Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

An intense light source radiates uniformly in all directions. At a distance of 5.0 m from the source, the radiation pressure on a perfectly absorbing surface is 9.0 \(\times\) 10\(^{-6}\) Pa. What is the total average power output of the source?

Short Answer

Expert verified
The power output of the source is approximately \( 8.5 \times 10^5 \) W.

Step by step solution

01

Understanding Radiation Pressure

Radiation pressure is the force exerted by electromagnetic radiation, such as light, on a surface. For a perfectly absorbing surface, the radiation pressure \( P \) can be described by the equation \( P = \frac{I}{c} \), where \( I \) is the intensity of the light and \( c \) is the speed of light in a vacuum, approximately \( 3.0 \times 10^8 \) m/s.
02

Calculate Intensity from Radiation Pressure

Given that the radiation pressure \( P \) is \( 9.0 \times 10^{-6} \) Pa, we can rearrange the equation \( P = \frac{I}{c} \) to find the intensity \( I \) as follows: \( I = P \cdot c = 9.0 \times 10^{-6} \times 3.0 \times 10^8 = 2.7 \times 10^{3} \) W/m².
03

Understanding Power and Intensity

Intensity \( I \) is defined as the power \( P_{source} \) per unit area. For a point source radiating uniformly in all directions, the area at a distance \( r \) from the source is the surface area of a sphere, calculated by \( A = 4\pi r^2 \).
04

Calculate the Surface Area at 5.0 m

The surface area of a sphere with a radius \( r = 5.0 \) m is \( A = 4\pi (5.0)^2 = 100\pi \) m².
05

Calculate the Total Power Output of the Source

Using the relationship between intensity and power, \( I = \frac{P_{source}}{A} \), where \( A = 100\pi \) m² and \( I = 2.7 \times 10^3 \) W/m², we solve for \( P_{source} \):\[P_{source} = I \cdot A = 2.7 \times 10^{3} \times 100\pi = 2.7 \times 10^{5}\pi \]which approximately equals \( 8.5 \times 10^5 \) W.

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 Vaia!

Key Concepts

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

Intensity of Light
The term "intensity of light" refers to the amount of energy that light delivers per unit area, per unit time. Think of it as how "strong" or "bright" the light is when it hits a surface. The mathematical expression used to define intensity, \( I \), is the power, \( P_{source} \), that a light source emits, divided by the area over which the light spreads. It's measured in watts per square meter (W/m²). For a perfectly absorbing surface, the relationship between intensity and radiation pressure is given by, \( P = \frac{I}{c} \), where \( c \) is the speed of light. This relationship shows that intensity directly affects the force applied by light on a surface. This force per unit area is what we measure as radiation pressure.
Electromagnetic Radiation
Electromagnetic radiation is a form of energy that travels through space as waves. It includes a spectrum of different types, such as light, radio waves, X-rays, and microwaves, to name a few. Light, in particular, is one type of electromagnetic radiation that is essential for understanding radiation pressure and intensity.
Light travels at approximately \( 3.0 \times 10^8 \) meters per second, also known as the speed of light. This speed is a constant when light moves through a vacuum. What makes electromagnetic radiation unique is its ability to exert pressure on surfaces. This pressure, although small, is measurable and plays a critical role in various scientific fields, including astrophysics and solar energy research.
Power Output
Power output refers to the total energy produced by a source per unit time. It's measured in watts (W), where one watt equals one joule of energy per second. In the context of light sources, power output is a measure of how much light energy is emitted by the source in total.
In our exercise, we determined the power output by relating it to intensity and the area over which the light is spread. Using the formula for intensity, \( I = \frac{P_{source}}{A} \), and modifying it to solve for \( P_{source} \), we found that the total power output of the source is approximately \( 8.5 \times 10^5 \) watts. Understanding power output is crucial for applications in designing lighting systems or assessing energy needs for different devices and environments.
Spherical Surface Area
When dealing with light sources that emit in all directions, it is essential to think in three-dimensional terms. A point source of light radiates uniformly in all directions, forming a spherical wavefront as it moves outward.
To calculate the surface area of this sphere at a certain distance, we use the formula \( A = 4\pi r^2 \), where \( r \) is the radius, or distance from the source. In our specific case, with \( r = 5.0 \) meters, the area is \( 100\pi \) m². This calculation allows us to understand how light spreads out, influencing both intensity and radiation pressure. Spherical surface area is an important concept because it affects how light interacts with objects, such as focusing on lenses or the amount of light reaching distant planets.

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

A circular wire loop has a radius of \(7.50 \mathrm{~cm}\). A sinusoidal electromagnetic plane wave traveling in air passes through the loop, with the direction of the magnetic field of the wave perpendicular to the plane of the loop. The intensity of the wave at the location of the loop is \(0.0275 \mathrm{~W} / \mathrm{m}^{2}\), and the wavelength of the wave is \(6.90 \mathrm{~m}\). What is the maximum emf induced in the loop?

For a sinusoidal electromagnetic wave in vacuum, such as that described by Eq. (32.16), show that the \(average\) energy density in the electric field is the same as that in the magnetic field.

An electromagnetic wave has an electric field given by \(\vec{E} (y, t)\) = (3.10 \(\times\) 10\(^5\) V/m) \(\hat{k}\) cos [ky - (12.65 \(\times\) 10\(^{12}\) rad/s)t]. (a) In which direction is the wave traveling? (b) What is the wavelength of the wave? (c) Write the vector equation for \(\vec{B} (y, t)\).

A satellite 575 km above the earth's surface transmits sinusoidal electromagnetic waves of frequency 92.4 MHz uniformly in all directions, with a power of 25.0 kW. (a) What is the intensity of these waves as they reach a receiver at the surface of the earth directly below the satellite? (b) What are the amplitudes of the electric and magnetic fields at the receiver? (c) If the receiver has a totally absorbing panel measuring 15.0 cm by 40.0 cm oriented with its plane perpendicular to the direction the waves travel, what average force do these waves exert on the panel? Is this force large enough to cause significant effects?

Consider electromagnetic waves propagating in air. (a) Determine the frequency of a wave with a wavelength of (i) 5.0 km, (ii) 5.0 mm, (iii) 5.0 nm. (b) What is the wavelength (in meters and nanometers) of (i) gamma rays of frequency 6.50 \(\times\) 10\(^{21}\) Hz and (ii) an AM station radio wave of frequency 590 kHz?

See all solutions

Recommended explanations on Physics Textbooks

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.

Sign-up for free