When we have to measure the colour transmitted through coloured bodies, we have to adopt a slightly different plan, which is extremely accurate. The first thing necessary is to make some arrangement whereby two beams of identical colour—that is, of the same wave-length—reach the screen, one of which passes through the transparent body to be measured, and the other unabsorbed. If we in addition have some means of equalizing the intensity of the two beams, we can then tell the amount cut off by the body through which one beam passes. The method that would be first thought of would be to use two spectra, from two sources of light; but should we adopt that plan there would be no guarantee that the spectra would not vary in intensity from time to time. The point then that had to be aimed at was to form two spectra from the same source of light, and with the same beam that passes through the slit of the collimator. Here we are helped by the property of Iceland spar, which is able to split up a ray into two divergent rays. By placing what is called a double-image prism of Iceland spar at the end of the collimator, we get two divergent beams of light falling on the prisms, and by turning the double-image prism we are able to obtain two spectra on the screen of the camera one above the other, and if the slit of the slide be sufficiently long two beams would issue through it of identical colour, and separated from one another by a dark space, the breadth of which depends on the length of the slit of the collimator. It is to be observed that by this arrangement we have exactly what we require: a light from one source passes through the same slit, is decomposed by the same prisms, and as the beams diverge in a plane passing through the slit of the collimator, the length of spectrum is the same. The problem to solve is how to utilize these two spectra now we have got them. We can make the light from the top spectrum pass through the coloured body by the following artifice. Let us place a right-angled prism in front of the top slit, reflecting say the beam to the right, and after it has travelled a certain distance, catch it by another right-angled prism, and thus reflect it on to the screen. Already in the path of the ray, issuing through the slit from the bottom spectrum, the lens L₄ is placed, forming a square patch on the screen. By placing a similar lens in the path of the other ray after reflection from the second right-angled prism, we can superpose a second patch of the same colour over the first patch, and by putting a rod in the path of the two beams we can have as before two shadows side by side, but this time each illuminated by the same colour. One shadow will be more strongly illuminated than the other, owing to the different intensities of beams into which the double-image prism splits up the primary ray. The two, however, can be equalized by placing a rotating apparatus in the path of one of the beams. When equalized the sector is read off, and tells us how much brighter one spectrum is than the other. Thus suppose in the direct beam the sectors had to be closed to an angle of 80°, to effect this, the bottom spectrum would be 180/80, or 2·25 times brighter than the bottom spectrum. It should be noted that as the two spectra are formed by the identical quality of light, this same ratio will hold good throughout their length. If it does not, it shows that the double-image prism is not in adjustment, and that the same rays are not coming through the slit in the slide, and it must be rotated till the readings throughout are the same. One of the most sensitive tests for adjustment is to form a patch with orange light, when the slightest deviation from adjustment will be seen by the two patches differing in hue.
We can now place the coloured transparent object in the path of the beam which is most convenient, and by again equalizing the shadows, measure the amount it cuts off; this we can do for any ray we choose. As both right-angled prisms can be attached to the card or slide which moves across the spectrum, nothing besides the card need be moved. In the following diagram we have the proportion of rays transmitted by the three different glasses, red, green, and blue, in terms of the unabsorbed spectrum. Take for instance on the scale of the spectrum the number 11. The curve shows that at that particular part of the spectrum which lies in the blue, the blue glass only allowed 4/100 or 1/25 of the ray to pass, whilst the green glass allowed 10/100 or 1/10 to pass. So at scale No. 4 in the orange, through the blue only 2% was transmitted, through the green glass 4%, and through the red 20%.
Fig. 17.—Absorption by Red, Blue, and Green Glasses.
Fig. 18.—Light reflected from Metallic Surfaces.
1. Vermilion 2. Carmine. 3. Mercuric Iodide. 4. Indian Red.
Fig. 19.