Page:Optics.djvu/72

 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. 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 1st refraction $$ERS,$$

$$\psi' $$ = angle of 2d 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 $$TVq=VRS + VSR= \phi - \phi' + \psi - \psi'.$$