Problem 25 - University Physics with Modern Physics

Question

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?

Step-by-Step Solution

Answer

The power output of the source is approximately 848 kW.

Step 1: Understanding the Problem

The problem involves finding the total average power output of a light source given its radiation pressure at a distance from the source. The radiation pressure on a perfectly absorbing surface is provided along with the distance.

Step 2: Identify Relevant Formulas

To find the power output, we need to use the formula for radiation pressure on a perfectly absorbing surface, which is given by: \( P = \frac{I}{c} \), where \( P \) is the pressure, \( I \) is the intensity of the radiation, and \( c \) is the speed of light \( 3 \times 10^8 \, \text{m/s} \). Also, intensity \( I \) is related to power \( S \) and area \( A \) by the equation \( I = \frac{S}{A} \).

Step 3: Calculate Intensity

From the radiation pressure formula \( P = \frac{I}{c} \), we can solve for intensity: \( I = P \times c \). Substitute the given values: \( I = 9.0 \times 10^{-6} \, \text{Pa} \times 3 \times 10^8 \, \text{m/s} = 2.7 \times 10^3 \, \text{W/m}^2 \).

Step 4: Calculate Area

Since the radiation is uniform and radiates in all directions, the surface area can be calculated using the formula for the surface area of a sphere: \( A = 4 \pi r^2 \), where \( r \) is the distance, 5.0 m. \( A = 4 \pi (5.0)^2 = 100 \pi \approx 314.16 \, \text{m}^2 \).

Step 5: Calculate Power Output

Use the relation \( I = \frac{S}{A} \) and solve for \( S \) (power output): \( S = I \times A \). Substitute the values: \( S = 2.7 \times 10^3 \, \text{W/m}^2 \times 314.16 \, \text{m}^2 \approx 848,232 \, \text{W} \).

Step 6: Final Result

The total average power output of the light source is approximately 848,232 watts (or 848 kW).

Key Concepts

Power Output Calculation

In this problem, we are tasked with calculating the total average power output of a light source. Power output is a measure of how much energy a source emits over a certain period of time. It's like asking how fast energy is leaving the source and is usually measured in watts (W). The process to find power output involves using the relationship between intensity, area, and power. In simpler terms, we know that the energy being radiated over a surface is related to its intensity, which is like asking how "bright" or "strong" the light is at a particular spot. Our goal is to manipulate these relationships to find the power output. Given the radiation pressure, which is the force the radiation exerts per unit area, we can determine the intensity of the radiation. Then, knowing the surface area over which this intensity is acting, we multiply to find the power output. This gives us a clearer picture of the energy the source emits.

Intensity and Power Relationship

Understanding the relationship between intensity and power is essential for solving this exercise. Intensity (\( I \)) is defined as the power per unit area, often measured in watts per square meter (\( ext{W/m}^2 \)). It essentially tells us how much energy is being transmitted through each square meter of area.To break it down:

• Intensity (\( I \)) = Power (\( S \)) divided by Area (\( A \)).

Using this formula, we see that as intensity increases, the power or energy flowing through the area also increases, and vice versa. So if you know the intensity of a source and the total area it spreads over, you can easily find out the total power output.In our problem, once we have the intensity from the previously calculated values, we multiply it by the area (given by the surface area of a sphere) to find the total power output. This ties directly back to understanding how light or energy spreads out from a point source.

Surface Area of a Sphere

The concept of the surface area of a sphere is critical in problems where light radiates uniformly in all directions. When light or any form of radiation emanates from a point source equally in every direction, it spreads out across the surface of an imaginary sphere centered on the source with a radius equal to the distance from the source.The formula for the surface area (\( A \)) of a sphere is:\[A = 4 \pi r^2\]Where \( r \) is the radius of the sphere. This equation helps determine the total area over which the light spreads. In our exercise, the distance from the source is the radius (\( 5 \text{ m} \)), and by plugging in this value into the formula, we calculate the sphere's surface area over which the light is spread. Knowing the surface area is essential because it allows you to connect to the intensity and find the total power output. As the light spreads out, knowing the area ensures that we can accurately determine how much energy is transmitted across that space.