Page:Optics.djvu/69

 ones, provided only the surfaces be parallel, the deviation on the whole is exactly the same.

Let $$QRSTVXY$$ (Fig. 58.) be the course of a beam of light through the media $$A, B, C, D, E, A: Q'V'X'Y'$$ that of another parallel to the former, at first, and therefore at last, passing only through the medium $$A, E, A$$.

Then since the refractions at $$X$$ and $$X'$$ are equal, the angles made by $$VX,$$ $$V'X',$$ with the final or the original courses of the rays, that is, the deviations, must be equal.

62. There is another thing to be observed here, which is rather remarkable. If the media $$B, C, D, E,$$ were separated from each other by sensible distances, that designated by $$A$$ pervading all the intermediate spaces, so that the ray came clear out of each before it entered the next, and if the surfaces were all as before parallel, it is evident that the last, like all the other emergent rays, would be parallel to the first incident. It appears then, that it is indifferent as to this phenomenon and its consequences, whether the surfaces touch or not, which is the more singular, as whatever be the nature of the action of substances on light, it certainly takes place only at insensible distances.

63. Given the direction in which a ray falls on a plane surface bounding a refracting medium; to find the direction of the refracted ray.

$$QR$$ (Fig. 59.) represents a ray refracted at $$R$$ into a direction $$RS,$$ which is produced backwards so as to intersect $$AQ,$$ a perpendicular to the surface through $$Q,$$ in $$q.$$

Let $$m$$ represent the ratio, sin inclination: sin refraction, sometimes called the index of refraction, or ratio of refraction.

$$\Delta ...................... AQ,$$

$$\Delta' ...................... Aq,$$

$$\theta ..................... \angle RQA,$$

$$\theta' ....................\angle RqA.$$

Then we have $$Aq=$$$$. \cot \theta' = AQ. \tan \theta. \cot \theta' ,$$

that is, $$\frac{\Delta'}{\Delta} = \frac{\tan \theta}{\tan \theta'}.$$

Errata