Radiation from a black body at the thermodynamic temperature T1 is measured by a small detector at distance d1 from it. When the temperature is increased to T, and the distance to d2, the power received by the detector is unchanged. What is the ratio d2 / d1 ?

Question

Radiation from a black body at the thermodynamic temperature T1 is measured by a small detector at distance d1 from it. When the temperature is increased to T, and the distance to d2, the power received by the detector is unchanged. What is the ratio d2 / d1 ? (A) (T2/T1) (B) ((T2/T1))2 (C) ((T1/T2))2 (D) ((T2/T1))4

Tardigrade Admin · Accepted Answer

Power received propto((E/d2)) propto (T4/d2) Here (T24/d22)=(T14/d12) ⇒ ((d2/d1))2=((T2/T1))4 or (d2/d1)=((T2/T1))2

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 $T,$ and the distance to $d_{2}$ , the power received by the detector is unchanged. What is the ratio $d_{2} / d_{1} ?$

$\frac{T _{2}}{T _{1}}$

$(\frac{T _{2}}{T _{1}})^{2}$

$(\frac{T _{1}}{T _{2}})^{2}$

$(\frac{T _{2}}{T _{1}})^{4}$

Power received $\propto (\frac{E}{d ^{2}}) \propto \frac{T ^{4}}{d ^{2}}$
Here $\frac{T _{2}^{4}}{d _{2}^{2}} = \frac{T _{1}^{4}}{d _{1}^{2}}$
$\Rightarrow (\frac{d _{2}}{d _{1}})^{2} = (\frac{T _{2}}{T _{1}})^{4}$ or $\frac{d _{2}}{d _{1}} = (\frac{T _{2}}{T _{1}})^{2}$

Q. Radiation from a black body at the thermodynamic temperature T1​ is measured by a small detector at distance d1​ from it. When the temperature is increased to T, and the distance to d2​, the power received by the detector is unchanged. What is the ratio d2​/d1​?

T1​T2​​

(T1​T2​​)2

(T2​T1​​)2

(T1​T2​​)4