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140
To make it as general as possible, let 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 | = | 12FEA=12{2π−(p+1)AEB} |
= | 12{2π−(p+1)(π−2φ′)} | |
= | (p+1)φ′−(p−1)π2; |
∴ θ | = | π−(π−φ)−(p+1)φ′+(p−1)π2 |
= | φ−(p+1)φ′+(p−1)π2. | |
Hence, dθdφ | = | 1−(p+1)dφ′dφ; and ∴ dφ′dφ, or cosφm·cosφ′=1p+1. |
We have then mcosφ′ | = | (p+1)cosφ; |
∴ m2cosφ′2 | = | (p+1)2cosφ2, |
and m2sinφ′2 | = | sinφ2; |
∴ m2 | = | (p+1)2cosφ2+sinφ2 |
= | (p2+2p)cosφ2+1; |
∴ cosφ=√m2−1p2+2p.
is of course found as before, 1, 2, 3, . . . . being put for 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
- ↑ 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