The incident intensity on a horizontal surface at sea level from the sun is about `1kW m^(-2)`.
Q. Find the ratio of this pressure to atmospheric pressure `p_0` (about `1xx10^(5)` Pa) at sea level

Options

A. `5xx10^(-11)`
B. `4xx10^(-8)`
C. `6xx10^(-12)`
D. `8xx10^(-11)`

Answer

A

Explanation

Pressure exerted by absorbed light `=(1)/(2)((S)/(c ))`
Pressure exerted by reflected light `=(1)/(2)((2S)/(c ))`
Total radiation pressure on the surface is
`P_(rad)=((3)/(2)S)/(c)=(1.5xx10^(3))/(3xx10^(8))=5xx10^(-6)` Pa
`(P_(rad))/(P_0)=(5xx10^(-6))/(1xx10^(5))=5xx10^(-11)`