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. `(T_(2))/(T_(1))`
B. `((T_(2))/(T_(1)))^(2)`
C. `((T_(1))/(T_(2)))^(2)`
D. `((T_(2))/(T_(1)))^(4)`

Answer

B

Explanation

We use `I = (p_(1))/(4 pi d_(1)^(2)) =(A sigma T_(1)^(4))/(4 pi d_(1)^(2))` …(1)
`rArr I=(p_(2))/(4 pi d_(2)^(2)) = (A sigmaT_(2)^(4))/(4 pi d_(2)^(2))` …(2)
From (1)& (2), we get
`(T_(1)^(4))/(d_(1)^(2)) =(T_(2)^(4))/(d_(2)^(2))`
`((d_(2))/(d_(1)))^(2) =((T_(2))/(T_(1)))^(4)`
`(d_(2)) /(d_(1)) =((T_(2))/(T_(1)))^(2)`