Page:Optics.djvu/163

 Then the angles $EAB$, $EBA$, $EBC$, $ECB$ being all equal, we have

Now when two successive emergent rays are parallel, $ATE=ABE−BAT=ABE−(EAT−EAB)$ remains unchanged, while $θ=φ′−(φ−φ′)=2φ′−φ$ becomes $θ$, or in other words, $φ$ is at a maximum or minimum, and $φ+dφ$;

But $θ$;

In order to find $dθ⁄dφ=0$ from this and the equation $∴2dφ′⁄dφ−1=0$, we must put for $dφ′⁄dφ=1⁄2$ the value that it has for any desired sort of homogeneous light refracted between air and water, and by the help of a table of natural sines and cosines, we shall obtain the angles $sinφ=m·sinφ′$ and $∴dφ′⁄dφ=cosφ⁄m·cosφ′$, and consequently $cosφ⁄m·cosφ′=1⁄2$, which will be the radius of the arc of that particular colour in the primary bow.

The investigation is very similar for the secondary bow, or indeed for any other.