Page:FizeauFresnel1859.pdf/14

 $u\left(\frac{v^{2}-v'^{2}}{v{}^{2}}\right).$|undefined

which is the amount by which the speed of propagation of the waves should be increased.

Also, if the speed of propagation in the state of rest is v', in the state of motion it will be

$v'\pm u\left(\frac{v^{2}-v'^{2}}{v{}^{2}}\right).$|undefined

Now, with the help of this expression, I am going to calculate the fringe displacement that should be observed in the experiment in question.

The speed of propagation of light in moving water, for each of the two rays that have to interfere, is one of the values expressed by the preceding formula. Using the same notation as in the previous example, the difference in velocities will be

$\Delta=E\left[\frac{v}{v'-u\left(\frac{v^{2}-v'^{2}}{v^{2}}\right)}-\frac{v}{v'+u\left(\frac{v^{2}-v'^{2}}{v^{2}}\right)}\right];$|undefined

By doing the calculations and some transformations, it is found

$\Delta=2E\frac{u}{v}\left[\frac{v^{2}-v'^{2}}{v'^{2}-u^{2}\left(\frac{v^{2}-v'^{2}}{v^{2}}\right)^{2}}\right].$|undefined

This expression may be simplified by taking into consideration that u is very small in relation to v, $$\left(\tfrac{u}{v'}=\frac{1}{33000000}\right)$$, and that the coefficient of $$u^2$$ is always smaller than unity, which permits the cancelation of the term in $$u^2$$ without appreciable error; m is the index of refraction, E is twice the length L of the tubes, and the final formula is

$\Delta=4L\frac{u}{v}\left(m^{2}-1\right),$