Two bodies A and B have thermal emissivities of 0.01 and 0.81 respectively. The outer surface areas of the two bodies are same. The two bodies emit total radiant power at the same rate. The wavelength `lambda_B` corresponding to maximum spectral radiancy from B is shifted from the wavelength corresponding to maximum spectral radiancy in the radiation from A by 1.0 `mum`. If the temperature of A is 5802 K, calculate (a) the temperature of B, (b) wavelength `lambda_B`.

Answer

`[1934 K, 1.5 mu m]`

Explanation

Given, `P_(1) =P_(2)` , So `A sigma epsilon_(1) T_(1)^(4)=A sigma epsilon_(2) T_(2)^(4)`
or `T_(2)=T_(1) ((epsilon_(1))/(epsilon_(2)))^(1//4)`
`=5802 ((0.01)/(0.81))^(1//4) =5802xx(1)/(3) =1934 K`
or `1-(1)/(lambda_(2)) =(T_(2))/(T_(1))= (1934)/(5802)`
or `(1)/(lambda_(2))= 1- (1934)/(5802)= (3868)/(5802)`
or `lambda_(2)= (5802)/(3868)= 1.5 mu m`