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MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics
8.02
Spring 2014
Practice Problem Set 11 Solutions
Problem 1: Electromagnetic Waves and the Poynting Vector
We have been studying one particular class of electric and magnetic fields solutions called plane
sinusoidal traveling waves. One special example is an electromagnetic wave traveling in the
positive x -direction with the speed of light c is described by the functions

⎛ 2π
⎞
E(x, y, z,t) = E y (x,t) ĵ = E y,0 sin ⎜ (x − ct)⎟ ĵ
⎝ λ
⎠

⎛ 2π
⎞
1
B(x, y, z,t) = Bz (x,t)k̂ = E y,0 sin ⎜ (x − ct)⎟ k̂
c
⎝ λ
⎠
Suppose the plane wave front (infinite yz -plane) traveling at speed c in the positive x -direction
passes through a rectangular volume of space that has area A perpendicular to the direction of
propagation and length cΔt , corresponding to the distance the electromagnetic wave travels in
time Δt .
Inside this volume of space the electric and magnetic fields store energy and that energy changes
in time as the wave passes through the volume.
The energy per time per area transported by this electromagnetic wave is called the Poynting
vector and is defined as
 1  
S= E×B.
µ0
The power transmitted by the Poynting vector through a surface is the flux given by the
expression

P(t) = ∫∫ S(t) ⋅ n̂ da .
surface
The flux of the Poynting vector on a surface describes the energy that ‘flows’ across the surface.
Question 1: What the Poytning vector associated with this wave?
Answer:
1
 1   1
⎛ 2π
⎞ 1
⎛ 2π
⎞
S = E×B =
E y,0 sin ⎜ (x − ct)⎟ ĵ × E y,0 sin ⎜ (x − ct)⎟ k̂
µ0
µ0
c
⎝ λ
⎠
⎝ λ
⎠
2
⎛ 2π
⎞
1 2
=
E y,0 sin ⎜ (x − ct)⎟ î
µ 0c
⎝ λ
⎠
.
So the energy per time per area is transported in the direction of propagation of the wave.
Question 2: What is the time-averaged Poynting vector field on the fixed plane x = 0 over one
period? The definition of a time average of a periodic function f (t) over one period is
f (t) =
1
T
T
∫ f (t) dt
0
Recall that c = 1 / µ 0 ε 0 .
Answer: Set x = 0 in your expression for the Poynting vector. Then

1 2
S(0,t) =
E y,0 sin 2 (−2πct / λ) î .
µ 0c
Note
that
sin 2 (−2πct / λ) = sin 2 (2πct / λ) .
Use
the
trigonometric
identity
that
sin 2 (2πct / λ) = (1/ 2)(1− cos(4πct / λ) .
The time averaged Poynting vector on the plane x = 0 is
T

1 1 2
S(0,t) = ∫
E (1/ 2)(1− cos(4πct / λ) dtî
T 0 µ 0 c y,0
=
2
E y,0
2µ 0 cT
T
2
E y,0
T
0
0
0
.
∫ dtî − 2µ cT ∫ cos(4πct / λ) dtî
The integration is straightforward,
T
2
2

E y,0
E y,0
T
sin(4πct / λ)
S(0,t) =
t î −
î
2µ 0 cT 0
2µ 0 cT (4πc / λ) 0
2
E y,0
2
E y,0
sin(4πcT / λ)
=
î −
î
2µ 0 c 2µ 0 cT (4πc / λ)
.
The second term vanishes because cT = λ and sin(4π) = 0 . Therefore
2
2

E y,0
S(0,t) =
î .
2µ 0 c
Recall that c 2 = 1/ (µ 0 ε 0 ) so 1/ (µ 0 c) = c / (µ 0 c 2 ) = ε 0µ 0 c / µ 0 = c ε 0 . In conclusion the time
averaged Poynting vector over one period on the plane x = 0 is given by

1
2
S(0,t) = c ε 0 E y,0
î
2
Because the integral was only over the time, it doesn’t matter what plane we choose to time
average the Poynting vector, our result holds for all planes and so

