## Assignment 3, solutions

Problem 1:

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, 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 = 1.00.  Therefore sinθcrit   =  1.33/1.52 , or θcrit  = 61.3o.  We thus have that α = 90o - θcrit = 28.7o.

Problem 2:

A beam of unpolarized light of intensity I0 passes through a series of ideal polarizing filters with their transmission axis turned to various angles, as shown in the figure.

(a)  What is the light intensity (in terms of I0) at points A, B, and C?
(b)  If we remove the middle filter, what will be the light intensity at point C?

Solution:

 Concepts: The law of Malus Reasoning: When unpolarized light passes through a polarizer, the intensity is reduced by a factor of ½.  The transmitted light is polarized along the axis of the polarizer. When polarized light of intensity Imax is incident on a polarizer, the transmitted intensity is given by I = Imaxcos2φ, where φ is the angle between the polarization direction of the incident light and the axis of the filter. Details of the calculation: For the second polarizer φ = 60o.  For the third polarizer φ = 90o – 60o = 30o. We then have that: (a)  At point A the intensity is I0/2 and the light is polarized along the vertical direction.  At point B the intensity is (I0/2)cos260o, = 0.125 I0, and the light is polarized along the axis of the second polarizer.  At point C the intensity is (0.125 I0)cos230o = 0.0938 I0. (b) If we remove the middle filter, for the last filter we now have that φ = 90o.  Thus I = 0.  It is important to visualize the fact that adding the middle filter increases the transmitted intensity!  This “paradoxical” effect is a signature of wave phenomena in general.

Problem 3:

Suppose light from a Helium-Neon laser (wavelength λ = 633 nm) is expanded and collimated into a beam with a diameter of 2 cm, and then split into two beams that intersect as shown in the drawing below.

(a) If the full angle between the two overlapping beams is 10o, how many fringes appear in the overlapping region?
(b) When the angle between the overlapping beams is increased to 30o, does the fringe spacing in the overlap region increase or decrease?  By how much?

Solution:

 Concepts: Interference Reasoning: In the regions where the beams overlap we have E = E1 + E2.  The intensity is proportional to . Details of the calculation: a) λ = 633 nm, beam diameter d = 2 cm. k1 = k2 = k = 2π/λ = 9.92*106 m-1.   k1x = kcosθ, k2x = kcosθ, k1y = ksinθ, k2y = -ksinθ. In the regions where the beams overlap we have E = E1 + E2.  Let E = Ek. E(r,t) = A1cos(k1∙r-ωt) + A2cos(k2∙r-ωt). E(r,t)2 = (A1cos(k1∙r-ωt))2 + (A2cos(k2∙r-ωt))2 + 2A1cos(k1∙r-ωt)A2cos(k2∙r-ωt). cosAcosB = (1/2)[cos(A+B) + cos(A-B)] cos(k1∙r-ωt+k2∙r-ωt) = cos(2kcosθ x - 2ωt). cos(k1∙r-ωt-k2∙r+ωt) = cos(2ksinθ y) E(r,t)2 = (A1cos(k1∙r-ωt))2 + (A2cos(k2∙r-ωt))2 + A1A2[cos(2kcosθ x - 2ωt)+cos(2k1sinθ y)] The average values of cos2(k1∙r-ωt) and cos2(k2∙r-ωt) are 1/2, and cos(2kcosθ x - 2ωt) is zero, when averaged over a large number of periods. = A12/2 +A22/2 + A1A2cos(2ksinθ y)] = + + 2()1/2cos(2ksinθ y)] The intensity therefore varies with y as C1 + C2cos((2π/λ')y) = C1 + C2cos(2ksinθ y). λ' = π/ksinθ. Φ = 90o - 2θ, w =b/cosΦ = b/sin(2θ), d/2 =wsinθ, d = 2bsinθ/sin(2θ) The maximum width of the overlap region is 4 cm*sinθ/sin(2θ) = 2 cm/cos(θ). When θ = 5o, then λ' = 3.63 10-6 m, we observe (0.04 m *sin5o)/(3.63 10-6 m sin10o) = ~5530 fringes. (b) When θ = 15o, then λ' = 1.22 10-6 m, we observe (0.04 m*sin15o)/(1.22 10-6 m sin30o) = ~16972 fringes. The fringe spacing decreases.

