Assignment 3, solutions

Problem 1:

A beam of monochromatic light with a wavelength of 500 nm is directed through an absorber having 5 equally narrow slits separated by 20 mm between adjacent slits.  The resulting diffraction pattern is observed on a screen that is perpendicular to the direction of light and 5 m from the slits.  The intensity of the central maximum is 1.3 W/m2.
(a)  What are the distances from the central maximum to the first and second principal maxima on the screen?
(b)  What will be the intensity of the central maximum if there are only 4 equally narrow slits (of the same width as in part a), separated by 20 mm between adjacent slits? 

Solution:

Concepts, principles, relations that apply to the problem:
Diffraction and interference
Why do they apply?
We are supposed to analyze the diffraction patterns produced by several slits.
How do they apply?
(a)  Constructive interference:
dsinq = ml,  m = 0, 1, 2, ….
sinq = z/(z2+L2)1/2 @ z/L if L >> z.

slitdiagram.gif (3469 bytes)
Here d = 20 mm, l  = 500nm, L = 5m
m = 1: sinq  = 2.5*10-2,  sinq = z/L,  z @ 12.5 cm.
m = 2: sinq  = 5*10-2,  sinq = z/L,  z @ 25 cm

(b)  The central intensity is proportional to the square of the number of sources.
If there are only 4 sources, then the intensity of the central maximum will be
1.3*16/25 W/m2 = 0.832 W/m2.

Details of the calculation:
None

Problem 2:

(a)  Determine the speed of light in water, which has a dielectric constant of 1.78.
(b)  An electromagnetic wave in vacuum has an electric field amplitude of 220 V/m.  Calculate the amplitude of the corresponding magnetic field.
(c)  What is the energy of a photon of blue light (l = 450nm) and of a photon of red light (l = 700nm) in units of eV.

Solution:

Concepts, principles, relations that apply to the problem:
The index of refraction,  v = c/n = (kmm0kee0)-1/2,
B = E/c (SI units),
Photon energy E = hn
Why do they apply?
We are asked to review some basic properties of EM waves.
How do they apply?
(a)  ke = k = 1.78, km = 1.
v = (km0e0)-1/2 = c/k1/2 = 2.23´108m/s.
(b)  B = E/c = (220N/C)/(3´108m/s) = (7.33´10-7N/Am) = 7.33´10-7T.
(c)  E = hc/l.
Blue light: E = (6.626´10-34Js)(3´108m/s)/(450´10-9m) = 4.4´10-19J = 2.76eV
Red light:  E = (6.626´10-34Js)(3´108m/s)/(700´10-9m) = 2.8´10-19J = 1.8eV
Details of the calculation:
None

Problem 3:

It has been proposed to drive a spacecraft remotely by directing an intense electromagnetic beam to the craft.
(a)  Is it more efficient to absorb the beam or reflect it from the craft?
(b)  If the beam carries 106 watts uniformly in an area of 5 m2 and the beam is reflected, how long would it take for a 1000 kg spaceship to reach a final velocity of 106 m/s, and how far would the craft travel in this time?

Solution:

Concepts, principles, relations that apply to the problem:
Energy and momentum flux for an electromagnetic wave, momentum conservation
Why do they apply?
Electromagnetic waves carry energy and momentum. When an EM wave is absorbed or reflected, momentum must be conserved.
How do they apply?
(a)  Approximately twice as much momentum is transferred by reflecting the beam than by absorbing the beam.
(b)  <S> = 2*105 W/m2 = energy flux.
<S>/c = momentum flux.
dp/dt = 2*<S>A/c = 2*106W/c = (2/3)*10-2 N.
dv/dt = (2/3)*10-5 m/s2.
t = 106s/((2/3)*10-5) = 1.5*1011s.
d = ½*(2/3)*10-5(1.5*1011)2m = 7.5*1016 m = 7.9 ly.
Details of the calculation:
None

Problem 4:

Fresnel's reflectance formulas are given by

R = |sin(qi-qr)/sin(qi+qr)|2   or   R = |tan(qi-qr)/tan(qi+qr)|2

depending on the polarization of the incident wave with respect to the plane of incidence, and with qi and qr the angles of incidence and refraction, respectively.
(a)  Specify the boundary conditions across an interface.
(b)  For simplicity assume m1 = m2 = m0, e1 = e0, e2 = e, 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, principles, relations that apply to the problem:
Maxwell's equations, boundary conditions for the electric and magnetic fields
Why do they apply?
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.
How do they apply?
(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 - wt)), 
E
r(r,t) =(y/y) Er exp(i(kr×r - wt)), 
E
t(r,t) =(y/y) Et exp(i(kt×r - wt)).
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(qi) = krsin(qr) = ktsin(qt).
ki = kr = (m0e0)1/2w = w/c.  kt = (m0e')1/2w = n2w/c.  Therefore
sin(qi) = sin(qr) = n2sin(qt).
This implies qi = qr (law of reflection) and sin(qr) = n2sin(qt) (Snell's law) for s-polarization.

