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 = f₀(v-v_o)/(v-v_s),
  f = observed frequency, f₀ = 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 non-zero φ.
(a)  Write down an expression |Ψ_total|² (proportional to the average intensity) at the detector.  Make the far-field 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 |Ψ₁(R) + Ψ₂(R)|²
• Details of the calculation:
  (a)  Ψ₁(R) = Aexp(i(kR - ωt))/R,  Ψ₂(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.
  |Ψ₁(R) + Ψ₂(R)|²
  = (2A²/R²) + (A²/R²) [exp(i(kR - ωt))exp(-i(kR - k·r - ωt + φ) + exp(-i(kR - ωt))exp(i(kR - k·r - ωt + φ))]
  = (2A²/R²) + (A²/R²) [exp(i(k·r - φ)) + exp(-i(k·r - φ))]
  = (2A²/R²) + (2A²/R²) 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)  |Ψ₁(R) + Ψ₂(R) + Ψ₃(R)|² = (3A²/R²) + (2A²/R²)[ 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 Ψ₁(R) + Ψ₂(R) + Ψ₃(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 point-like 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 = Ax² 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 y-axis.
  Use ray reversibility.  Rays incident parallel to the symmetry axis will be reflected through the focal point.
  For a ray incident parallel to the y-axis, find the intersection of the reflected ray with the y-axis.
  If this intersection does not depend on the distance x of the incident ray from the y-axis, then we have a focus.
  
  Let θ be the angle the slope at point (x,y) makes with the x-axis and the normal to the slope makes with the y-axis.
  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 - tan²θ).
  f = y + x(1 - tan²θ)/(2tanθ) = Ax² + x( 1 - 4A²x²)/(4Ax) = Ax² + 1/(4A) - Ax² = 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)  n₁ sinθ₁ = n₂ sinθ₂,  n₁ = 1, n₂ = sinθ₁/sinθ₂ = sin30^o/sin19.24^o = 1.517.
  (b)  λ = λ_free/n = 416.3 nm.
  (c)  f = c/λ_free = 4.74*10¹⁴ Hz.
  (d)  v = c/n = 1.97*10⁸ m/s.

Problem 5:

²³⁸Pu decays by α-emission with a half-life of 87 years.
²³⁸Pu --> ²³⁴U + α.
The half-life of ²³⁴U is much longer, 3.5*10⁵ years (ignore this decay).
The heat produced in this decay can be converted into useful electricity by radio-thermal 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:
⁴He:  4.002603 u,
²³⁴U: 234.040947 u
²³⁸Pu:  238.049555 u
Conversion:  1 u = 931.5 MeV

Solution:

• Concepts:
  Nuclear decay
• Reasoning:
  N = N₀exp(-λt),  R = R₀exp(-λt),  R₀ = λN₀.  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*10¹⁵/s = 2.53*10²³/year.
  R(t = 4 y) = 2.53*10²³/year = R₀exp(-7.97*10^-3*4).
  R₀ = 2.61*10²³/year.  N₀ = R₀/λ = 3.27*10²⁵.
  We need N₀/N_A = 54.4 mole of plutonium or 12.9 kg of plutonium at the beginning of the mission.