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Two electric lamps $A$ and $B$ radiate the same power. Their filaments have the same dimensions and have emissivities $e_A$ and $e_B$. Their surface temperatures are $T_A$ and $T_B$. The ratio $\frac{{{T_A}}}{{{T_B}}}$ will be equal to :-
${\left( {\frac{{{e_B}}}{{{e_A}}}} \right)^{\frac{1}{4}}}$
${\left( {\frac{{{e_B}}}{{{e_A}}}} \right)^{\frac{1}{2}}}$
${\left( {\frac{{{e_A}}}{{{e_B}}}} \right)^{\frac{1}{2}}}$
${\left( {\frac{{{e_A}}}{{{e_B}}}} \right)^{\frac{1}{4}}}$
Solution
The power radiated by a filament is $\mathrm{P}=\mathrm{e}\left(\sigma \mathrm{T}^{4}\right)$
(area), where $\mathrm{e}=$ emissivity
$\sigma=$ Stefan's constant,
$\mathrm{T}=$ surface temperature.
Here, $\mathrm{eT}^{4}=$ constant or $\mathrm{e}_{\mathrm{A}} \mathrm{T}_{\mathrm{A}}^{4}=\mathrm{e}_{\mathrm{B}} \mathrm{T}_{\mathrm{B}}^{4}$
