Question

The sun radiates electromagnetic energy at the rate of \[3.9 \times {10^{26}}{\text{ W}}\]. Its radius is \[6.96 \times {10^8}{\text{ m}}\]. The intensity of sunlight at the solar surface will be in \[\left( {\dfrac{W}{{{m^2}}}} \right)\]
A. \[1.4 \times {10^7}\]
B. \[2.8 \times {10^5}\]
C. \[64 \times {10^6}\]
D. \[5.6 \times {10^7}\]

Answer 1

Hint: In this question, we are given the power emitted by the sun and its radius and we need to find out the intensity of light at the solar surface. Assuming constant power is emitted by the sun by dividing the power of sunlight emitted by sun with its surface area.

Formula used:
\[I = \dfrac{P}{A}\]
Where \[P\] is the power emitted by the source
\[A\] is the area perpendicular to the Intensity

Complete step by step answer:
We have given,
Average Power of the electromagnetic radiation emitted by the sun, \[P = 3.9 \times {10^{26}}{\text{ W}}\]
The Radius of the sun, \[{\text{R = }}6.96 \times {10^8}{\text{ m}}\]
The Surface area of the sun, \[{\text{A = 4}}\pi {{\text{R}}^2} = 4\pi {\left( {6.96 \times {{10}^8}} \right)^2}{\text{ }}{{\text{m}}^2}\]
\[ \Rightarrow A = 4\pi \times 48.44 \times {10^{16}}\]
\[ \Rightarrow A = 60.9 \times {10^{17}}{\text{ }}{{\text{m}}^2}\]
Now, as we know intensity is the average power of a wave per unit area
Assuming the sun emits constant power from a single point in the core, we can use the formula \[I = \dfrac{P}{A}\] to find out the intensity
Thus, \[I = \dfrac{{{\text{Power emitted by sun}}}}{{{\text{Surface Area of sun}}}}\]
\[ \Rightarrow I = \dfrac{{3.9 \times {{10}^{26}}}}{{60.9 \times {{10}^{17}}}}\dfrac{W}{{{m^2}}}\]
On solving we will get,
\[ \Rightarrow I = 6.4 \times {10^7}{\text{ W/}}{{\text{m}}^2}\]
\[ \Rightarrow I = 64 \times {10^6}{\text{ W/}}{{\text{m}}^2}\]
Thus, Intensity of the sunlight at the solar surface is \[64 \times {10^6}{\text{ W/}}{{\text{m}}^2}\]
Therefore, the correct option is option C.

Additional information:
Intensity is a property of source and is independent of area, but it is dependent on distance from area.
For a point source, Intensity is inversely proportional to the square of the distance from the source
For line source or cylindrical source, Intensity is inversely proportional to the distance from the source
For planar source, Intensity is independent of the distance from the source

Note: In this question, we assume the sun as a point source emitting sunlight from the center of its core and the distance of the solar surface from the center of the sun is its radius. Intensity is a property of source and depends on strength and amplitude of a wave.