Page:Harold Dennis Taylor - A System of Applied Optics.djvu/31

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

Focal distance when entering rays are parallel.

and

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

If, on the other hand, $\scriptstyle {QA={\frac {r}{\mu -1}}}$,then $\scriptstyle {{\frac {\mu }{u'}}=0}$, and the rays of the refracted pencil are parallel.Refracted rays parallel.

We are now in a position to consider the cases of two spherical surfaces in succession enclosing glass between them and forming a lens. We will assume the axial thicknesses of such lenses to be negligible, the two spherical surfaces being brought to a sharp edge in the case of collective lenses and the diameter or aperture being very small So far, lenses assumed to have no central thickness. compared to the principal focal length, while in the case of dispersive lenses the two spherical surfaces may be supposed to touch one another on the lens axis, the axial thickness being zero. Let us take a case like Fig. 4a, wherein the rays after refraction at the first surface are convergent and ù, is positive. Let these convergent rays proceed through a second convex surface, as shown in Fig. 5a.

We saw that in the case of Fig. 4a the distance A..q or ù was given by the equation $\scriptstyle {{\frac {\mu }{u'}}={\frac {\mu -1}{r}}-{\frac {1}{u}}}$, from which we get $\scriptstyle {{\frac {1}{u}}={\frac {\mu -1}{r}}-{\frac {\mu }{u}}}$. We can apply this equation to the refraction, taken in the reverse direction, at the second surface, as shown in Fig. 5a, Plate II., wherein A₂..Q₂ corresponds to u, and A₂..q = ù; only in this case A₂..Q₂ may be better expressed as = v, and the radius of curvature as s, so that we get

$\scriptstyle {{\frac {\mu }{u'}}={\frac {\mu -1}{s}}-{\frac {1}{v}}}$

Course of rays at second surface considered reversed.

and

$\scriptstyle {{\frac {1}{v}}={\frac {\mu -1}{s}}-{\frac {\mu }{u'}}}$

But as the rays of the pencil are converging (left to right) into the second surface, and the distance ù becomes, relatively to the second surface, negative, therefore the above equation becomes

$\scriptstyle {{\frac {1}{v}}={\frac {\mu -1}{s}}+{\frac {\mu }{u'}}}$

But $\scriptstyle {\frac {\mu }{u'}}$ by the refraction at the first surface was shown to be $\scriptstyle {{\frac {\mu -1}{r}}-{\frac {1}{u}}}$. Substituting this value in the above equation we get

negative, therefore the above equation becomes

$\scriptstyle {{\frac {1}{v}}={\frac {\mu -1}{s}}+{\frac {\mu -1}{r}}-{\frac {1}{u}}}$

;