Force due to radiation pressure in oblique incidence

[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
(from: https://en.wikipedia.org/wiki/Radiation_pressure)

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.

 

[eljose79]

Great job on finding the solution! You have correctly proved that the statement is indeed false. Here is a more detailed explanation:

The formula for radiation pressure, P = 2 Ef/c * cos²(θ), takes into account the energy flux per unit area (Ef) and the angle of incidence (θ). This means that the pressure is dependent on both the intensity of the radiation (Power/A) and the angle at which it is incident on the surface.

When we convert this formula to force, we use the equation F = PA, where P is the pressure and A is the area. However, in this case, the area is not simply the surface area of the mirror, but rather the projected area of the incident radiation on the surface. This is because not all of the incident radiation is absorbed by the surface, as some of it may be reflected or transmitted. Therefore, the actual area on which the pressure is acting is reduced by a factor of cos(θ), which is the projection of the incident radiation's cross-sectional area onto the surface.

So, the correct formula for force due to radiation pressure is F = 2 Power/c * cos²(θ) * A/cos(θ). Simplifying this, we get F = 2 Power/c * cos(θ). Thus, the statement that one cos() is lost due to projection is false.

In summary, the statement is false because it does not take into account the projection of the incident radiation's cross-sectional area onto the surface. Your solution correctly shows how to account for this in the equation. Great work!