Question

A source of power 600 kw emits photon which is incident on a surface 2.5m away out of which 50 % reflected back,what is radiation pressure on the surface

Answer 1

P = radiation power coming from the source = 600 kW = 6 x 10⁵ watt

r = distance of the surface from source on which the radiation strikes = 2.5 m

intensity of the radiation at the surface is given as

I = P/(4πr²)

inserting the values

I = ( 6 x 10⁵)/(4π(2.5)²)

I = 7.64 x 10⁴

p = radiation pressure at the surface on which the radiation strike

c = speed of light = 3 x 10⁸ m/s

radiation pressure at the surface is given as

p = I/c

p = (7.64 x 10⁴)/(3 x 10⁸)

p = 2.55 x 10⁻⁴

Answer 2

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