Page:MillerTheory.djvu/7

 angles of reflexion are given by the equation

$\tan\frac{\phi'}{2}\ =\ \tan\frac{\phi}{2}\cdot\frac{\mathrm{V}-u}{\mathrm{V}+u}$

where u is the velocity of the surface perpendicular to its own plane; negative, if moving away from the approaching light. With the assumed adjustments, we therefore have for the reflexion of the unaccented system at D:

For the reflexion of the accented system we have:

By equation (3) of Dr. Hicks's paper, putting L1 for the perpendicular distance between wave-fronts of the light incident on D from the moving source,

$\frac{\lambda}{\mathrm L_{1}}\ =\ \frac{4-0.5}{4+0.5-2}=1.4$; and $\frac{\lambda'}{\mathrm L_{1}}=\frac{4-1}{4+1-4}\cdot\frac{4-0.5}{4+0.5+2}=\frac{21}{13}$.|undefined

Therefore

$\frac{\lambda}{\cos\delta}\ =\ \mathrm L_{1}1.4\div0.8=1.75\ \mathrm L_{1},$

and

$\frac{\lambda'}{\cos\delta'}\ =\ \mathrm L_{1}\frac{21}{13}\div\frac{12}{13}\ =\ 1.75\ \mathrm L$;

accordingly,

$\frac{\lambda'}{\cos\delta'}\ =\ \frac{\lambda}{\cos\delta}\ =\ l'\ =\ l$.

Therefore, if the intersection of k15 and i15 is on the line xy parallel to qT15, the intersection of i15, and h15 is also on the same line; that is, the phase-difference of the two sets of waves is constant along any line parallel to the axis of the observing telescope. The same thing may easily be proved for any one of eight equidistant points of the circumference commencing from the point where the motion of the apparatus is parallel to the axis of the telescope.