## Assignment 3, solutions

#### Problem 1:

A laser beam enters a prism from air.  For a symmetrical prism (one in which the apex angle lies at the top of an isosceles triangle), the total deviation angle δ of the beam is minimized when the light ray inside the prism travels parallel to the prism's base.
(a)  If the index of refraction of the glass is n and the apex angle is A, find total deviation angle δmin.
(b)  Give a numerical value for δmin if A = 60o and n = 1.5.

Solution:

• Concepts:
Snell's law, geometry
• Reasoning:
sinθ = n sinθglass,  θglass = A/2.
• Details of the calculation:
(a)  From geometry:
θ = sin-1(n sin(A/2)) is the incident angle for a light ray that travels inside the prism parallel to the prism's base.
δmin = 2θ - A = 2sin-1(n sin(A/2)) - A is the deviation angle.
(b)  A/2 = 30o.  δmin = 2sin-1(0.75) - 60o = 37.2o.

#### Problem 2:

A beam of monochromatic light of wavelength λ in vacuum is incident normally on a nonmagnetic dielectric film of refractive index n and thickness d.  Calculate the fraction of the incident energy that is reflected.

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:
Let the film lie between the planes z = 0 and z = d.
At the interfaces the tangential components of E and H are continuous.
Let the wave be incident from z = -∞  and the polarization direction be the x-direction.
Let k = 2π/λ,  ω = 2πc/λ.  Define the positive direction for E as shown.

ki = kr = kt = k,  k1 = k2  = nk.
For z < 0 we have
E(r,t) = i [Ei exp(i(kz - ωt)) - Er exp(i(-kz - ωt))],
B(r,t) = j [Ei exp(i(kz - ωt)) + Er exp(i(-kz - ωt))]/c,
H(r,t) = B(r,t)/μ0.
In the film for 0 < z < d we have
E(r,t) = i [E1 exp(i(nkz - ωt)) - E2 exp(i(-nkz - ωt))],
B(r,t) = j [E1 exp(i(nkz - ωt)) + E1 exp(i(-nkz - ωt))](n/c),   H(r,t) = B(r,t)/μ0.
For z > d we have
E(r,t) = i Et exp(i(kz - ωt)),
B(r,t) = j Et exp(i(kz - ωt))/c.   H(r,t) = B(r,t)/μ0.
The boundary conditions at z = 0 for the tangential component of E yield
Ei - Er = E1 - E2.
The boundary conditions at z = 0 for the tangential component of H yield
Hi + Hr = H1 + H2.  H = B/μ0 in all regions, B = En/c in the film, B = E/c outside the film.
Ei + Er  = n(E1 + E2).
Measuring all field strength in units of Ei we have express E1 and E2 in terns of Er.
Er = -E1 + E2 + 1,  Er = nE1 + nE2 - 1.  E2 = ((n + 1)Er - n + 1)/(2n),  E1 = (n + 1 - (n - 1)Er)/2n.
The boundary conditions at z = d for the tangential component of E yield
E1exp(inkd) - E2exp(-inkd) = Etexp(ikd).
The boundary conditions at z = d for the tangential component of H yield
H1exp(inkd) + H2exp(-inkd) = Htexp(ikd).
nE1exp(inkd) + nE2exp(-inkd) = Etexp(ikd).
-(n + 1)E2exp(-inkd) = (n - 1)E1exp(inkd).
Substituting E1 and E2 in terms of Er from above we get
-(n + 1)((n + 1)Er - n + 1)exp(-inkd) = (n - 1)(n + 1 - (n - 1)Er)exp(inkd).
Er[(n + 1)2exp(-inkd) - (n - 1)2exp(inkd)] = (n2 - 1)exp(-inkd) - (n2 - 1)exp(inkd).
Er[4n cos(nkd) + i2(n2 + 1) sin(kd)] = -i(n2 - 1)2sin(kd).
R = |Er|2 = 4(n2 - 1)2sin2(kd)/[16n2 cos2(nkd) + 4(n2 + 1)2 sin2(kd)].
R = (n2 - 1)2/[4n2 cot2(nkd) + (n2 + 1)2]
is the reflectance, i.e. the fraction of the incident energy that is reflected.

#### Problem 3:

(a)  Write down Maxwell's equations for a conducting medium with conductivity σ, permittivity ε0 and permeability μ0,
(b)  A plane wave of low frequency ω << σ/ε0 is propagating in the z-direction inside the conducting medium.
Let E = E0exp(i(kz - ωt)),  B = B0exp(i(kz - ωt)), where E0, B0, and k are complex.
Use Maxwell's equations to show that k = k1 + ik2, k1 ≈ k2 ≈ (μ0ωσ/2)½.
Calculate the ratio of the complex amplitude of the two fields, E0/B0 (magnitude and phase).
(c)  Calculate the energy flux (time averaged Poynting vector) in the conducting medium.

Solution:

• Concepts:
Maxwell's equations
• Reasoning:
In regions with  ρf = 0  and  jf = σE  Maxwell's equations can be used to show that both E and B satisfy the damped wave equation.
• Details of the calculation:
(a)  E = ρ/ε0×E = -∂B/∂t,  ∇B = 0,  ×B = μ0j + (1/c2)∂E/∂t.
Assume ε = ε0, μ = μ0, ρf = 0 and jf = σE in the conductor.  Then
×B = μ0σE + μ0ε0E/∂t.
×(×B) = (B) - 2B = μ0σ(×E) + μ0ε0∂(×E)/∂t.
2B - μ0σ∂B/∂t - μ0ε02B/∂t2 = 0.
Similarly:  ∇2E - μ0σ∂E/∂t - μ0ε02E/∂t2 = 0.
Both E and B satisfy the damped wave equation.

