==> \(P \propto {T^4}\) Also energy received per sec (p)\( \propto \frac{1}{{{d^2}}}\) (inverse square law)
==> \(P \propto \frac{{{T^4}}}{{{d^2}}}\)
==> \(\frac{{{P_1}}}{{{P_2}}} = {\left( {\frac{{{T_1}}}{{{T_2}}}} \right)^4} \times {\left( {\frac{{{d_2}}}{{{d_1}}}} \right)^2}\)
==> \(\frac{P}{{{P_2}}} = {\left( {\frac{T}{{2T}}} \right)^2} \times {\left( {\frac{{2d}}{d}} \right)^2} = \frac{1}{4}\) ==>\({P_2} = 4P.\)