Problem 4. To Find the Actinicity of Any Surface when Illuminated by a Light Source of Known Actinicity and Convergence. On a cloudless day lower the top sash of a window fronting the sky and close the lower part of the window with a black cloth so as to make a prac tically square opening in the top half. Arrange a white cloth vertically four widths of the open ing distant from it and about level with its lower edge. The low position of the cloth practically compensates for the square form of the opening which of course admits more light than would a circular opening of a diameter equal to the width of the window. The slight error involved is of no practical consequence. The opening thus subtends with practical exactness an f /4 aperture at each point of the cloth and the convergence is therefore 256 cone units. Now disregarding both the cloth and the opening and making a standard measurement of the sky with the meter its intensity is found to be 512 actinos. A meter measurement of the cloth under the conditions explained shows its actinicity to be 8 actinos. Now had the window subtended only one cone unit at the cloth instead of 256, it is evident that the actinicity of the white cloth would have been only of 8 actinos or of an actino. Furthermore had the intensity of the sky been only one actino instead of 512, then that of the cloth would have been only of L of an actino or of an actino.
It is found therefore that a unit light source, i.e., one of unit actinicity and unit cone value, will create on a white surface 35M of an actino of intensity. It is also evident that the intensity of a surface of a different color will be less than that of a white surface in the same ratio as the actinic factors of the two surfaces (p. 64). For this problem therefore is found the following rule.
Rule: To find the actinicity of a surface when illuminated by a light source of known actinicity and cone value; multiply ±, of an actino by the actinicity and again by the cone value of the light source; the product will be the actinicity of a white surface in actinos. To find the actin icity of a surface of another color multiply the actinicity of white by the actinic factor of that particular color.
Another method is as follows. It has already been ascertained (p. 48) that when measuring a maximum sky with a factor 8 medium in the f / 1 meter, the time is 1 second and its actinic ity is 512 actinos. It is also known that a one second incident light indicates 128 actinos for a white cloth. Therefore it is seen that when a white surface is illuminated by any other uniformly intense surface at a convergence of f / 1, the former has ; the intensity of that of the illuminating surface and also that a less convergence in the light source will cause a corresponding decrease of actinicity in the sur face lighted. The second rule for this problem
is therefore as follows.
Rule: Divide one fourth of the actinicity of the light source by the square of its f value; the quotient will be the actinicity of white sur face in that light. As given in the first rule, this number multiplied by the actinic factor of any particular color gives the actinicity of a surface of that color.
If the sky, at 512 actinos, illuminates a white cloth through an opening f / 4 in convergence, an actinicity equal to a< that of the sky or of 8 actinos, is created on the cloth. The whole hemisphere of sky whose convergence is approx imately 128 times that of an f/4 cone raises the actinicity of the surface mentioned to only 32 times 8 or to 256 actinos giving it one half the actinicity of the sky itself.
This apparent 'falling off, judged from the point of view of the law of inverse squares, so rightly and frequently employed for small cones of light, is due to the fact that when a light source approaches the solid angle dimension of a hemisphere in relation to a point which it illu minates, much of its light strikes the surface at a sharp angle and from a whole circle of direc tions and is again reflected off in such a manner as to make it impossible to measure the whole effect on any tinting medium.
In dealing with degrees of convergence much less than a complete hemisphere, as with f/1 and even 1/ 2 cones of light, still another ele ment of error enters. On doubling such rela tively large openings in a plane the equal area added is farther from the point of impingement than was the former opening. This applies only when a small area opposite the center of the opening is being illuminated as in the f / 1 meter.
It is for the sake of symmetry and clearness that the law of inverse squares is considered accurate for the calculation of light problems up to and including cones of f / 1 convergence.
To illustrate this problem suppose a blue cloth to be illuminated by the sun at 30° from the horizon, through a hole a of an inch in diameter and 32 inches distant from the cloth; what is the actinicity of the illuminated spot.
In this example the opening, being of less convergence than the sun itself, determines the form of the cone of light, which is seen to be 1/ 256 (32 4-D. Since the unit cone is f / 64 1/256 is equal to of a cone unit. It has already been found that the actinicity of the sun at is 16,000,000 or 16MM actinos. Now according to the first rule x 6 x 16MM = 64 and 64 x the factor for blue gives 32, the inten sity of the spot in actinos. Or according to the second rule; ÷ = 64. This intensity which is for a white surface, multiplied by !, the factor for blue, gives 32 actinos as before.