1
2
S(x,t) = c ε 0 E y,0
î .
2
Question 3: What is the time-average (over one period) of the energy density stored in the
electric and magnetic fields at the point (x, y, z,t) . Recall that uelec = (1/ 2)ε 0 E 2 and
umag = B 2 / 2µ 0 . How is that related to the magnitude of the time-averaged Poynting vector?
Answer: We begin by time averaging the electric field energy density
T
uelec
T
1
1 1
2
= ∫ uelec dt = ∫ ε 0 E y,0
sin 2 (2π(x − ct) / λ) dt .
T 0
T 02
Once again the integral
T
1
1
sin 2 (2π(x − ct) / λ) dt = .
∫
T 0
2
To see this explicitly at arbitrary plane with fixed values of x , make the change of variables
u = (2π(x − ct) / λ) . Then du = −2πcdt / λ ⇒ dt = (−λ / 2πc)du . The limits of the integral go
from u = 2πx / λ to u = (2π(x − cT ) / λ) = (2πx / λ − 2π) because cT = λ . Our time-averaged
integral is then
uelec = −
2
ε 0 E y,0
λ 2πx/λ−2π
4T πc
∫
2πx/λ
sin (u) du = −
2
2 2πx/λ−2π
ε 0 E y,0
4π
∫
sin 2 (u) du .
2πx/λ
Once again we have that
3
2πx/λ−2π
∫
2πx/λ
1
sin (u) du =
2
2
2πx/λ−2π
∫
2πx/λ
u=2πx/λ−2π
1 u=2πx/λ−2π sin(2u)
(1− cos(2u)) du = u u=2πx/λ −
2
2 u=2πx/λ
⎛ sin(2(2πx / λ − 2π) sin(2(2πx / λ) ⎞
1
= ((2πx / λ − 2π) − 2πx / λ) − ⎜
−
⎟⎠
2
2
2
⎝
⎛ sin(2(2πx / λ − 2π) sin(2(2πx / λ) ⎞
= −π − ⎜
−
⎟⎠ = −π
2
2
⎝
where we used the fact that sin(2(2πx / λ − 2π) = sin(2(2πx / λ) and therefore the second term in
the above expression is zero. So the time-averaged electric field energy density is
uelec = −
2
ε 0 E y,0
λ 2πx/λ−2π
4T πc
∫
sin (u) du =
2
2πx/λ
2
ε 0 E y,0
4
.
By a similar argument the time-averaged energy density stored in the electric and magnetic fields
2
umag
T
T
T
⎛ 2π
⎞
1
1 1 2
1
1
= ∫ umag dt = ∫
B dt = ∫
E y,0 2 sin ⎜ (x − ct)⎟ dt
2
T 0
T 0 2µ 0
T 0 2µ 0 c
⎝ λ
⎠
=
1
1
E y,0 2 = ε 0 E y,0 2
2
4
4µ 0 c
By a similar argument the time-averaged energy density stored in the electric and magnetic fields
is
1
utotal = uelec + umag = ε 0 E y,0 2 .
2
Recall that the time-averaged magnitude of the Poynting vector is

1
2
S(x,t) = c ε 0 E y,0
î .
2
Thus we see that time-averaged energy density is equal to the magnitude of the time-averaged
Poynting vector divided by the speed of light
utotal =

S(x,t)
c
Question 4: What is the time-averaged energy stored in the electric and magnetic fields in a
rectangular volume of cross-sectional area A and length cΔt , where the length cΔt is aligned

along the x -axis and cΔt << λ . The last assumption allows us to assume that S(x,t) is nearly
uniform across the box.
4
Answer:
1
U total = utotal ΔVvol = utotal AcΔt = c ε 0 E y0 2 AΔt
2
Question 5: What is the time-averaged rate of change of the total energy stored in the electric
and magnetic fields in the rectangular volume of cross-sectional area A and length cΔt ?
Answer:
d
U
dt total
1
c ε 0 E y0 2 AΔt
1
= lim 2
= c ε 0 E y0 2 A
Δt→0
Δt
2
Question 6: What is the time-averaged flow of power through the surface of area A located on
the surface given by x = x p , perpendicular to the direction of propagation? How does the power
that through the rectangular surface (this flux is negative since it flows into the volume) compare
to the time derivative of the energy stored in the fields inside the volume?
Answer: From the introductory remarks, the power through a surface (energy is flowing across
the surface) is given by

