We use Optical Character Recognition (OCR) during our scanning and processing workflow to make the content of each page searchable. You can view the automatically generated text below as well as copy and paste individual pieces of text to quote in your own work.
Text recognition is never 100% accurate. Many parts of the scanned page may not be reflected in the OCR text output, including: images, page layout, certain fonts or handwriting.
172 PROJECTION OF IMAGES OF OPAQUE OBJECTS [Cn. VII
of the lantern slide and the reflecting qualities of the opaque object (see § 274a).
§ 275. Aperture of the projection objective for transparencies and for opaque objects. — By comparing figures 90-91 it will be seen that for a transparency, relatively small aperture for the projection objective is sufficient. This also shows that if one were to use the same objective for both transparencies and for opaque objects, that the difference in brightness would be enormously exaggerated, if one used only the necessary aperture for the transparencies. If one used the proper objective for the opaque object, it would answer well for the transparency, but only a part of the aperture would be utilized. As the large aperture makes the objective very expensive, one wastes money by having the large aperture for transparencies In the best practice, an objective of moderate aperture is used for transparencies, and one of relatively very large aperture for opaque projection.
§ 276. As will be shown later (Ch. XIV, § 8 5 ya), with a given object and a given illumination, the brilliancy of the screen image depends upon the aperture of the objective and its distance from the screen. The larger the diameter of the lenses of an objective
§ 274a. Light flux getting through the objective with opaque projection. — •
It will be shown in § 857a that the light received from a perfectly white, perfectly diffusing surface is
>
vSin 26 d20 _ _ T B
— (i— cos 20)
o
(i — cos 20) lumens per square centimeter of the white reflecting
surface, where I is the intensity of illumination of the surface measured in meter candles, and 6 is the half angle of light subtended by the objective, or 26 is the angle of light subtended by the objective. The light received by the surface is I/io,ooo lumens and the proportion of light received by the surface
i — cos 2B which strikes the objective is then
In this problem the angle of light subtended by the objective is 20°, i. e. 26 = 20°. The proportion of light received by the objective is then (i — cos 20°)/2 = (i — -9397)/2 = .0603/2 = .0302 or about T,r"(.