Page:Optics.djvu/97

 The whole aberration is therefore

The angles $β=(∆″−r′)^{2}·(1⁄m∆′−1⁄∆″)v′$, $m∆″^{2}⁄∆′^{2}(∆′−r)^{2}(m⁄∆−1⁄∆′)v+(∆″−r′)^{2}·(1⁄m∆′−1⁄∆″)v′$ being very nearly the same, we may, without much error, establish that for a particular value of $AER$ the aberration varies as $Aer$ the $∆$ of $v$, that is, as the square of $versed sine$, the radius of the aperture.

Let us examine what kinds of value the aberration in a lens assumes in different cases. The aberration
 * 1) For the meniscus or concavo-convex lens we have ($AER$, $AR$, being both positive.)

Now suppose $r$, $r′$, $(A)={m·∆″^{2}⁄∆′^{2}·(∆′−r)^{2}(m⁄∆−1⁄∆′)+(∆″−r′)^{2}(1⁄m∆′−1⁄∆″)}v$, $m=3⁄2$ and therefiore

And if $r=1$ be the $r′=5⁄3$ of $∆=∞$ or $∆′=$, $3r=3$, nearly.

Note that the aberration is of a contrary sign to the focal distance, and therefore diminishes it.