238 
PROFESSOR FORBES ON THE EXTINCTION OF THE SOLAR RAYS 
39. v. Now let the ray of light be conceived to return in fig. 4. by the path O A A', 
or in fig. 5. through deb, then the angle of incidence is v' (fig. 4.) and the angle of 
refraction is (by fig. 5.) v' + or by eq. (4.) v' + & 0. Hence 
sin incidence sin?;' V\ + kq + k§ q 
sin refraction sin (v 1 + 8 6) V\ + kq ^ 
sin v' : sin {v' 6) \ Jc g -\- g :*/ \ + k g- 
And passing to differentials, 
sin v 1 : cos v' db — 1 + &p: — A 
V 5 2 VI + kq 
COS V 1 7 . k d q 
sin v' 2 (1 + kq) 
, . k do , 
dl, = - WTJi ) tanv <"•) 
40. vi. Let us now compare equations (1.) and (7.), being the differential values 
of extinction and of refraction. We observe first, that on the hypothesis of a uni- 
form temperature throughout the atmosphere, the logarithmic law which connects 
density with height gives us the relation 
— dg : § = dr : l, 
the logarithmic subtangent or height of the equiponderant column. Hence 
gdr = — Idg (8.) 
Substituting, equation (1.) becomes 
~ — d. logs = Qldg secy' (9.) 
Eq. (7 .) gives for the element of refraction 
. . kdq _ , 
rfl) =- 2(l + *;) tanK; 
dividing this by the former, 
dQ k . . 
QJ(TTh) s ' av ( 10 -) 
41. vii. It remains to determine the value of sin«/. 
By eq. (3.) dv' = dd — dv (11.) 
By fig. 4. A'F = rdv | , 
A' F = dr . tan v' f 
Equating the two last, 
(l 7* 
dv = — tan v' (13.) 
Substituting in eq. (11.) from (7.) and (13.) 
dv 1 f kdq dr 
tan v 1 ^ 2 ( 1 + A g) "t" r J ’ 
— cos v' dd dr kdq 
sinv' r ' 2 (1 + k q)' 
