Page:Optics.djvu/163

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

ATE=ABE−BAT=ABE−(EAT−EAB),
that is, θ=φ′−(φ−φ′)=2φ′−φ.

Now when two successive emergent rays are parallel, θ remains unchanged, while φ becomes φ+dφ, or in other words, θ is at a maximum or minimum, and dθ/dφ=0;

∴2dφ′/dφ−1=0, or dφ′/dφ=1/2.

But sinφ=m·sinφ′;

∴dφ′/dφ=cosφ/m·cosφ′,
and cosφ/m·cosφ′=1/2, or m·cosφ′=2cosφ;
∴ m²cosφ′²=4cosφ²,
but m²sinφ′²=sinφ²;

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∴ adding, m2	=	4cosφ2+sinφ2
	=	3cosφ2+1;

∴ cosφ=√m²−1/3.

In order to find θ from this and the equation θ=2φ′−φ, we must put for m 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 φ′ and θ, and consequently 2θ, 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.