A laser beam is used to levitate a metal disk against the...

A laser beam is used to levitate a metal disk against the force of Earth’s gravity. (a) Derive an equation giving the required intensity of light, I, in terms of the mass m of the disk, the gravitational acceleration g , the speed of light c, and the cross - sectional area of the disk A. Assume the disk is perfectly reflecting and the beam is directed perpendicular to the disk. (b) If the disk has mass 5.00 g and radius 4.00 cm, find the necessary light intensity. (c) Give two reasons why using light pressure as propulsion near Earth’s surface is impractical.

Key Concepts

Practical Limitations of Light Propulsion

The concept of using light pressure as a means of propulsion, although theoretically valid, is impractical near Earth’s surface due to several challenges. These include the necessity of extremely high light intensities, which are difficult and energy-inefficient to generate, and the susceptibility to losses such as atmospheric absorption and scattering. These factors limit the effectiveness of light-based propulsion methods in overcoming gravitational forces in real-world scenarios.

Force Balance / Equilibrium

Force balance or equilibrium refers to the state where the upward force provided by light’s radiation pressure equals the downward gravitational force acting on the object. Establishing this balance is essential for achieving levitation. The derivation of the required light intensity involves setting the radiation force equal to the gravitational force acting on the object.

Radiation Pressure

Radiation pressure refers to the pressure exerted upon any surface due to the exchange of momentum between the photons in the light and the surface. When a beam of light hits a surface, its photons transfer momentum, creating a force. In the case of a perfectly reflecting surface, the momentum change is doubled because the light is reflected rather than absorbed, which is crucial in deriving the necessary conditions for counteracting gravitational force with light.

Light Intensity

Light intensity is a measure of the power per unit area carried by a light beam. In problems involving radiation pressure, the intensity determines how much momentum is delivered to a surface per unit time. It is directly related to the force that the light can exert on an object, making it a vital parameter in balancing forces such as gravitational pull.

Momentum Transfer

Momentum transfer in the context of light involves the exchange of momentum between photons and a reflective surface. The change in momentum, especially in cases where the surface reflects the light, results in a force. This concept is key to understanding how a beam of light can exert enough force to overcome gravitational forces and levitate an object.

Transcript

00:01 To exert an upward force on the disk, this means that the laser beam should be aimed vertically upward, of course, striking the lower surface of the disk.

00:10 Now, the upward force exerted is going to be equal to the weight of the disc.

00:15 This would, of course, prove to be, this would have the system be in equilibrium and would actually levitate the disc.

00:23 And so we can say that from equation 21 .30, we have that the momentum, that electromagnetic radiation of intensity i, incident on a perfectly reflecting surface of area a, delivers to that surface in time delta t.

00:46 And so we can say that this is modeled by equation 21 .30.

00:50 And this would be delta p, equaling to, u divided by c or essentially 2 multiplied by the intensity times the area of the reflecting surface times the change in time divided by c the speed of light.

01:12 And so from the impulse momentum theorem, the average force exerted on the reflecting surface, f would be equaling delta p over delta t and this would be equaling to ia over c.

01:28 And so to just levitate the surface, this needs to be equal to, again, the weight of the disk.

01:35 And so the required intensity of light would be equalling mgc divided by two times the area of that reflecting surface.

01:46 And so, rather, this would be your answer for part a.

01:51 For part b, they simply want us to solve.

01:54 So that would be again mgc over 2a, and this would be equalling mgc over 2, and then pi r squared would be the area of the reflecting surface.

02:09 So this would be 0 .005 kilograms multiplied by 9 .80 meters per second squared, multiplied by 3 .00 times 10, 2 ,000...

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