Chapter 34
Questions: 9, 10, 11, 14, 17, 18, 21
Problems: 13, 17, 21, 29, 33, 35, 37
Questions
9. At that distance even a small change in the speed of light for di¤erent
frequencies would result in vastly di¤erent arrival times.
10. The optimum reception occurs when the electric …eld of the radiowave
(an electromagnetic wave) induces charge separation (and voltages) in the TV
antenna. This requires that the electric …eld of the radiowave being parallel
to the antenna rods. The radio wave you receive does not in general travel
horizontally, hence vertical rods will not interact well with the radio wave.
11. Polarizers allow for the component of the electric …eld of the EM
wave to pass through creating a polarization direction of the resulting EM
wave.. Hence two polarizers at right angles will block an EM wave. If you
place an additional polarizer between the two at right angles (at an angle
not parallel to either of them) then not only will some component of the E
…eld pass through, but the polarization direction of the resulting EM wave
will no longer be orthogonal to the …nal polarizer.
14. Hardly, millions of years of evolution has insured that the sensitivity
of our eyes (and all eyes of animals) match the peak in frequency distribution
of the energy emitted by the Sun.
17. The intensity is proportional to the square of the electric …eld, hence
the intensity is increased by a factor of 4.
18. No, the EM wave energy is distributed over a larger area.
21. The free electrons in a conducting metal interact with the electric
…eld of the EM wave. These electrons ‡ow in such a way to short out the
electric …eld.
Problems
13. The travel time is approximately
2h
2 36 106
d
=
=
= :24 sec
c
c
3 108
17. The electron beam travels 1cm in approximately
t=
10 4
=2
50
10 6 sec light travels a distance
t=
In 2
d = ct = 3
108
2
1
10
10
6
sec
6
= 600m
21. In an EM wave c =
wavelength is
f: If the frequency is f = 60Hz; then the
108 =60 = 5
= c=f = 3
106 m = 5000km
29. For polarized light the emitted intensity is
p
S = S0 cos2 ! cos = I=I0 :
If I=I0 = :2; then the angle is
= cos
1
p
:2 = 63:4
33. For unpolarized light the emitted intensity is
S=
S0
S0
cos2 =
cos2 35 = :336S0
2
2
35. For unpolarized light the emitted intensity is
S0
S0
4
cos8 20 = :304S0
cos2
=
2
2
37. The electric …eld of the polarized light makes an angle with the
polarizing axis of
(t) = !t = 2 f t = 20 t:
S=
Hence the transmitted intensity is
S (t) = S0 cos2 (t) = S0 cos2 20 t
2