Page:Optics.djvu/87

 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,

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

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