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 δ_{min}.

(b) Give a numerical value for δ_{min} if A = 60^{o} and
n = 1.5.

Solution:

- Concepts:

Snell's law, geometry - Reasoning:

sinθ = n sinθ_{glass}, θ_{glass}= 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.

δ_{min}= 2θ - A = 2sin^{-1}(n sin(A/2)) - A is the deviation angle.

(b) A/2 = 30^{o}. δ_{min}= 2sin^{-1}(0.75) - 60^{o}= 37.2^{o}.

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.k

_{i}= k_{r}= k_{t}= k, k_{1}= k_{2}= nk.

For z < 0 we have**E**(**r**,t) =**i [**E_{i }exp(i(kz - ωt)) -_{r }exp(i(**-**kz - ωt))],**B**(**r**,t) =**j [**E_{i }exp(i(kz - ωt)) +_{r }exp(i(**-**kz - ωt))]/c,**H**(**r**,t) =**B**(**r**,t)/μ_{0}.

In the film for 0 < z < d we have**E**(**r**,t) =**i [**E_{1 }exp(i(nkz - ωt)) -_{2 }exp(i(**-**nkz - ωt))],**B**(**r**,t) =**j [**E_{1 }exp(i(nkz - ωt))**+**E_{1 }exp(i(**-**nkz - ωt))](n/c),**H**(**r**,t) =**B**(**r**,t)/μ_{0}.

For z > d we have**E**(**r**,t) =**i**E_{t }exp(i(kz - ωt)),**B**(**r**,t) =**j**E_{t }exp(i(kz - ωt))/c.**H**(**r**,t) =**B**(**r**,t)/μ_{0}.

The boundary conditions at z = 0 for the tangential component of**E**yield

E_{i}- E_{r}= E_{1}- E_{2}._{}The boundary conditions at z = 0 for the tangential component of**H**yield

H_{i}+ H_{r}= H_{1}+ H_{2}. H = B/μ_{0}in all regions, B = En/c in the film, B = E/c outside the film.

E_{i}+ E_{r}= n(E_{1}+ E_{2}).

Measuring all field strength in units of E_{i}we have express E_{1}and E_{2}in terns of E_{r}.

E_{r}= -E_{1}+ E_{2}+ 1, E_{r}= nE_{1}+ nE_{2}- 1. E_{2}= ((n + 1)E_{r}- n + 1)/(2n), E_{1}= (n + 1 - (n - 1)E_{r})/2n.

The boundary conditions at z = d for the tangential component of**E**yield

E_{1}exp(inkd) - E_{2}exp(-inkd) = E_{t}exp(ikd).

The boundary conditions at z = d for the tangential component of**H**yield

H_{1}exp(inkd) + H_{2}exp(-inkd) = H_{t}exp(ikd).

nE_{1}exp(inkd) + nE_{2}exp(-inkd) = E_{t}exp(ikd).

-(n + 1)E_{2}exp(-inkd) = (n - 1)E_{1}exp(inkd).

Substituting E_{1}and E_{2}in terms of E_{r}from above we get

-(n + 1)((n + 1)E_{r}- n + 1)exp(-inkd) = (n - 1)(n + 1 - (n - 1)E_{r})exp(inkd).

E_{r}[(n + 1)^{2}exp(-inkd) - (n - 1)^{2}exp(inkd)] = (n^{2}- 1)exp(-inkd) - (n^{2}- 1)exp(inkd).

E_{r}[4n cos(nkd) + i2(n^{2}+ 1) sin(kd)] = -i(n^{2}- 1)2sin(kd).

R = |E_{r}|^{2}= 4(n^{2}- 1)^{2}sin^{2}(kd)/[16n^{2}cos^{2}(nkd) + 4(n^{2}+ 1)^{2}sin^{2}(kd)].

R = (n^{2}- 1)^{2}/[4n^{2}cot^{2}(nkd) + (n^{2}+ 1)^{2}]

is the reflectance, i.e. the fraction of the incident energy that is reflected.

(a) Write down Maxwell's equations for a conducting medium with conductivity σ,
permittivity ε_{0} and permeability μ_{0},

(b) A plane wave of low frequency ω << σ/ε_{0} is propagating in the
z-direction inside the conducting medium.

Let **E** = **E**_{0}exp(i(kz - ωt)),
**B** = **B**_{0}exp(i(kz
- ωt)), where **E**_{0}, **B**_{0}, and k are complex.

Use Maxwell's equations to show that k = k_{1} + ik_{2}, k_{1}
≈ k_{2} ≈ (μ_{0}ωσ/2)^{½}.

