Page:Optics.djvu/39

 {{center|$$ \begin{aligned} q\backprime=\frac{qf}{q+f-qv}&=\frac{qf}{q+f}\left(1-\frac{qv}{q+f}\right)^{-1}\\ &=\frac{qf}{q+f}\left\{1+\frac{qv}{q+f}+\left(\frac{qv}{q+f}\right)^2+....\right.\\ &=\frac{qf}{q+f}+\frac{q^2f.v}{(q+f)^2}+\frac{q^3fv^2}{(q+f)^3}+.... \end{aligned} $$}} 16. There is another series a little different from this, but which amounts nearly to the same thing.

If we take the letter $$\alpha$$ to represent the aberration, {{center|$$ \begin{aligned} \alpha=q\backprime-q\prime&=\frac{qf}{f+q\cos\theta}-\frac{qf}{f+q}\\ &=\frac{qf}{f+q}\left(\frac{f+q}{f+q\prime\cos\theta}-1\right)\\ &=\frac{qf}{q+f}.\frac{q-q\cos\theta}{f+q\cos\theta.}\\ &=\frac{qf}{q+f}.\frac{q\operatorname{ver~sin}\theta}{(f+q)\cos\theta+f\operatorname{ver~sin}\theta}\\ &=\frac{q^2f.v}{(q+f)^2.\cos\theta}.\left\{1+\frac{fv}{(q+f)\cos\theta}\right\}^{-1}\\ &=\frac{q^2f.v}{(q+f)^2.\cos\theta}\left\{1-\frac{fv}{(q+f)\cos\theta}+\frac{f^2v^2}{(q+f)^2\cos\theta^2}+\ldots\right.\\ &=\frac{q^2}{(q+f)^2}f.\frac{\operatorname{ver~sin}\theta}{\cos\theta}-\frac{q^2}{(q+f)^3}f^2.\frac{\operatorname{ver~sin}\theta^2}{\cos\theta^2}+.... \end{aligned} $$}} Now $$\frac{\operatorname{ver~sin}\theta}{\cos\theta}=\frac{1-\cos\theta}{\cos\theta}=\sec\theta-1$$; so that $$f$$ being half the radius, $$f\frac{\operatorname{ver~sin}\theta}{\cos\theta}$$ is the algebraical value of half the excess of the secant over the radius, and if $$RT$$ be drawn touching the surface at $$R$$, we shall have the following series in geometrical terms, $\alpha=\frac{QE^2}{QF^2}.\frac{AT}{2}-\frac{QE^3}{QF^3}.\frac{AT^2}{4QE}+\frac{QE^4}{QF^4}.\frac{AT^3}{8QE^2}-....$