Page:MichelsonRefractometer1882.djvu/5

Rh The effect is independent of the position of the glass plate, provided its surface is kept parallel with the corresponding mirror. Suppose, therefore, that it is in contact with the latter and let $$cd$$, fig. 4, represent the common surface. Let $$t=hi=$$ thickness of the glass, $$i=$$ angle of incidence, $$r=$$ angle of refraction, $$n=$$ index of refraction, $$\lambda=$$ wave-length of light. Let $$ef$$ represent the image of the other mirror, and put $$n_{0}=\tfrac{hk}{t}$$.

It can be readily demonstrated that the path of the rays in the instrument is equivalent to that given in the figure, where one of the rays follows the path $$qnmh$$, and the other the path $$rolh$$. Suppose the mirrors $$cd$$ and $$ef$$ parallel. Then as has been previously shown, the curves of interference are concentric circles, formed at an infinite distance. Therefore the rays $$qn$$, $$ro$$, whose path is to be traced, are parallel, and from the point $$h$$ they coincide. Their difference of path is $$2nm-2hl-op$$, and their difference of phase is

Let it be proposed to find the value of $$n_0$$ which renders any particular ring achromatic. The condition of achromatism, as given by Cornu, is $$\tfrac{d\varphi}{d\lambda}=0$$, which gives

$\varphi+2t\left(\cos r\frac{dn}{d\lambda}-n\sin r\frac{dr}{dn}\cdot\frac{dn}{d\lambda}\right)=0$

We have

$n=\frac{\sin i}{\sin r}$ whence $\frac{dr}{dn}=-\frac{\sin^{2}r}{\sin i\cos r}$, whence

$\varphi+\frac{2t}{\cos r}\cdot\frac{dn}{d\lambda}=0$

By Cauchy's formula we have $$n=a_{1}+\tfrac{\alpha_{2}}{\lambda^{2}}$$ whence $$\tfrac{dn}{d\lambda}=-\tfrac{2\alpha_{2}}{\lambda^{3}}$$.