784 Prof. J. H. Jeans on the 



between the two media these expressions must be equal. 

 Now we have, from equation (17), 



sin 1 sin 6 2 sin 6 2 



and hence 



cos 6 1 sin 0, d$i cos 6* sin 6 2 d6 2 



~r? ~" ~v? • 



Since R 12 is the same function of 6 1 as is E 2 i o£ 2 , it 

 follows that 



yj I (1 — R I2 )cos^ 1 sin^ 1 ^ l = ^- 2 i (1 — R 21 )cos0 2 sin0 2 d0 2 , 



.... (23) 

 so that expressions (21) and (22) will be equal if 



EjV^E^ 3 (24) 



This is a known result of which we have now obtained 

 a proof by electrodynamical, without thermodynamical, 

 principles. 



16. The energy in a cavity in a body of any kind being 

 of the form 



E,=/(T,/0. (25) 



it follows that the energy in the interior of the body itself 

 must be of the form 



E 3 =(^) 3 /(T );J ), (26) 



so that the partition of energy inside matter can be deter- 

 mined in terms of black-body radiation. Using this value 

 for E 2 , the total emission per unit area per unit time from 

 the surface of the body, by equation (22), 



= dpf (T, p) cy^-2 \ (1 — R 21 ) cos 6 2 sin 6 2 d$ 2 



= dpf(T, p) iVj ( (1-R ]2 ) cos 6 l sin <V0 l5 



by equation (23). Since 6 1 is the angle the issuing radiation 

 makes with the normal, the limits for ± are from to 



^ . The coefficient of absorption A for light incident on the 



