Page:Optics.djvu/75

51

Also $\phi '+\psi '=\iota ;\quad \therefore d\phi '+d\psi '=0$.

Eliminating therefore $d\phi '$ and $d\psi '$, we find

$\cos \psi '.\cos \phi .d\phi +\cos \phi '.\cos \psi .d\psi =0;$

$\therefore d\psi =-{\frac {\cos \phi }{\cos \phi '}}.{\frac {\cos \psi '}{\cos \psi }}d\phi ,$

and $d\delta =d\phi \left\{1-{\frac {\cos \phi }{\cos \phi '}}.{\frac {\cos \psi '}{\cos \psi }}\right\}$.

Now this will clearly be equal to nothing, when $\phi$ and $\psi$ are equal, as their cosines and those of $\phi '$ and $\psi '$ will also be equal, and therefore ${\frac {\cos \phi }{\cos \psi }}.{\frac {\cos \psi '}{\cos \phi '}}=1$.

It appears then, that the deviation is a minimum, when the incident and emergent ray make equal angles with the sides of the prism.

72.Prop.A thin conical pencil of rays pass nearly perpendicularly through both sides of a very thin prism; required the focus of the emergent rays.

Let $Q$, (Fig. 64.) be the focus of the incident rays, $QO$ a perpendicular to the surface. The focus of the rays in their passage through the prism will be $x$, a point in $OQ$, such that

$Ox=m\times OQ,$ (Art. 63.).

From $x$ draw $xy$ perpendicular to the second side of the prism, then, as $x$ is the focus of rays incident on that surface at $S$, their focus after the second refraction will be a point $q$, such that

$yx=m\times yq$.

73.To find experimentally the refracting power of any given substance, we may form a piece of it, if solid, into a prism, and observe an object through it. See Fig. 65, where $P$ is the place of the eye, $Q$ the object observed. Let the angles at $P,\ Q$ be measured; also the angle $PAH$ or $PAI,$ in order to have the angle of incidence at $A$.