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Chapter 12: Problem 25

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

Short Answer

Expert verified
The total average power output of the source is approximately 848 watts.

Step by step solution

01

Understand Radiation Pressure

Radiation pressure (p al) is exerted by light on a surface. For a perfectly absorbing surface, the pressure can be calculated using the formula \( P = \frac{I}{c} \), where \( I \) is the intensity of light and \( c \) is the speed of light (\(3 \times 10^8 \text{ m/s}\)).
02

Calculate Intensity

Rearrange the formula \( P = \frac{I}{c} \) to find intensity: \( I = Pc \). Substitute the given values: \( P = 9.0 \times 10^{-6} \text{ Pa} \) and \( c = 3 \times 10^8 \text{ m/s} \). Thus, \( I = 9.0 \times 10^{-6} \text{ Pa} \times 3 \times 10^8 \text{ m/s} = 2.7 \text{ W/m}^2 \).
03

Use Intensity to Find Power

The intensity \( I \) at a distance \( r \) from a point source is related to the power output \( P_t \) by the formula \( I = \frac{P_t}{4 \pi r^2} \). Rearrange this to find \( P_t \): \( P_t = I \times 4 \pi r^2 \). Substitute \( I = 2.7 \text{ W/m}^2 \) and \( r = 5.0 \text{ m} \).
04

Calculate Power Output

Substitute the known values into the formula \( P_t = I \times 4 \pi r^2 \): \( P_t = 2.7 \text{ W/m}^2 \times 4 \pi (5.0 \text{ m})^2 \). Calculate \( r^2 = 5.0^2 = 25 \) and \( 4 \pi \approx 12.57 \), so the calculation is \( P_t = 2.7 \times 12.57 \times 25 \approx 848 \text{ W} \).
05

Conclusion

The calculation shows that the total average power output of the source is approximately 848 watts.

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Key Concepts

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

Intensity of Light
The intensity of light refers to the amount of light energy that reaches a surface per unit area. It is typically measured in watts per square meter (W/m虏). Intensity becomes a crucial factor when dealing with radiation pressure, as seen in the original exercise. The formula to calculate intensity, when dealing with light and radiation pressure, is derived from the equation for radiation pressure on a perfectly absorbing surface:
  • Radiation Pressure: \( P = \frac{I}{c} \)
  • Where \( I \) is the intensity and \( c \) is the speed of light (approximately \(3 \times 10^8\) m/s).
This relationship shows that the intensity of light depends directly on the radiation pressure exerted and the speed of light. By rearranging the formula to solve for intensity \( I = P \times c \), we can calculate how intense the light must be if a certain pressure is observed on the absorbing surface. This calculation is paramount for understanding how much light energy is actually impacting a surface at a given point.
Power Output Calculation
When exploring the concept of power output from a light source, we focus on how much energy is dispensed by the source in all directions. The total average power output is calculated using the relationship between intensity and distance from the source. In the context of the exercise, the power output \( P_t \) is connected to intensity \( I \) and distance \( r \) by the formula:
  • Intensity Formula: \( I = \frac{P_t}{4 \pi r^2} \)
  • Rearranged for Power: \( P_t = I \times 4 \pi r^2 \)
This equation takes into account how the light spreads as a sphere expands outward from the source. It is critical to understand that as the distance \( r \) increases, the intensity \( I \) decreases, indicating that the light's energy is dispersed over a larger area.
By calculating \( P_t \), using both the derived intensity and the specific distance, we establish how much total power the source is outputting, ignoring energy loss due to other factors like absorption.
Absorbing Surface
An absorbing surface in physics, specifically in optical studies, is a hypothetical or real surface that completely absorbs all the light that hits it, without reflecting any of it. This is essential in theoretical exercises because it simplifies the calculation of radiation pressure. For the task at hand, knowing that the surface is perfectly absorbing means that:
  • All light energy is transferred to the surface,
  • The full effect of radiation pressure depends solely on the intensity of light.
A perfectly absorbing surface simplifies the scenario by focusing on the absorption's effect on the light's energy, allowing us to use the formula \( P = \frac{I}{c} \).
This concept highlights how light interacts with materials, whether they absorb, reflect, or transmit the light. In practical terms, understanding absorbing surfaces aids in studying phenomena such as solar cells and sensors that utilize the full energy of absorbed light for efficiency. Learning how absorbing surfaces work provides a foundational skill for those interested in material science and optical engineering.

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Most popular questions from this chapter

The electric-field component of a sinusoidal electromagnetic wave traveling through a plastic cylinder is given by the equation \(E=(5.35 \mathrm{~V} / \mathrm{m}) \cos \left[\left(1.39 \times 10^{7} \mathrm{rad} / \mathrm{m}\right) x-(3.02 \times\right.\) \(\left.\left.10^{15} \mathrm{rad} / \mathrm{s}\right) t\right] .\) (a) Find the frequency, wavelength, and speed of this wave in the plastic. (b) What is the index of refraction of the plastic? (c) Assuming that the amplitude of the electric field does not change, write a comparable equation for the electric field if the light is traveling in air instead of in plastic. Fn. AE

Electromagnetic waves propagate much differently in conductors than they do in dielectrics or in vacuum. If the resistivity of the conductor is sufficiently low (that is, if it is a sufficiently good conductor), the oscillating electric field of the wave gives rise to an oscillating conduction current that is much larger than the displacement current. In this case, the wave equation for an electric field \(\overrightarrow{\boldsymbol{E}}(x, t)=E_{y}(x, t) \hat{\boldsymbol{j}}\) propagating in the \(+x\)-direction within a conductor is $$ \frac{\delta^{2} E_{y}(x, t)}{\delta x^{2}}=\frac{\mu}{\rho} \frac{\delta E_{y}(x, t)}{\delta t} $$ where \(\mu\) is the permeability of the conductor and \(\rho\) is its resistivity. (a) A solution to this wave equation is $$ E_{y}(x, t)=E_{\max } e^{-k_{C} x} \cos \left(k_{C} x-\omega t\right) $$ where \(K_{C}=\sqrt{\omega \mu / 2 \rho}\). Verify this by substituting \(E_{y}(x, t)\) into the above wave equation. (b) The exponential term shows that the electric field decreases in amplitude as it propagates. Explain why this happens. (Hint: The field does work to move charges within the conductor. The current of these moving charges causes \(i^{2} R\) heating within the conductor, raising its temperature. Where does the energy to do this come from?) (c) Show that the electric-field amplitude decreases by a factor of \(1 / e\) in a distance \(1 / k_{\mathrm{C}}=\sqrt{2 \rho / \omega \mu}\), and calculate this distance for a radio wave with frequency \(f=\) \(1.0 \mathrm{MHz}\) in copper (resistivity \(1.72 \times 10^{-8} \Omega \cdot \mathrm{m}\); permeability \(\mu=\mu_{0}\) ). Since this distance is so short, electromagnetic waves of this frequency can hardly propagate at all into copper. Instead, they are reflected at the surface of the metal. This is why radio waves cannot penetrate through copper or other metals, and why radio reception is poor inside a metal structure.

(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 from earth to Sirius in kilometers?

A small helium-neon laser emits red visible light with a power of \(4.60 \mathrm{~mW}\) in a beam that has a diameter of \(2.50 \mathrm{~mm}\). (a) What are the amplitudes of the electric and magnetic fields of the light? (b) What are the average energy densities associated with the electric field and with the magnetic field? (c) What is the total energy contained in a \(1.00-\mathrm{m}\) length of the beam?

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