P(t) = ∫∫ S(x,t) ⋅ n̂ da .
surface
Therefore the time-averaged power is on the surface of area A located on surface given by
x = xp ,
P(t) =
∫∫
surface


1
S(x p ,t) ⋅ n̂ da = S(x p ,t) A = c ε 0 E y0 2 A .
2
For simplicity (to avoid complicated integrals), let’s assume the wave at time t has just entered
the box and the box is empty. In time Δt later, the wave has not left the volume, so no power
flows out. Thus we conclude that the power that flows into the rectangular volume (this flux is
negative since it flows into the volume) equals the time derivative of the energy stored in the
fields inside the volume

 d
−
S(x p ,t) ⋅ d A =
U
.
∫
∫
dt total
closed
surface
5
Problem 2: Poynting Vector and Radiation Pressure:
A plane electromagnetic wave transports energy in the direction of propagation of the wave. The
power per square area is given by the Poynting vector
 
 E×B
.
S=
µ0
The power that flows into the rectangular volume cross-sectional area A and length cΔt appears
as rate of change of the energy stored in the fields inside the volume.

d
Ppower = S A =
U
.
dt total
The electromagnetic wave also transports momentum, and hence can exert a radiation pressure
on a surface due to the absorption and reflection of the momentum.
The momentum carried by an electromagnetic wave is related to the energy of the wave
according to

U =cp
If the plane electromagnetic wave is completely absorbed by a surface of cross-sectional area A

then the momentum Δp delivered to the surface in a time Δt is given by
 ΔU
Δp =
.
c
The force that the wave exerts on the surface is then the rate of change of the momentum in time
F = lim
Δt→0

Δp
Δt
1 ΔU 1 dU
=
.
Δt→0 c Δt
c dt
= lim
Since the rate of change of energy is related to the power flowing across the surface, the force is
F=
1 dU 1
1 
= Ppower =
S A.
c dt
c
c
The radiation pressure Ppressure is then defined to be the force per area that the wave exerts on the
surface
abs
Ppressure
≡
F 1 
=
S ,
A c
perfectly absorbing.
When the surface completely reflects the wave, then the change in momentum is twice the
absorbing case since the wave completely reverses direction,
6
ΔU

Δp = 2
.
c
Therefore the radiation pressure of a wave on a perfectly reflecting surface is
ref
Ppressure
≡
F
1 
=2 S ,
A
c
perfectly reflecting.
Suppose a space station is placed in orbit the same distance away from the sun as the earth but on
the opposite side of the sun from the earth in a circumpolar orbit (the orbit plane is perpendicular
to the plane of the earth’s orbit around the sun). The average earth-sun distance is
re,s = 1.50 × 1011 m . An astronaut of mass m = 100 kg on a space walk loses contact with the
space station. The spacesuit has a cross sectional area of approximately A = 2 m 2 . The mass of
the sun is ms = 1.99 × 1030 kg . The gravitational constant is G = 6.67 × 10−11 N ⋅ m 2 ⋅ kg −2 .
a) What is the radiation pressure of the sun on the astronaut’ spacesuit. Assume it is perfectly
reflecting?
Answer: We need to know the magnitude of the Poynting vector. We can use the fact that the
time averaged power output of the sun is Psun = 4 × 1026 W and hence

Psun
3.9 × 1026 W
S =
=
= 1.4 × 103 W ⋅ m -2 .
2
11
2
4πres
(4π)(1.50 × 10 m)
Therefore
ref
pressure
P
1 
1 
2(1.4 × 103 W ⋅ m −2 )
=2 S =2 S =
= 9.2 × 10−6 Pa
8
−1
c
c
(3.0 × 10 m ⋅ s )
b) What is the force on the astronaut?
Answer:
F=P
ref
pressure
1 
2(1.4 × 103 W ⋅ m −2 )(2.0 m 2 )
A= 2 S A=
= 1.8 × 10−5 N
8
−1
c
(3.0 × 10 m ⋅ s )
c) How does this force compare to the force of gravity?
Answer: The gravitational force is
Fgrav = G
mms
2
re,s
=
(6.67 × 10−11 N ⋅ m 2 ⋅ kg −2 )(100 kg)(1.99 × 1030 kg)
= 6 × 10−1 N
(1.50 × 1011 m)2
So the gravitational force is more than ten thousand times stronger.
7
Problem 3: Intensity of the Sun At the upper surface of the earth’s atmosphere, the timeaveraged magnitude of the Poynting vector, referred to as the solar constant, is given by