Problem 4:

Fresnel's reflectance formulas are given by

R = |sin(θi - θr)/sin(θi + θr)|2   or   R = |tan(θi - θr)/tan(θi + θr)|2

depending on the polarization of the incident wave with respect to the plane of incidence, and with θi and θr the angles of incidence and refraction, respectively.
(a)  Specify the boundary conditions across an interface.
(b)  For simplicity assume μ1 = μ2 = μ0, ε1 = ε0, ε2 = ε, and derive the reflectance formulas.
(c)  For polarization parallel to the plane of incidence, find the Brewster angle for an index of refraction of n2 = 1.50, and comment on the polarization of the reflected radiation for a wave of mixed polarization incident on a plane interface at the Brewster angle.

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:
(a)  At a dielectric-dielectric interface Maxwell's equation yield the boundary conditions

 (SI units)

(b)  In the figure below, the y-direction is the direction of the polarization for s-polarization.

Ei(r,t) =(y/y) Ei exp(i(ki∙r - ωt)),
E
r(r,t) =(y/y) Er exp(i(kr∙r - ωt)),
E
t(r,t) =(y/y) Et exp(i(kt∙r - ωt)).
The tangential component of E is continuous across the boundary.
The y-direction is tangential to the interface.
Therefore Ei exp(i(ki∙r)) + Er exp(i(kr∙r)) = Et exp(i(kt∙r)) for all r in the z = 0 plane.
This is only possible if ki∙r = kr∙r = kt∙r for all r in the z = 0 plane.
Let r = x/x, then kix = krx = ktx,   or   kisin(θi) = krsin(θr) = ktsin(θt).
ki = kr = (μ0ε0)1/2ω = ω/c.  kt = (μ0e')1/2ω = n2ω/c.  Therefore
sin(θi) = sin(θr) = n2sin(θt).
This implies θi = θr (law of reflection) and sin(θr) = n2sin(θt) (Snell's law) for s-polarization.

The tangential component of B = μ0H is continuous across the boundary.  The x-direction is tangential to the interface.
B = k×E/ω,  Bi = (1/ω)(ki×Ei)∙x = (1/ω)(Ei × x)∙ki =(1/ω)Eikiz = (1/ω)Eikicos(θi).
Similarly Br∙x = -(1/ω)Erkrcos(θr),  Bt∙x = (1/ω)Etktcos(θt).
We then have Eikicos(θi) - Erkrcos(θr) = Etktcos(θt),
or  Eicos(θi) - Ercos(θr) = n2Etcos(θt).
Combining this equation with Ei + Er = Et and θi = θr from above yields
Er/Ei = (cosθi - n2cosθt)/(cosθi + n2cosθt)
= (cosθi - (sinθi/sinθt)cosθt)/(cosθi + (sinθi/sinθt)cosθt)
= (1 - (tanθi/tanθt))/(1 + (tanθi/tanθt))
= (tanθt - tanθi)/(tanθt + tanθi)
= sin(θt - θi)/sin(θt + θi).
R =|Er/Ei|2 = |sin(θt - θi)/sin(θt + θi)|2.
[sin(a - b)/sin(a + b) = (sina cosb - cosa sinb)/(sina cosb + cosa sinb)
= sina cosb/(sina cosb + cosa sinb) - cosa sinb/(sina cosb - cosa sinb)
= 1/(1 + tanb/tana) - 1/(1 + tana/tanb)
= (tana - tanb)/(tana + tanb)]

In the figure below, the direction of B is tangential to the interface for p-polarization.

