Light with an enargy flux of `25xx10^(4) Wm^(-2)` falls on a perfectly reflecting surface at normal incidence. If the surface area is `15 cm^(2)`, the average force exerted on the surface is

Options

A. `3.0 xx 10^(-6) N`
B. `1.25 xx 10^(-6)N`
C. `2.50 xx 10^(-6)N`
D. `1.20 xx 10^(-6)N`

Answer

3

Explanation

Radiation pressure = `(2I)/(C )`
`(I to "Intensity", C to "speed of light")`
so average exerted
`F = ((2I)/(C))xxA = (2xx25xx10^(4))/(3xx10^(8))xx15xx10^(-4)`
`F_(av) = 2.50 xx10^(-6) "Newton"`