Page:Popular Science Monthly Volume 28.djvu/193

Rh the telescope is α, but it will appear to an eye looking into the eye-piece to lie at an angular distance β from the axis. The magnifying power of the telescope is therefore equal to the angle β divided by the angle α.

The distance A of the focus of the converging waves from the axis is very small, and will equal zero when the luminous point is on the axis, when F will equal the focal length of the object-glass and f of the eye-piece. Extremely small angles being proportional to their tangents, the diagram shows the following expression to be true:

Magnifying power of telescope $$=\frac{\beta}{\alpha}=\frac{tan. \beta}{tan. \alpha}=\frac{\frac{A}{f}}{\frac{A}{F}}=\frac{F}{f}$$, proving

that the magnifying power of a telescope equals the focal length of its object-glass divided by the focal length of its eye-piece.

We have just seen, by similar triangles in Diagram 6, that the focal lengths of the object-glass and eye-piece are proportional to the diameters of the cylinders of plane wave-fronts entering the object-glass and emerging from the eye-piece; it follows, therefore, that the magnifying power of a telescope equals the diameter of the entering cylinder of light divided by the diameter of the emerging cylinder of light.

The easiest way to measure the magnifying power of a telescope is to divide the diameter of the clear aperture of the object-glass by the diameter of the little circle of light seen in the center of the eye-piece when the telescope is pointed at the bright sky, it being assumed that it is in focus for an infinitely distant object. This small circle of light Been in the center of the eye-piece is really an image of the object-glass formed by the eye-piece; but, when the light-waves emerge with plane fronts, the size of this image is exactly equal to the size of the emerging cylinder of plane wave-fronts, so that this method of finding the magnifying power is strictly accurate.

We have seen that, with an eye-piece not exceeding two and a half inches in focal length, luminous points appear through the telescope as many times brighter than they do to the naked eye as the area of the object-glass exceeds the area of the pupil of the eye; and it also follows directly from what has been already stated that, with this eye-piece, the apparent angular distance between two luminous points is proportional to the focal length of the object-glass used. A curious thing following from this is, that surfaces having sensible areas appear no brighter through large telescopes than they do to the naked eye; and it can be stated generally that, using a two-and-a-half-inch eye-piece, which gives the brightest image of an object with any sized object-glass, the surface will appear equally bright, whether seen by the naked eye or through a telescope of any size. The apparent dimensions of the surface, however, will increase directly with the