Assignment 3, solutions

Problem 1:

A spherical, shiny holiday decoration ball is acting as a convex mirror.  The sphere has a radius of 4 cm.  Your eye is 10 cm from the mirror.  How much bigger or smaller is the image of your eye than the actual size of your eye?  Is the image real or virtual, upright or inverted?


The mirror equation
f = R/2,   1/f = 1/so + 1/si,  M = -si/so.
bulletDetails of the calculation:
Here R = -4 cm, f = -2 cm, so = 10 cm, si = -10/6 cm,  M = 1/6.
The image is 6 times smaller.  It is located behind the mirror, it is a virtual image, it is upright.

Problem 2:

For a symmetrical prism (one in which the apex angle lies at the top of an isosceles triangle), the total deviation angle ϕ of a light ray is minimized when the ray inside the prism travels parallel to the prism’s base.  
Assume that a beam of light passes through a glass equilateral prism with refractive index 1.5.  The prism is in air and is mounted on a rotation stage, as shown in the figure.  When the prism is rotated, the angle by which the beam is deviated changes.  What is the minimum angle ϕ bywhich the beam is deflected?


Snell’s law, geometry
sinθ = n sinθglass.
At the angle of minimum deviation, the refraction at each interface is the same, so the beam inside the prism is parallel to the base of the prism.  At both interfaces, the internal angle between the beam and the normal is thus 30o.
bulletDetails of the calculation
By Snell’s Law, the external angle between the beam and the normal is thus α,
where 1.0*sinα =1.5*sin30o.
α = sin-1(1.5*sin30o) = sin-1(0.75) = 48.6o.
From geometry:

The angle of minimum deviation is ϕ = 2α + 120o – 180o.
ϕ = 2α – 60o= 2 sin-1(0.75) – 60o = 37.2o.

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.


Maxwell's equations, boundary conditions for the electric and magnetic fields
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.
bulletDetails of the calculation:
The incident sinusoidal plane wave is of the form  E = E0i ei(kz - ωt)B = B0jei(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
ki = kr = kt = ω/c,  k1 = k2 = nω/c,  n = (ε/ε0)½ = 2.
The boundary condition at z = 0 for the tangential component of E yields
Ei - Er = E1 - E2.
The boundary conditionsat z = 0 for the tangential component of H yields
Hi + Hr = H1 + H2.  H = B/μ0 in all regions.
B = (1/v)(k/k)×EBi = (y/y)Ei/c, Br = (y/y)Er/c, B1 = (y/y)E1n/c, B2 = (y/y)E2n/c.
Ei + Er  = n(E1 + E2) = 2(E1 + E2).
Measuring all field strength in units of Ei we canexpress E1 and E2 in terns of Er.
Er = -E1 + E2 + 1,  Er = 2E1 + 2E2 - 1.  E2 = (3Er -1)/4,  E1 = (3 - Er)/4.

The boundary conditions at z = z0 for the tangential component of E yield
E1exp(ik1z0) - E2exp(-ik2z0) = Etexp(iktz0).
E1exp(i2ωz0/c) - E2exp(-i2ωz0/c) = Etexp(iωz0/c).
The boundary conditions at z = z0 for the tangential component of H yield
H1exp(i2ωz0/c) + H2exp(-i2ωz0/c) = Htexp(iωz0/c).
2E1exp(i2ωz0/c) + 2E2exp(-i2ωz0/c) = Etexp(iωz0/c).
-3E2exp(-i2ωz0/c) = E1exp(i2ωz0/c).

Substituting E1 and E2 in terms of Er from above we get
3(3Er - 1)exp(-i2ωz0/c) + (3 - Er)exp(i2ωz0/c) = 0.
Er(9 exp(-i2ωz0/c) - exp(i2ωz0/c)) = 3 exp(-i2ωz0/c) - 3exp(i2ωz0/c).
Er(8 cos(2ωz0/c) - 10i sin(2ωz0/c)) = -6i sin(2ωz0/c).
1/Er = (-8 cos(2ωz0/c) + 10i sin(2ωz0/c))/(6i sin(2ωz0/c)).
1/R = 1/|Er|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.

Problem 4:

An electromagnetic wave with circular frequency ω propagates in a medium of dielectric constant ε, magnetic permeability μ, and conductivity σ.
(a)  Show that there is a plane wave solution in which the amplitude of the E and B fields decreases exponentially along the direction of propagation, and find the characteristic decay length.
(b)  Simplify by assuming that σ is great enough so that σ/(εω) >> 1.


Maxwell's equations
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.
bulletDetails of the calculation:
(a)  E = ρ/ε,  ×E = -∂B/∂t,  ∇B = 0,  ×B = μj + (1/c2)∂E/∂t.
Assume ρf = 0 and jf = σE in the conductor.  Then
×B = μσE + με∂E/∂t.
×(×B) = (B) - 2B = μσ(×E) + με∂(×E)/∂t.
2B - μσ∂B/∂t - με∂2B/∂t2 = 0.
Similarly:  ∇2E - μσ∂E/∂t - με∂2E/∂t2 = 0.
Both E and B satisfy the damped wave equation.

Try solutions of the form E(r,t) = E0 exp(i(kr - ωt)).
Then k2 = iμσω + μεω2 = μεω2(1 + iσ/(εω)).
k2 = μεω2(1 + σ2/(εω)2)½ e,  tanφ = σ/(εω).
k = (με)½ω(1 + σc2/(εω)2)¼ eiφ/2 = |k|eiφ/2.
k is a complex number, k = β + iα/2.
To find the real and imaginary parts we have to find cos(φ/2) and sin(φ/2) in terms of tanφ.
cos(φ/2) = ((1 + cosφ)/2)½,  sin(φ/2) = ((1 - cosφ)/2)½,  cosφ = (1 + tan2φ) in the first quadrant.
Therefore cos(φ/2) = 2(1 + (1 + (σ/(εω))2))½
sin(φ/2) = 2(1 - (1 + (σ/(εω))2))½.
k = (με/2)½ω[(1 + (σ/(εω))2)½ + 1]½ + i(με/2)½ω[(1 + (σ/(εω))2) - 1]½.
β = (με/2)½ω[(1 + (σ/(εω))2)½ + 1]½,  α/2 = (με/2)½ω[(1 + (σ/(εω))2)½ - 1]½.
The skin depth δ is the distance it takes to reduce the amplitude by a factor of 1/e.
δ = 2/α. 
(b)  Let σ/(εω) >> 1.
Then α/2 ~ (με/2)½ω(σ/(εω))½ = (μσω/2)½,
β ~ α/2 ~ (μσω/2)½ = 1/δ.

Problem 5:

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.


Diffraction and interference
Double slit interference can be observed at angles where single slit diffraction provides non-zero intensity.
bulletDetails 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)  40 mm/5 mm = 8 interference fringes will be seen within the central maximum in the diffraction pattern.