Radiation from a black body at the thermodynamic temperature T_1 is measured by a small detector at distance d

Q: 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 $d_2/d_1$ ?

b We use $I_{1}=\frac{P_{1}}{4 \pi d_{1}^{2}}=\frac{A \sigma T_{1}^{4}}{4 \pi d_{1}^{2}}$ $I_{2}=\frac{P_{2}}{4 \pi d_{2}^{2}}=\frac{A \sigma T_{2}^{4}}{4 \pi d_{2}^{2}}$ From $( 1)$ and $(2),$ We get $\frac{T_{l}^{4}}{d_{1}^{2}}=\frac{T_{l}^{4}}{d_{2}^{2}}$ $\left(\frac{\mathrm{d}_{2}}{\mathrm{d}_{1}}\right)^{2}=\left(\frac{\mathrm{T}_{2}}{\mathrm{T}_{1}}\right)^{4}$ $\frac{{{d_2}}}{{{d_1}}} = {\left( {\frac{{{T_2}}}{{{T_1}}}} \right)^2}$

SECTION - A [PHYSICS - MCQ]

GSEB - English Medium

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 $d_2/d_1$ ?

A$\frac{{{T_2}}}{{{T_1}}}$
B${\left( {\frac{{{T_2}}}{{{T_1}}}} \right)^2}$
C${\left( {\frac{{{T_1}}}{{{T_2}}}} \right)^2}$
D${\left( {\frac{{{T_2}}}{{{T_1}}}} \right)^4}$

Medium

STD 11 Science
NEET
STD 11 - 10.2. transmission of heat
Physics

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