S = 1.35 × 103 W ⋅ m −2 .
a) If you assume that the sun’s electromagnetic radiation is a plane sinusoidal wave, what are the
magnitudes of the electric and magnetic fields?
Answer: The magnitude of the time-averaged Poynting vector is related to the amplitude of the
electric field by

1
2
S = c ε 0 E y,0
.
2
So
E y,0 =

2 S
cε 0
=
2(1.35 × 103 W ⋅ m −2 )
= 1.01 × 103 V ⋅ m −1 .
8
−1
−12
2
−1
−2
(3.0 × 10 m ⋅ s )(8.85 × 10 C ⋅ N ⋅ m )
The amplitude of the magnetic field is
E0 1.01 × 103 V ⋅ m −1
B0 =
=
= 3.4 × 10−6 T .
8
-1
c
3.0 × 10 m ⋅ s
The associated magnetic field is less than one tenth the earth’s magnetic field.
b) What is the time-averaged power radiated by the sun? The mean sun-earth distance is
res = 1.50 × 1011 m .
Answer: The time-averaged power radiated by the sun at the distance re, s that the earth lies from
the sun is given by
Psun =
∫∫


S(t) ⋅ n̂ da = S
surface
−2
r = res
4πres 2
.
= (1.35 × 10 W ⋅ m )(4π)(1.50 × 10 m) = 3.8 × 10 W
3
11
2
26
8
Problem 4: Intensity of Double Slit Interference
When coherent monochromatic laser light falls on two slits separated by a distance d , the
emerging light will produce an interference pattern on a viewing screen a distance D from the
center of the slits. The geometry of the double slit interference is shown in the figure below.
Suppose the slits are located at the plane x = − D . The light that emerges from slit 1 and slit 2 at
time t are in phase. Suppose a screen is placed at the plane x = 0 . Suppose the transverse
component of the electric field of the wave from slit 1 at the point P is given by
E1 = E0 sin (ωt ) .
Let’s assume that the plane wave from slit 2 has the same amplitude E0 as the wave from slit 1.
Since the plane wave from slit 2 has to travel an extra distance to the point P equal to the path
length, this wave will have a phase shift φ relative to the wave from slit 1,
E2 = E0 sin (ωt + φ ) .
Question 1: Show that the phase shift φ , the wavelength λ , the distance between the slits, and
the angle related θ by
φ=
2π
d sin θ .
λ
As a hint determine a relationship between, the ratio of the phase shift φ to 2π and the ratio of
the path length Δr = d sin θ to wavelength λ ; i.e. how are φ / 2π and Δr / λ related?
Answer: The ratio of the phase shift φ to 2π is the same as the ratio of the path length
Δr = d sin θ to wavelength λ ,
9
φ Δr
=
.
2π λ
Therefore the phase shift φ is given by
φ=
2π
d sin θ .
λ
Question 2: The total electric field at the point P is the superposition of the these two fields
Etotal = E1 + E2 = E0 (sin (ωt ) + sin (ωt + φ )).
Use the trigonometric identity
⎛ A+ B ⎞
⎛ A− B ⎞
sin A + sin (B ) = 2sin ⎜
⎟ cos ⎜
⎟.
⎝ 2 ⎠
⎝ 2 ⎠
To show that the total transverse component of the electric field is
φ⎞
⎛
⎛φ⎞
Etotal = E1 + E2 = 2 E0 sin ⎜ ωt + ⎟ cos ⎜ ⎟ .
2⎠
⎝
⎝ 2⎠
Answer: We first use the superposition principle to add the component sof the two electric fields
yielding
⎛ ωt + ωt + φ ⎞
⎛ ωt − (ωt + φ) ⎞
Etotal = E0 sin ωt + sin ωt + φ = 2E0 sin ⎜
cos ⎜
⎟
⎟⎠
2
2
⎝
⎠
⎝
( ( )
(
))
Thus the total transverse component of the electric field is (noting that cos(−φ / 2) = cos(φ / 2)
φ⎞
⎛
⎛φ⎞
Etotal = E1 + E2 = 2 E0 sin ⎜ ωt + ⎟ cos ⎜ ⎟ .
2⎠
⎝
⎝ 2⎠
Question 3: The intensity of the light is equal to the time-averaged Poynting vector

