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or ${\frac {1}{\rho }}={\frac {1}{b}}+{\frac {1}{f}}.$

Now $b$ is the radius of the object, and $f$ is that of the image of a straight line at the vertex; moreover the curvature of a line is aptly measured by the reciprocal of its radius of curvature. It appears then that the curvature of the image of a small portion of a sphere, is equal to that of the object, together with that of the conoidal image of a small plane object, at the same place.^[1]

51.With regard to the magnitude of the image produced by a spherical mirror, it is easy to see that as it subtends the same angle at the centre of the mirror, that the object does, if we suppose them to be plane, an hypothesis which agrees very well with ordinary cases of experiment, the linear magnitudes, that is, the lengths or breadths, of the object and image, will be in direct proportion to their distances from the centre, so that, if we put $L,\ l,$ for the lengths of the object and images, $q_{0}',\ q_{0}'$ for the distances $EP,\ Ep;$

$l=L.{\frac {q_{0}'}{q_{0}}}.$

But $q_{o}=c$, and ${\frac {1}{q_{o}'}}={\frac {2}{r}}+{\frac {1}{c}};\quad \therefore q'={\frac {cr}{2c+r}};$

$\therefore \ l=L.{\frac {r}{2c+r}}=L\div {\Bigl (}1+{\frac {2c}{r}}{\Bigr )},$

or if $f={\frac {r}{2}}$, being the principal focal distance,

$l=L.{\frac {f}{c+f}}=L\div {\Bigl (}1+{\frac {c}{f}}{\Bigr )}.$

52.It might be expected that we should treat of images produced by reflexion at surfaces not spherical, but the subject is in

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1. We here suppose that the curvatures of the object and mirror are opposed to each other, that is, are of the same kind; if the radii lie in the same direction, so that the one be convex, and the other concave, the curvature of the image will be less than in the case of a right line, the expression being then ${\frac {1}{\rho }}={\frac {1}{f}}-{\frac {1}{b}}$

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