Page:Optics.djvu/49

 To find the form of the caustic, when parallel rays are reflected by a spherical mirror. (Fig. 26.)

Taking as usual a section of the mirror, let $Pp$ be one of the reflected rays, touching the caustic in $p$. Then, since we know that in this case $Pp$ is one quarter of the chord, if $EP$ be bisected in $O$, and $Op$ be joined, $OpP$ will be a right angle, and if a circle be described through the points $PpO$, $OP$ will be the diameter of it. Let $R$ be the centre, join $Rp$. Then since $OPp=EPQ=PEA$, if a circle be described with centre $E$ and radius $EO$, cutting the axis $EA$ in $F$, the principal focus, the arc $OF$ which measures the angle $PEA$ to the radius $EF$, must be equal to the arc $Op$, which measures twice the angle $OPp$ to the radius $OR$, which is half of the other radius $EO$.

It is plain therefore that the curve $CpF$ must be an epicycloid described by the revolution of a circle equal to $PpO$, on that of which $FO$ is a part: and that there must be a similar epicycloid on the other side of the axis; moreover that if the other part of the circle, $CBc$, represent a convex mirror, there will be a similar pair of epicycloids formed by the intersection of the reflected rays, considering them to extend behind the mirror without limit as all straight lines are supposed to do in the higher analysis.

To describe the caustic given by a spherical reflector, when the radiating point is at the extremity of the diameter.

We shall see that the section of the caustic consists again of two epicycloids.

Let $Q$, (Fig. 28.) be the radiating point. $Pp$ a reflected ray touching the caustic in $p$.

It appears from the equation $1⁄u+1⁄v=2⁄rcosφ$, that $Pp$ is in this case one-third of the chord, since $u=2rcosφ$; and therefore if $E$, $G$ be joined, and $PE$ be trisected in the points $R$, $O$, we shall have the triangle $PRp$ similar to $PEG$, two sides of which being radii, $PR$ must be equal to $Rp$. With centre $R$ let a circle be described through the points $P$, $p$, $O$; and with centre $E$ and radius $EO$, another circle which will of course cut $EC$ in $V$, the focus of rays reflected at points infinitely near $A$, $AV$ being one