Question

If temperature of sun is $T=6000K$, radius of sun is $R=7.2\times {{10}^{5}}km$, radius of Earth ${{R}_{e}}=6000km$ and distance between earth and sun $d=15\times {{10}^{7}}km$. Find the intensity of the light on Earth.
$\begin{aligned} & A.19.2\times {{10}^{16}} \\\ & B.12.2\times {{10}^{16}} \\\ & C.18.3\times {{10}^{16}} \\\ & D.9.2\times {{10}^{16}} \\\ \end{aligned}$

Answer 1

First of all find out the energy emitted by the sun. The total energy emitted by the sun is given as,
${{E}_{S}}=\sigma {{T}^{4}}\times 4\pi {{R}^{2}}$
Then find out the energy received by the earth. The total energy received by earth is given as,
${{E}_{e}}=\dfrac{\sigma {{T}^{4}}\times 4\pi {{R}^{2}}}{4\pi {{d}^{2}}}\times \pi {{R}_{e}}^{2}$
The intensity of light received on earth is obtained by substituting the values in it. In this way, we can find out the answer.

Complete step by step answer
First of all, let us look at the quantities mentioned in the question.
The temperature of the sun is given as,
$T=6000K$
The radius of the sun is mentioned as,
$R=7.2\times {{10}^{5}}km$
The radius of the earth can be written as,
${{R}_{e}}=6000km$
At last, the distance between the sun and the earth is given as,
$d=15\times {{10}^{7}}km$
Using these parameters, we have to find the intensity of light on the earth.
For that, first of all we have to calculate the energy emitted by the sun,
Therefore the total energy radiated by the sun is given as,
${{E}_{S}}=\sigma {{T}^{4}}\times 4\pi {{R}^{2}}$
Then the energy received by the earth can be written as,
${{E}_{e}}=\dfrac{\sigma {{T}^{4}}\times 4\pi {{R}^{2}}}{4\pi {{d}^{2}}}\times \pi {{R}_{e}}^{2}$
Simplifying this equation will give,
${{E}_{e}}=\dfrac{\sigma {{T}^{4}}\times {{R}^{2}}}{{{d}^{2}}}\times \pi {{R}_{e}}^{2}$
As we all know the value of the constant,
$\sigma =5.67\times {{10}^{-8}}$
Substituting the values in it will make the equation as,
${{E}_{e}}=\dfrac{5.67\times {{10}^{-8}}\times {{\left( 6000 \right)}^{4}}\times 4\pi {{\left( 7.2\times {{10}^{5}}km \right)}^{2}}}{4\pi {{\left( 15\times {{10}^{7}}km \right)}^{2}}}\times \pi {{\left( 6000km \right)}^{2}}$
Simplifying this equation will give the intensity of the light on earth,
$I=19.2\times {{10}^{16}}$
Therefore the correct answer is option A.

Note: We can see that as the distance is increasing, the light is getting spread out over a huge surface and the brightness of the surface decreases according to a one-over r squared relationship. This reduction goes as r squared since the area up to which the light is spreading out is proportional to the square of the distance.