Calculate the ratio of the complex amplitude of the two fields, E_{0}/B_{0}
(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 ρ_{f }= 0 and**j**= σ_{f }**E**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_{0},**∇**×**E**= -∂**B**/∂t**, ∇**∙**B**= 0,**∇**×**B**= μ_{0}**j**+ (1/c^{2})∂**E**/∂t.Assume ε = ε_{0}, μ = μ_{0}, ρ_{f }= 0 and**j**= σ_{f }**E**in the conductor. Then

**∇**×**B**= μ_{0}σ**E**+ μ_{0}ε_{0}∂**E**/∂t.

**∇**×(**∇**×**B**) =**∇**(**∇**∙**B**) -**∇**^{2}**B**= μ_{0}σ(**∇**×**E**) + μ_{0}ε_{0}∂(**∇**×**E**)/∂t.

∇^{2}**B**- μ_{0}σ∂**B**/∂t - μ_{0}ε_{0}∂^{2}**B**/∂t^{2}= 0.

Similarly: ∇^{2}**E**- μ_{0}σ∂**E**/∂t - μ_{0}ε_{0}∂^{2}**E**/∂t^{2}= 0.

Both**E**and**B**satisfy the damped wave equation.

(b) Let**E**(**r**,t) =**E**_{0 }exp(ikz)exp(-iωt). We have a plane wave propagating into the z-direction.

Then k^{2}= iμ_{0}σω + μ_{0}ε_{0}ω^{2}= μ_{0}ε_{0}ω^{2}(1 + iσ/(ε_{0}ω)).

k^{2}= μ_{0}ε_{0}ω^{2}(1 + σ^{2}/(ε_{0}ω)^{2})^{½ }e^{iφ}. tanφ = σ/(ε_{0}ω).

k = (μ_{0}ε_{0})^{½}ω(1 + σ^{2}/(ε_{0}ω)^{2})^{¼ }e^{iφ/2}= |k|e^{iφ/2}.

Assume σ/(ε_{0}ω) >> 1. tanφ --> ∞. φ -- > 90^{o}. φ/2 --> 45^{o}, and (1 + σ^{2}/(ε_{0}ω)^{2})^{¼}--> (σ/(ε_{0}ω))^{½}.

k = (μ_{0}σω)^{½ }e^{iπ/4 }= (μ_{0}σω)^{½ }(cos(π/4) + i cos(π/4)) = (μ_{0}σω/2)^{½}(1 + i) = k_{1}+ ik_{2},

k_{1}≈ k_{2}≈ (μ_{0}ωσ/2)^{½}.×

∇**E**= i**k**×**E**= -∂**B**/∂t.

**B**(**r**,t) =**B**(**r**)_{ }exp(-iωt). Therefore i**k**×**E**= iω**B**. ik(**z**/z)×**E**= iω**B**.

E_{0}/B_{0}= ω/k = ω/|k| e^{-iπ/4}= (ω/(μ_{0}σ))^{½}e^{-iπ/4}.

With a convenient choice of the zero of t we have

Re(E) =|E_{0}| exp(-k_{2}z) cos(k_{1}z - ωt), Re(B) = |B_{0}| exp(-k_{2}z) cos(k_{1}z - ωt + π/4). "B lags behind E" when plotting E and B for fixed z as a function of time.

(c)**S**= (1/μ_{0})[Re(**E**) × Re(**B**)] =**k**|E_{0}|^{2}(σ/(ωμ_{0}))^{½}exp(-2k_{2}z) cos(k_{1}z - ωt) cos(k_{1}z - ωt + π/4)

=**k**|E_{0}|^{2}(σ/(2ωμ_{0}))^{½}exp(-2k_{2}z) (cos^{2}(k_{1}z - ωt) - cos(k_{1}z - ωt) sin(k_{1}z - ωt)).

<**S**> =**k**| ½E_{0}|^{2}(σ/(2ωμ_{0}))^{½}exp(-(2μ_{0}σω)^{½}z).

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.

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 n_{a}and n_{b}, respectively. With this convention, Snell's law reads n_{a}sinθ_{a}= n_{b}sinθ_{b}. Both angles are measured from the surface normal. For n_{a}> n_{b}. we have that θ_{a}> θ_{b}; For θ_{a}> θ_{crit}no light is refracted into medium b. This corresponds to the condition that θ_{b}= 90^{o}, sinθ_{b}= 1. We thus have that sinθ_{crit}= n_{b}/n_{a}. - Details of the calculation:

In our problem the light has an angle of incidence of 0^{o}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 + θ

_{crit}= 90^{o}.

(a) For a glass-air interface, n_{a}= 1.52 and n_{b}= 1.00.

We have that sinθ_{crit}= 1/1.52. Thus, θ_{crit}= 41.1^{o}. We conclude that α = 90^{o}- θ_{crit}= 48.9^{o}.

(b) In this case the refraction problem corresponds to glass -->water, for which n_{b}= 1.33. Therefore sinθ_{crit}= 1.33/1.52 , or θ_{crit}= 61.3^{o}. We thus have that α = 90^{o}- θ_{crit}= 28.7^{o}.