1  
I= S =
E×B .
µ0
Because the amplitude of the magnetic field is related to the amplitude of the electric field by
B0 = E0 c . The intensity of the light is proportional to the time-averaged square of the electric
field,
I  Etotal 2 ,
10
where the time-averaged value of the square of the sine function is
φ⎞
1
⎛
sin 2 ⎜ ωt + ⎟ = .
2⎠
2
⎝
Determine the time averaged total electric field,
(E + E )
1
2
2
.
Answer: The intensity of the electric field time-averaged square of the electric field is
(E + E )
1
2
2
⎛ φ⎞
⎛
⎛ φ⎞
φ⎞
= 4E0 2 cos 2 ⎜ ⎟ sin 2 ⎜ ωt + ⎟ = 2E0 2 cos 2 ⎜ ⎟ .
2⎠
⎝ 2⎠
⎝
⎝ 2⎠
Let I max be the amplitude of the intensity. Then the intensity of the light at the point P is
⎛φ⎞
I = I max cos 2 ⎜ ⎟
⎝ 2⎠
Question 4: Show that the intensity is maximal when
d sin θ = mλ, m = 0, ± 1, ± 2, ± 3, ... .
Answer: The intensity has a maximum when the argument of the cosine is an integer number of
2π
d sin θ , we have that
multiples of π , φ 2 = ± mπ . Since the phase shift is given by φ =
λ
π
d sin θ = ± mπ . Thus we have the condition for constructive interference,
λ
d sin θ = ± mλ, m = 0, ± 1, ± 2, ± 3, ...
Question 5: Graph the intensity pattern on the screen as a function of distance y from the point
O for the case that D >> d and d >> λ .
11
Question 6: Determine the average over all phases of the time averaged total electric field,
(E + E )
1
2
2
phase
average
=
2
1 2mπ
E1 + E2 ) dφ .
(
∫
2mπ 0
Answer: The intensity of the electric field time-averaged square of the electric field is
(E + E )
1
2
2
⎛ φ⎞
= 2E0 2 cos 2 ⎜ ⎟ .
⎝ 2⎠
If we now average this over all phases,
2 E0 2
E0 2 2 mπ
E0 2
1 2 mπ 2 ⎛ φ ⎞
2π
cos
d
φ
=
(1
+
cos
φ
)
d
φ
=
(2mπ + sin φ 0 ) = E0 2 .
⎜ ⎟
∫
∫
0
0
2mπ
mπ
2mπ
⎝ 2⎠
Question 7: Do intensities add, i.e. is it true that
(E + E )
1
2
2
= E12 + E2 2 ?
Or do intensities add when we average over all phases,
(E + E )
1
2
2
phase
average
= E12 + E2 2 ?
Answer: The time-average of the individual electric fields are given by
E12 = E0 2 sin 2 (ωt ) = E0 2 2 ,
E2 2 = E0 2 sin 2 (ωt + φ ) = E0 2 2 .
If you only time average the total electric fields then
(E + E )
1
2
2
≠ E12 + E2 2 .
But if you also average over all possible phases then
(E + E )
1
2
2
phase
average
= E12 + E2 2 .
12
Problem 5: Superposition of Two Travelling Waves
Suppose the electric field of an electromagnetic wave is given by the superposition of two waves

E = E0 cos(kz − ω t) î + E0 cos(kz + ω t) î . You may find the following identities useful
cos(kz ± ω t) = cos(kz)cos(ω t)  sin(kz)sin(ω t) .

