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?
Solution:
Concepts: The mirror equation  
Reasoning: f = R/2, 1/f = 1/s_{o} + 1/s_{i}, M = s_{i}/s_{o}.  
Details of the calculation: Here R = 4 cm, f = 2 cm, s_{o} = 10 cm, s_{i} = 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?
Solution:
Concepts: Snell’s law, geometry  
Reasoning: sinθ = n sinθ_{glass}. 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 30^{o}.  
Details 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*sin30^{o}. α = sin^{1}(1.5*sin30^{o}) = sin^{1}(0.75) = 48.6^{o}. From geometry: The angle of minimum deviation is ϕ = 2α + 120^{o} – 180^{o}. ϕ = 2α – 60^{o}= 2 sin^{1}(0.75) – 60^{o} = 37.2^{o}. 
Problem 3:
The region 0 ≤ z ≤ z_{0} is filled with a dielectric material of
permittivity ε = 4ε_{0 }and
permeability μ = μ_{0}.
A
linearly polarized wave of amplitude E = E_{0}i
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)csc^{2}(2ωz_{0}/c)]^{1}.
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: The incident sinusoidal plane wave is of the form E = E_{0}i_{ }e^{i(kz  ωt)}. B = B_{0}je^{i(kz  ωt)}. Maxwell's equation yield the boundary conditions (E_{2}  E_{1})∙t = 0, (H_{2}  H_{1})∙t = 0 for dielectricdielectric interfaces.

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.
Solution:
Concepts: Maxwell's equations  
Reasoning: In regions with ρ_{f }= 0 and j_{f }= 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/ε, ∇×E = ∂B/∂t, ∇∙B = 0, ∇×B = μj + (1/c^{2})∂E/∂t. Assume ρ_{f }= 0 and j_{f }= σE in the conductor. Then ∇×B = μσE + με∂E/∂t. ∇×(∇×B) = ∇(∇∙B)  ∇^{2}B = μσ(∇×E) + με∂(∇×E)/∂t. ∇^{2}B  μσ∂B/∂t  με∂^{2}B/∂t^{2} = 0. Similarly: ∇^{2}E  μσ∂E/∂t  με∂^{2}E/∂t^{2} = 0. Both E and B satisfy the damped wave equation. Try solutions of the form E(r,t) = E_{0 }exp(i(k∙r  ωt)). Then k^{2} = iμσω + μεω^{2} = μεω^{2}(1 + iσ/(εω)). k^{2} = μεω^{2}(1 + σ^{2}/(εω)^{2})^{½ }e^{iφ}, tanφ = σ/(εω). k = (με)^{½}ω(1 + σ_{c}^{2}/(εω)^{2})^{¼ }e^{iφ/2} = ke^{iφ/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 + tan^{2}φ)^{½} in the first quadrant. Therefore cos(φ/2) = 2^{½}(1 + (1 + (σ/(εω))^{2})^{½})^{½}, sin(φ/2) = 2^{½}(1  (1 + (σ/(εω))^{2})^{½})^{½}. k = (με/2)^{½}ω[(1 + (σ/(εω))^{2})^{½} + 1]^{½} + i(με/2)^{½}ω[(1 + (σ/(εω))^{2})^{½}  1]^{½}. β = (με/2)^{½}ω[(1 + (σ/(εω))^{2})^{½} + 1]^{½}, α/2 = (με/2)^{½}ω[(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)^{½}ω(σ/(εω))^{½} = (μσω/2)^{½}, β ~ α/2 ~ (μσω/2)^{½} = 1/δ. 
Problem 5:
In a doubleslit 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 nonzero 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) 40 mm/5 mm = 8 interference fringes will be seen within the central maximum in the diffraction pattern. 