Page:Optics.djvu/91

 This is shown in Figs. 90, 91, 92, 93, where $$QRST$$ is the course of the light, and it will easily be seen that the spherical surfaces at $$R,\ S,$$ can be parallel only when the radii $$ER,\ E'S,$$ are so.

The point $$O,$$ where the refracted ray $$RS$$ cuts the axis, is called by some writers, the centre of the lens; it is within the lens in the cases of double-concave and double-convex lenses, but without, in the meniscus and concavo-convex.

The point $$O$$ is invariably the same at whatever angle the parallel radii be drawn, for

$$EO = EE' \cdot \frac{ER}{ER \mp E'S}=(r \mp r' \pm t). \frac{r}{r \mp r'}. $$

The point $$m$$ where the incident ray cuts the axis is easily found: we have only to put the value of $$AO$$ for $$\Delta'$$ in the equation, for the single refracting surface, and find $$\Delta.$$

$$AO = EO - AE = (r \mp r' \pm t) \frac{r}{r \mp r'}-r = \frac{rt}{r \mp r'}. $$

Then $$\frac{1}{\Delta} = \pm \frac{m-1}{mr} + \frac{1}{m \Delta'} = \pm \frac{m-1}{mr}+ \frac{r \mp r'}{mrt} $$

$$= \pm \frac{(m-1) t + r \mp r'}{mrt}; $$

$$\therefore \Delta = \frac{mrt}{r \mp r' \pm (m-1)\ t}. $$

If the thickness of the lens be supposed inconsiderable, $$QRST$$ may be taken as a straight line, and $$A,\ O,$$ as one point.

It appears from this, that when a pencil of rays enters a thin lens obliquely, that ray which passes through the centre is not refracted at all: it serves as an axis to the pencil, and the focus of refracted rays lies on it at the same distance from the centre of the lens as when the axis of the pencil coincides with that of the lens, though the refraction is not quite accurate.

94. To return to the simple approximate formula of the lens.

Let $$\phi$$ represent the distance hitherto called $$\Delta''.$$

Then, $$\frac{1}{\phi}=\frac{1}{F}+\frac{1}{\Delta}$$ or $$= \frac{m-1}{\rho} + \frac{1}{\Delta},$$ (vid. Note, p. 62.)