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Chapter 31: Problem 35

A monochromatic point source of light emits \(1.5 \mathrm{~W}\) of electromagnetic power uniformly in all directions. Find the Poynting vector at a point situated at each of the following locations: a) \(0.30 \mathrm{~m}\) from the source b) \(0.32 \mathrm{~m}\) from the source c) \(1.00 \mathrm{~m}\) from the source

Short Answer

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Question: Calculate the Poynting vector (energy flux) at points with different distances from a point source of light that emits 1.5 W of electromagnetic power uniformly in all directions: a) 0.30m, b) 0.32m, and c) 1.00m. Answer: a) At 0.30m, the Poynting vector is 1.326 W/m虏. b) At 0.32m, the Poynting vector is 1.161 W/m虏. c) At 1.00m, the Poynting vector is 0.119 W/m虏.

Step by step solution

01

Find the area at 0.30m distance

First, we need to find the surface area for a sphere whose radius is 0.30m. We'll use the formula for the surface area of a sphere: $$A = 4\pi r^2$$ With \(r = 0.30 \mathrm{~m}\): $$A_{0.30m} = 4\pi (0.30)^2 = 4\pi (0.09) = 1.131\mathrm{~m虏}$$
02

Calculate the Poynting vector at 0.30m distance

Now, we'll use the formula for the Poynting vector to find the energy flux at 0.30m distance from the source: $$\vec{S}=\frac{P}{A}$$ With \(P = 1.5 \mathrm{~W}\) and \(A = 1.131\mathrm{~m虏}\): $$\vec{S}_{0.30m}=\frac{1.5}{1.131} = 1.326\mathrm{~W/m虏}$$
03

Find the area at 0.32m distance

As we did in Step 1, we need to find the surface area for a sphere whose radius is 0.32m: $$A_{0.32m} = 4\pi (0.32)^2 = 4\pi (0.1024) = 1.292\mathrm{~m虏}$$
04

Calculate the Poynting vector at 0.32m distance

Now, we'll use the formula for the Poynting vector to find the energy flux at 0.32m distance from the source: $$\vec{S}_{0.32m}=\frac{P}{A}$$ With \(A = 1.292\mathrm{~m虏}\): $$\vec{S}_{0.32m}=\frac{1.5}{1.292} = 1.161\mathrm{~W/m虏}$$
05

Find the area at 1.00m distance

Finally, we need to find the surface area for a sphere whose radius is 1.00m: $$A_{1.00m} = 4\pi (1.00)^2 = 4\pi = 12.566\mathrm{~m虏}$$
06

Calculate the Poynting vector at 1.00m distance

We'll use the formula for the Poynting vector to find the energy flux at 1.00m distance from the source: $$\vec{S}_{1.00m}=\frac{P}{A}$$ With \(A = 12.566\mathrm{~m虏}\): $$\vec{S}_{1.00m}=\frac{1.5}{12.566} = 0.119\mathrm{~W/m虏}$$ To summarize, the Poynting vector for the given distances is: a) \(0.30\mathrm{~m}\) : \(\vec{S}_{0.30m}=1.326\mathrm{~W/m虏}\) b) \(0.32\mathrm{~m}\) : \(\vec{S}_{0.32m}=1.161\mathrm{~W/m虏}\) c) \(1.00\mathrm{~m}\) : \(\vec{S}_{1.00m}=0.119\mathrm{~W/m虏}\)

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Most popular questions from this chapter

Quantum theory says that electromagnetic waves actually consist of discrete packets-photons-each with energy \(E=\hbar \omega,\) where \(\hbar=1.054573 \cdot 10^{-34} \mathrm{~J} \mathrm{~s}\) is Planck's reduced constant and \(\omega\) is the angular frequency of the wave. a) Find the momentum of a photon. b) Find the angular momentum of a photon. Photons are circularly polarized; that is, they are described by a superposition of two plane-polarized waves with equal field amplitudes, equal frequencies, and perpendicular polarizations, one-quarter of a cycle \(\left(90^{\circ}\right.\) or \(\pi / 2\) rad \()\) out of phase, so the electric and magnetic field vectors at any fixed point rotate in a circle with the angular frequency of the waves. It can be shown that a circularly polarized wave of energy \(U\) and angular frequency \(\omega\) has an angular momentum of magnitude \(L=U / \omega .\) (The direction of the angular momentum is given by the thumb of the right hand, when the fingers are curled in the direction in which the field vectors circulate. c) The ratio of the angular momentum of a particle to \(\hbar\) is its spin quantum number. Determine the spin quantum number of the photon.

Determine the distance in feet that light can travel in vacuum during \(1.00 \mathrm{~ns}\).

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