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]

The correct answer is (Cos(θ))^2*Pr[p]

I'm really lost. I don't know how the cos function gets squared. Please help. Thanks in advance, MrMoose

 

[NascentOxygen]

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

 

[haruspex]

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

The area of the plate is fixed. As the beam is tilted away from the normal, the cross sectional area of the beam that hits the plate reduces. So the intensity of the beam on the plate reduces. At the same time, the radiation pressure per unit of intensity reduces because the momentum is not completely reversed. Both reductions are as cos theta.

 

[Nickie]

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Hello MrMoose,

Your approach to the problem is correct, but there is a small mistake in your final answer. The correct expression for radiation pressure at angle θ given perpendicular pressure is Pr[θ] = (Cos(θ))^2 * Pr[p].

To understand why the cosine function is squared, let's look at the derivation of the expression. The total radiation pressure Pr[p] is given by the equation Pr[p] = 2*I/c, where I is the intensity of the laser beam and c is the speed of light. This pressure is exerted perpendicular to the surface.

Now, when the beam is reflected at an angle θ, the pressure exerted on the surface will be less than Pr[p] because the beam is not perpendicular to the surface anymore. To find the pressure at this angle, we need to consider the component of Pr[p] that is perpendicular to the surface, which is given by Pr[p] * Cos(θ). However, this only gives us the pressure in the y-direction (as shown in your diagram), so we need to multiply by another factor of Cos(θ) to get the total pressure at angle θ, which is given by Pr[θ] = Pr[p] * (Cos(θ))^2.

I hope this helps clarify why the cosine function is squared in the final expression. Keep up the good work in your studies!