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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 -W light bulb on a 2 -mm-diameter spot 5 m away e) All of the above exert the same pressure.

Short Answer

Expert verified
a) a 1-mW laser on a 2-mm-diameter spot 5 m away b) a 200-W light bulb on a 4-mm-diameter spot 5 m away c) a 100-W light bulb on a 2-mm-diameter spot 5 m away d) a 200-W light bulb on a 2-mm-diameter spot 5 m away Answer: The largest radiation pressure is exerted by option d) a 200-W light bulb on a 2-mm-diameter spot 5 m away.

Step by step solution

01

Calculate the area of the illuminated spots

For each option, we need to find the area of the illuminated spot using the formula for the area of a circle. We will use the diameter to find the radius and then calculate the area. a) Radius = \(\frac{2\,\text{mm}}{2} \times 10^{-3}\,\text{m}\) = \(10^{-3}\,\text{m}\). Area = \(\pi (10^{-3})^2 = \pi\times10^{-6}\,\text{m}^2\). b) Radius = \(\frac{4\,\text{mm}}{2} \times 10^{-3}\,\text{m}\) = \(2\times10^{-3}\,\text{m}\). Area = \(\pi (2\times10^{-3})^2 = 4\pi\times10^{-6}\,\text{m}^2\). c) Radius = Same as (a), Area = same as (a) which is \(\pi\times10^{-6}\,\text{m}^2\). d) Radius = Same as (a), Area = same as (a) which is \(\pi\times10^{-6}\,\text{m}^2\).
02

Calculate the intensity of light for each option

Using the power of the light source and the area calculated in step 1, calculate the intensity of light for each option. a) Intensity = \(\frac{1\,\text{mW}}{\pi\times10^{-6}\,\text{m}^2} = 1000/(\pi\times10^{-6}\,\text{m}^2) = \frac{10^6}{\pi}\,\text{W/m}^2\) b) Intensity = \(\frac{200\,\text{W}}{4\pi\times10^{-6}\,\text{m}^2} = \frac{200}{4\pi\times10^{-6}\,\text{m}^2} = \frac{50\times10^6}{\pi}\,\text{W/m}^2\) c) Intensity = \(\frac{100\,\text{W}}{\pi\times10^{-6}\,\text{m}^2} = \frac{10^8}{\pi}\,\text{W/m}^2\) d) Intensity = \(\frac{200\,\text{W}}{\pi\times10^{-6}\,\text{m}^2} = \frac{2\times10^8}{\pi}\,\text{W/m}^2\)
03

Calculate the radiation pressure for each option

Using the intensity calculated in step 2, calculate the radiation pressure (in pascals) for each option. The speed of light (c) is approximately \(3\times10^8\,\text{m/s}\). a) Pressure = \(\frac{10^6/\pi}{3\times10^8} = \frac{1}{3\pi}\times10^{-2}\,\text{Pa}\) b) Pressure = \(\frac{50\times10^6/\pi}{3\times10^8} = \frac{50}{3\pi}\times10^{-2}\,\text{Pa}\) c) Pressure = \(\frac{10^8/\pi}{3\times10^8}=\frac{1}{3\pi}\times10^{-1}\,\text{Pa}\) d) Pressure = \(\frac{2\times10^8/\pi}{3\times10^8}=\frac{2}{3\pi}\times10^{-1}\,\text{Pa}\)
04

Compare the radiation pressures

Compare the radiation pressures calculated in step 3 to find which option exerts the largest radiation pressure: a) \(P = \frac{1}{3\pi}\times10^{-2}\,\text{Pa}\) b) \(P = \frac{50}{3\pi}\times10^{-2}\,\text{Pa}\) c) \(P = \frac{1}{3\pi}\times10^{-1}\,\text{Pa}\) d) \(P = \frac{2}{3\pi}\times10^{-1}\,\text{Pa}\) Since the radiation pressure in option d) is the largest among all the options, the correct answer is option d) a 200-W light bulb on a 2-mm-diameter spot 5 m away.

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Key Concepts

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

Understanding Intensity of Light
Let's unravel the concept of intensity of light, a pivotal factor in determining radiation pressure. The intensity of light can be conceptualized as the amount of energy delivered by light waves to a certain area over a specific time period. It's expressed in the units of Watts per square meter (W/m2).