The tangential component of B = m0H is continuous across the boundary.  The x-direction is tangential to the interface.
B = k´E/wBi = (1/w)(ki´Ei)×x = (1/w)(Ei ´ x)×ki =(1/w)Eikiz = (1/w)Eikicos(qi).
Similarly Br×x = -(1/w)Erkrcos(qr),  Bt×x = (1/w)Etktcos(qt).
We then have Eikicos(qi) - Erkrcos(qr) = Etktcos(qt),
or  Eicos(qi) - Ercos(qr) = n2Etcos(qt).
Combining this equation with Ei + Er = Et and qi = qr from above yields
Er/Ei = (cosqi - n2cosqt)/(cosqi + n2cosqt)
= (cosqi - (sinqi/sinqt)cosqt)/(cosqi + (sinqi/sinqt)cosqt)
= (1 - (tanqi/tanqt))/(1 + (tanqi/tanqt))
= (tanqt - tanqi)/(tanqt + tanqi)
= sin(qt - qi)/sin(qt + qi).
R =|Er/Ei|2 = |sin(qt - qi)/sin(qt + qi)|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)]

Details of the calculation:
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 - wt)), 
B
r(r,t) = -(y/y) Br exp(i(kr×r - wt)), 
B
t(r,t) = -(y/y) Bt exp(i(kt×r - wt)).

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(qi)  - Er exp(i(kr×r))cos(qr) = Et exp(i(kt×r))cos(qt) 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(qi) = krsin(qr) = ktsin(qt).
ki = kr = (m0e0)1/2w = w/c.  kt = (m0e')1/2w = n2w/c.  Therefore
sin(qi) = sin(qr) = n2sin(qt).
This implies qi = qr (law of reflection) and sin(qr) = n2sin(qt) (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/w,  Bi = (1/w)Eiki, = Ei/c. Br = (1/w)Erkr = Er/c, Bt = (1/w)Etkt = n2Et/c.
We then have Ei + Er = n2Et.
Combining this equation with Ei cos(qi)  - Er cos(qr) = Et cos(qt) and qi = qr  fom above yields
Er/Ei = (n2cosqi - cosqt)/(n2cosqi + cosqt)
= ((sinqi/sinqt)cosqi - cosqt)/((sinqi/sinqt)cosqi + cosqt)
= (sinqicosqi - cosqtsinqt)/(sinqicosqi + cosqtsinqt)
= tan(qt - qi)/tan(qt + qi).
R =|Er/Ei|2 = |tan(qt - qi)/tan(qt + qi)|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 qt + qi = p/2, the reflectance is zero for p-polarization.
This happens when qi = qB, n2sinqB  = n2sin((p/2)-qB), tanqB = n2/n1.
qB is called the Brewster angle.
When qi = qB, then only light with s-polarization is reflected.  For a wave incident with mixed polarization incident with qi = qB, the reflected light is polarized.
With n1 = 1, n2 = 1.5 we have qB = 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 w is incident on the plasma.
Let the electric field component of the plane wave be E = E0exp(i(kx - wt)).
(a)  Solve the equation of motion for a single charge and find the current density j and the conductivity s of the plasma as a function of w.
(b)  Assume a plane wave of the form E = E0exp(i(kx - wt)) propagate in the plasma with conductivity s.  Find the dispersion relation —the relation between k and w— for the electromagnetic wave in the plasma and the index of refraction as a function of w.

Solution:

Concepts, principles, relations that apply to the problem:
Newton's 2nd law, charge and current density, plane waves in a conducting medium
Why do they apply?
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.
How do they apply?
(a)  Equation of motion for a single charge: md2r/dt2 = -eE = -eE0exp(i(kx - wt)).
r(t) = (eE0/(mw2))exp(i(kx - wt)).  v(t) = (eE0/(imw))exp(i(kx - wt)) = (e/(imw))E
j(t) = -nrv(t) = -(ne2/(imw))E(t). j = sEs = (ine2/(mw)).

(b)  Ñ´B = m0sE + m0e0E/t.
Ñ´(Ñ´B) = Ñ(Ñ×B) - Ñ2B = m0s(Ñ´E) + m0e0(Ñ´E)/t.
Ñ2B - m0B/t - m0e02B/t2 = 0.
Similarly:  Ñ2E - m0E/t - m0e02E/t2 = 0.
Plane wave solutions of the form E = E0exp(i(kx - wt)) have to satisfy
-k2 + iwm0s + w2/c2 = 0,  k2 = w2/c2 - (m0ne2/m ) = (1/c2)(w2wp2),
k = (1/c)( w2wp2)1/2,
where wp = (m0ne2c2/m)= (ne2/(e0m)).
c/n = w/k.  n = ck/w = (1 – wp2/w2)1/2.  For wp > w n is imaginary, the wave is absorbed.

Details of the calculation:
None

Problem 6:

Consider an electromagnetic traveling wave with electric and magnetic fields given by

Ex = E0cos(kz – wt + f),

and

By = B0cos(kz – wt + f).

Using Maxwell’s equations show that B0 can be written in terms of E0.

Solution:

Concepts, principles, relations that apply to the problem:
Maxwell's equations
Why do they apply?
Maxwell's equations in free space  yield the wave equation for both E and B.   They can also be used to show that E ^ B,  E ^ k,  B ^ k,  B = (m0e0)1/2(k/k)´E.
How do they apply?
From Maxwell’s equations:
Ñ
´E = -B/t.
Ñ
´E = Ex/z j = -kE0sin(kz – wt + f) j,  -B/t = -wB0sin(kz – wt + f) j.
Therefore kE = wB0,  B0 = (k/w)E0.
In free space k/w = 1/c.

Details of the calculation:
None

Problem 7:

Light of wavelength 640nm 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, principles, relations that apply to the problem:
Double slit interference
Why do they apply?
Destructive interference occurs when the optical path length difference is ml/2, with m = odd integer.
How do they apply?
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 = l/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 = (2pn/l - 2p/l)x = (2p/l)(n - 1)x.
For the minimum thickness we need  (2p/l)(n - 1)x = p, x = l/(2(n-1))  = l/1.2 = 533 nm.  The minimum thickness is 533 nm.
Details of the calculation:
None