Solution Set 1 – Phys 4510 Optics – Fall 2014

Due date: Tu, September 9, in classScoring rubric4 points/sub-problem, total: 40 points3: Small mistake in calculation or formula2: Correct formula but calculation error1: Partial credit depend on the answer

Reading: Hecht, skim Ch. 2, thoroughly read Ch 3 introduction, 3.1, 3.2, 3.3.1, 3.3.2, 3.4.1, 3.4.3.

1. Tell me in a few words about your main interests in physics, your current or possible plan to doresearch in optics, your plan to go to graduate school or industry or other after graduating, etc.This will help me fine tune topics we cover.

2. We derived the wave-equation for the electric field E(y, t) which reads for the 1D case:

∂2E(y, t)

∂y2− 1

c2∂2E(y, t)

∂t2= 0 (1)

a) Show that the ansatz E(y, t) = E0cos(ωt − ky + ϕ) is a solution of the wave equation, withwavevector k, angular frequency ω, and phase ϕ. What are the units of k, ω, ϕ, and c? Whatis the relationship between k and ω (it is called dispersion relation)?

Solution:

A straightforward substitution demonstrates that the ansatz is valid as long as the dispersionrelation k(ω) = ω/c holds. The argument of the cosine function requires as its input radians,hence we have for the dimensions of k, ω, ϕ, and c

[k] =radians

length,

[ω] =radians

time,

[ϕ] = radians,

[c] =length

time.

b) What is the spatial and temporal evolution of the plane wave? Discuss, whether and how realphysical waves can be represented by plane waves.

Solution:

Clearly at a fixed position in space, the wave evolves sinusoidally in time. At a fixed time, thewave has a sinusoidal spatial variation. Due to the infinite extent in space and time, a planewave carries an infinite amout of energy and extends infinitely in both space and time and ishence non-physical.

1

ky

Figure 1: Plot of the spatial distribution of the normalized electric field at a time t (blue) and a time ∆tlater (red).

c) Sketch the wave at t = 0 and t = t′ for 0 < t′ < 2π/ω. What can you say about the propaga-tion direction of the wave? What is the velocity with which the crest of the wave propagatesalong the y-direction?

Solution:

The sketch is shown in Fig. 1. Now, consider a point on the wave: when the time increases by∆t, the position increases by ∆y. The condition that we stay at the same place on the wave isequivalent to saying that the cosine takes on the same value. This means that ∆(ωt − ky) =ω∆t − k∆y = 0. Hence we have ∆t/∆y = ω/k = c. Thus the wave speed is c. Furthermore,since ω and k have different signs in the argument of the cosine function, if ∆t > 0 we musthave ∆y > 0 so that any point on the wave is traveling to the right or positive direction.

3. Consider a electromagnetic plane wave propagating in the x-direction, and polarized in the y-direction.

a) Show from Maxwell’s equations that

∂Ey

∂x= −∂Bz

∂t(2)

Solution:

We can take Faraday’s law

−∂B

∂t= ∇× E = (

∂Ez

∂y− ∂Ey

∂z)x+ (

∂Ex

∂z− ∂Ez

∂x)y + (

∂Ey

∂x− ∂Ex

∂y)z

Since the direction of propagating wave is x, non-x derivatives are 0.

∇× E = −∂Ez

∂xy +

∂Ey

∂xz

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In this case, Ez must be 0 because polarization direction of propagating wave is y. So

∇× E =∂Ey

∂xz

Now we know that −∂B∂t only have components in z-direction. So

∂Ey

∂x= −∂Bz

∂t

b) Let us now assume a specific harmonic wave of the form Ey(x, t) = Ey,0cosω(t− x/c).What is the corresponding expression for the magnetic fieldBz(x, t)? Also show that Ey = cBz.

Solution:

∂Ey

∂x=

ω

cEy,0sin(ωt− ωx/c) = −∂Bz

∂t

So we can derive expression of magnetic field

Bz(x, t) =1

cEy,0cos(ωt− ωx/c)

And we know the relationship iscBz = Ey

c) Sketch in a single 3D graph Ey and Bz as they propagate in the x-direction. What is thephase relationship between Ey and Bz? Lastly calculate the corresponding Poynting vector(component) Sx.

Solution:

The electric and magnetic field oscillate in phase, but their direction of oscillation is perpen-dicular (Sketch is shown in Fig. 2).

