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Chapter 21: Problem 56

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 \(5.00 \mathrm{~g}\) and radius \(4.00 \mathrm{~cm}\), find the necessary light intensity. (c) Give two reasons why using light pressure as propulsion near Earth's surface is impractical.

Short Answer

Expert verified
The necessary light intensity to levitate the disk of given parameters is obtained by substituting the respective values into the derived equation. The two reasons why using light pressure as propulsion near the Earth's surface is impractical are high energy requirements and the need for perfect alignment and reflection conditions.

Step by step solution

01

Derive a Formula for Required Light Intensity

The force required to levitate the disk can be calculated using the equation \( F = m \cdot g \). For a light beam of intensity \( I \) incident perpendicular to the surface of the disk, the pressure \( P \) is given by \( P = 2I/c \). Equating the force exerted by light to the force of gravity, the required light intensity can be derived: \( m \cdot g = 2I \cdot A/c \) => \( I = (m \cdot g \cdot c) / (2 \cdot A) \).
02

Calculate the Required Light Intensity

Substitute the given values for the disk mass \( m = 5.00 \) g (or 0.005 kg), gravitational acceleration \( g = 9.81 \) m/s虏, speed of light \( c = 3.00 \times 10^8 \) m/s, and the cross-sectional area \( A = \pi \times (4.00 \times 10^{-2} m)^2 \) into the formula. The calculated value gives the needed light intensity for levitation: \( I = (0.005 kg \cdot 9.81 m/s虏 \cdot 3.00 \times 10^8 m/s) / (2 \cdot 3.14 \cdot (4.00 \times 10^{-2} m)^2) \).
03

Describing Impracticality of Light Propulsion near Earth's Surface

The reasons why using light pressure as propulsion near the Earth's surface is impractical are: (1) The intensity of light required is high, which requires a high amount of energy. (2) The pressure exerted by light is small compared to the gravitational pull by the Earth. Therefore, it requires perfectly reflective materials and perfect alignment of the light beam to have a practical application of light pressure, which in practice can be highly complex and cost-intensive.

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Key Concepts

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

Intensity of Light
Intensity of light is a measure of the power per unit area that a light beam conveys. This concept is closely related to the energy of the light beam and its capacity to exert force, like lifting a disk against gravity. In our exercise, the intensity is required to counteract the gravitational force pulling the disk downwards. To calculate the intensity of light necessary to levitate a disk, we use the formula derived from equating light pressure to gravitational force.
  • Intensity formula: \[ I = \frac{m \cdot g \cdot c}{2 \cdot A} \]
This formula indicates that the intensity depends on
  • the mass of the object, \(m\)
  • gravitational acceleration, \(g\)
  • the speed of light, \(c\)
  • and the area over which the light is spread, \(A\)
It is essential to understand that the intensity must be sufficient to generate enough pressure to balance out the gravitational pull.
Gravitational Force
Gravitational force is a natural phenomenon by which all things with mass are brought towards one another, including objects like the metal disk and the planet Earth. This force is described by the formula:
  • Gravitational force formula: \[ F = m \cdot g \]
Here, \(m\) represents the mass of the object and \(g\) is the gravitational acceleration (approximately \(9.81 \, \text{m/s}^2\) on Earth). This concept is essential in our exercise as it dictates the force that the light beam needs to counterbalance. In simpler terms, the more massive an object, the greater the gravitational pull, and consequently, the more intense the light must be to levitate it.
Reflective Materials
Reflective materials are crucial when dealing with light pressure. When light strikes a surface, some of its energy can be transferred to the surface, exerting a force. However, the effectiveness of this force largely depends on the material's ability to reflect light. Using perfectly reflective materials, as assumed in our exercise, optimizes this process.
  • They reflect nearly all incident light, maximizing the pressure exerted.
  • This results in twice the force compared to an absorbing surface, as the light is reversed in direction.
The ideal application would require materials with little to no energy loss to completely reflect back the light energy and ensure efficient use of the intensity applied.
Speed of Light
The speed of light is a fundamental constant of nature, \(c = 3 \times 10^8 \text{ m/s} \). It is one of the core parameters in calculating light pressure intensity because it determines how fast light pressure can respond to exert force. In the context of our equation, \(c\) governs how efficiently energy from the light is transferred into pressure. High speed means light can quickly exert momentum onto a surface.Facts about the speed of light:
  • It is the fastest speed at which energy, matter, or information can travel.
  • In our formula, it serves to balance out the pressure exerted by the light beam against the gravitational force.
Thus, using the speed of light highlights how efficient and rapid this force transfer can be, enabling the theoretical concept of using light pressure for levitation or propulsion.

