台大生的作業－山客的部落格

18. From the equation immediately preceding Eq. 33-12, we see that the maximum value of ¶B/¶t is wB_m. We can relate B_m to the intensity:

and relate the intensity to the power P (and distance r) using Eq. 33-27. Finally, we relate w to wavelength l using w = kc = 2pc/l. Putting all this together, we obtain

14. (a) The power received is

(b) The power of the source would be

30. Eq. 33-27 suggests that the slope in an intensity versus inverse-square-distance graph (I plotted versus r ^-2) is P/4p. We estimate the slope to be about 20 (in SI units) which means the power is P = 4p(30) » 2.5 ×10² W.

22. (a) The radiation pressure produces a force equal to

(b) The gravitational pull of the Sun on Earth is

which is much greater than F_r.

28. The mass of the cylinder is where D is the diameter of the cylinder. Since it is in equilibrium

We solve for H:

31. We shall assume that the Sun is far enough from the particle to act as an isotropic point source of light.

(a) The forces that act on the dust particle are the radially outward radiation force and the radially inward (toward the Sun) gravitational force . Using Eqs. 33-32 and 33-27, the radiation force can be written as

where R is the radius of the particle, and is the cross-sectional area. On the other hand, the gravitational force on the particle is given by Newton’s law of gravitation (Eq. 13-1):

where is the mass of the particle. When the two forces balance, the particle travels in a straight path. The condition that implies

which can be solved to give

(b) Since varies with and varies with if the radius R is larger, then , and the path will be curved toward the Sun (like path 3).

34. After passing through the first polarizer the initial intensity I₀ reduces by a factor of 1/2. After passing through the second one it is further reduced by a factor of cos² (p – q₁ – q₂) = cos² (q₁ + q₂). Finally, after passing through the third one it is again reduced by a factor of cos² (p – q₂ – q₃) = cos² (q₂ + q₃). Therefore,

Thus, 0.045% of the light’s initial intensity is transmitted.

35. Let I₀ be the intensity of the unpolarized light that is incident on the first polarizing sheet. The transmitted intensity is and the direction of polarization of the transmitted light is q₁ = 40° counterclockwise from the y axis in the diagram. The polarizing direction of the second sheet is q₂ = 20° clockwise from the y axis, so the angle between the direction of polarization that is incident on that sheet and the polarizing direction of the sheet is 40° + 20° = 60°. The transmitted intensity is

and the direction of polarization of the transmitted light is 20° clockwise from the y axis. The polarizing direction of the third sheet is q₃ = 40° counterclockwise from the y axis. Consequently, the angle between the direction of polarization of the light incident on that sheet and the polarizing direction of the sheet is 20° + 40° = 60°. The transmitted intensity is

Thus, 3.1% of the light’s initial intensity is transmitted.

33. The angle between the direction of polarization of the light incident on the first polarizing sheet and the polarizing direction of that sheet is q₁ = 70°. If I₀ is the intensity of the incident light, then the intensity of the light transmitted through the first sheet is

The direction of polarization of the transmitted light makes an angle of 70° with the vertical and an angle of q₂ = 20° with the horizontal. q₂ is the angle it makes with the polarizing direction of the second polarizing sheet. Consequently, the transmitted intensity is

40. We note the points at which the curve is zero (q₂ = 0° and 90°) in Fig. 33-45(b). We infer that sheet 2 is perpendicular to one of the other sheets at q₂ = 0°, and that it is perpendicular to the other of the other sheets when q₂ = 90°. Without loss of generality, we choose q₁ = 0°, q₃ = 90°. Now, when q₂ = 30°, it will be Dq = 30° relative to sheet 1 and Dq¢ = 60° relative to sheet 3. Therefore,

36. We examine the point where the graph reaches zero: q ₂ = 160º. Since the polarizers must be “crossed” for the intensity to vanish, then q₁ = 160º – 90º = 70º. Now we consider the case q ₂ = 90º (which is hard to judge from the graph). Since q₁ is still equal to 70º, then the angle between the polarizers is now Dq =20º. Accounting for the “automatic” reduction (by a factor of one-half) whenever unpolarized light passes through any polarizing sheet, then our result is

cos²(Dq) = 0.442 » 44%.

48. (a) For the angles of incidence and refraction to be equal, the graph in Fig. 33-50(b) would consist of a “y = x” line at 45º in the plot. Instead, the curve for material 1 falls under such a “y = x” line, which tells us that all refraction angles are less than incident ones. With q₂ < q₁ Snell’s law implies n₂ > n₁.

(b) Using the same argument as in (a), the value of n₂for material 2 is also greater than that of water (n₁).

