Page:Optics.djvu/72

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If the ray after emergence from the refracting substance, that we have just been considering, meet with another with surfaces parallel to those of the former, whose thickness is $T'$ and for which the ratio of refraction is $m',$ its direction will, after the second emergence, be of course parallel to the former, and its intersection with $AQ$ will be removed by an additional distance ${\frac {m'-1}{m'}}T',$ so that the whole deviation will be ${\frac {m-1}{m}}T+{\frac {m'-1}{m'}}T',$ and if there were more refracting substances of the same description, their effects would, in like manner, be all added together.

67. Prop.⁠To determine the refraction which a ray experiences in passing through a medium bounded by planes not parallel, for example, a triangular prism of glass.

We will suppose the incidence to take place in a plane perpendicular to the axis of the prism, in which case a transverse section of the prism such as $HIK,$ (Fig. 61.) will contain all the lines necessary for the figure.

Let $QR$ be the incident ray, refracted at $R$ into the direction $RS,$ and again at $S$ into the direction $ST.$

⁠Let $\iota$ = the angle of the prism $HIK,$

⁠$\phi$ = angle of incidence $QRx,$

⁠$\phi '$= angle of 1^st refraction $ERS,$

⁠$\psi '$ = angle of 2^d refraction $ESR,$

⁠$\psi$ = angle of emergence $TSy.$

Then $\sin \phi =m\sin \phi ',\quad \sin \psi =m\sin \psi '.$Now $IRS+ISR+RIS=180^{o};$

that is, ${\frac {\pi }{2}}-\phi '+{\frac {\pi }{2}}-\psi '+\iota =\pi ,\quad \therefore \phi '+\psi '=\iota .$

From these equations, knowing $\iota$ and $\phi ,$ we may find successively $\phi ',\psi ',\mathrm {and} \ \psi .$

The deviation $TVq=VRS+VSR=\phi -\phi '+\psi -\psi '.$