JEE Main & Advanced Physics Transmission of Heat Kirchoff's Law

According to this law the ratio of emissive power to absorptive power is same for all surfaces at the same temperature and is equal to the emissive power of a perfectly black body at that temperature. Hence \[\frac{{{e}_{1}}}{{{a}_{1}}}=\frac{{{e}_{2}}}{{{a}_{2}}}=...\,\,{{\left( \frac{E}{A} \right)}_{\text{Perfectly black body}}}\]

But for perfectly black body \[A=1\] i.e. \[\frac{e}{a}=E\]

If emissive and absorptive powers are considered for a particular wavelength l, \[\left( \frac{{{e}_{\lambda }}}{{{a}_{\lambda }}} \right)={{({{E}_{\lambda }})}_{\text{black}}}\]

Now since \[{{({{E}_{\lambda }})}_{black}}\] is constant at a given temperature, according to this law if a surface is a good absorber of a particular wavelength it is also a good emitter of that wavelength.

This in turn implies that a good absorber is a good emitter (or radiator)