Page:Optics.djvu/59

35

Then supposing the object $PQ$ to be beyond $E$ the centre of the mirror,

Let $r$ represent the radius $EA,$

$\theta$. . . . . . . . . . . . angle $PEQ,$

$c$. . . . . . . . . . . . . . line $EP,$

$q{\Bigl (}={\frac {c}{\cos \theta }}{\Bigr )}$. . . . . . $EQ,$

$q'$ . . . . . . . . . . . . . . . .Eq^{[errata 1]}.

Then, referring to page 8, we find

${\frac {1}{q'}}={\frac {2}{r}}+{\frac {1}{q}}={\frac {2}{r}}+{\frac {1}{c}}\cos \theta .$

Now the polar equation to an ellipse referred to the near focus is

${\frac {1}{\rho }}={\frac {1}{a(1-e^{2})}}+{\frac {e}{a(1-e^{2})}}\cos \theta .$

which coincides with the above, provided $a(1-e^{2})={\frac {r}{2}},$ and

${\frac {a(1-e^{2})}{e}}=c;\quad \mathrm {that~is,~} e={\frac {r}{2c}},$

and $a={\frac {r}{2(1-e^{2})}}={\frac {r}{2{\Bigl (}1-{\frac {r^{2}}{4c^{2}}}{\Bigr )}}}={\frac {2c^{2}r}{4c^{2}-r^{2}}}.$

It is also necessary, that $e$ be less than unity, that is, $c>{\frac {r}{2}}$.

44.There are some things to be remarked in this elliptic image. We found the quantity $a\ (1-e^{2})$ to be constantly equal to the half radius. Now $a\ (1-e^{2})$ is half the latus rectum of the ellipse, which is also the radius of curvature at the vertex. It appears then, that although the place of the vertex $p,$ and the magnitude of the ellipse, depend on the place of the point $P,$ yet the curvature of the image at the vertex is invariable, as is the

Errata

1. Original: EQ was amended to Eq: detail