Page:Optics.djvu/64

40

Let $AEOP,$ (Fig. 54,) be the line joining the centres, which we will consider as the axis of the mirror. $PQ$ a part of the object; $pq$ its image.

Let $OP=b;\quad EO=c,\quad EQ=q,\quad Eq=q',\quad QEP,\mathrm {or~} pEq=\theta ,\quad EF=f,\quad Ep=p.$

Then we know that $q=c.\cos \theta +{\sqrt {b^{2}+c^{2}\sin \theta ^{2}}},$ and if to simplify the problem, we suppose $PQ$ and $\theta$ to be very small, we may put for this

$q=c{\Bigl (}1-{\frac {\theta ^{2}}{2}}{\Bigr )}+b-{\frac {c^{2}}{2b}}\theta ^{2}=(b+c).{\Bigl (}1-{\frac {c\theta ^{2}}{2b}}{\Bigr )}.$

Then ${\frac {1}{q'}}={\frac {1}{f}}+{\frac {1}{q}}={\frac {1}{f}}+{\frac {1}{b+c}}\left\{1+{\frac {c\theta ^{2}}{2b}}\right\}$

${\frac {1}{p}}={\frac {1}{f}}+{\frac {1}{b+c}}.$

It is not easy to determine any thing about the form of the curve from this, but we may deduce one rather remarkable conclusion.

The diameter of curvature of $pq$, is equal to ${\frac {pq^{2}}{qr}}$ if $pr$ be a tangent at $p.$ Now this is equal to ${\frac {p^{2}\theta ^{2}}{p\sec \theta -q'}}.$

Hence, calling the radius of curvature $\rho$, we have

${\frac {1}{2\rho }}={\frac {p\sec \theta -q'}{p^{2}\theta ^{2}}}={\frac {p-q'}{p^{2}\theta ^{2}}}+{\frac {p.{\frac {1}{2}}\theta ^{2}}{p^{2}\theta ^{2}}}$

$={\frac {p-q'}{pq'\theta ^{2}}}+{\frac {1}{2p}},$ nearly

$={\Bigl (}{\frac {1}{q'}}-{\frac {1}{p}}{\Bigr )}{\frac {1}{\theta ^{2}}}+{\frac {1}{2p}}$

$={\frac {c}{2b(b+c)}}+{\frac {1}{2f}}+{\frac {1}{2(b+c)}}$

$={\frac {1}{2b}}+{\frac {1}{2f'}},$