Simple plane waves

Problem:

An electromagnetic wave with frequency 6.98*10¹⁴ Hz propagates with a speed of 2.22*10⁸ m/s in a certain piece of glass.
(a)  Find the wavelength of the wave in the glass.
(b)  What is the wavelength of a wave with the same frequency propagating in vacuum?
(c)  What is the index of refraction n of the glass for an EM wave of this frequency?
(d)  What is the largest incident angle (with respect to the normal) for light of this frequency to be able to travel from glass into vacuum?
(Give numerical answers!)

Solution:

• Concepts:
  EM waves
• Reasoning:
  We are supposed to recall some basic properties of EM waves.
• Details of the calculation:
  (a)  λ = v/f = 318 nm.
  (b)  λ = c/f = 430 nm.
  (c)  v = c/n, n = 1.35.
  (d)  critical angle θ_c: sinθ_c = 1/n,  θ_c = 47.8^o.

Problem:

An electromagnetic wave in vacuum has an electric field given by

E = j (1200 V/m) sin((4*10¹⁵/s)*t - kx).

(a)  What is the direction of motion of this electromagnetic wave?
(b)  What is the wavelength?
(c)  Write a formula for the magnetic field vector B.
(d)  Find the intensity of this electromagnetic wave.

Solution:

• Concepts:
  Electromagnetic plane waves, the Poynting vector
• Reasoning:
  The electric field of a plane wave is given and we are asked about the magnetic field, the wavelength, the direction of propagation, and the intensity of the wave.
• Details of the calculation:
  (a)  The wave travels into the positive x-direction.
  (b)  λ = 2πc/ω = 2πc/(4*10¹⁵) = 4.7*10^-7 m.
  (c)  B = (E/c) k in SI units.  Here k is the unit vector in the z-direction.
  (d) The magnitude of the Poynting vector S = (1/μ₀)E×B relates the intensity of an electromagnetic wave to the field strength.  For a sinusoidal electromagnetic wave we have <S> = (|E₀²|/2)(ε₀/μ₀)^½.
  <S> = 720000/(μ₀c) = 1910W/m.

Problem:

(a)  Determine the speed of light in water, which has a dielectric constant of 1.78.
(b)  An electromagnetic wave in vacuum has an electric field amplitude of 220 V/m.  Calculate the amplitude of the corresponding magnetic field in SI units.
(c)  What is the energy of a photon of blue light (λ = 450 nm) and of a photon of red light (λ = 700 nm) in units of eV.

Solution:

• Concepts:
  The index of refraction,  v = c/n = (k_mμ₀k_eε₀)^-½,
  B = E/c (SI units),
  Photon energy E = hf
• Reasoning:
  We are asked to review some basic properties of EM waves.
• Details of the calculation:
  (a)  k_e = 1.78, k_m = 1.
  v = (k_eμ₀ε₀)^-½ = c/k_e^½ = 2.23*10⁸ m/s.
  (b)  B = E/c = (220 N/C)/(3*10⁸ m/s) = (7.33*10^-7 N/Am) = 7.33*10^-7 T.
  (c)  E = hc/λ.
  Blue light: E = (6.626*10^-34 Js)(3*10⁸ m/s)/(450*10^-9 m) = 4.4*10^-19 J = 2.76 eV.
  Red light:  E = (6.626*10^-34 Js)(3*10⁸ m/s)/(700*10^-9 m) = 2.8*10^-19 J = 1.8e V.

Problem:

An electromagnetic plane wave with wavelength λ propagates in vacuum along the direction indicated by the vector  2i + j.  The electric field has amplitude E₀, and is oriented along the vertical z direction.  Write the expression for the electric and magnetic fields knowing that at the origin, at time t = 0, the electric field points in the negative z-direction with magnitude equal to 0.5 E₀, and its magnitude is increasing.

Solution:

• Concepts:
  Electromagnetic plane waves, the Poynting vector
• Reasoning:
  E(r, t) = (z/z)E₀cos(k_xx + k_yy - ωt + φ).
  (We write z/z to not confuse the unit vector with the wave vector k.)
  E(0, 0) and the direction of propagation are given.
• Details of the calculation:
  k_x = 2k_y,  k_x² + k_y² = k²,  k = 2π/λ,  ω = 2πf = 2πc /λ .
  k_x = 2k/√5, k_y = k/√5,
  E(r, t) = (z/z) E₀cos(4πx/(λ√5) + 2πy/(λ√5)  - 2πft + φ)
  = (z/z)  E₀cos((2π/λ)(2x/√5 + y/√5  - ct + λφ/(2π)).
  E(0, 0) = -0.5 E₀(z/z) --> cos(φ) = -0.5, φ = ±2π/3
  Since the magnitude of E is increasing at t = 0, φ = -2π/3.
  E(r, t) =  (z/z)E₀cos((2π/λ)(2x/√5 + y/√5  - ct - λ/3).
  |B| = |E|/c.
  The direction of B is
  (z/z) × k/|k| = (z/z)   × ((2/√5)i + (1/√5)j) = (2/√5)j - (1/√5)i.
  B(r, t) =  ((E₀/(c√5)) (-i + 2j) cos((2π/λ)(2x/√5 + y/√5  - ct - λ/3).