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Chapter 27:

- 2: for visible range of 400nm - 700nm, λ in crown glass is 263nm - 460nm, for 380-760 nm it is 250-500 nm in crown glass
- 8: d = 1.22 μm
- 11: 577 nm
- 13: m=62 (set Sinθ=1, solve for m=62.5, but the biggest integer value of m=62)
- 23: 8990 lines/cm (solve for d, which is meters/line, convert to cm/line, and invert for lines/cm)
- 25: 5000 lines/cm gives d = 2.00 μm, λ = 707 nm
- 45: a) D=1.35 μm, b) θ=69.7
^{o} - 51: a) You can't use the small angle approximation, because the angles are 17.1306
^{o}and 17.1455^{o}, 0.0150^{o}different, b) And you cannot use the small angle approx to determine the separation, because it depends on the angles themselves. Find each y-value, then take the difference to get 0.273 mm - 58: 107 m
- 62: a) 2.237 x 10
^{-4}radians, which is 0.0128^{o}, b)5.8 km, c) 179 μm d) that is narrower than a pencil line, about the width of two hairs. - 86: 0.046 W/m
^{2} - 89: Final intensity is 1/4 the original intensity. If middle polarizer is removed, final intensity is zero.

Chapter 26:

- 2: 50.3 diopters
- 4: calculate the magnification, get 5.3 x 10
^{-6}, then calc the distance = 3.75 km - 6: 2.0 m
- 9: Original power = 51 diopters. so 51 = 1/x + 1/0.02, giving x = 1.0 m
- 16: -5 diopters is the prescription for contacts
- 17: -5.4 diopters is the prescription for glasses. (-5.6 D for glasses at 2 cm from eye)

Chapter 25:

- 2: Show that the change of direction equals 180 degrees
- 5: 2.25 x 10
^{8}m/s and 2.04 x 10^{8}m/s - 8: ice, n=1.31
- 13: n=1.457, so probably fused quartz (which is a fairly commonly used material
- 14: the time delay for two trips through the atmosphere is 5.86 x 10
^{-8}s, which is 2.29 x 10^{-6}% of the round trip time. - 22: 66.3
^{o} - 29: θ
_{red}= 46.5^{o}, θ_{violet}= 46.0^{o} - 36: 20.0 diopters
- 41: Screen is 3.43 m away. Mag = 33.3, so image is 80 cm x 120 cm.
- 42: a) -135 cm, that is, 135 cm from lens on the same side as the mole, b) m = 10, c) 50.0 mm image size
- 57: f = 36.0 cm
- 59: a) m = 0.111, b)image distance = -0.334 cm (behind the cornea), c) f = -0.376 cm, so R =2f = -0.752 cm

Chapter 24:

- 3: 1.50 x 10
^{5}V/m - 7: a) wavelengths from 188 to 556 m for AM, and 2.78 to 3.41 m for FM
- 12: λ = 0.25 μm, which is about the size of the smallest detail you could see with the UV microscope.
- 18: a) 9.46 x 10
^{15}m, b) 1.9 x 10^{22}m, c) 1.1 x 10^{26}m - 23: a) 193 nm has f = 1.55 x 10
^{15}Hz, the resolution ratio will be the ratio of the wavelengths, so for 400 nm light, 400/193 ≈2 two times better resolution. - 26: in 0.1 ns light goes 3 cm, which is the round trip uncertainty. Half of that is 1.5 cm, the uncertainty of the distance to the moon. b) 3.9 x 10
^{-9}percent - 32: Intensity, I = Power/area, 318 W/m
^{2}, b) B = 1.6 μT, c) E = 490 V/m - 42: a) 1.50 μm IR light, 200 W, in a 25.0 cm dia circle on your shoulder. Intensity, I=P/A = 4074 W/m
^{2}, b) E = 1750 V/m, c) B = 5.84 x 10^{-6}T, d) to heat 4.00 kg of shoulder requires 2.78 x 10^{4}J, and power = energy/time, so t = E/P = 139 seconds.

