MR B. STEWART ON RADIANT HEAT. 17 
27. 2d Proposition. 
1st Case.—If AB represent a surface (the substance being below AB), and CF a 
surface of lamp-black indefinitely extended (as in Art. 25), 
from which rays fall on AB through a small angle CAD; = a 
then, if AE be drawn perpendicular to GB, the heat that peo RRO 
falls on AB will = a constant x AE, whatever be the 
value of the angle CAB. 
For, since the angle CAD is exceedingly small, CD may 
be considered very small in comparison with CF or CG; therefore the heat which 
impinges on AB through the angle CAD may be taken to be that which radiates 
from OG in directions between CA and DA; but, since the radiative power of lamp- 
black in any direction varies as the sine of the angle which that direction makes 
with the surface, this will = const.x AE. Hence, if Rx CAD be the quantity 
of heat which falls on AB, when AB is perpendicular to GB, that which falls on 
it when GB makes any angle GBA with AB, will be Rx CAD sinGBA. 
If 2 denote the angle which GB makes with the perpendicular to AB, then the 
heat impinging on AB will be Rcos7 x CAD. 
2d Case.—If the substance be above AB, and the rays falling on AB originate 
in the substance, the same formula will hold; for it has been shown, in Prop. 1st, 
that in this case, the heat falling on AB through the small angle CAD = that 
which falls on AE through the same small angle; but, since the radiation from 
the interior of the substance is the same in all directions (each particle radiating 
independently and equally in all directions), the amount falling on AE will not 
be affected by the angle which AE makes with the surface; hence the heat fall- 
ing on AB=const. x AE=const. x sinGBA. 
If R’ x CAD=quantity which falls on AB when AB is perpendicular to GB, that 
which falls on it when GB makes any angle GBA with AB, will be R’ x CAD sin 
GBA; also the expression corresponding to R.coszx CAD will be R’ cosz’ x CAD. 
28. 3d Proposition. 
Let a ray strike the surface of a medium, at an angle of incidence =i; and 
another ray at an angle of incidence i+ 0i, it is required to find the difference 
between the two angles of refraction. 
Let pu be the index of refraction, then, 

sini=p sin? 
Hence, ) (sin 7)= po (sin 2’) 
cos i Oi= ficos VOoVv 
; COS2 9g. 
Hence, ov = Tae oi 

29. I shall also make the following supposition with regard to the laws of re- 
flection and refraction. 
VOL. XXII. PART I. E 
