Problem 28 - University Physics with Modern Physics

Question

In the \(25-\) fil Space Simulator facility at NASA's Jet Propulsion Laboratory, a bank of overhead arc lamps can produce light of intensity 2500 \(\mathrm{W} / \mathrm{m}^{2}\) at the floor of the facility. (This simulates the intensity of sunlight near the planet Venus.) Find the average radiation pressure (in pascals and in atmospheres) on (a) a totally absorbing section of the floor and (b) a totally reflecting section of the floor. (c) Find the average momentum density (momentum per unit volume) in the light at the floor.

Step-by-Step Solution

Answer

Radiation pressure is 8.33×10⁻⁶ Pa (8.22×10⁻¹¹ atm) for absorption and 1.67×10⁻⁵ Pa (1.65×10⁻¹⁰ atm) for reflection; momentum density is 2.78×10⁻¹⁴ Ns/m³.

Step 1: Understand the Problem

We need to find the average radiation pressure for two scenarios: (a) when the floor is totally absorbing and (b) when it is totally reflecting. Additionally, we need to find the average momentum density of the light at the floor.

Step 2: Radiation Pressure on Absorbing Surface

For a totally absorbing surface, the radiation pressure is given by the formula: \[ P = \frac{I}{c} \] where \( I = 2500 \, \text{W/m}^2 \) is the intensity of light and \( c = 3 \times 10^8 \, \text{m/s} \) is the speed of light. Substituting the values, we get: \[ P = \frac{2500}{3 \times 10^8} \, \text{Pa} \approx 8.33 \times 10^{-6} \, \text{Pa}. \]

Step 3: Radiation Pressure on Reflecting Surface

For a totally reflecting surface, the radiation pressure is twice that of a totally absorbing surface, because the momentum change is doubled when light reflects. Thus, \[ P = \frac{2I}{c} = \frac{2 \times 2500}{3 \times 10^8} \, \text{Pa} \approx 1.67 \times 10^{-5} \, \text{Pa}. \]

Step 4: Convert Pressure to Atmospheres

To convert pressure from pascals to atmospheres, use: \[ 1 \, \text{atm} = 1.013 \times 10^5 \, \text{Pa}. \] So, for absorbing surface: \[ P = \frac{8.33 \times 10^{-6}}{1.013 \times 10^5} \, \text{atm} \approx 8.22 \times 10^{-11} \, \text{atm}. \] For reflecting surface: \[ P = \frac{1.67 \times 10^{-5}}{1.013 \times 10^5} \, \text{atm} \approx 1.65 \times 10^{-10} \, \text{atm}. \]

Step 5: Calculate Momentum Density

The average momentum density \( p \) can be calculated using \( p = \frac{I}{c^2} \). Substituting the values, we get: \[ p = \frac{2500}{(3 \times 10^8)^2} \, \text{Ns/m}^3 \approx 2.78 \times 10^{-14} \, \text{Ns/m}^3. \]

Key Concepts

Absorbing Surface

When light hits an absorbing surface, it transfers all its energy and momentum to that surface, causing a change in pressure. This is what we refer to as radiation pressure. For a totally absorbing surface, the formula for calculating radiation pressure is simple: \[P = \frac{I}{c}\] where:

• \( P \) represents the radiation pressure.
• \( I \) is the intensity of light, measured in watts per square meter \( \text{W/m}^2 \).
• \( c \) is the speed of light in a vacuum, approximately \( 3 \times 10^8 \, \text{m/s} \).

By substituting actual values into this formula, you can calculate how much pressure is exerted on the floor by the overhead arc lamps.

Reflecting Surface

Unlike absorbing surfaces, reflecting surfaces bounce the incoming light back. This reflection doubles the change in momentum compared to absorption. Therefore, the radiation pressure is twice as high for a reflecting surface.The radiation pressure for a totally reflecting surface can be determined through the formula:\[P = \frac{2I}{c}\]This equation tells us that if a surface reflects all light unhindered, each photon's momentum change is doubled, maximizing the pressure exerted on the surface.This concept is extremely relevant in fields such as solar sailing, where spacecraft maneuver using light pressure from solar sails.

Momentum Density

Momentum density refers to the amount of momentum present per unit volume of light. This concept is particularly intriguing because it highlights the momentum-carrying capacity of light itself. Momentum density \( p \) is given by the formula:\[p = \frac{I}{c^2}\]using the same variables as before. In essence, momentum density is a small theoretical value when calculating for areas like the facility's floor, yet it crucially impacts the behavior of radiation and its pressure in real-world applications. This value is important for understanding how forces are distributed by light across different surfaces.

Intensity of Light

The intensity of light is a measure of how much energy the light waves carry over a particular area per unit time. This is particularly important in analyzing radiation pressure because the intensity directly correlates with how much pressure the light can exert on a surface.The formula for intensity used is:\[I = \text{power per unit area}\]In the case of the NASA facility's lamps, they produce a constant intensity, 2500 \( \text{W/m}^2 \), which simulates conditions similar to planet Venus. Understanding light intensity allows us to understand how energy delivered to the surface changes, which then affects radiation pressure and the resulting momentum transfer.

Speed of Light

The speed of light, denoted by \( c \), is a universal constant utilized in various physics equations, including those concerning radiation pressure. It holds a constant value of approximately \( 3 \times 10^8 \, \text{m/s} \).In calculations related to light pressure and momentum density, \( c \) plays a crucial role:

• In the radiation pressure formulas, it's the constant that normalizes the pressure exerted by a given intensity.
• In momentum density, \( c^2 \) is used in the denominator to contextualize how much momentum the light holds at specific intensities.

Understanding the role of \( c \) is vital for comprehending how light interacts with materials and affects various physical outcomes.