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Most popular questions from this chapter

A microwave oven is powered by an electron tube called a magnetron that generates electromagnetic waves of frequency \(2.45 \mathrm{GHz}\). The microwaves enter the oven and are reflected by the walls. The standing-wave pattern produced in the oven can cook food unevenly, with hot spots in the food at antinodes and cool spots at nodes, so a turntable is often used to rotate the food and distribute the energy. If a microwave oven is used with a cooking dish in a fixed position, the antinodes can appear as burn marks on foods such as carrot strips or cheese. The separation distance between the burns is measured to be \(6.00 \mathrm{~cm}\). Calculate the speed of the microwaves from these data.

Operation of the pulse oximeter (see previous problem). The transmission of light energy as it passes through a solution of light-absorbing molecules is described by the Beer-Lambert law $$ I=I_{0} 10^{-c a} \text { or } \log _{10}\left(\frac{l}{I_{0}}\right)=-\epsilon C L $$ which gives the decrease in intensity \(I\) in terms of the distance \(L\) the light has traveled through a fluid with a concentration \(C\) of the light- absorbing molecule. The quantity \(\epsilon\) is called the extinction coefficient, and its value depends on the frequency of the light. (It has units of \(\mathrm{m}^{2} / \mathrm{mol}\).) Assume the extinction coefficient for \(660-\mathrm{nm}\) light passing through a solution of oxygenated hemoglobin is identical to the coefficient for \(940-\mathrm{nm}\) light passing through deoxygenated hemoglobin. Also assume \(940-\mathrm{nm}\) light has zero absorption \((\epsilon=0)\) in oxygenated hemoglobin and \(660-\mathrm{nm}\) light has zero absorption in deoxygenated hemoglobin. If \(33 \%\) of the energy of the red source and \(76 \%\) of the infrared energy is transmitted through the blood, what is the fraction of hemoglobin that is oxygenated?

An AC: source operating at \(60 \mathrm{~Hz}\) with a maximum voltage of \(170 \mathrm{~V}\) is connected in series with a resistor \((R=\) \(1.2 \mathrm{k} \Omega\) ) and an inductor \((L=2.8 \mathrm{H})\). (a) What is the maximum value of the current in the circuit? (b) What are the maximum values of the potential difference across the resistor and the inductor? (c) When the current is at a maximum, what are the magnitudes of the potential differences across the resistor, the inductor, and the AC source? (d) When the current is zero, what are the magnitudes of the potential difference across the resistor, the inductor, and the AC source?

Assume the solar radiation incident on Earth is \(1340 \mathrm{~W} / \mathrm{m}^{2}\) (at the top of Earth's atmosphere). Calculate the total power radiated by the Sun, taking the average separation between Earth and the Sun to be \(1.49 \times 10^{11} \mathrm{~m}\).

A transformer on a pole near a factory steps the voltage down from \(3600 \mathrm{~V}\) (rms) to \(120 \mathrm{~V}\) (rms). The transformer is to deliver \(1000 \mathrm{~kW}\) to the factory at \(90 \%\) efficiency. Find (a) the power delivered to the primary, (b) the current in the primary, and (c) the current in the secondary.
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