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Chapter 33: Problem 30

A small laser emits light at power \(5.00 \mathrm{~mW}\) and wavelength \(633 \mathrm{~nm}\). The laser beam is focused (narrowed) until its diameter matches the \(1266 \mathrm{~nm}\) diameter of a sphere placed in its path. The sphere is perfectly absorbing and has density \(5.00 \times\) \(10^{3} \mathrm{~kg} / \mathrm{m}^{3}\). What are (a) the beam intensity at the sphere's location, (b) the radiation pressure on the sphere, \((\mathrm{c})\) the magnitude of the corresponding force, and (d) the magnitude of the acceleration that force alone would give the sphere?

Short Answer

Expert verified
(a) Intensity: \( 3.97 \times 10^9 \text{ W/m}^2 \); (b) Pressure: \( 13.2 \text{ N/m}^2 \); (c) Force: \( 1.66 \times 10^{-11} \text{ N} \); (d) Acceleration: \( 3.13 \times 10^6 \text{ m/s}^2 \).

Step by step solution

01

Calculate the Beam Intensity

Intensity is the power per unit area. First, we calculate the area of the sphere which acts as a cross-sectional area for the beam. The diameter of the sphere is given as \(1266 \text{ nm}\), thus the radius \(r\) is \(633 \text{ nm} = 633 \times 10^{-9} \text{ m}\). The area \(A\) of the circle is given by: \[ A = \pi r^2 = \pi (633 \times 10^{-9})^2 \approx 1.26 \times 10^{-12} \text{ m}^2 \] The intensity \(I\) is the power \(P = 5 \times 10^{-3} \text{ W}\) divided by the area \(A\):\[ I = \frac{P}{A} = \frac{5 \times 10^{-3}}{1.26 \times 10^{-12}} \approx 3.97 \times 10^9 \text{ W/m}^2 \]
02

Calculate the Radiation Pressure

For perfectly absorbing materials, the radiation pressure \( p \) is given by \( p = \frac{I}{c} \) where \( c \) is the speed of light \( 3 \times 10^8 \text{ m/s} \). We already have \( I = 3.97 \times 10^9 \text{ W/m}^2 \): \[ p = \frac{3.97 \times 10^9}{3 \times 10^8} \approx 13.2 \text{ N/m}^2 \]
03

Determine the Force on the Sphere

The force \( F \) exerted by the radiation pressure on the sphere is given by \( F = pA \), where \( A \) is the cross-sectional area calculated previously. So, \[ F = 13.2 \times 1.26 \times 10^{-12} \approx 1.66 \times 10^{-11} \text{ N} \]
04

Compute Acceleration of the Sphere

The acceleration \( a \) due to the force \( F \) on the sphere can be found using Newton's second law, \( F = ma \), where \( m \) is the mass of the sphere. To find the mass \( m \), we use the formula: \[ m = \text{density} \times \text{Volume} = 5 \times 10^3 \text{ kg/m}^3 \times \frac{4}{3}\pi(633 \times 10^{-9})^3 \approx 5.3 \times 10^{-18} \text{ kg} \] By substituting \( F \) and \( m \) into \( F = ma \), we get \[ a = \frac{F}{m} = \frac{1.66 \times 10^{-11}}{5.3 \times 10^{-18}} \approx 3.13 \times 10^6 \text{ m/s}^2 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Beam Intensity
When dealing with a laser beam, understanding beam intensity is crucial. The **intensity** of a beam refers to the power per unit area. It is a measure of how much energy the beam carries over a given area. More specifically, the intensity (I) is calculated using the formula:
  • \( I = \frac{P}{A} \)
where \( P \) is the power of the laser, and \( A \) is the cross-sectional area that the beam spreads over.

For example, in the exercise, the laser operates at a power of 5 mW. To find the area, consider the sphere in the laser's path. The sphere's diameter matches the beam's diameter, allowing us to use the sphere's cross-sectional area for calculations.