Bi(r,t) = -(y/y) Bi exp(i(ki∙r - ωt)),
B
r(r,t) = -(y/y) Br exp(i(kr∙r - ωt)),
B
t(r,t) = -(y/y) Bt exp(i(kt∙r - ωt)).

The tangential component of E is continuous across the boundary.
The x-direction is tangential to the interface.
Therefore the x-component of E is continuous across the boundary.
Ei exp(i(ki∙r))cos(θi)  - Er exp(i(kr∙r))cos(θr) = Et exp(i(kt∙r))cos(θt) for all r in the z = 0 plane.
This is only possible if ki∙r = kr∙r = kt∙r for all r in the z = 0 plane.
Let r = x/x, then kix = krx = ktx,   or   kisin(θi) = krsin(θr) = ktsin(θt).
ki = kr = (μ0ε0)1/2ω = ω/c.  kt = (μ0e')1/2ω = n2ω/c.  Therefore
sin(θi) = sin(θr) = n2sin(θt).
This implies θi = θr (law of reflection) and sin(θr) = n2sin(θt) (Snell's law) for p-polarization.

The tangential component of B = mH is continuous across the boundary.
The y-direction is tangential to the interface.
Bi + Br = Bt.
B = k×E/ω,  Bi = (1/ω)Eiki, = Ei/c. Br = (1/ω)Erkr = Er/c, Bt = (1/ω)Etkt = n2Et/c.
We then have Ei + Er = n2Et.
Combining this equation with Ei cos(θi)  - Er cos(θr) = Et cos(θt) and θi = θr  fom above yields
Er/Ei = (n2cosθi - cosθt)/(n2cosθi + cosθt)
= ((sinθi/sinθt)cosθi - cosθt)/((sinθi/sinθt)cosθi + cosθt)
= (sinθicosθi - cosθtsinθt)/(sinθicosθi + cosθtsinθt)
= tan(θt - θi)/tan(θt + θi).
R =|Er/Ei|2 = |tan(θt - θi)/tan(θt + θi)|2.
[(sina cosa - cosb sinb)/(sina cosa + cosb sinb)
= (sina cosa(sin2b + cos2b) - cosb sinb(sin2a + cos2a))/(sina cosa(sin2b + cos2b) + cosb sinb(sin2a + cos2a))
= (sin(a-b)cos(a+b)/[sin(a+b)cos(a-b)] = tan(a-b)/tan(a+b)]

(c)  When θt + θi = п/2, the reflectance is zero for p-polarization.
This happens when θi = θB, n2sinθB  = n2sin((п/2)-θB), tanθB = n2/n1.
θB is called the Brewster angle.
When θi = θB, then only light with s-polarization is reflected.  For a wave incident with mixed polarization incident with θi = θB, the reflected light is polarized.
With n1 = 1, n2 = 1.5 we have θB = 56.31o.

Problem 5:

A “tenuous” plasma consists of free electric charges of mass m and charge –e (where e is positive).  There are n charges per unit volume.  Assume that the density is uniform and that the interactions between the charges may be neglected.  Also assume that the charges can be treated classically.
A linearly-polarized electromagnetic wave of frequency ω is incident on the plasma.
Let the electric field component of the plane wave be E = E0exp(i(kx - ωt)).
(a)  Solve the equation of motion for a single charge and find the current density j and the conductivity σ of the plasma as a function of ω.
(b)  Assume a plane wave of the form E = E0exp(i(kx - ωt)) propagate in the plasma with conductivity σ.  Find the dispersion relation —the relation between k and ω— for the electromagnetic wave in the plasma and the index of refraction as a function of ω.

