# I [Solved] Force due to radiation pressure in oblique incidence

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1. Jun 7, 2016

### Uchida

The radiation pressure for a perfect mirror is given by:

P = 2 Ef/c * cos²(θ)

Where Ef is the energy flux per area (Power/A), and θ is the angle of incidence

Since force due to pressure is F = PA, force due to radiation pressure become:

F = 2 Power/c * cos²(θ) ??

The problem is with the cos²(θ), I'm unable to prove this statement from a fellow:
"When converting radiation pressure formula to force (such as the one from Wikipedia), One will find out that one cos() is lost because of projection."

Which should give
F = 2 Power/c * cos(θ)

How can I prove this? Or is the statement false?

------------------------- EDIT --------------------------

I've found the solution:

For a oblique incidence, the area of incidence on the surface is A/cos(θ), due to projection of the beam cross section area.
from Ef = Power/Area, with the cross section area of the beam A, and F = Pressure x Area, with projected area A/cos(θ), and we have

F = 2 (Power/A)/c * cos²(θ) * A/cos(θ) = 2 Power/c * cos(θ)

This topic can be closed.

Last edited: Jun 7, 2016