Problem 1:
A sonar device emits a sound with a frequency of 30 kHz as it approaches a
wall and measures the frequency of the echo as 30.9 kHz.
If the speed of sound is 340 m/s, what is the speed with which the device
approaches the wall?
Solution:
 Concepts:
The Doppler effect
 Reasoning:
f = f_{0}(vv_{o})/(vv_{s}),
f = observed frequency, f_{0} = frequency of source,
v = wave velocity, v_{o} = velocity of observer,
v_{s} = velocity of source.
v_{o} and v_{s} are not the speeds, but the components of the
observer's and the source's velocity in the direction of the velocity of the
sound reaching the observer.
 Details of the calculation:
The wall receives sound with frequency f' = 30kHz(340 m/s)/(340 m/s
 u).
Here u is the speed with which the device approaches the wall.
The wall is the source of the reflected wave with frequency f'.
The device measures the frequency of the echo as 30.9 kHz = f'(340 m/s + u)/(340
m/s).
30.9 kHz = (340 m/s + u)/(340 m/s  u). u = 5.02 m/s.
Problem 2:
Two point sources emit spherical waves with wave number k and frequency ω
into 3D space. One source is located at the origin and one source a small
distance r away from the origin. A detector is located far
away at R, with R >> r.
For each source, let Ψ_{i} = A exp(i(kr_{i}  ωt
+ φ_{i}))/r_{i},
where r_{i} is the distance of the detector from the source and A is real. For the source at the
origin choose φ = 0, while for the source at r choose a nonzero φ.
(a) Write down an expression Ψ_{total}^{2} (proportional
to the average intensity) at the detector. Make the farfield approximation, i.e.
lines from each source to the detector are parallel to each other.
(b) Let k = kR/R. What condition must hold for k·r,
so that the average intensity at the detector is maximized?
(c) What happens if another point source is added at position 2r
with phase 2φ.
(d) For case (c), what condition must hold for k·r,
so that the average intensity at the detector zero?
Solution:
 Concepts:
Wave interference
 Reasoning:
The average intensity at R is proportional to Ψ_{1}(R) + Ψ_{2}(R)^{2}
 Details of the calculation:
(a) Ψ_{1}(R) = Aexp(i(kR  ωt))/R, Ψ_{2}(R) =
Aexp(i(kR  k·r  ωt + φ))/R  r.
Since R >> r we can neglect the difference between R and R  r
in the denominator, but not in the phase.
Ψ_{1}(R) + Ψ_{2}(R)^{2}
= (2A^{2}/R^{2}) + (A^{2}/R^{2}) [exp(i(kR  ωt))exp(i(kR
 k·r  ωt + φ) + exp(i(kR  ωt))exp(i(kR 
k·r  ωt + φ))]
= (2A^{2}/R^{2}) + (A^{2}/R^{2}) [exp(i(k·r
 φ)) + exp(i(k·r  φ))]
= (2A^{2}/R^{2}) + (2A^{2}/R^{2}) cos(k·r
 φ).
(b) We need cos(k·r  φ) = 1 for maximum
intensity.
We need k·r  φ = n2π, of k·r
= n2π + φ, n = 0, 1, 2, ... .
(c) Ψ_{1}(R) + Ψ_{2}(R) + Ψ_{3}(R)^{2}
= (3A^{2}/R^{2}) + (2A^{2}/R^{2})[ 2cos(k·r
 φ) + cos(2(k·r  φ))].
We need cos(k·r  φ) = 1 and cos(2(k·r
 φ)) = 1 for maximum intensity.
This still happens when k·r = n2π + φ, n =
0, 1, 2, ... .
(d) We need Ψ_{1}(R) + Ψ_{2}(R) + Ψ_{3}(R)
= 0.
exp(i(kR  ωt)) + exp(i(kR
 k·r  ωt + φ) + exp(i(kR
 2k·r  ωt + 2φ) = 0.
1 + exp(i(k·r  φ)) + exp(i(2k·r 
2φ)) = 0.
Let x = k·r  φ.
1 + e^{ix} + e^{i2x} = e^{ix}(1 + e^{ix} + e^{ix})
= e^{ix}(1 + 2 cosx) = 0.
(1 + 2 cosx) = 0, cosx = ½, x = 2π/3 + n2π or x = 4π/3 + n2π.
We need k·r  φ = 2π/3 + n2π or 4π/3 + n2π for zero
intensity.
Problem 3:
It is known that if a pointlike source light is placed in the focus F of a concave
