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Chapter 15: Problem 2145

Light with an energy flux of \(18 \mathrm{w} / \mathrm{m}^{2}\) or \(\mathrm{Wm}^{-2}\) falls on a non-reflecting surface at normal to surface. If the surface has an area of \(20 \mathrm{~m}^{2}\). The average force exerted on the surface during 30 minutes is (A) \(6.48 \times 10^{5} \mathrm{~N}\) (B) \(3.60 \times 10^{2} \mathrm{~N}\) (C) \(1.2 \times 10^{-6} \mathrm{~N}\) (D) \(2.16 \times 10^{-3} \mathrm{~N}\)

Short Answer

Expert verified
The short version of the answer is: The average force exerted on the surface during 30 minutes is \(2.16 \times 10^{-3} \mathrm{~N}\) (option (D)).

Step by step solution

01

Calculate the total energy received by the surface

First, we need to find the total energy received by the non-reflecting surface. To do this, we multiply the energy flux (18 W/m²) by the surface area (20 m²), and then by the time interval in seconds: Total energy = Energy flux × Surface area × Time Total energy = 18 W/m² × 20 m² × (30 min × 60 s/min)
02

Calculate the momentum of the photons absorbed by the surface

To find the momentum of the photons absorbed by the surface, we will use the relationship between energy (E) and momentum (p) of photons: \[E = pc\] where p = momentum of photons c = speed of light ≈ \(3 \times 10^8\) m/s Therefore, we can find the momentum as follows: \[p = \frac{E}{c}\]
03

Calculate the average force exerted on the surface

Now, we should find the average force exerted on the surface. Force (F) is defined as the change in momentum (∆p) divided by the time interval (∆t): \[F = \frac{\Delta p}{\Delta t}\] Since the momentum before the light hit the surface was zero, the change in momentum is equal to the momentum of absorbed photons. Thus: \[F = \frac{p}{\Delta t}\] Finally, substitute the values of p and ∆t to find the average force exerted on the surface.
04

Choose the correct answer

Based on the calculated average force value, choose the appropriate answer from the given options (A), (B), (C), or (D).

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