Question

If temperature of Sun $=6000$K, radius of Sun is $7.2\times {10}^5$ Km, radius of Earth $=6000$km & distance between earth and Sun $=15\times {10}^7$ Km. Find intensity of light on Earth.

• A. $19.2\times {10}^{16}$
• B. $12.2\times {10}^{16}$
• C. $18.3\times {10}^{16}$
• D. $9.2\times {10}^{16}$

Answer 1

Answer: (A)

$19.2\times {10}^{16}$

Given that,

Temperature of sun$T=6000K$

Radius of sun $R=7.2\times {10}^{5}Km$

Radius of earth $E=6000Km$

The distance between earth and sun $d=15\times {10}^{7}Km$

We know that,

Total the energy emitted by Sun =$\sigma T^4\times 4\pi R^2$

Energy received by earth =$\frac{\sigma T^4\times 4\pi R_s^2}{4\pi d^2}\times \pi R_e^2$

=$\frac{\sigma T^4.R_s^2}{d^2}\times \pi R_e^2$

=$\frac{5.67\times {10}^{-8}\times (6000{)}^4\times (7.2\times {10}^8)2\times 3.14\times (6000\times {10}^3)}{(1.5\times {10}^{1}1{)}^2}$

=$\frac{5.67\times 36\times 36\times 7.2\times 7.2\times 36\times 3.14}{2.25}\times {10}^{1}0$

=$19.2\times {10}^{16}$

So, The intensity of the light on the earth is $19.2\times {10}^{16}$.

Hence, Option A is correct.