More Problems

Problem 1:

A radio telescope consists of two antennas separated by a distance of 200 m.  Both antennas are tuned to a particular frequency, such as 20 MHz.  The signals from each antenna are fed into a common amplifier, but one signal first passes through a phase adjuster that delays its phase by a chosen amount so that the telescope can look in different directions.  When the phase delay is zero, plane radio waves that are incident vertically on the antennas produce signals that add constructively at the amplifier.  What should the phase delay be so that signals coming from an angle θ = 10^o with the vertical (in the plane formed by the vertical and the line joining the antennas) will add constructively at the amplifier?

Solution:

• Concepts:
  Interference
• Reasoning:
  The path difference is Δ = d sinθ.  The phase difference therefore is kΔ
• Details of the calculation:
  kΔ = (ω/c)dsinθ = (2πf/c)dsinθ = 14.55 rad.
  In the diagram, the signal arriving at the left antenna in the diagram should be delayed.
  
  

Problem 2:

Consider a highly excited He atom.  One electron is in a state with n₁ = 4, l₁ = 3 and the other in a state with n₂ = 5,  l₂ = 4
(a)  Find the possible values of l (total orbital angular momentum quantum number) for the system.
(b)  Find the possible values of s (total spin angular momentum quantum number) for the system.
(c)  Find the possible values of j (total angular momentum quantum number) for the system.
(d)  How many distinct angular momentum states with j = 1 are there?

Solution:

• Concepts:
  Addition of angular momentum
• Reasoning:
  We are supposed to add the orbital and spin angular momentum.
• Details of the calculation:
  (a)  The quantum numbers associated with the total orbital angular momentum will range from a maximum value found by adding the individual quantum numbers together to a minimum value found by taking the absolute value of the difference of the two numbers in integer steps.
  l_max = l₁ + l₂ = 3 + 4 = 7.
  l_min = |l₁ - l₂| = |3 - 4| = 1.
  The possible values for L are 1, 2, 3, 4, 5, 6, and 7.
  (b)  The quantum numbers associated with the total spin angular momentum will range from a maximum value found by adding the individual quantum numbers together to a minimum value found by taking the absolute value of the difference of the two numbers in integer steps.
  s_max = s₁ + s₂ = ½ + ½ = 1.
  s_min = |s₁ - s₂| = |½ - ½| = 0.
  The possible values for s are 0 and 1.
  (c)  The largest possible value for j will be found by adding together the maximum values for both l and s. The minimum value for j will be found by subtracting the largest possible value of s from the smallest possible value for l.
  j_smax = l_max + s_max = 7 + 1 = 8.
  j_min = |l_min - s_max| = |1 - 1| = 0.
  j = 0, 1, 2, 3, 4, 5, 6, 7, 8.
  (d)  A state with quantum number j = 1 can have l = 1, s = 0, or l = 2, s = 1, or l = 1, s = 1
  Each of these states can have m_j = -1, 0, +1.
  So there are 9 distinct states with j = 1.

Problem 3:

The region 0 ≤ z ≤ z₀ is filled with a dielectric material of permittivity ε = 4ε₀ and permeability μ = μ₀.  A linearly polarized wave of amplitude E = E₀i and angular frequency ω is incident normally on the interface at z = 0 from the region z < 0.  Show that the ratio of the reflected intensity to the incident intensity in the z < 0 region is
[1 + (16/9)csc²(2ωz₀/c)]^-1.

Solution:

• Concepts:
  Maxwell's equations, boundary conditions for the electric and magnetic fields
• Reasoning:
  The intensities of the reflected and transmitted waves are proportional to the squares of the corresponding electric field amplitudes.  We find the ratio of these amplitudes to the incident amplitude by applying boundary conditions.
• Details of the calculation:
  The incident sinusoidal plane wave is of the form  E = E₀i e^{i(kz - ωt)}.  B = B₀je^{i(kz - ωt)}.
  Maxwell's equation yield the boundary conditions (E₂ - E₁)∙t = 0,  (H₂ - H₁)∙t = 0 for dielectric-dielectric interfaces.
   

  
  If we set up the problem as shown in the figure then
  k_i = k_r = k_t = ω/c,  k₁ = k₂ = nω/c,  n = (ε/ε₀)^½ = 2.
  The boundary condition at z = 0 for the tangential component of E yields
  E_i - E_r = E₁ - E₂.
  The boundary conditions at z = 0 for the tangential component of H yields
  H_i + H_r = H₁ + H₂.  H = B/μ₀ in all regions.
  B = (1/v)(k/k)×E.  B_i = (y/y)E_i/c, B_r = (y/y)E_r/c, B₁ = (y/y)E₁n/c, B₂ = (y/y)E₂n/c.
  E_i + E_r  = n(E₁ + E₂) = 2(E₁ + E₂).
  Measuring all field strength in units of E_i we can express E₁ and E₂ in terns of E_r.
  E_r = -E₁ + E₂ + 1,  E_r = 2E₁ + 2E₂ - 1.  E₂ = (3E_r -1)/4,  E₁ = (3 - E_r)/4.
  
  The boundary conditions at z = z₀ for the tangential component of E yield
  E₁exp(ik₁z₀) - E₂exp(-ik₂z₀) = E_texp(ik_tz₀).
  E₁exp(i2ωz₀/c) - E₂exp(-i2ωz₀/c) = E_texp(iωz₀/c).
  The boundary conditions at z = z₀ for the tangential component of H yield
  H₁exp(i2ωz₀/c) + H₂exp(-i2ωz₀/c) = H_texp(iωz₀/c).
  2E₁exp(i2ωz₀/c) + 2E₂exp(-i2ωz₀/c) = E_texp(iωz₀/c).
  -3E₂exp(-i2ωz₀/c) = E₁exp(i2ωz₀/c).
  
  Substituting E₁ and E₂ in terms of E_r from above we get
  3(3E_r - 1)exp(-i2ωz₀/c) + (3 - E_r)exp(i2ωz₀/c) = 0.
  E_r(9 exp(-i2ωz₀/c) - exp(i2ωz₀/c)) = 3 exp(-i2ωz₀/c) - 3exp(i2ωz₀/c).
  E_r(8 cos(2ωz₀/c) - 10i sin(2ωz₀/c)) = -6i sin(2ωz₀/c).
  1/E_r = (-8 cos(2ωz₀/c) + 10i sin(2ωz₀/c))/(6i sin(2ωz₀/c)).
  1/R = 1/|E_r|² = (64 cos²(2ωz₀/c) + 100 sin²(2ωz₀/c))/(36 sin²(2ωz₀/c))
  = ( 64 + 36 sin²(2ωz₀/c))/(36 sin²(2ωz₀/c)) = (64 csc²(2ωz₀/c))/36 + 1).
  R = (1 + (16/9)csc²(2ωz₀/c))^-1.