Page:Optics.djvu/99

75

100.Some writers treat of another aberration arising from that which we have been investigating: it is the distance $qz,$ (Fig. 97.) $Aq$ being the ultimate focal distance and $qz$ perpendicular to $Aq.$

This distance $qz$ is called the lateral aberration, $qo$ the longitudinal,

$qz=qv.{\frac {aS}{av}}=qv{\frac {AR}{Av}},$ nearly.

Since $qv$ varies as the square of $AR,$ it appears that $qz$ varies as its cube.

101.It is important, particularly when a lens is used as a burning-glass to determine whereabouts all the refracted rays are collected within the least space, that is, technically speaking, to find the least circle of aberration or diffusion.

Let $ST$ (Fig. 98.) be the extreme refracted ray on one side: sv^{[errata 1]} a ray on the other side intersecting with this in $n,\ nm$ perpendicular to the axis. Now it is plain that at the maximum state of $mn$, if it has one, all the refracted rays on the same same side with $Sv$ will pass through it, and passing from the section to the actual lens, the circle having $mn$ for its radius will just contain all the rays, so that it will be the circle we seek.

In order to find $mn$ we must know the extreme aberration $qT:$ let this be called $a.$

Let AR	=K which must be measured,
Ar	=k (unknown, or variable,)
Tm	=x,
AT	=T, Av=t.

Since the aberration (longitudinal) varies as the square of the radius of the aperture,

$qv=a.{\frac {k^{2}}{K^{2}}};\quad \therefore vT=a.{\frac {K^{2}-k^{2}}{K^{2}}}..........(1),$

$mn=Tm.{\frac {AR}{AT}}=x.{\frac {K}{T}};\quad \therefore mn\propto x,$ and they are at a maximum together.

Errata

1. Original: Sv was amended to sv: detail