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  • The incident intensity on a horizontal surface at sea level from the sun is about <img alt="1 \mathrm{kWm}^{-2}" src="https://entrancecorner.oncodecogs.com/gif.latex?1%20%5Cmathrm%7BkWm%7D%5E%7B-2%

The incident intensity on a horizontal surface at sea level from the sun is about 1 \mathrm{kWm}^{-2}. Assuming that 50 percent of this intensity is reflected and 50 per cent is absorbed, determine the radiation pressure on this horizontal surface (in pascals).

Option: 1

8.2 \times 10^{-2}


Option: 2

5 \times 10^{-6}


Option: 3

3 \times 10^{-5}


Option: 4

6 \times 10^{-5}


Answers (1)

best_answer

Given: \mathrm{I}=1000 \mathrm{Wm}^{-2}


As 50 percent of light is reflected, thus \mathrm{e}=0.5
Radiation pressure, \mathrm{P}= (1+e) I\mathrm{c}
\therefore \mathrm{P}=\frac{(1+0.5) \times 1000}{3 \times 10^8}=5 \times 10^{-6} \mathrm{~Pa}

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