Solution:

 Concepts: Newton's 2nd law, charge and current density, plane waves in a conducting medium Reasoning: The “tenuous” plasma is a conducting medium.  We can find its conductivity by solving for the motion of the free charges using Newton's second law. Details of the calculation: (a)  Equation of motion for a single charge: md2r/dt2 = -eE = -eE0exp(i(kx - ωt)). r(t) = (eE0/(mω2))exp(i(kx - ωt)).  v(t) = (eE0/(imω))exp(i(kx - ωt)) = (e/(imω))E j(t) = -nev(t) = -(ne2/(imω))E(t). j = σE.  σ = (ine2/(mω)).(b)  ∇×B = μ0σE + μ0ε0∂E/∂t. ∇×(∇×B) = ∇(∇∙B) - ∇2B = μ0σ(∇×E) + μ0ε0∂(∇×E)/∂t. ∇2B - μ0σ∂B/∂t - μ0ε0∂2B/∂t2 = 0. Similarly:  ∇2E - μ0σ∂E/∂t - μ0ε0∂2E/∂t2 = 0. Plane wave solutions of the form E = E0exp(i(kx - ωt)) have to satisfy -k2 + iωμ0σ + ω2/c2 = 0,  k2 = ω2/c2 - (μ0ne2/m ) = (1/c2)(ω2 – ωp2), k = (1/c)( ω2 – ωp2)1/2, where ωp = (μ0ne2c2/m)= (ne2/(ε0m)). c/n = ω/k.  n = ck/ω = (1 – ωp2/ω2)1/2.  For ωp > ω n is imaginary, the wave is absorbed.

Problem 6:

A (locally) plane electromagnetic wave in vacuum is propagating in the positive z-direction.  The angular frequency of the wave is ω.  The wave generator slowly decreases the amplitude of the wave.  At the position z = 0 the amplitude of the wave is E0(1 – at) for times between t = 0 and t = 1/a, with a << ω.
Consider a cylindrical surface as shown.

(a)  Find the average outward energy flux from the cylinder.
(b)  Find the field energy enclosed by the surface, and show that it decreases at a rate equal to the flux found in part (a).

Solution:

 Concepts: The Poynting vector, energy density Reasoning: The Poynting vector represents the energy flux. Details of the calculation: (a)  At z = 0 and t = t0 we have E(0,t0) = E0(1 – at0) cos(ωt0). The field at z = L and t = t0 is the same as the field at z = 0 and t = t0 - L/c. E(L,t0) = E0(1 – a(t0 – L/c) cos(ω(t0 – L/c)). S = (1/μ0)(E×B) = k (1/μ0)E2 = k ε0c E2. S points in the z-direction, so the integral of S∙n over the surface of the cylinder reduces to an integral over the ends of the cylinder. Since a <<  ω, averaging over one period of the oscillation yields (b)  The energy density is u = (1/(2μ0))B2 + (ε0/2)E2 = ε0E2. The energy density averaged over one period is =  (ε0/2)E2. The average energy inside the cylinder at t = t0  therefore is W decreases at a rate At t = t0, the rate at which W decreases equals the average outward energy flux from the cylinder.

Problem 7:

Light of wavelength 640 nm is shone on a double-slit apparatus and the interference pattern is observed on a screen.  When one of the slits is covered with a very thin sheet of plastic of index of refraction n = 1.6, the center point on the screen becomes dark.  What is the minimum thickness of the plastic?

Solution:

 Concepts: Double slit interference Reasoning: Destructive interference occurs when the optical path length difference is mλ/2, with m = odd integer. Details of the calculation: The geometrical path length from each slit to the center of the diffraction pattern is the same.  Let the thickness of the plastic sheet be x.  The optical path length through the plastic sheet is nx and the optical path length difference is therefore (n - 1)x.  We need (n - 1)x = λ/2, for the minimum thickness n. Alternatively: Reaching the slits, the waves passing the slits are in phase.  Upon exiting the slits they have a phase difference (k' - k)x = (2πn/λ - 2π/λ)x = (2π/λ)(n - 1)x. For the minimum thickness we need  (2π/λ)(n - 1)x = π, x = λ/(2(n-1))  = λ/1.2 = 533 nm.  The minimum thickness is 533 nm.