parabolic mirror the reflected rays are parallel to the optical axis. Suppose the parabolic mirror is formed by the rotation of
a parabola y = Ax^{2} around
the y axes. Find the focal distance of the concave parabolic mirror.
Solution:
 Concepts:
The law of reflection
 Reasoning:
We find F using geometry.
 Details of the calculation:
Set up a coordinate system as shown. We have rotational symmetry about the
yaxis.
Use ray reversibility. Rays incident parallel to the symmetry axis will be
reflected through the focal point.
For a ray incident parallel to the yaxis, find the intersection of the
reflected ray with the yaxis.
If this intersection does not depend on the distance x of the incident ray from
the yaxis, then we have a focus.
Let
θ be the angle the slope at point (x,y) makes with the xaxis and the normal to
the slope makes with the yaxis.
tanθ = dy/dx = 2Ax.
The law of reflection: θ_{i} = θ_{r} = θ for a ray
incident on the point (x,y).
Geometry: f = y + x/tan(2θ). tan(2θ) = 2tanθ/(1  tan^{2}θ).
f = y + x(1  tan^{2}θ)/(2tanθ) = Ax^{2} + x( 1  4A^{2}x^{2})/(4Ax)
= Ax^{2} + 1/(4A)  Ax^{2} = 1/(4A).
f = 1/(4A).
A has dimension of length^{1}.
We have verified that f does not depend on x.
We have a focus.
Problem 4:
A laser beam is incident at an angle of 30.0^{o} to the vertical onto
a solution of corn syrup in water. If the beam is refracted to 19.24^{o}
to the vertical,
(a) What is the index of refraction of the syrup solution?
Suppose that the light is red, with vacuum wavelength 632.8 nm. Find its
(b) wavelength,
(c) frequency, and
(d) speed in the solution.
Give numerical answers.
Solution:
 Concepts:
Snell's law, simple periodic waves
 Reasoning:
We apply Snell's law at the interface.
In the medium f stays the same, λ = λ_{free}/n, v = c/n, λf = v.
 Details of the calculation:
(a) n_{1
}sinθ_{1 }= n_{2 }sinθ_{2}, n_{1}
= 1, n_{2} = sinθ_{1}/sinθ_{2} = sin30^{o}/sin19.24^{o}
= 1.517.
(b) λ = λ_{free}/n = 416.3 nm.
(c) f = c/λ_{free} = 4.74*10^{14} Hz.
(d) v = c/n = 1.97*10^{8} m/s.
Problem 5:
^{238}Pu decays by αemission with a halflife of 87 years.
^{238}Pu > ^{234}U + α.
The halflife of ^{234}U is much longer,
3.5*10^{5} years (ignore this decay).
The heat produced in this decay
can be converted into useful electricity by radiothermal generators (RTG's). The Voyager 2 space probe, which was launched in August 1977, flew past four
planets, including Saturn, which it reached in August 1981.
How much plutonium would an RTG on Voyager 2 with 5.5 % efficiency have to
carry at the start to deliver at least 395 W of electric power when the probe
flies past Saturn?
Masses:^{
4}He: 4.002603 u, ^{
234}U: 234.040947 u^{
238}Pu: 238.049555 u
Conversion: 1 u = 931.5 MeV
Solution:

Concepts:
Nuclear decay

Reasoning:
N = N_{0}exp(λt), R = R_{0}exp(λt),
R_{0} = λN_{0}. R = decay rate.
λ = ln2/t_{½}
= 7.97*10^{3}/year.

Details of the calculation:
Each decay releases E = (238.049555  234.040947  4.002603) * 931.5 MeV
= 8.96*10^{13} J of energy.
To release 395 W with 5.5% efficiency we need a decay rate of
395/(0.055 * 8.96*10^{13})/s = 8.01*10^{15}/s = 2.53*10^{23}/year.
R(t = 4 y) = 2.53*10^{23}/year = R_{0}exp(7.97*10^{3}*4).
R_{0} = 2.61*10^{23}/year. N_{0} = R_{0}/λ
= 3.27*10^{25}.
We need N_{0}/N_{A} = 54.4 mole of plutonium or 12.9 kg of
plutonium at the beginning of the mission.