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.
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
(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)),
Er(r,t) =(y/y) Er exp(i(kr×r
- wt)),
Et(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/w,
Bi = (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)),
Br(r,t) = -(y/y) Br exp(i(kr×r
- wt)),
Bt(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:
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:
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 |
