Page:Optics.djvu/64

 Let $$AEOP,$$ (Fig. 54,) be the line joining the centres, which we will consider as the axis of the mirror. $$PQ$$ a part of the object; $$pq$$ its image.

Let $$OP=b; \quad EO=c, \quad EQ=q, \quad Eq=q', \quad QEP, \mathrm {or~} pEq= \theta, \quad EF=f, \quad Ep=p.$$

Then we know that $$q = c. \cos \theta + \sqrt{b^2+c^2 \sin \theta^2},$$ and if to simplify the problem, we suppose $$PQ$$ and $$\theta$$ to be very small, we may put for this

Then $$\frac{1}{q'}=\frac{1}{f}+\frac{1}{q}=\frac{1}{f} +\frac{1}{b+c} \left \{ 1+\frac{c \theta^2}{2b} \right \}$$

$$\frac{1}{p}=\frac{1}{f}+\frac{1}{b+c}.$$

It is not easy to determine any thing about the form of the curve from this, but we may deduce one rather remarkable conclusion.

The diameter of curvature of $$pq$$, is equal to $$\frac{pq^2}{qr}$$ if $$pr$$ be a tangent at $$p.$$ Now this is equal to $$\frac{p^2 \theta^2}{p \sec \theta - q'}.$$

Hence, calling the radius of curvature $$\rho$$, we have

$$\frac{1}{2 \rho} = \frac{p \sec \theta - q'}{p^2 \theta^2} = \frac{p - q'}{p^2 \theta^2} + \frac{p. \frac{1}{2} \theta^2}{p^2 \theta^2}$$

$$= \frac{p-q'}{p q' \theta^2} + \frac{1}{2p},$$ nearly

$$= \Bigl( \frac{1}{q'} - \frac{1}{p} \Bigr) \frac{1}{\theta^2} + \frac{1}{2p}$$

$$= \frac{c}{2b (b+c)} + \frac{1}{2f} + \frac{1}{2(b+c)}$$

$$= \frac{1}{2b} + \frac{1}{2f'},$$