Radiation from a black body at the thermodyna | vedclass

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(B) According to the Stefan-Boltzmann law,the total power $P$ radiated by a black body of surface area $A$ at temperature $T$ is given by $P = A \sigm

Vedclass · Accepted Answer

$(\frac{T_2}{T_1})^2$

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(B) According to the Stefan-Boltzmann law,the total power $P$ radiated by a black body of surface area $A$ at temperature $T$ is given by $P = A \sigma T^4$.The power received by a small detector at distance $d$ is proportional to the intensity $I$,which follows the inverse square law: $I = \frac{P}{4 \pi d^2} = \frac{A \sigma T^4}{4 \pi d^2}$.Given that the power received by the detector remains unchanged when the temperature changes from $T_1$ to $T_2$ and the distance changes from $d_1$ to $d_2$,we have:$\frac{A \sigma T_1^4}{4 \pi d_1^2} = \frac{A \sigma T_2^4}{4 \pi d_2^2}$Canceling the common constants $A, \sigma,$ and $4 \pi$,we get:$\frac{T_1^4}{d_1^2} = \frac{T_2^4}{d_2^2}$Rearranging the terms to find the ratio $d_2/d_1$:$(\frac{d_2}{d_1})^2 = (\frac{T_2}{T_1})^4$Taking the square root of both sides:$\frac{d_2}{d_1} = (\frac{T_2}{T_1})^2$

Radiation from a black body at the thermodynamic temperature $T_1$ is measured by a small detector at distance $d_1$ from it. When the temperature is increased to $T_2$ and the distance to $d_2$,the power received by the detector is unchanged. What is the ratio $d_2/d_1$?

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