Page:Optics.djvu/43

This page has been validated.

19

and if we multiply the first by $v$ , and the second by $u$ , and subtract, we shall have

$0=2 uv -(u + v) r cos φ$ ;
and $∴ v = ur cos φ / 2 u - r cos φ$ , or $1 / u + 1 / v = 2 / r cos φ$ ,

and if we put $2 f$ for $r$ the expressions will be rather simpler,

$v = uf cos φ / u - f cos φ$ , or $1 / u + 1 / v = 1 / f cos φ$ .

23. Here we may observe, that if $u$ be infinite, that is, the incident rays parallel, we have simply $v = f cos φ$ , which referred to the geometrical figure, shows that $Aq$ is in that case one-fourth of the chord of the osculating circle so that if $Ry$ be a tangent, and $Qy$ perpendicular to it, calling $Qy$ , $p$ , we have in general

$1 / u + 1 / v = 2 dp / pdu$ , and $v = pudu / 2 udp - pdu$	$= du / 2 dp / p - du / u$
	$= du / dl p 2 / u$ .

24. Prop. Given the radiant point, and the reflecting surface, to describe the caustic.

In order to determine the nature of the caustic curve, that is, the section of the superficial caustic, we may consider it as a spiral having $Q$ for its pole.

Let $Qx$ (Fig. 18.) be drawn perpendicular on $Rq$ which is of course a tangent to the caustic at $q$ . Call $Qq$ , $u'$ ; and $Qx$ , $p'$

$u' 2$	$= u 2 + v 2 -2 uv cos2 φ$ ,
$p'$	$= u sin2 φ$ ,
$cos φ$	$= p / u$ ; $∴ cos2 φ = 2 p 2 / u 2 -1$ ; $sin2 φ = 2 p / u \sqrt 1- p 2 / u 2$ ;
$∴ u' 2$	$= u 2 + v 2 -2 uv (2 p 2 / u 2 -1)=(u + v) 2 - 4 p 2 v / u$ ,