(c) It’s easiest to examine the topmost point of each curve. With q₂ = 90º and q₁ = ½(90º), and with n₂ = 1.33 (Table 33-1) we find n₁ = 1.9 from Snell’s law.

(d) Similarly, with q₂ = 90º and q₁ = ¾(90º), we obtain n₁ = 1.4.

47. The angle of incidence for the light ray on mirror B is 90° – q. So the outgoing ray r' makes an angle 90° – (90° – q) = q with the vertical direction, and is antiparallel to the incoming one. The angle between i and r' is therefore 180°.

54. (a) From n₁sinq₁ = n₂sinq₂ and n₂sinq₂ = n₃sinq₃, we find n₁sinq₁ = n₃sinq₃. This has a simple implication: that q₁ =q₃ when n₁ = n₃. Since we are given q₁ = 40º in Fig. 33-56(a) then we look for a point in Fig. 33-56(b) where q₃ = 40º. This seems to occur at n₃ = 1.6, so we infer that n₁ = 1.6.

(b) Our first step in our solution to part (a) shows that information concerning n₂ disappears (cancels) in the manipulation. Thus, we cannot tell; we need more information.

50. (a) A simple implication of Snell’s law is that q₂ = q₁ when n₁ = n₂. Since the angle of incidence is shown in Fig. 33-52(a) to be 30º, then we look for a point in Fig. 33-52(b) where q₂ = 30º. This seems to occur when n₂ = 1.7. By inference, then, n₁ = 1.7.

(b) From 1.7sin(60º) = 2.4sin(q₂) we get q₂ = 38°.

51. Consider a ray that grazes the top of the pole, as shown in the diagram that follows. Here q₁ = 90° – q = 35°, and The length of the shadow is x + L. x is given by

According to the law of refraction, n₂ sin q₂ = n₁ sin q₁. We take n₁ = 1 and n₂ = 1.33 (from Table 33-1). Then,

L is given by

The length of the shadow is 0.35 m + 0.72 m = 1.07 m.

67. (a) A ray diagram is shown below.

Let q₁ be the angle of incidence and q₂ be the angle of refraction at the first surface. Let q₃ be the angle of incidence at the second surface. The angle of refraction there is q₄ = 90°. The law of refraction, applied to the second surface, yields n sin q₃ = sin q₄ = 1. As shown in the diagram, the normals to the surfaces at P and Q are perpendicular to each other. The interior angles of the triangle formed by the ray and the two normals must sum to 180°, so q₃ = 90° – q₂ and

According to the law of refraction, applied at Q, The law of refraction, applied to point P, yields sin q₁ = n sin q₂, so sin q₂ = (sin q₁)/n and

Squaring both sides and solving for n, we get

(b) The greatest possible value of sin² q₁ is 1, so the greatest possible value of n is

(c) For a given value of n, if the angle of incidence at the first surface is greater than q₁, the angle of refraction there is greater than q₂ and the angle of incidence at the second face is less than q₃ (= 90° – q₂). That is, it is less than the critical angle for total internal reflection, so light leaves the second surface and emerges into the air.

(d) If the angle of incidence at the first surface is less than q₁, the angle of refraction there is less than q₂ and the angle of incidence at the second surface is greater than q₃. This is greater than the critical angle for total internal reflection, so all the light is reflected at Q.

63. When examining Fig. 33-62, it is important to note that the angle (measured from the central axis) for the light ray in air, q, is not the angle for the ray in the glass core, which we denote q ' . The law of refraction leads to

assumingThe angle of incidence for the light ray striking the coating is the complement of q ', which we denote as q'_comp and recall that

In the critical case, q'_comp must equal q_c specified by Eq. 33-47. Therefore,

which leads to the result: With n₁ = 1.58 and n₂ = 1.53, we obtain

71. (a) The first contribution to the overall deviation is at the first refraction: The next contribution to the overall deviation is the reflection. Noting that the angle between the ray right before reflection and the axis normal to the back surface of the sphere is equal to q_r, and recalling the law of reflection, we conclude that the angle by which the ray turns (comparing the direction of propagation before and after the reflection) is The final contribution is the refraction suffered by the ray upon leaving the sphere: again. Therefore,

(b) We substitute into the expression derived in part (a), using the two given values for n. The higher curve is for the blue light.

(c) We can expand the graph and try to estimate the minimum, or search for it with a more sophisticated numerical procedure. We find that the q_dev minimum for red light is 137.63°137.6°, and this occurs at q_i = 59.52°.

(d) For blue light, we find that the q_dev minimum is 139.35°139.4°, and this occurs at q_i = 59.52°.

(e) The difference in q_dev in the previous two parts is 1.72°.