a) What is the associated magnetic field B(x, y, z,t) ?
Answer: We treat each contribution to the electric field separately. Then we have that
 E
E
B = 0 cos(kz − ω t) ĵ − 0 cos(kz + ω t) ĵ ,
c
c
where we used that fact that that the amplitude of the magnetic field is related to the amplitude of
the electric field by B0 = E0 / c , and the direction of the field satisfy
 

dir(E × B) = dir(propagation) . The contribution E1 = E0 cos(kz − ω t) î is propagating in the

positive k̂ -direction, and the contribution E2 = E0 cos(kz + ω t) î is propagating in the negative
k̂ -direction.
b) Use the identity cos(a)sin(a) = (1/ 2)sin(2a) to show that the energy per unit area per
unit time (the Poynting vector) transported by this wave is

S = (E0 2 / cµ0 )sin(2kz)sin(2ω t)k̂ .
Answer: We first use the identities cos(kz ± ω t) = cos(kz)cos(ω t)  sin(kz)sin(ω t) , to rewrite the
electric field as

E = E0 cos(kz − ω t) î + E0 cos(kz + ω t) î = 2E0 cos(kz)cos(ω t) î .
The corresponding expression for the magnetic field is
 E
E
2E0
B = 0 cos(kz − ω t) ĵ − 0 cos(kz + ω t) ĵ =
sin(kz)sin(ω t) ĵ .
c
c
c
The Poynting vector is then
 1  
⎡ 2E
⎤
1
S=
E × B = ⎡⎣ 2E0 cos(kz)cos(ω t) î ⎤⎦ × ⎢ 0 sin(kz)sin(ω t) ĵ⎥
µo
µo
⎣ c
⎦
 4E 2
S= 0 cos(kz)cos(ω t)sin(kz)sin(ω t)k̂
c µ0
.
13
1
Now use the identity cos(a)sin(a) = sin(2a) to rewrite the Poynting vector as
2
 E2
S = 0 sin(2kz)sin(2ω t)k̂ .
c µ0
c) What is the time-average of the Poynting vector? Briefly explain your answer.
Answer: Note the time average of the Poynting vector is given by
T

1 
S ≡ ∫ Sdt .
T 0
Therefore
T

1 E2
S = ∫ 0 sin(2kz)sin(2ω t) dtk̂
T 0 µ0 c
T
E0 2 1
=
sin(2kz) ∫ sin(2ω t) dtk̂
µ0 c T
0
.
The integral is then
T
T
1
1
sin(2ω t)dt = −
cos(2ω t) 0 = 0 ,
∫
20
2ω
where we used the fact that 2ω T = 4π and cos(2ω T ) = cos(4π ) = 1
Thus


