# Determine Radiation pressure at angle given Perpendicular Pressure

MrMoose

## Homework Statement

A laser beam of intensity I reflects from a flat, totally reflecting surface of area A whose normal makes an angle θ with the direction of the beam. Write an expression for the radiation pressure Pr[θ] exerted on the surface, in terms of the pressure Pr[p] that would be exerted if the beam were perpendicular to the surface.

## Homework Equations

Radiation Pressure equation for total reflection back along the original path.

Pr[p] = 2*I/c

## The Attempt at a Solution

See attached picture.

Assuming F[θ] = F[p] = I*A / c

Use geometry to find F[θ]y

F[θ]y = Cos(θ) * F[θ]

Divide the equation by A to find pressure:

Pr[θ]y = Cos(θ) * Pr[θ]

Assuming Pr[θ] = Pr[p]

Pr[θ]y = Cos(θ) * Pr[p]

Since the beam is totally reflected, multiply by 2,

Pr[θ]y = 2 * Cos(θ) * Pr[p]

#### Attachments

• q43.jpg
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Staff Emeritus
Because the plate is angled away from the light, it intercepts fewer light rays than when it is head-on.

MrMoose
Thanks Nascent Oxygen, I'm having a lot of trouble visualizing this problem and radiation pressure in general. I know that Pr for a beam that is totally reflected along the original path is just dependent on the Intensity and speed of light:

Pr = 2*I / c

It's twice what it would be for a beam that is totally absorbed.

I'm having trouble understanding your statement, "Because the plate is angled away from the light, it intercepts fewer light rays than when it is head-on."

Mathematically, does that mean that the area intercepted by the beam becomes longer as theta increases, and as a result intensity is less (since I = Power/A)? Sorry, I'm still trying to understand the concept. Thanks