Page:Optics.djvu/71

47

It appears then that $\Delta '$ is in general greater than its ultimate value by the quantity ${\frac {m^{2}-1}{2m}}\Delta .\tan \theta ^{2},$ supposing the angle $\theta$ to be small.

65. Prop. Supposing that a ray passes through a refracting substance bounded by two parallel plane surfaces; it is required to determine its direction after emergence.

Let $AR,A'R'$ (Fig. 60.) represent the two surfaces; $R'S$ the course of the ray after emergence, which, produced backwards, cuts $AQ$ in $V.$ We must observe, that as the refraction at $R'$ is contrary to that at $R,$ if in the latter case ${\frac {\sin \mathrm {inc^{e}.} }{\sin \mathrm {ref^{n}.} }}=m,$ in the former we must have ${\frac {\sin \mathrm {inc^{e}.} }{\sin \mathrm {ref^{n}.} }}={\frac {1}{m}}.$

Then we have $Aq=m.AQ,A'V={\frac {1}{m}}.A'q;$ that is, if we call $AA',$ the thickness of the substance, $T,$ and $AV,\Delta '',$

$\Delta ''+T={\frac {1}{m}}(T+\Delta ')={\frac {1}{m}}(T+m\Delta );$

and $\therefore \quad \Delta ''=\Delta -{\frac {m-1}{m}}T,$

whence it appears that $VQ={\frac {m-1}{m}}AA'.$

It appears from this, that a pencil of rays passing nearly perpendicularly through a refracting medium bounded by parallel planes, suffers no alteration as to convergency or divergency, only that the point of concourse of the rays is brought nearer to the surface of the medium.

66. If we take into consideration the aberrations, we shall find that the distance $Aq$ being taken too small in the calculation we have just made, the point $R'$ is rather too high, (that is, too far from $A',$) which has the effect of throwing $V$ too far out from $A',$ that is, making $A'V$ too great; but again, on account of the aberration at $R',A'V$ is too small; so that the two aberrations correct each other.