Two bodies $A$and $B$ have thermal emissivities of $0.01$ and $0.81$ respectively. The outer surface areas of the two bodies are the same. The two bodies emit total radiant power at the same rate. The wavelength ${\lambda _B}$ corresponding to maximum spectral radiancy in the radiation from $B$ is shifted from the wavelength corresponding to maximum spectral radiancy in the radiation from $A$, by $1.00\mu m$. If the temperature of $A$ is $5802\;K$
IIT 1994, Diffcult
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(d) According to Stefan’s law
$E = eA\sigma {T^4} \Rightarrow {E_1} = {e_1}A\sigma T_1^4$ and ${E_2} = {e_2}A\sigma T_2^4$
${E_1} = {E_2}$ $\therefore $${e_1}T_1^4 = {e_2}T_2^4$
$ \Rightarrow $${T_2} = {\left( {\frac{{{e_1}}}{{{e_2}}}T_1^4} \right)^{\frac{1}{4}}} = {\left( {\frac{1}{{81}} \times {{(5802)}^4}} \right)^{\frac{1}{4}}}$$ \Rightarrow $${T_B} = 1934\;K$
And, from Wein’s law ${\lambda _A} \times {T_A} = {\lambda _B} \times {T_B}$
$ \Rightarrow \frac{{{\lambda _A}}}{{{\lambda _B}}} = \frac{{{T_B}}}{{{T_A}}}$$ \Rightarrow $$\frac{{{\lambda _B} - {\lambda _A}}}{{{\lambda _B}}} = \frac{{{T_A} - {T_B}}}{{{T_A}}}$
$ \Rightarrow $$\frac{1}{{{\lambda _B}}} = \frac{{5802 - 1934}}{{5802}} = \frac{{3968}}{{5802}} \Rightarrow {\lambda _B} = 1.5\;\mu m$
$E = eA\sigma {T^4} \Rightarrow {E_1} = {e_1}A\sigma T_1^4$ and ${E_2} = {e_2}A\sigma T_2^4$
${E_1} = {E_2}$ $\therefore $${e_1}T_1^4 = {e_2}T_2^4$
$ \Rightarrow $${T_2} = {\left( {\frac{{{e_1}}}{{{e_2}}}T_1^4} \right)^{\frac{1}{4}}} = {\left( {\frac{1}{{81}} \times {{(5802)}^4}} \right)^{\frac{1}{4}}}$$ \Rightarrow $${T_B} = 1934\;K$
And, from Wein’s law ${\lambda _A} \times {T_A} = {\lambda _B} \times {T_B}$
$ \Rightarrow \frac{{{\lambda _A}}}{{{\lambda _B}}} = \frac{{{T_B}}}{{{T_A}}}$$ \Rightarrow $$\frac{{{\lambda _B} - {\lambda _A}}}{{{\lambda _B}}} = \frac{{{T_A} - {T_B}}}{{{T_A}}}$
$ \Rightarrow $$\frac{1}{{{\lambda _B}}} = \frac{{5802 - 1934}}{{5802}} = \frac{{3968}}{{5802}} \Rightarrow {\lambda _B} = 1.5\;\mu m$
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