S =1

µ0E ×B

=1

µ0c(Ey,0cos(ωt− ωx/c)y)× (Ey,0cos(ωt− ωx/c)z)

=(Ey,0)

2

µ0ccos2(ω(t− x/c))x

4. This problem shall give you a feeling for intensities, field amplitudes, and frequencies of light in thevisible and infrared. Consider two cw laser sources, (1) a green, e.g., HeNe laser (λ = 543 nm), and(2) an IR, e.g., CO2 laser (λ ∼ 10µm), both with a power of 10 mW each.

a) Calculate for each laser light in vacuum the frequency, the wavevector, the photon energy (ineV, and in multiples of kT at room temperature), and the intensity in number of photons persecond. Note on units: quantum energy is most conveniently given in eV, wavevector in cm−1).

Solution:

The relationship between λ, ν, and c is given by λν = c. From this it is easy to calculate thefrequencies resulting in

νHe:Ne = 5.52× 1014Hz,

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x

y

z

Figure 2: Sketch of the electromagnetic field propagation.

νCO2 = 3.00× 1013Hz.

For the photon energy, we have another plug-and-chug relation: Eγ = hν where h is Planck’sconstant. Using the value of this constant h = 6.58× 10−16eV · s we find

EHe:Ne = 2.28eV,

ECO2 = 0.124eV.

It is interesting to compare these energies to a thermal energy unit kbT ≈ 0.0259eV at roomtemperature. Note, we thermally emit infrared light whose energy roughly, say within a factorof 2 or 3, corresponds approximately to kbTbody ≈ 0.0267eV giving λ ≈ 46µm. Calculating thewave vector is a simple task using the relation k = 2π/λ giving (of course after appropriatechanges in units)

kHe:Ne = 1.16× 105cm−1,

kCO2 = 6.28× 103cm−1.

Finally we want to calculate the intensity in number of photons per second. We are given thebeam power in the problem, which is the amount of energy crossing a plane perpendicularto the beam axis each second. Thus if we knew the energy for one photon, we could dividethe power by this number to get the number of photons per second passing through a planeperpendicular to the beam. We have already calculated this number for both of our beams.Applying the result we find

IHe:Ne = 2.73× 1016photons

s,

ICO2= 5.03× 1017

photons

s.

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b) Assume you focus the light using appropriate optics down to a spot size of 10µm diameter.What is the irradiance (aka intensity) [most convenient in W/cm2]? What is the electric fieldamplitude? What is the corresponding magnetic field amplitude?

Solution:

The spot radius is 5µm, thus the power per unit area or intensity I = P/A is

I = 12732W

cm2.

The electric field amplitude can be calculated using I = cnϵ02 |E0|2 = n

2Z0|E0|2. Inputing the

approprate values for free space and sloving for E0 we find

E0 = 3.09× 105V

m.

And the corresponding magnetic field amplitude is given by B0 = E0

c

B0 = 1.03× 10−3T.

Note that interatomic fields are roughly of the order 1011V/m, so with a .1mW laser we won’tbe displacing the electrons far from their equilibrium position. As a result, the electron ex-periences a linear restoring force so that we are doing linear optics as in problem 5. Withmuch stronger fields of the order of the interatomic field, we will begin to see the effects ofnonlinearities in the restoring force which leads into topics of nonlinear optics where effectssuch as harmonic generation can occur.

c) Now we use a 100 W light bulb converting say 2% of its electric power into light. Distributethe radiation evenly over a solid angle Ω = 1 sterad by a reflector. Calculate the light intensityand electric field of the light field at a distance of 1m from the bulb.

Solution:

Our light produces 2W of light all of which is collected into a solid angle of 1sterad. Solidangle is given by Ω = S/r2. where S is the surface area of a spherical cap of radius r. So wewish to find S at r = 100cm which is just Ωr2 thus

I = 2× 10−4 W

cm2.

E0 = 38.8V

m.

d) Discuss how intensity and field strength compare if you now were to use sunlight collectedwith a 5 cm diameter lens, still focused to a spot size of 10µm diameter (For this problemstart from your knowledge of the luminosity of the sun of L = 3.9× 1026 W, and the distanceof the earth from the sun of 1.5× 108 km; neglect atmospheric effects).

Solution:

We can use proportional relation to derive luminosity of the sun in the lens

3.9× 1026 : 4π(1.5× 1013)2 = P : π(2.5)2

5

P = 2.708W

Then we can simply claculate intensity

I =P

A=

2.708

π(5× 10−4)2= 3.44793× 106

W

cm2

And we know I = cnϵ02 |E0|2 = n

2Z0|E0|2, thus

E0 = 5.09523× 106V

m.

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