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A point source of light emits isotropically with a power of \(200 \mathrm{~W}\). What is the force due to the light on a totally absorbing sphere of radius \(2.0 \mathrm{~cm}\) at a distance of \(20 \mathrm{~m}\) from the source?

Short Answer

Expert verified
The force is approximately \(1.67 \times 10^{-12} \, \text{N}\).

Step by step solution

01

Understand Light Intensity Formula

The intensity (\(I\) of light at a distance \(r\) from a point source is given by the formula:\[I = \frac{P}{4\pi r^2}\]where \(P\) is the power emitted by the source. Here, \(P = 200\, \text{W} \) and \(r = 20\, \text{m}\).
02

Calculate Light Intensity

Substitute the values into the intensity formula:\[I = \frac{200}{4\pi (20)^2}\]Calculate:\[I = \frac{200}{4\pi \times 400} = \frac{200}{1600\pi} = \frac{1}{8\pi} \approx 0.0398 \, \text{W/m}^2\].
03

Find Area of Sphere

The area \(A\) of the sphere is needed to calculate the force and is given by:\[A = \pi R^2\]where \(R = 2 \times 10^{-2}\, \text{m}\). Calculate:\[A = \pi (0.02)^2 = \pi \times 0.0004 = 0.0004\pi \approx 0.001257 \, \text{m}^2\].
04

Calculate Radiant Force

The force \(F\) due to light on a totally absorbing surface is given by:\[F = IA \times \frac{1}{c}\]where \(c\) is the speed of light \( \approx 3 \times 10^8 \, \text{m/s} \). Substitute the known values:\[F = \left(0.0398\right) \times \left(0.001257\right) \times \frac{1}{3 \times 10^8}\].Calculate:\[F \approx 1.67 \times 10^{-12} \, \text{N}\].

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Key Concepts

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

Light Intensity
Light intensity essentially measures how much power is carried per unit area by a wave or a beam of light. This is a crucial concept in understanding how light propagates from a point source into the surrounding environment. When a light source emits energy, this energy spreads out equally in all directions (isotropy). As a result, the intensity decreases with distance from the source because the same amount of energy is distributed over a larger area.

Understanding the mathematical form of light intensity is key. It's given by the formula:- \[I = \frac{P}{4\pi r^2}\]
Where:
  • \(I\) is the light intensity (in watts per square meter, \( ext{W/m}^2\))
  • \(P\) is the power of the source (in watts, \( ext{W}\))
  • \(r\) is the distance from the light source (in meters, \( ext{m}\))
By plugging in the values of power and distance given in the exercise, we find the intensity of light at that specific location from the point source.
Radiant Force
Radiant force relates to the momentum carried by light. Even though light has no mass, it carries momentum and can exert a force when absorbed or reflected by surfaces. This force, called the radiation pressure, is particularly interesting when dealing with highly sensitive devices or specific physical scenarios such as spacecraft propulsion.

The force experienced by an absorbing surface is linked to the light intensity and the area of the surface that the light hits. For a totally absorbing sphere, where all incoming light intensity is absorbed, the force is calculated through the formula:- \[F = IA \times \frac{1}{c}\]
Where:
  • \(F\) is the force (in newtons, \( ext{N}\))
  • \(I\) is the light intensity (in \( ext{W/m}^2\))
  • \(A\) is the cross-sectional area of the absorbing sphere (in \( ext{m}^2\))
  • \(c\) is the speed of light \(\approx 3 \times 10^8 \, \text{m/s}\)
This formula helps us quantify the tiny, yet fascinating effect of light on the sphere.
Isotropic Emission
Isotropic emission describes a scenario where a point light source emits light uniformly in all directions. This assumption simplifies calculations because the intensity at any given point depends solely on the distance from the source.

In isotropic emission, no energy is lost in translation from one direction to another. The energy flows symmetrically in all directions, resulting in a spherical wavefront of light. When we consider the total power emitted, it's essential to note that this power is distributed over the full surface area of a sphere surrounding the light source.

The spherical distribution of power is key because it forms the foundation of the light intensity formula \(I = \frac{P}{4\pi r^2}\). Understanding this distribution helps in accurately computing how much energy reaches a specific distance from the source.
Absorbing Sphere
An absorbing sphere is a theoretical concept used to understand how objects interact with light. It assumes the object in question fully absorbs all the light that hits it without reflecting any back into the environment. Such an assumption is simplifying yet very useful to model light-matter interactions.

For calculations involving a totally absorbing sphere, the sphere's cross-sectional area is critical. Instead of considering the entire surface area, we look at the area that directly faces the light source, which can be simplified to \(A = \pi R^2\) for a sphere with radius \(R\).

This radius is easy to determine from the problem’s parameters and allows you to compute how much light energy is effectively "captured" by the sphere. When combined with the intensity of light, this helps calculate the radiant force exerted on the sphere, mirroring real-world scenarios you might see in physics labs or advanced technology applications.

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Most popular questions from this chapter

Project Seafarer was an ambitious program to construct an enormous antenna, buried underground on a site about \(10000 \mathrm{~km}^{2}\) in area. Its purpose was to transmit signals to submarines while they were deeply submerged. If the effective wavelength were \(1.0 \times 10^{4}\) Earth radii, what would be the (a) frequency and (b) period of the radiations emitted? Ordinarily, electromagnetic radiations do not penetrate very far into conductors such as seawater, and so normal signals cannot reach the submarines.

(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.)

Someone plans to float a small, totally absorbing sphere \(0.500 \mathrm{~m}\) above an isotropic point source of light, so that the upward radiation force from the light matches the downward gravitational force on the sphere. The sphere's density is \(19.0 \mathrm{~g} / \mathrm{cm}^{3}\), and its radius is \(2.00 \mathrm{~mm}\). (a) What power would be required of the light source? (b) Even if such a source were made, why would the support of the sphere be unstable?

At a beach the light is generally partially polarized due to reflections off sand and water. At a particular beach on a particular day near sundown, the horizontal component of the electric field vector is \(2.3\) times the vertical component. A standing sunbather puts on polarizing sunglasses; the glasses eliminate the horizontal field component. (a) What fraction of the light intensity received before the glasses were put on now reaches the sunbather's eyes? (b) The sunbather, still wearing the glasses, lies on his side. What fraction of the light intensity received before the glasses were put on now reaches his eyes?

An isotropic point source emits light at wavelength \(500 \mathrm{~nm}\), at the rate of \(200 \mathrm{~W}\). A light detector is positioned \(400 \mathrm{~m}\) from the source. What is the maximum rate \(\partial B / \partial t\) at which the magnetic component of the light changes with time at the detector's location?

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