Page:Optics.djvu/38

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

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

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

$$\theta$$ 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 $$v,$$

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

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

and so on, whence

This in geometrical terms amounts to

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