Page:Optics.djvu/82

 When $$Q$$ is between $$A$$ and $$E$$, $$q$$ is between $$Q$$ and $$E.$$ This may easily be seen from the geometrical construction, (Fig. 77.) or it may be shown from the formula: for

which shows that $$\Delta$$ is greater or less than $$\Delta'$$ according as it is greater or less than $$r.$$

When $$Q$$ comes to $$A$$, $$q$$ coincides with it.

By differentiating the equation $$\frac{1}{\Delta'}=\frac{m-1}{mr}+\frac{1}{m \Delta},$$ we find

which shows that the distance $$Qq$$ is at a maximum in the space between $$A$$ and $$E$$ when $$mr^2=(\overline{m-1} \Delta + r)^2     $$, or

for when $$\Delta - \Delta'$$ is at a maximum $$d \Delta - d \Delta' = 0;$$ and $$\therefore \frac{d \Delta'}{d \Delta} = 1.$$

If we place $$Q$$ on the other side of $$A,$$ (Fig. 78.) or make $$\Delta$$ negative, we shall have

whence we collect that as long as $$\Delta < \frac{r}{m-1},$$ $$\Delta'$$ is negative and increasing: that when $$\Delta = \frac{r}{m-1},$$ or $$Q$$ is at $$f$$, $$\Delta'$$ is infinite, and that afterwards it becomes positive, or that $$q$$ goes to the other side of $$A.$$

Obs. It will probably have occurred to the reader, that by placing $$Q$$ within the denser medium, we have virtually passed from the first case to the fourth, with the only difference that the places of $$Q$$and $$q$$ are inverted. I have, however, purposely placed $$Q$$ in all possible positions, in order to illustrate the connexion between the