Page:FizeauFresnel1859.pdf/11

 The motion takes place in a relative direction which is opposed to each of the rays.

If we call $$\Delta$$ the difference in speeds being sought, E the length of the column of water traversed by the rays, by applying the principles of the theory of interferences, we find that

$\Delta=E\left(\frac{v}{v'-u}-\frac{v}{v'+u}\right),$

or also,

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

Since u is so small in relation to v, $$\left(\tfrac{u}{v'}=\tfrac{1}{33000000}\right)$$, this expression may be reduced, without appreciable error, to

$2E\frac{u}{v}\frac{v^{2}}{v'^{2}}$|undefined

Replacing the expression $$\tfrac{v}{v'}$$ by m, the refractive index of water, in the above formula we have

$\Delta=2E\frac{u}{v}m^{2}$

Each ray traverses the tubes twice, and so the length E is twice the actual length of the tubes. By calling L the actual length of each tube (1.4875 meters), the previous expression becomes

$\Delta=4L\frac{u}{v}m^{2}.$

Numerical calculation yields:

$\Delta$ = 0.0002418 millimeters

This is the difference in distance that should exist between the two rays if the first hypothesis were true.