| Physics Demystified, 2nd edition |
| Stan Gibilisco |
| Explanations for Quiz Answers in Chapter 14 |
| 1. We know that light of any wavelength (including the wavelength for green) propagates through a vacuum at 3.00 x 108 m/s. If we want to know the speed at which a ray of light will propagate through a medium whose index of refraction equals 1.80, we divide 3.00 x 108 m/s by 1.80 to get 1.67 x 108 m/s. The correct choice is C. |
| 2. We need a converging lens to collimate (make parallel) the rays of light that come from a nearby point source. A plano-convex lens behaves as a converging lens. The answer is A. Choices B and C are both absolutely wrong! Lenses of those types would cause the already divergent rays from a point source to spread out even more. |
| 3. Total internal reflection with visible light can only occur when a ray encounters a boundary between media having two different refractive indices (or, if you prefer, indexes), and the second medium has a refractive index that's lower than that of the first medium. Of the three choices given here, only C describes a situation of that sort. Water has a higher refractive index than air, so if we aim a visible laser beam upward toward the surface (with the laser device located underwater), we'll observe total internal refraction if the angle of incidence at the surface is larger than or equal to the critical angle. |
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4. Let's use the formula for the critical angle xc from pages 440 and 441. When a ray of light passes through a flat boundary from a medium having a relatively higher refractive index r to a medium having a relatively lower refractive index s, then xc = Arcsin (s/r) We're told that r = 1.75 (for the clear, solid material) and s = 1.33 (for the fresh water). Therefore xc = Arcsin (1.33 / 1.75) The correct choice is B. |
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5. Let's refer to the formula for numerical aperture Ao in terms of the refractive index rm of a medium M and the angle q as shown in Fig. 14-15: Ao = rm sin q This information appears on pages 462 and 463. When we compare Fig. 14-15 with Fig. 14-16 (page 466), we see that in the situation at hand, rm = 1.50 and q = 25º. When we input these values into the formula, we get Ao = 1.50 sin 25º We round the answer off to two significant figures because we only know the measure of q to that level of accuracy. The answer is C. |
| 6. The diameter of the objective lens determines the resolving power of a refracting telescope. Remember, resolving power expresses the extent to which the telescope allows us to distinguish between two distant objects that appear "close together," no matter how great the magnification. In practical situations, the resolving power determines the amount of detail that a telescope can render. The eyepiece diameter, eyepiece focal length, and objective focal length affect various other parameters, but not the resolving power of the entire instrument. The correct choice is D. |
| 7. A first-surface concave paraboloidal mirror with a focal length of 80 cm will make an ideal objective for a Newtonian reflecting telescope (although, if we decide to build the telescope, we'll get a fairly small instrument, suitable only for amateur astronomy). The answer is A. |
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8. To calculate the magnification factor m for a telescope, we divide the focal length of the objective by the focal length of the eyepiece. In this case, the objective has a focal length of 1.50 m while the eyepiece has a focal length of 30 mm = 0.030 m. Therefore m = 1.50 / 0.030 The correct choice is B. |
| 9. The clear, solid material has a refractive
index of 1.05 while the water has a refractive index of 1.33. Before we
proceed further, we had better make certain that the angle of incidence,
66º, is less than the critical angle as the ray encounters the first
boundary between the water and the solid material. (If the angle of
incidence equals or exceeds the critical angle, then the ray will undergo
total internal reflection at the boundary.) Let's remember the
formula for the critical angle xc. When a ray passes
from from a medium having a relatively higher refractive index
r into a medium having a relatively lower refractive index
s, then
xc = Arcsin (s/r) In this case, we have r = 1.33 (for the water) and s = 1.05 (for the solid material). Therefore xc = Arcsin (1.05 / 1.33) The angle of incidence exceeds the critical angle, so the ray will not, in fact, enter the solid material. Instead, it will reflect back into the water. The correct choice is D. |
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10. When we want to approximate the magnification m for a microscope, we can multiply the magnification of the objective by the magnification of the eyepiece. In this case, the objective has a magnification of 60x and the eyepiece has a magnification of 15x. Therefore m = 60 x 15 The answer is A. |