Chapter 23:

- 4: a) ccw, b) cw c) cw
- 9: a) 3.04 mV, b) pretty low power, not much temperature increase. Even if R = 0.001 Ω the power is only 9 mW for a short time.
- 12: B = 0.425 T
- 17: 0.63 V, not much
- 27: a) I=0, b) I = cw, c) I = 0, d) I = ccw, e) I=0
- 50: a) 2.24, b) 0.447 (which is 1/2.24), c) loss is 20% of what it was before (0.447 squared)
- 67: L = 200 H, b) temperature increases by 5
^{o}C

Chapter 22:

- 2: a) right, b) Perp. out of page, c) down, to bottom of page, d) 0, e) left, f) up, to top of page
- 3: a) right, b) perp. into page, c) down, to bottom of page
- 10: 10.1
^{o}or 169.9^{o} - 14: 0.261 T, easily achieved even by permanent magnets.
- 32: a) left, b) perp. out of page, c) up to top of page
- 37: B = 1.80 T
- 48: τ = 0.471 N m, viewed from above, rotation will be cw.
- 57: F = 7.5 x 10
^{-5}N toward the bottom of the page, in line with the right side of the square. - 68: Need I
_{2}going down. I_{1}/I_{2}= 1/π - 84: need F
_{B}= mg, so you need to figure out m. I looked up the density of copper, which I won't tell you so you also have the pride of that accomplishment. Ultimately, I got I = 1730 A. If earth contributes a field of 3 x 10^{-5}T, this is 11% of the field contributed by the lower wire, so decrease its current by 11% to 1540 A.

Chapter 21:

- 3: Series gives the max R
_{s}=786 Ω, parallel gives the min,R_{p}=20.3 Ω - 6: a) I=0.400 A, P
_{24}=3.84 W, P_{96}=15.4 W, b) R_{p}=19.2 Ω, I=2.50 A, 2.0 A through R_{24}and 0.5 A through R_{96} - 16: 2.9994 V
- 23: a) 200 A, b) 10.0 V, c) 2000 W, d) 80 A, V
_{mot}= 4.0 V, P_{mot}= 320 W - 26: b) 1.67 A, c) 27.9 kW, d) 20 MΩ

For Resistors

(A) R

Note: R

(B) R

Note: R

(C) R

If instead they are capacitors

(A) C

(B) C

(C) C

Chapter 20:

- 4: 4.00 mA
- 9: 0.12 C, 7.5 x 10
^{17}electrons - 23: 0.20 mA
- 25: 0.322 Ω
- 27: the Al will be 1.24 times the diameter of the Cu wire
- 40: P = 2 x 10
^{12}W. But not for very long. - 49: beam power is 1500 W
- 56: $3.94 billion
- 69: a) = 2.08 x 10
^{5}A, for that current, b) the waste heat will be I^{2}R= 4.3 x 10^{10}W. c) but that is greater than the power output of the generator! So it cannot happen. d) Don't confuse the idea of how much power you want to get to the destination with how much power you need to produce at the source. - 73: 480 V
- 76: R = 576 Ω, power during melt down is 100 W.

Chapter 19:

- 2: eV=1/2 mv
^{2}, so v=1.2 x 10^{8}m/s - 5: 1940 V
- 7: a) 2 GJ, b) 765 kg of water can evaporate (mcΔT + mL
_{f}... remember?) - 13: yup, they are equivalent
- 17: a) E=2.5 MV, so no breakdown (no spark), b) at d=1.7 mm you get a big enough E-field.
- 29: Q = +0.833 μC
- 38: sketch so equipotentials are perp to E-field. Higher potential is in direction opposite the arrows of E-field
- 50: 20,000 V
- 61: C
_{net}= 11.4 μF. (a couple of intermediate combinations give 2.06 and 1.37 μF) - 66: series: 14.2 μC and 63.8 μJ, parallel: 84.6 μC and 381 μJ

Chapter 18:

- 1: a) 1.25 x 10
^{10}electrons b) 3.13 x 10^{12}electrons - 6: a) 1.88 x 10
^{6}electrons short of a full load, b) 1.88 x 10^{-10} - 16: 14.4 N away from Qa, (or in 1 sig fig, 10 N)
- 17: 3.45 x 10
^{16}m/s^{2} - 20: 1.04 nC
- 25: 85.9 cm right of negative charge, or, if both positive 10.9 cm left of +3 μC
- 30: 69.4 nC and 6.25 N/C
- 43: a) 4 x 10
^{7}N/C, b) 7 cm, in the middle, c)No, there is no suitable place 0-8 cm, d) The net charge in fig (a) is zero, but the net charge in fig (b) is +1, so E goes to zero faster in the case of fig. (a), and e) Add the electric fields due to the three charges for some point right of 11 cm. This is ugly algebra, which is why they suggest looking at the graph of the sum of the E-fields (if you have some way to make a graph of a function), then the graph of the sum of the E-fields crosses the zero at x=30.6 cm - 48: Vector addition! All the x-components cancel, and the net vector is 0.102 N in the negative y-direction
- 57: remember circular motion? F=ma=mv
^{2}/r, and v=rω, but F=kqQ/r^{2}, put it all together and get ω = 4.12 x 10^{16}rad/sec (and from v=rω get v=2.2 x 10^{6}m/s, if you are interested)

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