We calculate using:
  • Area \( A = \pi r^2 \)
Where \( r \) is the radius of the sphere, determined from its diameter. After finding \( A \), substitute back into \( I = \frac{P}{A} \) to compute the beam's intensity. This value tells us how much power is delivered per square meter at the sphere's location.
Force on Sphere
Radiation pressure results in a force exerted on an object, such as a sphere in the pathway of a laser beam. This force can be determined by analyzing radiation pressure and the cross-sectional area involved.

**Radiation pressure** arises when electromagnetic radiation like light impinges on a surface. For a perfectly absorbing surface, like the sphere, the formula to calculate pressure is:
  • \( p = \frac{I}{c} \)
where \( I \) is the beam intensity and \( c \) is the speed of light (~3 脳 10鈦� m/s). This pressure manifests as a force pushing against the object.

To find the force \( F \) on the sphere, use:
  • \( F = pA \)
The cross-sectional area \( A \) aligns with the area used for intensity calculations. The force calculated contributes to determining how the sphere reacts physically to the light pressure, further affecting its acceleration.
Acceleration Calculation
The last part of understanding the sphere's response to the laser involves calculating its acceleration. To compute this, we apply Newton's second law:
  • \( F = ma \)
where \( F \) is the force, \( m \) the mass of the sphere, and \( a \) the acceleration.

First, determine the sphere's mass. Its mass depends on its volume and density. Given the sphere's diameter, we can calculate its volume with:
  • \( ext{Volume} = \frac{4}{3} \pi r^3 \)
Multiply the volume by the density (5 脳 10鲁 kg/m鲁 in our case) to find the mass.

Finally, insert the force and the mass into the formula \( a = \frac{F}{m} \) to find the acceleration. This calculation reveals how quickly the sphere will speed up in response to the force exerted by the laser beam. Understanding this acceleration helps in analyzing motion and impact in such scenarios.

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Most popular questions from this chapter

A plane electromagnetic wave, with wavelength \(3.0 \mathrm{~m}\), travels in vacuum in the positive direction of an \(x\) axis. The electric field, of amplitude \(300 \mathrm{~V} / \mathrm{m}\), oscillates parallel to the \(y\) axis. What are the (a) frequency, (b) angular frequency, and (c) angular wave number of the wave? (d) What is the amplitude of the magnetic field component? (e) Parallel to which axis does the magnetic field oscillate? (f) What is the time-averaged rate of energy flow in watts per square meter associated with this wave? The wave uniformly illuminates a surface of area \(2.0 \mathrm{~m}^{2}\). If the surface totally absorbs the wave, what are \((\mathrm{g})\) the rate at which momentum is transferred to the surface and (h) the radiation pressure on the surface?

(a) How long does it take a radio signal to travel \(150 \mathrm{~km}\) from a transmitter to a receiving antenna? (b) We see a full Moon by reflected sunlight. How much earlier did the light that enters our eye leave the Sun? The Earth-Moon and Earth-Sun distances are \(3.8 \times 10^{5} \mathrm{~km}\) and \(1.5 \times 10^{8} \mathrm{~km}\), respectively. (c) What is the round-trip travel time for light between Earth and a spaceship orbiting Saturn, \(1.3 \times 10^{9} \mathrm{~km}\) distant? (d) The Crab nebula, which is about 6500 light-years (ly) distant, is thought to be the result of a supernova explosion recorded by Chinese astronomers in A.D. 1054 . In approximately what year did the explosion actually occur? (When we look into the night sky, we are effectively looking back in time.)

Prove, for a plane electromagnetic wave that is normally incident on a flat surface, that the radiation pressure on the surface is equal to the energy density in the incident beam. (This relation between pressure and energy density holds no matter what fraction of the incident energy is reflected.)

A small spaceship with a mass of only \(1.5 \times 10^{3} \mathrm{~kg}\) (including an astronaut) is drifting in outer space with negligible gravitational forces acting on it. If the astronaut turns on a \(10 \mathrm{~kW}\) laser beam, what speed will the ship attain in \(1.0\) day because of the momentum carried away by the beam?

A certain helium-neon laser emits red light in a narrow band of wavelengths centered at \(632.8 \mathrm{~nm}\) and with a "wavelength width" (such as on the scale of Fig. \(33-1\) ) of \(0.0100 \mathrm{~nm}\). What is the corresponding "frequency width" for the emission?
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