The spectral emissive power $E_\lambda $ for a body at temperature $T_1$ is plotted against the wavelength and area under the curve is found to be $A$. At a different temperature $T_2$ the area is found to be $9A$. Then $\lambda _1/\lambda _2 =$
Diffcult
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The spectral emissive power is directly proportional to the fourth power of the absolute temperature.
$E_{\lambda}=k T^{4}$
$E_{1}=k T_{1}^{4}=A$
$E_{2}=k T_{2}^{4}=9 A$
$\frac{E_{2}}{E_{1}}=\left(\frac{T_{2}}{T_{1}}\right)^{4}=9$
$\frac{T_{2}}{T_{1}}=\sqrt{3}$
$T_{2}=\sqrt{3} T_{1}$
According to Wien's displacement law.
$\lambda T=b=$ constant
$\lambda_{1} T_{1}=\lambda_{2} T_{2}$
$\frac{\lambda_{1}}{\lambda_{2}}=\frac{T_{2}}{T_{1}}=\sqrt{3}$
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