Page:MichelsonRefractometer1882.djvu/3

Rh $$\vartheta$$. Draw $$P'D$$ parallel to $$cf$$, and $$P'C$$, at right angles, and complete the rectangle $$BDP'C$$. Let $$P'PC=i$$ and $$DPP'=\theta$$. Let $$PP'=P$$, and the distance between the surfaces at $$P'=2t_{0}$$. We have then

We see that in general $$\Delta$$ has all possible values, and therefore all phenomena of interference would be obliterated. If, however, we observe the point $$P$$ through a small aperture, $$ab$$, the pupil of the eye, for instance, the light which enters the eye from the surfaces will be limited to the small cone whose angle is $$bPa$$, and if the aperture be sufficiently small the differences in $$\Delta$$ may be reduced to any required degree.

It is proposed to find such a distance $$P$$, that with a given aperture these differences shall be as small as possible, which is equivalent to finding the distance from the mirrors at which the phenomena of interference are most distinct. The change of $$\Delta$$ for a change in $$\theta$$, is

The change of $$\Delta$$ for a change in $$i$$, is

For $$\tfrac{\delta\Delta}{\delta\theta}=0$$ we have $$\theta=0$$ (or $$\Delta=0$$).

For $$\frac{\delta\Delta}{\delta i}=0$$ we have $$\left(1+\tan^{2}i+\tan^{2}\theta\right)P\tan\varphi=\left(t_{0}+P\tan\varphi\tan i\right)\tan i$$, or

$\left(1+\tan^{2}\theta\right)P\tan\varphi=t_{0}\tan i$, whence $P=\frac{t_{0}}{\tan\varphi}\tan i\cos^{2}\theta$|undefined

Hence the fringes will be most distinct when $$\theta=0$$ and when

This condition coincides nearly with that found by Feussner.

If the thickness of the film is zero, or if the angle of incidence is zero, the fringes are formed at the surface of the mirrors. If the film is of uniform thickness, the fringes appear at infinity. If at the same time $$\varphi=0$$, and $$t_{0}=0$$, or $$i=0$$ and $$\varphi=0$$, the position of the fringes is indeterminate. If $$i$$ has the