|Physics Demystified, 2nd edition|
|Explanations for Quiz Answers in Chapter 12|
1. Let's use the formula for photon energy in terms of wavelength, given on page 391. If we let e represent the energy that a single photon contains (in joules) and we let λ represent the wavelength (in meters), then
e = 1.9865 x 10-25 / λ
We're told that λ = 1300 nm = 1.300 x 10-6 m. Therefore
(1.9865 x 10-25) / (1.300 x 10-6)
The correct choice is B.
|2. The particle theory, in which we imagine radiant energy as comprising discrete photons (energy "packets"), does a good job of explaining the physical pressure that a ray of radiant energy produces when it strikes a tiny object such as a grain of dust. The particle theory doesn't satisfactorily explain interference patterns, refraction, or variable propagation speeds in different media, so we can rule out choices A, B, and C. The only reasonable choice is D.|
3. Let's remember the formula relating the speed of light c (in meters per second) to the frequency f (in hertz) and the wavelength λ (in meters) for an electromagnetic (EM) wave in free space. That formula is
c = f λ
We can consider c = 3.00 x 108 m/s. Rearranging the above formula with a little algebra tells us that
f = 3.00 x 108 / λ
In this case, λ = 21 cm = 0.21 m, so we have
f = 3.00 x 108 / 0.21
The correct choice is A.
|4. All three of the wave types described here can propagate in the transverse mode, meaning that the motions of the particles in the medium move back and forth at right angles to the direction of propagation. The answer is D.|
5. When two waves having different frequencies beat together, we observe new waves at the sum and difference frequencies. In this case, the original waves have frequencies of 775 Hz and 995 Hz. Let's call these frequencies f and g, respectively. We therefore observe beat notes at frequencies of
g + f = 995 + 775
g - f = 995 - 775
Both choices A and C answer this question correctly, so we should select D, "More than one of the above."
6. To solve this problem, we need the formula relating the propagation speed c (in meters per second) to the wavelength λ (in meters) and the period T (in seconds) for a general wave disturbance. That formula is
λ = c T
We're told that c = 500 m/s for acoustic waves in the atmosphere of the planet on which we find ourselves. We want to know the frequency of a wave with T = 0.155 s. When we input these two values to the formula given above, we get
500 x 0.155
The correct choice is A.
|7. Most musical instruments produce irregular (or complex) acoustic waves. The correct choice is B. Specialized electronic musical instruments can generate rectangular (choice A) or sine (choice C) waves, but these systems are exceptions, not the rule. As for choice D, we might produce a transverse sound wave if we could play a musical instrument inside a solid medium such as steel, but as far as I know, no one has ever done it! In the atmosphere where people normally play musical instruments, the acoustic waves propagate in the longitudinal (or compression) mode, not in the transverse mode.|
8. As we did in the solution to Question 3, let's use the formula relating the speed of light, 3.00 x 108 m/s, to the frequency f (in hertz) and the wavelength λ (in meters) for an EM wave in free space:
f = 3.00 x 108 / λ
In this case, λ = 150 m, so
f = 3.00 x 108 / 150
The correct choice is C.
|9. The speed with which transverse waves will travel down a dangling rope depends on the type of rope (thick and heavy, or thin and light), and also on the mass of any object that we might attach to the bottom end. We don't have enough information to answer this question as stated. We must choose D.|
|10. Diffraction occurs easily when the wavelength of a disturbance is comparable to, or greater than, the diameter of an obstruction or the "sharpness" of a corner. When the wavelength is short compared to the diameter of an obstruction or the "sharpness" of a corner, diffraction doesn't readily take place. Of the four choices offered here, A represents the "worst-case scenario" for diffraction, because visible-light waves are far shorter than the diameter of a typical utility pole. In the other three cases, the wavelengths greatly exceed the diameters of the obstructions or the "sharpness" of the corners. We should choose A as our answer.|