Page:Optics.djvu/78

54

(3) A rarer medium with a concave surface, Fig. 68.

(4) A rarer medium with a convex surface, Fig. 69.

In all these figures $QR$ is the incident ray cutting the axis $AE$ in $Q$; $RS$ the refracted ray cutting $AE$ in $q$. $E$ is the centre of the surface.

Let $r=EA{\mbox{ or }}ER$,

$\theta =\angle REA,$

$\phi =\angle ERQ{\mbox{ or }}GRQ,{\mbox{ the angle of incidence}},$

$\phi '=\angle ERq{\mbox{ or }}GRq,{\mbox{ the angle of refraction}},$

$\Delta =AQ,$

$\Delta '=Aq.$

Then $m={\frac {\sin \phi }{\sin \phi '}}={\frac {\sin \phi }{\sin \theta }}\cdot {\frac {\sin \theta }{\sin \phi '}}={\frac {QE}{QR}}\cdot {\frac {Rq}{Eq}}.$

76. It is not possible to obtain a general, simple, algebraical expression from this, but if merely the ultimate value of $Aq$ be required, that is, the limit of its value when the angle $RQA$ is diminished sine limite, we may put $AQ$, for $RQ$, $Aq$ for $Rq$^[1]; we have then

that is, (Case1, 3.) $m={\frac {\Delta -r}{\Delta }}\cdot {\frac {\Delta '}{\Delta '-r}};$

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1. There is one case in which $Aq$ has but one value, all the rays being refracted accurately so as to meet in a single point ; it is when

   $AQ=(m+1)AE{\mbox{ or }}EQ=m.EA.$ (Fig. 66.)

   Then since $ER=EA,{\frac {QE}{ER}}=m={\frac {QE}{QR}}\cdot {\frac {Rq}{Eq}};$

   $\therefore {\frac {QR}{RE}}={\frac {Rq}{qE}}.$

   The triangles $QRE,RqE$ are therefore similar, and $Eq=ER.{\frac {RE}{EQ}}={\frac {EA^{2}}{EQ}}$, and is consequently the same for all points $R$.