Page:Optics.djvu/76

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Then $QPA=\theta ,$

$PAB=\pi -\phi +\phi ',$

$ABQ=\pi -\psi +\psi ',$

$BQP=\eta ,$

${\overline {2\pi =2\pi -\phi +\phi '-\psi +\psi '+\theta +\eta ;}}$

$\therefore \ \psi =\theta +\eta -\phi +\phi '+\psi '$

$=\theta +\eta +\iota -\phi$ . . . . . . . . . . . . . . . . . . . . (1).

We have now to find $\psi ',$

$\sin \phi =m\sin(\iota -\psi '),$${\displaystyle }$

$\sin \psi =m\sin \psi ',$

${\frac {\sin \psi -\sin \phi }{\sin \psi +\sin \phi }}={\frac {\sin \psi '-\sin(\iota -\psi '}{\sin \psi '-\sin(\iota -\psi '}}$

$\therefore \tan {\frac {\psi -\phi }{2}}.\cot {\frac {\psi +\phi }{2}}=\tan {\frac {2\psi '-\iota }{2}}\cot {\frac {\iota }{2}},$

$\tan {\Bigl (}\psi '-{\frac {\iota }{2}}{\Bigr )}=\tan {\frac {\iota }{2}}.\tan {\frac {\psi -\phi }{2}}.\cot {\frac {\psi +\phi }{2}};$

$\therefore \psi '={\frac {\iota }{2}}+\tan ^{-1}{\Bigl (}\tan {\frac {\iota }{2}}.\tan {\frac {\psi -\phi }{2}}.\cot {\frac {\psi +\phi }{2}}{\Bigl )}$

Having found $\psi$ and $\psi ',$ we have only to divide $\sin \psi$ by $\sin \psi '$ to get $m,$ the index of the refracting power required^[1].

This process furnishes an easy method of verifying the third law, by calculating the values of $m$ from different observations, changing the position of the eye or object, and prism; they will be found all to agree.

74.In order to find the refracting power of a liquid substance, we have only to put some of it into a hollow prism of glass, having the surfaces of its sides ground very true and parallel. The refractions of the ray in passing through these will not change the, direction of the emergent ray, as far as regards the angles it makes

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1. The experiment fails, when $\psi$ and $\psi '$ are equal to nothing, but in that case
   $\sin \phi =m\sin \iota ;\quad \therefore m={\frac {\sin \phi }{\sin \iota }}.$