Page:Optics.djvu/82

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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

${\frac {1}{\Delta }}-{\frac {1}{\Delta '}}={\bigl (}1-{\frac {1}{m}}{\bigr )}{\frac {1}{\Delta }}-{\frac {m-1}{mr}}={\frac {m-1}{m}}{\bigl (}{\frac {1}{\Delta }}-{\frac {1}{r}}{\bigr )},$

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

${\frac {d\Delta '}{d\Delta }}={\frac {1}{m}}\cdot {\frac {\Delta '^{2}}{\Delta ^{2}}}={\frac {mr^{2}}{({\overline {m-1}}\Delta +r)^{2}}}$

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

$\Delta ={\frac {\surd m-1}{m-1}}r,$

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

${\frac {1}{\Delta '}}={\frac {m-1}{mr}}-{\frac {1}{m\Delta }},$

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