S = 0.
This result may seem surprising at first. If we rewrite the z -component of the Poynting vector as
S z (t) = A(t)sin(2kz) ,
where
A(t) =
E0 2
sin(2ω t)
µ0 c
is the time dependent amplitude. Then the spatial dependence sin(2kz) remains fixed with
respect to time i.e. there is no propagation. Only the amplitude A(t) varies sinusoidally with
time.
E2
In the graph below, we plot S z (t) = 0 sin(2ω t)sin(2kz) vs. z for various values of
µ0 c
2ω t = 0, ± π / 12, ± π / 6, ± π / 4, ± π / 3, ± π / 2 .
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We see that the nodal structure does not change. If we time average S z (t) we see that the
sinusoidal amplitude averages to zero over one cycle.
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Problem 6: Diffraction
A monochromatic light with a wavelength of λ = 600 nm passes through a single slit which has a
width of 0.800 mm.
Question1: What is the distance between the slit and the screen if the first minimum in the
diffraction pattern is at a distance 1.00 mm from the center of the screen?
Answer: The general condition for destructive interference is
sin θ = m
λ
a
m = ±1, ± 2, ± 3, ...
For small θ , we employ the approximation sin θ ≈ tan θ = y / L , which yields
y
λ
≈m
L
a
The first minimum corresponds to m = 1 . If y1 = 1.00 mm , then
ay1 (8.00 ×10− 4 m)(1.00 ×10− 3 m)
L=
=
= 1.33 m
mλ
(1)(600 ×10− 9 m)
Question 2: Calculate the width of the central maximum.
Answer: The width of the central maximum is (see figure below)
w = 2 y1 = 2(1.00 ×10− 3 m) = 2.00 mm
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Problem 7 Second-Order Bright Fringe
A monochromatic light is incident on a single slit of width 0.800 mm , and a diffraction pattern
is formed at a screen that is 0.800 m away from the slit. The second-order bright fringe is at a
distance 1.60 mm from the center of the central maximum. What is the wavelength of the
incident light?
Solution:
The general condition for destructive interference is
sin θ = m
λ y
 .
a L
where small-angle approximation has been made. Thus, the position of the m-th order dark fringe
measured from the central axis is
λL
ym = m
.
a
Let the second bright fringe be located halfway between the second and the third dark fringes.
That is,
1
1 ⎛ λL
λL⎞ 5 λL
.
y2b = ( y1 + y2 ) = ⎜ 2
+3 ⎟ =
2
2⎝ a
a ⎠ 2 a
The approximate wavelength of the incident light is then
2 y2b a (2)(0.800 × 10−3 m)(1.60 × 10−3 m)
λ=
=
= 6.40 × 10−7 m .
5L
(5)(0.800 m)
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Problem 8: Phase Difference
In the double-slit interference experiment
shown in the figure, suppose d = 0.100 mm
and L = 1.20 m , and the incident light is
monochromatic with a wavelength λ=600 nm.
(a) What is the phase difference between
the two waves arriving at a point P on
the screen when θ = 0.800° ?
Answer:
δ
d sin θ
φ = 2π = 2π
λ
λ
(1.00 × 10 −4 m)(sin 0.8°)
= 2(3.14)
= 14.6 rad.
6.00 × 10 −7 m
(b) What is the phase difference between the two waves arriving at a point P on the screen when
y = 4.00 mm ?
Answer: We use the approximation that sin θ  y / L . Then
φ=
2π d sin θ d y

λ
λL
= 2(3.14)
(1.00 × 10 −4 m)(4.00 × 10 −3 m)
= 3.49 rad .
(6.00 × 10 −7 m)(1.20 m)
(c) If the phase difference between the two waves arriving at point P is φ=1/3 rad, what is the
value of θ ?
Answer:
φ=
1
d sin θ
rad = 2π
3
λ
⎛ λφ ⎞
⇒ θ = sin −1 ⎜
= 3.18 × 10−4 rad = 0.0182° .
⎟
⎝ 2π d ⎠
(d) If the path difference is δ = λ / 4 , what is the value of θ ?
Answer:
⎛δ ⎞
⎛ λ ⎞
δ = d sin θ ⇒ θ = sin −1 ⎜ ⎟ = sin −1 ⎜
= 1.50 × 10 −3 rad = 0.0860° .
⎟
⎝ d⎠
⎝ 4d ⎠
(e) In the double-slit interference experiment, suppose the slits are separated by d = 1.00 cm
and the viewing screen is located at a distance L = 1.20 m from the slits. Let the incident
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light be monochromatic with a wavelength λ = 500 nm . Calculate the spacing between
the adjacent bright fringes on the viewing screen.
Answer:
Because yb = mλ L / d , the spacing between adjacent bright fringes is
Δyb = yb (m + 1) − yb (m) = (m + 1)
=
λL
λL λL
−m
=
d
d
d
(5.00 × 10 −7 m)(1.20 m)
= 6.00 × 10 −5 m = 60.0 µ m.
(1.00 × 10 −2 m)
(f) What is the distance between the third-order fringe and the center-line on the viewing screen?
Answer:
λL
λL
−0= 3
d
d
−7
(5.00 × 10 m)(1.20 m)
=3
= 1.80 × 10 −4 m = 180 µ m.
−2
(1.00 × 10 m)
Δyb = yb (3) − yb (0) = (3)
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Problem 9: Generating Infinite Plane Waves.
Open the following applet that we worked with during Week 14D1 in class.
http://web.mit.edu/8.02t/www/802TEAL3D/visualizations/light/planewaveapp/planewaveapp.htm
Try pulling the sheet and observe the fields that are generated. In the problem below we will
calculate the electric and magnetic fields associated with an oscillating infinite charged sheet.
Consider an infinite charged sheet that has a uniform charge density σ and is being pulled in the

