Elements of natural philosophy (Volume 2-3) . uation (66), expresses microscope; L , tlie apparent magnitude of an object referred to that at the limit of distinct vision, taken as unity; and what-ever has been demonstrated of the powers of lenses gen-erally, is truetof magnifying powers. Thus, in Equation(31), we have the magnifying power of any combinationof lenses equal to the algebraic sum of the magnifyingpowers taken separately. Should any of the individualsof the combination be concave, they will enter with signscontrary to those of the opposite The power of a single m
Elements of natural philosophy (Volume 2-3) . uation (66), expresses microscope; L , tlie apparent magnitude of an object referred to that at the limit of distinct vision, taken as unity; and what-ever has been demonstrated of the powers of lenses gen-erally, is truetof magnifying powers. Thus, in Equation(31), we have the magnifying power of any combinationof lenses equal to the algebraic sum of the magnifyingpowers taken separately. Should any of the individualsof the combination be concave, they will enter with signscontrary to those of the opposite The power of a single microscope is, Equation (66), magnifying power ofa8ingie equal to the limit of distinct vision divided by its jprinci-microscope; pai focal distance, and the numerical value of the powerwill be greater as the refractive index and curvature aregreater. § 74:. To obtain a general expression for the visual an-gle under which the image of an object formed by alens, and having any position in reference to the eye, ELEMENTS OF OPTICS. 241 Fig. is seen, let Q P, be an object in front of a concave lens. From P, draw through the optical centre E, the line P E ; from P, draw the extreme ray P M, and from J/draw M S, making with P Jf produced the angle SMT equal to the power of the lens; then will, § 47, MS be the corresponding deviated ray, and its intersection p, with the ray P E, through the optical centre, will be a point in the image ; from p, draw p q, parallel to P Q, till it is cut by the ray Q E, through the other extreme of the object and optical centre; p q will be the image. Let 0, be the optical centre of the eye; then denoting the visual angle p 0 q by J., we have, To find thovisual angleunder which animage formed bya lens is seen; A=qp- qp Oq Eq-OE Value of visualangle; and representing the distances Q E by /, Eqhj /,and EO by d, we find, • qp = QFr. Eq-EO=f-d;and hence a- Qp f ? Same in otherterms; and denoting the visual angle P EQ by A, A_A f f-d\ d_ f Ratio of v
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