two disks g, r will be coloured complementarity, the contrast 

 between the green and the red being stronger in proportion to the quantity 

 of polarized light in c d. 



The action of this arrangement will be readily evident. Let it be sup- 

 posed, in the first place, that the two sources of light, D and C, are exactly 

 equal. They will each be divided by H into two disks d' d and c c\ and 

 the two polarized rays of which c d is compounded will also be absolutely 

 equal in intensity, and will neutralize each other, and form common light, 

 no trace of free polarization being present. In this case the two disks of 

 light, g, r, will be colourless. Let it now be supposed that one source 

 of light (D, for instance) is stronger than the other (C). It follows that 

 the two images d', d will be more luminous than the two images c, c', and 

 that the vertically polarized ray d will be stronger than the horizontally 

 polarized ray c. The compound disk c d will therefore shine with partially 

 polarized light, the amount of free polarization being in exact ratio with 

 the photometric intensity of D over C. In this case the image of the 

 selenite plate in front of the aperture I will be divided by K into a red 

 and a green disk. 



Fig. 2 shows the instrument fitted up. A is the eyepiece (shown in 

 enlarged section at fig. 3) ; G B is a brass tube, blacked inside, having a 



