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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 $TEA$	$=$	$1 / 2 FEA = 1 / 2 {2 π -(p +1) AEB}$
	$=$	$1 / 2 {2 π -(p +1)(π -2 φ')}$
	$=$	$(p +1) φ'-(p -1) π / 2$ ;

$∴ θ$	$=$	$π -(π - φ)-(p +1) φ'+(p -1) π / 2$
	$=$	$φ -(p +1) φ'+(p -1) π / 2$ .
Hence, $dθ / dφ$	$=$	$1-(p +1) dφ' / dφ$ ; and $∴ dφ' / dφ$ , or $cos φ / m \cdotcos φ' = 1 / p +1$ .

We have then $m cos φ'$	$=$	$(p +1)cos φ$ ;
$∴ m 2 cos φ' 2$	$=$	$(p +1) 2 cos φ 2$ ,
and $m 2 sin φ' 2$	$=$	$sin φ 2$ ;

$∴ m 2$	$=$	$(p +1) 2 cos φ 2 +sin φ 2$
	$=$	$(p 2 +2 p)cos φ 2 +1$ ;

$∴ cos φ = \sqrt m 2 -1 / p 2 +2 p$ .

$\theta$ 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

↑ 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

↑ Original: innermost, or was amended to innermost arc: detail

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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.

[2] Original: innermost, or was amended to innermost arc: detail