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.

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 \(A_4 = A_1 = \pi \times 10^{-6} \,\text{m}^2\).

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) \(I_4 = \frac{200\,\text{W}}{\pi \times 10^{-6} \,\text{m}^2} = \frac{2 \times 10^5}{\pi} \,\text{W m}^{-2}\)

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.