(b)  Let E(r,t) = E0 exp(ikz)exp(-iωt).  We have a plane wave propagating into the z-direction.
Then k2 = iμ0σω + μ0ε0ω2 = μ0ε0ω2(1 + iσ/(ε0ω)).
k2 = μ0ε0ω2(1 + σ2/(ε0ω)2)½ e.  tanφ = σ/(ε0ω).
k = (μ0ε0)½ω(1 + σ2/(ε0ω)2)¼ eiφ/2 = |k|eiφ/2.
Assume σ/(ε0ω) >> 1.  tanφ --> ∞.  φ -- > 90o.  φ/2 --> 45o,  and (1 + σ2/(ε0ω)2)¼ --> (σ/(ε0ω))½.
k = (μ0σω)½ eiπ/4 = (μ0σω)½ (cos(π/4) + i cos(π/4)) = (μ0σω/2)½(1 + i) = k1 + ik2,
k1 ≈ k2 ≈ (μ0ωσ/2)½.
×E = ik×E = -∂B/∂t.
B(r,t) = B(r) exp(-iωt).  Therefore  ik×E  = iωB.  ik(z/z)×E  = iωB.
E0/B0 = ω/k = ω/|k| e-iπ/4 = (ω/(μ0σ))½ e-iπ/4.
With a convenient choice of the zero of t we have
Re(E) =|E0| exp(-k2z) cos(k1z - ωt),  Re(B) = |B0| exp(-k2z) cos(k1z - ωt + π/4).  "B lags behind E" when plotting E and B for fixed z as a function of time.

(c)  S = (1/μ0)[Re(E) × Re(B)] = k |E0|2(σ/(ωμ0))½ exp(-2k2z) cos(k1z - ωt) cos(k1z - ωt + π/4)
= k |E0|2(σ/(2ωμ0))½ exp(-2k2z) (cos2(k1z - ωt) -  cos(k1z - ωt) sin(k1z - ωt)).
<S> = k | ½E0|2(σ/(2ωμ0))½ exp(-(2μ0σω)½z).

#### Problem 4:

In a double-slit experiment with 500 nm light, the slits each have a width of 0.1 mm.
(a)  If the interference fringes are 5 mm apart on a screen which is 4 m from the slits, determine the separation of the slits.
(b)  What is the distance from the center of the pattern to the first diffraction minimum on one side of the pattern?
(c)  How many interference fringes will be seen within the central maximum in the diffraction pattern?  Draw a sketch of the pattern.

Solution:

• Concepts:
Diffraction and interference
• Reasoning:
Double slit interference can be observed at angles where single slit diffraction provides non-zero intensity.
• Details of the calculation:
(a)  dsinθ = mλ.  Small angle approximation:  sinθ = tanθ = θ = m*5*10-3/4.
d = (m*5*10-7 m)/( m*5*10-3/4) = 4*10-4 m is the slit separation.
(b)  wsinθ = λ,  θ = x/L = λ/w = (5*10-7 m)/(10-4 m) = 5*10-3,
x = (4 m)* 5*10-3 = 2 cm is the distance from the center of the pattern to the first diffraction minimum on one side of the pattern.
(c)  There is an interference maximum at 0.5 cm, 1 cm, and 1.5 cm on ether side of the central maximum.  The interference maximumat 2 cm is missing since it falls on the first minimum in the diffraction pattern.  Therefore 7 interference fringes (including the center fringe) will be seen within the central maximum in the diffraction pattern.

#### Problem 5:

Light is incident along the normal on face AB of a glass  prism of index of refraction 1.52, as shown in the figure.

Find the largest value the angle α can have  such that there is no light refracted out of the prism at face AC if
(a)  the prism is immersed in air.
(b)  the prism is immersed in water.

Solution:

• Concepts:
Total internal reflection
• Reasoning:
We use the general notation that the light is incident to the interface from medium a, and is refracted into medium b.  The corresponding indices of refraction are denoted by na and nb, respectively.  With this convention, Snell's law reads nasinθa = nbsinθb.  Both angles are measured from the surface normal.  For na >  nb. we have that θa > θb;  For   θa > θcrit  no light is refracted into medium b.  This corresponds to the condition that θb = 90o, sinθb = 1.  We thus have that sinθcrit = nb/na.
• Details of the calculation:
In our problem the light has an angle of incidence of 0o at the AB surface, so that it enters the glass without being bent, as shown in the figure.

At the AC face, under the condition for total internal reflection we have that α + θcrit = 90o.
(a)  For a glass-air interface, na = 1.52  and nb = 1.00.
We have that sinθcrit = 1/1.52.  Thus, θcrit = 41.1o.  We conclude that α = 90o - θcrit = 48.9o.
(b)  In this case the refraction problem corresponds to glass -->water, for which nb = 1.33.  Therefore sinθcrit = 1.33/1.52 , or θcrit = 61.3o.  We thus have that α = 90o - θcrit = 28.7o.