Problem 4 - University Physics with Modern Physics

Question

Which of the following exerts the largest amount of radiation pressure? a) a \(1-\mathrm{mW}\) laser pointer on a \(2-\mathrm{mm}\) -diameter spot \(1 \mathrm{~m}\) away b) a 200-W light bulb on a 4 -mm-diameter spot \(10 \mathrm{~m}\) away c) a 100 -W light bulb on a 2 -mm-diameter spot 4 m away d) a 200 - \(\mathrm{W}\) light bulb on a 2 -mm-diameter spot \(5 \mathrm{~m}\) away e) All of the above exert the same pressure.

Step-by-Step Solution

Answer

The light sources that exert the largest amount of radiation pressure are (b) a 200-W light bulb on a 4-mm-diameter spot 10 meters away and (d) a 200-W light bulb on a 2-mm-diameter spot 5 meters away.

Step 1: Calculate the area for each scenario

First, we need to find the area of the spot on which the light is incident for each scenario. The area can be calculated using the formula for the area of a circle, \(A = \pi r^2\), where r is the radius of the circle. a) For a \(2\,\text{mm}\)-diameter spot, the radius is \((2 \times 10^{-3})/2\,\text{m} = 1 \times 10^{-3}\,\text{m}\). The area is \(A_1 = \pi \cdot (1 \times 10^{-3})^2 = \pi \times 10^{-6} \,\text{m}^2\). b) For a \(4\,\text{mm}\)-diameter spot, the radius is \((4 \times 10^{-3})/2\,\text{m} = 2 \times 10^{-3}\,\text{m}\). The area is \(A_2 = \pi \cdot (2 \times 10^{-3})^2 = 4\pi \times 10^{-6} \,\text{m}^2\). c) For a \(2\,\text{mm}\)-diameter spot, the area is the same as in part a, so \(A_3 = A_1 = \pi \times 10^{-6} \,\text{m}^2\). d) For a \(2\,\text{mm}\)-diameter spot, the area is the same as in part a, so \(A_4 = A_1 = \pi \times 10^{-6} \,\text{m}^2\).

Step 2: Calculate the intensity for each scenario

Now we can calculate the intensity for each scenario. The intensity is defined as the power of the effect divided by the area, \(I=P/A\). a) \(I_1 = \frac{1\,\text{mW}}{\pi \times 10^{-6} \,\text{m}^2} = \frac{10^{-3} \,\text{W}}{\pi \times 10^{-6} \,\text{m}^2} = \frac{10^3}{\pi} \,\text{W m}^{-2}\) b) \(I_2 = \frac{200\,\text{W}}{4\pi \times 10^{-6} \,\text{m}^2} = \frac{2 \times 10^5}{\pi} \,\text{W m}^{-2}\) c) \(I_3 = \frac{100\,\text{W}}{\pi \times 10^{-6} \,\text{m}^2} = \frac{10^5}{\pi} \,\text{W m}^{-2}\) d) \(I_4 = \frac{200\,\text{W}}{\pi \times 10^{-6} \,\text{m}^2} = \frac{2 \times 10^5}{\pi} \,\text{W m}^{-2}\)

Step 3: Compare the intensities

Now that we have the intensity for each scenario, we can compare them to see which has the largest amount of radiation pressure, keeping in mind that pressure is proportional to intensity. a) \(I_1 = \frac{10^3}{\pi} \,\text{W m}^{-2}\) b) \(I_2 = \frac{2 \times 10^5}{\pi} \,\text{W m}^{-2}\) c) \(I_3 = \frac{10^5}{\pi} \,\text{W m}^{-2}\) d) \(I_4 = \frac{2 \times 10^5}{\pi} \,\text{W m}^{-2}\) Comparing the intensities, we can see that \(I_2 = I_4 > I_3 > I_1\). Thus, the light source that exerts the largest amount of radiation pressure is (b) and (d) - a 200-W light bulb on a 4-mm-diameter spot \(10\,\text{m}\) away and a 200-W light bulb on a 2-mm-diameter spot \(5\,\text{m}\) away.

Key Concepts

Intensity of Light

The intensity of light helps us understand how much light energy is hitting a specific area. To find this, we use the formula:

• Intensity, \(I\), is calculated as the Power, \(P\), of the light source divided by the Area, \(A\), it shines on. Mathematically, this is \(I = \frac{P}{A}\).

For example, if a laser pointer has a power of 1 milliwatt and strikes a spot of a known size, the intensity will tell us how concentrated the light is at that spot.
Understanding intensity is crucial for calculating other properties like radiation pressure.

Power and Area Relationship

The power of a light source and the area it illuminates are directly related to the intensity of the light. Here's how these elements interact:

• The power is the total light energy emitted per second from a source.
• The area is the surface over which the light spreads.

When you calculate intensity, higher power or smaller area results in higher light intensity. For instance, a bright 200-watt light bulb focused on a tiny spot has a much greater intensity than the same bulb shining over a large area.
Hence, concentrating light power into a smaller spot increases the intensity.

Radiation Pressure Calculation

Radiation pressure is the force exerted by light when it hits a surface. The pressure depends on the intensity of the light:

• Radiation pressure \(P_\text{rad}\) is proportional to light intensity \(I\).
• Therefore, higher intensity leads to more radiation pressure.

This concept is useful in scenarios like solar sails for space travel, where light pushes a large, thin surface, moving the spacecraft through space.
In our exercise, comparing light sources' intensity tells us which exerts more pressure.

Laser Pointer

A laser pointer is a small hand-held device emitting a narrow, intense beam of light. Lasers have specific characteristics:

• They are monochromatic, meaning they produce light of a single color.
• Their beams are coherent, with photons moving in parallel, amplifying the intensity.

When you shine a laser pointer on a small spot, the intensity is quite high due to the concentrated beam, even if the power is relatively low.
This makes laser pointers useful in various applications from presentations to scientific measurements.

Light Bulb

Light bulbs are a common source of artificial light, available in different power ratings. Key points about light bulbs include:

• They emit light due to electrical energy converting into visible light energy.
• Their power rating (e.g., 100 watts) indicates how much energy they use per second.

Though not as intense as lasers, light bulbs can be very powerful when directed on a small area. This concentration can lead to high radiation pressure, essential in scenarios like our exercises to compare the effects of different lighting setups.
Understanding the relationship between power, area, and intensity helps explain the behavior of light bulbs in various applications.