For example, when comparing two lightbulbs of equal wattage, the bulb illuminating a smaller area would have a greater intensity since the same amount of energy is concentrated over a smaller surface. This is crucial in the study of radiation pressure as a higher intensity signifies a stronger force being exerted on the illuminated surface. In effect, more energy per unit area translates to greater radiation pressure exerted.
Area of Illumination and Its Effect on Radiation Pressure
When discussing radiation pressure, the area of illumination holds great significance. It refers to the size of the surface area that light is shining upon. As we know, radiation pressure is the force exerted by the photon momentum of light on any surface it hits. However, the extent of this force is inversely proportional to the area being illuminated.

For instance, focusing a laser beam to a pinpoint creates a high intensity of light and subsequently a high radiation pressure on that tiny spot. In contrast, spreading the same amount of light over a larger area reduces the intensity and thus the radiation pressure. This explains why, in the exercise provided, the options with illumination over smaller areas (with everything else being equal) will exert a higher radiation pressure. We calculate the area using the formula for the area of a circle, taking care to convert diameters to radii, and the result directly influences the resulting light intensity and hence the radiation pressure.
Pascals as the Pressure Unit in Physics
To understand the output of our calculations for radiation pressure, we need to be familiar with Pascals (Pa), the SI unit of pressure. One Pascal is defined as one newton per square meter (N/m2). It's a small unit, often used in conjunction with prefixes such as kilopascal (kPa) or megapascal (MPa) for practical applications.

Pascals measure not just mechanical pressure but also stress and tensile strength. In the context of radiation pressure, the values are often very small, which is why we see exponents like -2 or -1 following the numbers in our calculations. These small values don't negate their significance though; for highly sensitive instruments, even the tiniest force exerted by radiation pressure can have measurable effects. This brings practical relevance to understanding the Pascal when studying phenomena such as radiation pressure in physics.

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Most popular questions from this chapter

Two polarizers are out of alignment by \(30.0^{\circ} .\) If light of intensity \(1.00 \mathrm{~W} / \mathrm{m}^{2}\) and initially polarized halfway between the polarizing angles of the two filters passes through both filters, what is the intensity of the transmitted light?

What is the electric field amplitude of an electromagnetic wave whose magnetic field amplitude is \(5.00 \cdot 10^{-3} \mathrm{~T} ?\)

During the testing of a new light bulb, a sensor is placed \(52.5 \mathrm{~cm}\) from the bulb. It records a root-mean-square value of \(9.142 \cdot 10^{-7} \mathrm{~T}\) for the magnetic field of the radiation emitted by the bulb. What is the intensity of that radiation at the sensor's location?

A continuous-wave (cw) argon-ion laser beam has an average power of \(10.0 \mathrm{~W}\) and a beam diameter of \(1.00 \mathrm{~mm} .\) Assume that the intensity of the beam is the same throughout the cross section of the beam (which is not true, as the actual distribution of intensity is a Gaussian function). a) Calculate the intensity of the laser beam. Compare this with the average intensity of sunlight at Earth's surface ( \(1400 . \mathrm{W} / \mathrm{m}^{2}\) ). b) Find the root-mean-square electric field in the laser beam. c) Find the average value of the Poynting vector over time. d) If the wavelength of the laser beam is \(514.5 \mathrm{nm}\) in vacuum, write an expression for the instantaneous Poynting vector, where the instantaneous Poynting vector is zero at \(t=0\) and \(x=0\). e) Calculate the root-mean-square value of the magnetic field in the laser beam.

A microwave operates at \(250 .\) W. Assuming that the waves emerge from a point source emitter on one side of the oven, how long does it take to melt an ice cube \(2.00 \mathrm{~cm}\) on a side that is \(10.0 \mathrm{~cm}\) away from the emitter if \(10.0 \%\) of the photons that strike the cube are absorbed by it? How many photons of wavelength \(10.0 \mathrm{~cm}\) hit the ice cube per second? Assume a cube density of \(0.960 \mathrm{~g} / \mathrm{cm}^{3}\)

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