Detailed answers
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2.1
(a) A sound wave is not polarized - it is a longitudinal wave (wave
that transfers energy in the same direction as that of propagation)
that works by compression and expansion of molecules along the path of
the wave. A transverse wave can be polarized (energy is transferred
perpendicular to the path of propagation).
http://www.physics.isu.edu/~hackmart/waves100.PDF
(b) Amazingly, if you put a polarizer at a 45 degree angle between two
orthogonal polarizers, you find that light is transmitted. This is
because a polarizer is not a passive element. It selects out a
particular polarization from a light wave - that is, it resolves that
particular polarization.
The first polarizer selects a particular component of the light
wave...say the y-direction. If you put polarizer at 45 degrees to this
you transmit 1/2 the intensity of the original wave (the square of the
electric field, or 1/(root 2) squared). Now, if you put a polarizer in
the x-direction, you resolve another 1/2 of the wave that is polarized
in the 45 degree orientation, or 1/4 of the original intensity.
(c) (see the Stokes vector section - this will do.
(d) If everything is emitting with the same color distribution and
intensity (called isotropic), then you can't see any features. It all
looks the same.
(e) Kirchhoff's law will fail when the molecules don't emit as though
they are blackbodies at the local temperature. This is most pronounced
at very high altitudes (> 60 km) where collisional quenching is too
slow to mix up all the energies, so you see specific, preferred,
emission lines.
(f) The sun can vary in intensity. It is called the solar constant
because it looks like a constant in an equation that we use for
balancing incoming and outgoing radiation, but it actually varies in
time
(to first order, at any one time it doesn't vary with position in
orbit, so long as the distance to the sun is the same).
(g) Stars emit like blackbodies, planets reflect light. But if there
weren't a light source like a star, the planet would look like a
blackbody (albeit, a cold one), so I am not sure I like this problem!
2.2
(a) Circular, rotating counterclockwise in x/y plane
(b) Elliptical, out of phase by 45 degrees, rotating clockwise
(c) Linear, 45 degree
2.3
E = 1016 kg m s-2 Co-1, H = 3.39 x 10-6
T
2.4
E = 34.7 kg m s-2 Co-1,
H = 1.16 x 10-7 T
2.5
E = Eo cos(kx
- wt) j
H = Eo/c cos
(kx - wt) k
Eo = 8.68 kg m s-2 Co-1
w = 3.77 x 1015 s-1
k = 1.26 x 107 m-1
2.6
v = 0.004c (or 1.2 x 106 m s-1) away from
Earth
2.7
(a) as in class notes
(b) 7.3 s-1 per mi hr-1
2.8
0.01
0.1
1.0 10 (m s-1)
0.69 2.9x104 2.9x105
2.9x106 2.9x107
1.04 1.9x104 1.9x105
1.9x106 1.9x107
10.6 1.9x103 1.9x104
1.9x105 1.9x106
3 cm 0.67
6.7
67 670
5 0.40
4.0
40
402
10 0.20
2.0
20
201
2.9
(b) [5/4, -3/4, 1, 0]
(d) [5/4, -3/4, 0, -1]
(e) [1, 0, 1/root(2), 1/root(2)]
2.10
(a) [1, 0, 0, -1]
(b) [1, 0, 1/root(2), -1/root(2)]
(c) [1, 0, 1, 0]
2.11
299 K (the problem should say 0.98 x 104 erg cm-2
micrometer-1 s-1
2.12
(a) 0.05
(b) 0.046
(c) 1.43 x 10-20
(d) exp(-456) - essentially zero, 300 K is too low a temperature to
produce any UV
2.13
To solve this, differentiate the two equations 2.47 and 2.44, set equal
to zero, and find the
non-zero or non-infinite root. You will end up with two transcendental
equations of the form:
(A) x = 3(1 - e-x)
(B) x = 5(1 - e-x)
These can be solved iteratively - start with x = 1.
(A) new x = 1.9, 2.55, 2.77, 2.81, 2.82, 2.821, 2.8214, etc.
(B) new x = 3.16, 4.79, 4.96, 4.965, 4.9651, etc.
Ratio = 4.9651/2.8214 = 1.76
2.14
Use taylor series expansion for exponential term, since
frequency goes to zero as wavelength goes to infinity
2.15
6.61 micrometers