Page:Optics.djvu/36

 language as applied to Geometry. This is however more clearly distinguished in the case of the other formula, which here becomes for, as before,

$$\frac{1}{\Delta} - \frac{1}{\Delta'}=-\frac{2}{r},$$

whence $$\Delta r - 2 \Delta \Delta' - r \Delta'=0,$$ and dividing by $$\Delta \Delta' r,$$

$$\frac{1}{\Delta'} - \frac{2}{r} - \frac{1}{\Delta} = 0,$$

or $$\frac{1}{\Delta} - \frac{1}{\Delta'} = -\frac{2}{r}$$ as above.

Here then $$\Delta'$$ and $$r$$ become both negative, which the Figure plainly shows.

Making $$\Delta$$ infinite, or $$\frac{1}{\Delta}=0,$$ we find $$\frac{1}{\Delta'} = \frac{2}{r}$$, or $$\Delta' = \frac{r}{2} $$ as before, so that the principal focus is here at the bisection of the radius of the reflector, and of course behind it. In fact, as long as $$\Delta$$ is positive, or $$Q$$ is in front of the mirror, so long must $$q$$ necessarily be behind it, and the reflected rays diverge.

If, however, the incident rays converge to a point behind the mirror, that is, $$\Delta$$ be negative, the formula will become

$$-\frac{1}{\Delta} - \frac{1}{\Delta'} = -\frac{2}{r},$$ or $$\frac{1}{\Delta'} = \frac{2}{r} - \frac{1}{\Delta},$$

and $$\Delta'$$ may be positive, provided $$\frac{1}{\Delta} > \frac{2}{r},$$ that is, $$\Delta < \tfrac{1}{2} r,$$ so that if the focus of the incident rays lie between the principal focus and the back of the mirror, the reflected rays will converge. Of course, if the incident rays converge to the principal focus, the reflected rays are parallel.

,