Page:Optics.djvu/87

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In the double-concave lens $r'$ is negative,

⁠${\frac {1}{F}}=(m-1)\left\{{\frac {1}{r}}+{\frac {1}{r'}}\right\}.$

In the double-convex $r$ is negative,

⁠${\frac {1}{F}}=-(m-1)\left\{{\frac {1}{r}}+{\frac {1}{r'}}\right\}.$

In the plano-concave either $r'$ is infinite, or $r$ is infinite, and $r'$ negative; therefore putting $r$ for the single radius

⁠${\frac {1}{F}}={\frac {m-1}{r}},\quad F={\frac {r}{m-1}}.$

In the plane-convex,

⁠${\frac {1}{F}}=-{\frac {m-1}{r}},\quad F={\frac {r}{m-1}}.$

When in the double-concave, or double-convex lens the radii are equal,

⁠${\frac {1}{F}}=\pm (m-1).{\frac {2}{r}},\quad \mathrm {or} \quad F=\pm {\frac {r}{2(m-1)}}.$^[1]

88.⁠It appears from all this, that the place of the principal focus is the same, whichever side of a lens is turned towards the incident light, and that

The concavo-convex[2]	$\scriptstyle {\left.{\begin{matrix}\ \\\\\ \\\ \ \end{matrix}}\right\}\,}$	make parallel rays diverge.
the double-concave
and the plano-concave
The meniscus	$\scriptstyle {\left.{\begin{matrix}\ \\\\\ \\\ \ \end{matrix}}\right\}\,}$	make parallel rays converge.
the double-convex
and the plano-convex

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1. If $m={\frac {3}{2}}$ which is nearly the case in glass, $f=\pm r,$ or the principal focal length is equal to the radius of sphericity.
2. See Fig. 87, for the relation between these different kinds of lens. Those placed together are equivalent.