Elements of natural philosophy (Volume 2-3) . pe; The rule for finding the magnifying power of this Magnifyinginstrument is the same as in the former case: for we pov;er fo™d analytically; have, which in Equation (67), after making/, and d, negative,gives A 11- (73) Ratio °f visual angles; and for parallel rays, A F, U m . . (74) Same for parallelrays; The second member being positive, shows that objects objects appearseen through the Galilean telescope appear erect. § 78. If we divide both numerator and denominator ofEquation (72), by Fu. (F„), it becomes, A A m i~ F. n Magnifyingpower in ter
Elements of natural philosophy (Volume 2-3) . pe; The rule for finding the magnifying power of this Magnifyinginstrument is the same as in the former case: for we pov;er fo™d analytically; have, which in Equation (67), after making/, and d, negative,gives A 11- (73) Ratio °f visual angles; and for parallel rays, A F, U m . . (74) Same for parallelrays; The second member being positive, shows that objects objects appearseen through the Galilean telescope appear erect. § 78. If we divide both numerator and denominator ofEquation (72), by Fu. (F„), it becomes, A A m i~ F. n Magnifyingpower in termsof the powers ofthe lenses; and denoting by Z, the power of the field, and by l% thatof the eye lens, we have A^A L • • • • (75) Ratio of visualangles; 243 NATURAL PHILOSOPHY. Rule for magnifying power. that is, the magnifying power of the astronomical tele-scope is equal to the quotient arising from dividing thepower of the eye lens hy that of the field lens. Fig. 51. PL Geometrical J? illustration of the *field of view; Q. Generalexplanation; Q § 79. If E, be the optical centre of the field, and 0that of the eye lens of an astronomical telescope, theline E 0, passing through the points E and 6>, is calledthe axis of the instrument. Let Qr P be any objectwhose centre is in this axis, and q p> its image. Now,in order that all points in the object may appear equallybright, it is obvious from the figure, that the lens mustbe large enough to embrace as many rays from the pointsP and Q\ as from the intermediate points. It is notso in the figure; a portion, if not all the rays from thosepoints will be excluded from the eye, and the object, inconsequence, appear less luminous about the exteriorthan towards the centre, the brightness increasing to acertain boundary, within which all points will appearequally bright. The angle subtended at the centre of thefield lens, by the greatest line that can be drawn withinthis boundary, is called the field of view. To find
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