More Problems

Problem 1:

The quantum numbers l₁ and l₂ of the orbital momenta of particle A and particle B are 1 and 2, respectively.  Find the 15 possible 'kets' in the coupled representation (notation |l₁,l₂; L,M_L>) where L represents the quantum number of the total orbital momentum.

Solution:

• Concepts:
  Addition of angular momentum
• Reasoning:
  We are supposed to add two angular momenta l₁ and l₂  and find the possible values for L and M.
• Details of the calculation:
  l₁ = 1 and l₂ = 2.  The possible values for L are L = 3, 2, 1.
  L = 1:  Possible values M are 1, 0, -1.
  Corresponding kets: |1,2; 1,1>, |1,2; 1,0>, |1,2; 1,-1>.
  (notation: |l₁,l₂; L,m>)
  L = 2:  Possible values M are 2, 1, 0, -1, -2.
  Corresponding kets: |1,2; 2,2>, |1,2; 2,1>, |1,2; 2,0>, |1,2; 2,-1>, |1,2; 2,-2>.
  L = 3:  Possible values M are 3, 2, 1, 0, -1, -2, -3.
  Corresponding kets: |1,2; 3,3>, |1,2; 3,2>, |1,2; 3,1>, |1,2; 3,0>, |1,2; 3,-1>, 
  |1,2; 3,-2>, |1,2; 3,-3>.
  We have listed 15 kets.

Problem 2:

An s-polarized electromagnetic plane wave with wavelength λ is propagating in vacuum along the direction indicated by the vector cos(30^o)e_z + sin(30^o)e_x, after emerging from an interface perpendicular to the z-axis. 

At t = 0, the electric field at the origin is zero and its y-component is increasing.  Write the expression for the electric and magnetic fields in terms of λ and the electric field amplitude E₀.

Solution:

• Concepts:
  Electromagnetic plane waves, the Poynting vector
• Reasoning:
  E(r, t) = e_y E₀sin(k_xx + k_zz - ωt),
  since E(0, 0) and the direction of propagation are given.
• Details of the calculation:
  k_x = kcos(30^o), k_z = ksin(30^o),  k = 2π/λ,  ω = 2πf = 2πc /λ .
  E(r, t) = e_y E₀sin(kx/2 + √3kz/2 - 2πct/λ)
  =  e_y E₀sin((2π/λ)(x/2 + √3z/2 - ct)).
  |B| = |E|/c.
  The direction of B is the direction of
  (cos(30^o)e_z + sin(30^o)e_x)×e_y = -√3e_x/2+ e_z/2.
  B(r, t) =  E₀sin((2π/λ)(x/2 + √3z/2 - ct))(-√3e_x/2+ e_z/2).

Problem 3:

A mirror creates an image of an object which is 3 times larger than the object.  The object is then moved to a new location, and the image is 4 times larger than the object.
(a)  Is this a convex or concave mirror?
(b)  What is the ratio of the distance the object has been moved to the radius of curvature of the mirror?

Solution:

• Concepts:
  The mirror equation
• Reasoning:
  1/x_o + 1/x_i = 1/f,  M = -x_i/x_o.
• Details of the calculation:
  (a)  Only concave mirrors produce images larger than the object.  This is a concave mirror.
  (b)  There are several solutions. 
  (i)  Both images are real and inverted.
  real image:  M = -3,  x_i = 3x_o.  x_o = 4f/3.
  real image:  M = -4,  x_i = 4x_o.  x_o = 5f/4.
  Distance d the object has been moved:  d = f/12. 
  The radius of curvature R = 2f, d/R = 1/24.
  
  (i)  Both images are virtual and not inverted.
  virtual image:  M = 4, x_i = -4x_o.  x_o = 3f/4.
  virtual image:  M = 3, x_i = -3x_o.  x_o = 2f/3.
  Distance d the object has been moved:  d = f/12,  d/R = 1/24.
  
  (iii)  One of the images is a real, inverted image and one of the images is a virtual, upright image.
  real image:  M = -4,  x_i = 4x_o.  x_o = 5f/4.
  virtual image:  M = 3, x_i = -3x_o.  x_o = 2f/3.
  Distance d the object has been moved:  d = 7f/12,  d/R = 7/24.
  
  or
  real image:  M = -3,  x_i = 3x_o.  x_o = 4f/3.
  virtual image:  M = 4, x_i = -4x_o.  x_o = 3f/4.
  Distance d the object has been moved:  d = 7f/12,  d/R = 7/24.