Page:Light waves and their uses.djvu/148

130 The resolving power, as shown in one of the preceding lectures, depends on the size of the diffraction rings which are produced about the image of a star. It was also shown that the smallest angle which a telescope could resolve was that subtended at the center of the lens by the radius of the first dark ring, and this angle is equal to the ratio of the length of the light wave to the diameter of the objective. For example, if we consider a 4-inch glass, the length of the light wave being of an inch, this angle would be. If the lens were a 40-inch glass, the angle would be something like, which can be represented by the angle subtended by a dime at the distance of fifteen miles. Hence, if we had two such dimes placed side by side, the largest glass would scarcely separate them.

Fig. 90 is an actual photograph of the image of a point of light taken with an aperture smaller than that of a telescope, but otherwise under the same conditions under which a telescope is used. It is easy to see that, surrounding the point of the image, there is a more or less defined white disc, and beyond this a dark ring. Outside of this dark ring there are a bright ring and another dark ring. Theoretically, there are a great number of those rings; practically, we see only one or two under the most favorable conditions.

This figure represents the appearance of the image of one of Jupiter's satellites as it would be observed in one of the largest telescopes under the most favorable conditions. If it be required to measure the diameter of one of these very