Q. If temperature of Sun = 6000K, radius of Sun is 7.2 × 105 Km, radius of Earth = 6000 Km & distance between earth and Sun = 15 × 107 Km. Find intensity of light on Earth.
AIIMSAIIMS 2019
Solution:
Total energy emitted by Sun = $\sigma$T4 × 4$\pi$R2
Energy received by earth $=\frac{\sigma T^{4}4\pi R^{^2}_{S}}{4\pi d^{2}}\times\pi\left(Re\right)^{2}$
$=\frac{\sigma T^{4}R^{^2}_{_s}}{d^{2}}\times\pi\left(Re\right)^{2}$
$=\frac{5.67\times10^{-8}\times\left(6000\right)^{4}\times\left(7.2\times10^{8}\right)^{2}\times3.14\times\left(6000\times10^{3}\right)}{\left(1.5\times10^{11}\right)^{2}}$
$=\frac{5.67\times36\times36\times7.2\times7.2\times36\times3.14}{2.25}\times10^{10}=19.2\times10^{16}$
