Radiation pressure?

Question

A laser pointer directs a narrow beam of light that can be used by a speaker to illuminate locations on a projected image during a lecture. If a 3.7 mW laser pointer illuminates a circular area with a radius of 1.2 mm on a projection screen with an albedo of 0.86, find the pressure the laser exerts on the illuminated portion of the screen (in N/m2).

Answer 1

The energy of a photon is related to its momentum by:

E = pc

Take the time derivative of both sides and you see that the power delivered is going to be related to the force by:

P = Fc

So if all the photons landed on the screen and stuck, the pressure would be:

pressure = F / area = P / (c*area)

But the photons don't land and stick.  A percentage of them (alpha, the albedo) bounce off and deliver more momentum and hence more force.  And they don't just bounce straight back.  They go in all directions.  The perpendicular component of their speed/momentum (which is what we care about) is proportional to the cosine of the angle they bounce at.  So if they bounce at all angles, the average value of cos(theta) is 1/2*.

So if ALL the photons bounced straight back, they'd deliver double momentum, and the pressure would be:

pressure = F / area =

2 P / (c*area)

If only a fraction alpha of the photons bounced straight back, the pressure would be:

pressure = F / area =

(1 + alpha) P / (c*area)

But since those reflected photons bounce all over and the average recoil momentum is half the original momentum, the answer is:

pressure = F / area

= (1 + alpha/2) P / (c*area)

= (1 + alpha/2) P / (c pi r^2)

and the rest is plugnchug.

*EDIT--After reading Jim's soln, realized that my average value of cos (theta) was off because I was doing the integral in one-D.  The correct answer is:

Avg = integral of cos(theta) d(cos(theta)) from cos(theta) = 1 to zero

/ integral of d(cos(theta)) over the same range

= 1/2, same as he gets

So replaced my factor of 2/pi (the average of cos over a semi-circle) by 1/2 (the average over a sphere).

Answer 2

You can use conservation of momentum in this problem.

The momentum of radiation is energy/c. When the 3.8 mW beam hits the screen, a fraction of 0.86 is reflected back, the rest is absorbed (we can assume that the absorbed radiation heats the screen and that the heat partially in the form of infrared radiation is radiated away in random directions).

Conservation of momentum implies that the change in momentum per unit time of the photons in the laser beam must be compensated for by the screen plus rest of the world (the screen is fixed). The absorbed momentum by the screen per unit time is the force exerted on the screen by the photons. If you divide this by the area of the illuminated point you get the pressure.

Let's first compute the change in momentum of the photons that are reflected back. Let's call the direction toward the screen the negative z direction, and the direction away for the screen the positive z-direction. Then the reflected photons can have any momentum in the positive z-direction.

Note that the screen does not reflect the light back in one direction like a perfect mirror. If it did do that, then the illuminated dot on the screen would not be visible to everyone in the audience.

So, let's assume for simplicity that the reflected photons are scattered back with equal probability per unit solid angle in the positive z-direction. We can then evaluate the average momentum of a reflected photon.

sin(theta) dtheta d phi

is a solid angle element in terms of the spherical coordinates. The z-component of the photon's momentum is cos(theta) P, where P is the magnitude of the momentum of a single reflected photon. The total solid angle the reflected photon can be in is 2 pi. So, the average momentum is:

P/(2 pi) Integral of sin(theta) cos(theta)d theta

d phi from theta = 0 to pi/2 and phi= 0 to 2 pi =

P Integral of sin(theta) cos(theta)d theta

d phi from theta = 0 to pi/2 = P/2

(in the z-direction)

So, the change in momentum is when a photon of momentum P hits the screen and is reflected is 3/2 P in the z-direction. The change in momentum when a photon get's absorbed is P in the z-direction, because the heat that is radiated away is assumed to be isotropic and thus has no momentum.

If we take the power of the laser beam of 3.8 mW and divide this by c, we get the total momentum per unit time in the form of photons in the beam. A fraction 0.86 of this will be reflected back, which means that the change in momentum per unit time due to the photons that are reflected back is:

0.86*3/2* 3.7 mW/c

The rest of the beam is absorbed, the change in momentum of the beam per unit time due to the absorption of photons is:

(1-0.86)* 3.7* mW/c

Total momentum change of photons in the beam per unit time is thus:

0.86*3/2* 3.7 mW/c + 0.14* 3.7 mW/c =

1.76*10^(-11) N

This is in the z-direction. Total momentum is conserved, so the change in momentum per unit time of the screen is equal in magnitude to this but in the opposite direction (the negative z-direction, which is the direction from the laser pointer to the screen)

If we divide this force by the area of the illuminated circle, we find that the pressure is given by 1.56*10^(-5) Pa

Answer 3

radiation pressure = F/A=(change in momentum)/ area =energy density / speed of light

Albedo = 0.86

You can use reflection or absorption; I chose absorption.

86% of 3.7 mW is reflected = 3.096 mW reflected

So 14% is absorbed 14% 0f 3.7 mW = 0.518 mW absorbed

Area of 1.2mm diameter circle = pi(0.6)^2 = 1.131 x 10^-6 m^2

Area=1.131 x 10^-6 m^2

The Intensity of the beam = I

I = Power/Area= 3.7 x 10^-3 W / 1.131 x 10^-6 m^2

I=3.2714412 x 10^3 watts/m^2 =incident intensity

Reflected intensity =3.096 x 10^-3W/1.131 x 10^-6

m^2=2.73740053 watts/m^2

Absorbed intensity =0.518 x 10^-3 W /1.131 x 10^-6

m^2=4.58001768 x 10^-4 watts /m^2

If you know the intensity, I, of the radiation, then it is most convenient to use a formula for the pressure, P, (force per unit area) exerted by the radiation. The radiation pressure depends on whether the radiation is absorbed or reflected. If it is reflected, the object which reflects it must recoil with enough momentum to stop the incoming wave, and then send it back out again, while if it is absorbed, the object must only stop it. Hence, the reflection formulas give answers that are twice the size of those given by the absorption formulas. The radiation pressure formula for absorption is

P(pressure)= I/c (for absorption)

Pressure = (4.58 x 10^-4 watts /m^2)/ (3 x 10^8m/s)=1.53 x 10^-12 Pascals