Page:Optics.djvu/38

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$\left.{\begin{aligned}\mathrm {Let} ~&q&~\mathrm {represent} ~&EQ\\&q'&~\ldots \ldots \ldots \ldots \ldots ~&Eq\\&f&~\ldots \ldots \ldots \ldots \ldots ~&EF\end{aligned}}\right\}~\mathrm {as~before,}$

and $v$ being the actual intersection of the reflected ray and axis,

let $q\backprime =Ev$.

Then since $q'$ is the limiting value of $q\backprime$ when $\theta =0$, we must have

$q\backprime =q'+{\frac {dq\backprime }{d\theta ^{2}}}.\theta +{\frac {1}{1.2}}{\frac {d^{2}q\backprime }{d\theta ^{2}}}.\theta ^{2}+....$

$\theta$ being made equal to $0$ in every differential coefficient, or if we consider $q\backprime$ as a function of the versed sine of $\theta$, which we will call $v,$

$q\backprime =q'+\left({\frac {dq\backprime }{dv}}\right)v+{\frac {1}{1.2}}\left({\frac {d^{2}q\backprime }{dv^{2}}}\right)v^{2}+....$

The brackets indicating that $v$ is made $=0$ in each coefficient.

${\begin{aligned}\mathrm {Now} ~q\backprime &={\frac {qf}{f+q\cos \theta }}={\frac {qf}{f+q(1-\operatorname {ver~sin} \theta )}}={\frac {qf}{q+f-qv}}\\{\frac {dq\backprime }{dv}}&={\frac {q^{2}f}{(q+f-qv)^{2}}}~\mathrm {which~when} ~v=0,~\mathrm {becomes} ~{\frac {q^{2}f}{(q+f)^{2}}}\\{\frac {1}{1.2}}{\frac {d^{2}q\backprime }{dv^{2}}}&={\frac {1}{1.2}}{\frac {2q^{3}f}{(q+f-qv)^{3}}}..................{\frac {q^{3}f}{(q+f)^{3}}},\end{aligned}}$

and so on, whence

$q\backprime ={\frac {qf}{q+f}}+{\frac {q^{2}f.v}{(q+f)^{2}}}+{\frac {q^{3}f.v^{2}}{(q+f)^{3}}}+{\frac {q^{4}.fv^{3}}{(q+f)^{4}}}+....$

This in geometrical terms amounts to

$Eq+{\frac {QE^{2}}{QF^{2}}}.{\frac {AN}{2}}+{\frac {QE^{3}}{QF^{3}}}.{\frac {AN^{2}}{4EF}}+....$

The Aberration is represented by this series without its first term; and when the angle $\theta$, and a fortiori its versed sine, are but small, the second term of the series will give a near approximate value.

Note. The above is perhaps the neatest way of obtaining the series for the aberration: it is sometimes done by a method simpler in its principle