/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 68 \(\bullet$$\bullet\) A powerful ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 23: Problem 68

\(\bullet$$\bullet\) A powerful searchlight shines on a man. The man's cross- sectional area is 0.500 \(\mathrm{m}^{2}\) perpendicular to the light beam, and the intensity of the light at his location is 36.0 \(\mathrm{kW} / \mathrm{m}^{2}\) . He is wearing black clothing, so that the light incident on him is totally absorbed. What is the magnitude of the force the light beam exerts on the man? Do you think he could sense this force?

Short Answer

Expert verified
The force is \( 6.0 \times 10^{-5} \text{ N} \); it is too small to be sensed.

Step by step solution

01

Understand the Problem Statement

A searchlight is shining on a man who is wearing black clothing, absorbing all the incident light. We need to determine the force exerted on him by the light beam, given his cross-sectional area as 0.500 \( \text{m}^2 \) and the intensity of the light as 36.0 \( \text{kW/m}^2 \).
02

Recall the Relationship between Light, Force, and Pressure

The light exerts a pressure on the man because it carries momentum. When light is absorbed, the force exerted is equal to the momentum change (pressure) per unit area times the area. This relationship is given by \( F = P \times A \).
03

Calculate the Pressure Exerted by the Light

For light, the pressure \( P \) is related to the intensity \( I \) by \( P = \frac{I}{c} \), where \( c \) is the speed of light in a vacuum (approximately \( 3 \times 10^8 \text{ m/s} \)). Substituting the given intensity \( I = 36.0 \times 10^3 \text{ W/m}^2 \), we have:\[ P = \frac{36.0 \times 10^3}{3 \times 10^8} = 1.2 \times 10^{-4} \text{ N/m}^2 \]
04

Calculate the Force Exerted on the Man

Using the formula \( F = P \times A \), where \( A = 0.500 \text{ m}^2 \) and we found \( P = 1.2 \times 10^{-4} \text{ N/m}^2 \), the force \( F \) is:\[ F = 1.2 \times 10^{-4} \times 0.500 = 6.0 \times 10^{-5} \text{ N} \]
05

Consider the Sensation of the Force

The calculated force \( 6.0 \times 10^{-5} \text{ N} \) is extremely small, many orders of magnitude below what a human can physically sense. Therefore, the man would not be able to feel the force exerted by the light beam.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Momentum Change
When light interacts with objects, it transfers momentum, even though it has no mass. This is crucial to understanding why light can exert a force. The momentum of light is defined by the equation \( p = \frac{E}{c} \), where \( E \) is the energy of the light and \( c \) is the speed of light. In this case, the energy of light striking an object and being absorbed leads to a change in momentum, resulting in a force. The force exerted by light is related to how much momentum is transferred, which depends on the light's intensity and the area it strikes.
For an absorbed light, the change in momentum is directly linked to the light's pressure on the surface. The force a surface feels is the rate of momentum change per unit time. This means the greater the intensity and the larger the area exposed to light, the more considerable the momentum change, resulting in a larger force.
Intensity of Light
The intensity of light is a measure of the energy the light carries per unit area per unit time. It is expressed in watts per square meter (\( ext{W/m}^2 \)). In this scenario, the intensity is key to determining the pressure the light exerts when absorbed. The relationship between intensity and pressure is described by the equation \( P = \frac{I}{c} \), where \( I \) is the intensity and \( c \) is the speed of light.
This formula shows that the pressure exerted by the light is directly proportional to its intensity. Hence, higher intensity means more exerted pressure, leading to a more substantial force for a given area. In practical terms, when you know the intensity of light at a certain point, you can predict how much pressure it will apply if the light is completely absorbed.
Cross-Sectional Area
The cross-sectional area is a crucial factor when calculating the force exerted by light. It refers to the surface area perpendicular to the direction of the incident light. In this example, it's the area of the man that the light directly shines upon, which is 0.500 \( ext{m}^2 \).
This area determines how much of the light's energy and momentum are absorbed. The larger the area, the more light is absorbed, leading to a greater influence of pressure and a higher force exerted on the surface. To compute the force, we use the equation \( F = P \times A \), where \( F \) is the force, \( P \) is the pressure, and \( A \) is the cross-sectional area. Thus, both the intensity of light and the cross-sectional area are essential to understanding and calculating the impact of light pressure.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

\(\bullet\) The refractive index of a certain glass is \(1.66 .\) For what angle of incidence is light that is reflected from the surface of this glass completely polarized if the glass is immersed in (a) air or (b) water?

\(\bullet$$\bullet\) A thin beam of light in air is incident on the surface of a lanthanum flint glass plate having a refractive index of 1.80 . What is the angle of incidence, \(\theta_{a}\) of the beam with this plate, for which the angle of refraction is \(\theta_{a} / 2 ?\) Both angles are measured relative to the normal.

\(\bullet\) Radio station \(\mathrm{WCCO}\) in Minneapolis broadcasts at a frequency of 830 \(\mathrm{kHz}\) . At a point some distance from the transmitter, the magnetic-field amplitude of the electromagnetic wave from \(\mathrm{WCCO}\) is \(4.82 \times 10^{-11} \mathrm{T}\) . Calculate (a) the wavelength, (b) the wave number, (c) the angular frequency, and (d) the electric-field amplitude.

\(\bullet\) (a) How much time does it take light to travel from the moon to the earth, a distance of \(384,000 \mathrm{km} ?\) (b) Light from the star Sirius takes 8.61 years to reach the earth. What is the distance to Sirius in kilometers?

\(\bullet$$\bullet\) Heart sonogram. Physicians use high-frequency \((f=\) 1 MHz to 5 MHz) sound waves, called ultrasound, to image internal organs. The speed of these ultrasound waves is 1480 \(\mathrm{m} / \mathrm{s}\) in muscle and 344 \(\mathrm{m} / \mathrm{s}\) in air. We define the index of refraction of a material for sound waves to be the ratio of the speed of sound in air to the speed of sound in the material. Snell's law then applies to the refraction of sound waves. (a) At what angle from the normal does an ultrasound beam enter the heart if it leaves the lungs at an angle of \(9.73^{\circ}\) from the normal to the heart wall? (Assume that the speed of sound in the lungs is 344 \(\mathrm{m} / \mathrm{s} .\) ) (b) What is the critical angle for sound waves in air incident on muscle?
See all solutions

Recommended explanations on Physics Textbooks

Fundamentals of Physics

Read Explanation

Wave Optics

Read Explanation

Linear Momentum

Read Explanation

Engineering Physics

Read Explanation

Physical Quantities And Units

Read Explanation

Energy Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.