1.

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

Answer»

<P>`8.2xx10^(-2)`
`5xx10^(-6)`
`6XX10^(-12)`
`8xx10^(-11)`

Solution :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)`


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