Page:Harold Dennis Taylor - A System of Applied Optics.djvu/30

holds good if we put u for A..Q, r for the radius A..C, and ù for A..q, and this formula interprets itself for all cases, provided the following conventions are strictly adhered to, viz.:—

Convention as to signs of focal distancesThe radii of all surfaces, whether convex or concave, to be considered intrinsically positive with respect to the conjugate distances whose signs are to be assessed.

Then for convex surfaces—

Rays of incident pencil diverging, then Q..A or u is positive.
Figs. 4a and 4e.
Rays of incident pencil converging, then A..Q or u is negative.
Figs. 4c and 4g.
Rays of refracted pencil converging, then A..q or ù is positive.
Figs. 4a, 4c, and 4g.
Rays of refracted pencil diverging, then q..A or ù is negative.

Fig. 4e.

And for concave surfaces—

Rays of incident pencil converging, then A..Q or u is positive.
Figs. 4b and 4f.
Rays of incident pencil diverging, then Q..A or u is negative.
Figs. 4d and 4h.
Rays of refracted pencil diverging, then q..A or ù is positive.
Figs. 4b, 4d, and 4h.
Rays of refracted pencil converging, then A..q or ù is negative.

Fig. 4f.

Thus, in the case of Fig. 4c, A..Q is convergent and therefore u is negative, and

$\textstyle {{\frac {\mu }{u'}}={\frac {\mu -1}{r}}-{\frac {1}{u}}}$

becomes Rays entering convex surface convergent.

$\textstyle {{\frac {\mu }{u'}}={\frac {\mu -1}{r}}+{\frac {1}{u}}}$

And, again, in a case where Q..A in Fig. 4a becomes less than $\textstyle {\frac {r}{\mu -1}}$, then of course $\textstyle {{\frac {\mu }{u'}}={\frac {\mu -1}{r}}-{\frac {1}{u}}}$ Rays leaving convex surface divergent. gives a negative result, and the refracted pencil is shown to be divergent, as in Fig. 4e.

If the rays of the incident pencil are parallel and therefore

$\textstyle {{\frac {1}{Q..A}}=zero}$

therefore