Page:Optics.djvu/164

140

To make it as general as possible, let $p$ represent the number of reflexions within the drop; (Fig. 209.)

θ	=	ATE the half radius, as before,
ATE	=	π−EAT−TEA.

Now EAT=π−TAM=π−φ,

and TEA	=	1/2FEA=1/2{2π−(p+1)AEB}
	=	1/2{2π−(p+1)(π−2φ′)}
	=	(p+1)φ′−(p−1)π/2;

∴ θ	=	π−(π−φ)−(p+1)φ′+(p−1)π/2
	=	φ−(p+1)φ′+(p−1)π/2.
Hence, dθ/dφ	=	1−(p+1)dφ′/dφ; and ∴ dφ′/dφ, or cosφ/m·cosφ′=1/p+1.

We have then mcosφ′	=	(p+1)cosφ;
∴ m2cosφ′2	=	(p+1)2cosφ2,
and m2sinφ′2	=	sinφ2;

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∴ m2	=	(p+1)2cosφ2+sinφ2
	=	(p2+2p)cosφ2+1;

∴ cosφ=√m²−1/p²+2p.

$\theta$ is of course found as before, 1, 2, 3, . . . . being put for $p,$ according as the question relates to the primary, secondary, or tertiary bow.

In this manner the radius^[1] of the innermost arc^{[errata 1]} of the lower bow, is found to be 40° 17′, that of the outermost 42° 2′. And the extreme values for the second bow, are 50° 57′, and 54° 7′.

182. It is easy to verify these results by observation, for as the center of the bows is in the line joining the center of the Sun

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1. These arcs are considered as parts of small circles of the celestial sphere, and the radius is the distance of each from its pole.

Errata

1. Original: innermost, or was amended to innermost arc: detail