Page:Optics.djvu/70

46

Now, if we suppose that the incidence of the ray $QR$ takes place nearly perpendicularly to the surface, the angles $RQA,RqA$ may be considered so small, that their sines and tangents are to all sense the same: we may therefore substitute the ratio of the sines for that of the tangents, and remembering that $RQA,RqA$ are equal to their alternate angles $QRN,qRN$ which are those of incidence and refraction, we find

${\frac {\Delta '}{\Delta }}={\frac {\sin \theta }{\sin \theta '}}=m,\quad \mathrm {or} \quad \Delta '=m\Delta .$

It follows from this, that a thin pencil of rays proceeding from a point at the distance $\Delta$ from the plane surface of a refracting medium, appears after refraction to proceed from a point $\mathrm {q}$ at the distance $\mathrm {m\Delta '.}$ In common glass, this latter distance is about half as great again as the former.

64.⁠Now let us examine the amount of the error we commit in substituting the ratio of the sines for that of the tangents, or in other words, let us see what is the difference between the ultimate value of $\Delta '(Aq)$ and that which it actually bears when the angle $RQA$ though small, is taken into consideration: this difference, from its analogy with one of the phænomena of reflexion at curved surfaces, we will call the Aberration.

Let us then take instead of $AQ,Aq,$ the tangents of the angles $RQA,RqA$ to the radius $RA,RQ,$ and $Rq,$ which are to each other as the sines of the angles $RqA,RQq,$ that is, as $\sin \theta ',$ and $\sin \theta .$

We have then $Rq=$ M x RQ^{[errata 1]}, that is, if $AR=v,$ squaring both sides,

⁠$\Delta '^{2}+v^{2}=m^{2}(\Delta ^{2}+v^{2})$

⁠$\therefore \quad \Delta '^{2}=m^{2}\Delta ^{2}+(m^{2}-1)v^{2}$

⁠$=m^{2}\Delta ^{2}+(m^{2}-1)\Delta ^{2}\tan \theta '$

⁠$=m^{2}\Delta ^{2}{\Bigl (}1+{\frac {m^{2}-1}{m^{2}}}\tan \theta ^{2}{\Bigr )};$

⁠$\therefore \quad \Delta '=m\Delta {\Bigl (}1+{\frac {m^{2}-1}{2m^{2}}}\tan \theta ^{2}{\Bigr )}.$

Errata

1. Original: mRQ was amended to M x RQ: detail