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)`?

Options

A. (a)`(T_(2))/(T_(1))`
B. (b)`(T_(2))/(T_(1)^(2))`
C. (c)`((T_(1))/(T_(2)))^(2)`
D. (d)`((T_(2))/(T_(1)))^(2)`

Answer

B

Explanation

Intensity in first case
`I_(1) = (P_(1))/(4piR_(1)^(2)) = (sigmaAT_(1)^(4))/(4pid_(1)^(2))`
Intensity in second case
`I_(2) = (P_(2))/(4piR_(2)^(2)) = (sigmaAT_(2)^(4))/(4pid_(2)^(2))`
Given `I_(1)= I_(2) implies (sigmaAT_(1)^(4))/(4pid_(1)^(2)) = (sigmaAT_(2)^(4))/(4pid_(2)^(2)) implies (d_(2))/(d_(1)) = ((T_(2))/(T_(1)))^(2)`