positive y -direction at a velocity v (see two views of the moving sheet below).
a) What is a vector description of the surface current density?
b) Use Ampere’s Law to find the direction and magnitude of the magnetic field that is
generated by the current sheet.
c) If the sheet position oscillates as y(t) = y0 sin(ω t) , the magnetic field generated by the
surface current will propagate along the x-axis. What is the direction and magnitude of
the magnetic field as a function of (x,t) , for both regions x > 0 and x < 0 ?
d) What is the direction and magnitude of the electric field as a function of (x,t) , for both
regions x > 0 and x < 0 ?
e) What is the Poynting vector associated with this infinite plane wave in the regions
regions x > 0 and x < 0 ?
f) What is total time-average powerd per unit area radiated away?
g) What is the source of energy associated with the radiation?
h) Calculate the time-averaged power per unit area generated by the electric force on the
moving charges to determine the efficiency of the power generation in the
electromagnetic radiation.
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Solution:
a) What is a vector description of the surface current density?


The surface current density is given by K = σ v = σ v y ĵ .
b) Use Ampere’s Law to find the direction and magnitude of the magnetic field that is
generated by the current sheet.
We draw an Amperian loop as shown in the rotated view of the infinite sheet with the surface
current density pointing out of the page.
So we need to do Ampere’s law around the loop. The current enclosed is the product of the
surface current density K with the length of the Amperian loop l . Therefore Ampere’s Law
becomes
 
B
∫ ⋅ d s = µ0 Ienc ⇒ 2Bl = µ0 Kl = µ0σ v y l ⇒
B = µ0σ v y / 2
The magnetic field is then
⎧ µoσ v y
 ⎪⎪− 2 k̂; x > 0
B=⎨
⎪ µoσ v y
⎪⎩ 2 k̂; x < 0
c) If the sheet position oscillates as y(t) = y0 sin(ω t) , what are is the direction and
magnitude of the magnetic field as a function of (x,t) ?
Suppose the sheet is now oscillating according to y(t) = y0 sin(ω t) . Then the y-component of the
velocity of the sheet is given by the expression v y (t) = ω y0 cos(ω t) . Then the magnetic field
propagates according to
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⎧ µ0σω y0 cos(ω t − kx)
k̂; x > 0
⎪⎪−

2
B(x,t) = ⎨
⎪ µ0σω y0 cos(ω t + kx) k̂; x < 0
⎪⎩
2
d) What is the direction and magnitude of the electric field as a function of (x,t) , for both
regions x > 0 and x < 0 ?
The associated electric field is then
⎧ cµ0σω y0 cos(ω t − kx)
ĵ; x > 0
⎪⎪−

2
E(x,t) = ⎨
.
⎪− cµ0σω y0 cos(ω t + kx) ĵ; x < 0
⎪⎩
2
e) What is the Poynting vector associated with this infinite plane wave in the regions
regions x > 0 and x < 0 ?
⎧ cµ0σ 2ω 2 y0 2 cos 2 (ω t − kx)


î; x > 0

E(x,t) × B(x,t) ⎪⎪
4
S(x,t) =
=⎨
2 2
2
2
µ0
⎪− cµ0σ ω y0 cos (ω t + kx) î; x < 0
2
⎩⎪
f) What is total time-averaged power per unit area radiated away?
Because the power is radiated in both positive a negative x-directions, the total time averaged
power per unit area radiated away is
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2 2
2
cµ σ 2ω 2 y0 2
1 cµ0σ ω y0
= 2 S =2
= 0

Area
2
4
4
Ptotal
two sides
g) Where is the source of energy associated with the radiation?
It is coming from the energy required to move the charged sheet.
h) Calculate the time-averaged power per unit area generated by the electric force on the
moving charges to determine the efficiency of the power generation in the
electromagnetic radiation.
The electric field exerts a force on the charges, and they are moving, so
P
A
=
Fv
A
=
q Ev
A
= σ Ev = σ
cµ0σω 2 y0 2 cos 2 (ω t − kx)
2
cµ0σ 2ω 2 y0 2
=
4
This is the same as the time-averaged power radiated, so this is 100% efficient process!
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