homework 18 – KIM, JI – Due: May 5 2008, 4:00 am
1
Question 1, chap 33, sect 3.
part 1 of 1
10 points
An unpolarized light beam with intensity
of
I
0
passes through 2 polarizers shown in the
picture.
I
0
33
◦
63
◦
I
If the first polarizer is rotated at an angle
of 33
◦
with respect to vertical and the sec
ond at an angle 63
◦
with respect to vertical,
what is the beam intensity
I
after the second
polarizer?
1.
I
=
7
16
I
0
2.
I
=
1
8
I
0
3.
I
=
1
4
I
0
4.
I
=
3
8
I
0
correct
5.
I
=
5
8
I
0
6.
I
=
5
16
I
0
7.
I
=
1
2
I
0
8.
I
=
9
16
I
0
9.
I
=
3
16
I
0
10.
I
=
1
16
I
0
Explanation:
Let :
θ
1
= 33
◦
and
θ
2
= 63
◦
.
The beam intensity after the first polarizer
is
I
1
=
I
0
2
,
since the initial beam is unpolarized.
We use the formula for the intensity of the
transmitted (polarized) light. Thus the beam
intensity after the second polarizer is
I
=
I
1
cos
2
(Δ
θ
)
=
I
1
cos
2
(
θ
2

θ
1
)
=
I
1
cos
2
(63
◦

33
◦
)
=
I
0
2
cos
2
(30
◦
)
=
I
0
2
parenleftBigg
√
3
2
parenrightBigg
2
=
3
I
0
8
.
Question 2, chap 33, sect 5.
part 1 of 1
10 points
At what distance from a 80 W electromag
netic wave point source (like a light bulb) is
the amplitude of the electric field 13 V
/
m?
All quantities given here are in the SI units.
Correct answer:
5
.
32939
m (tolerance
±
1
%).
Explanation:
Let :
P
= 80 W
,
E
max
= 13 V
/
m
,
and
μ
0
c
= 376
.
991 Ω
.
The wave intensity (power per unit area) is
given by
I
=
S
av
=
E
2
max
2 (
μ
o
c
)
=
(13 V
/
m)
2
2 (376
.
991 Ω)
= 0
.
224143 W
/
m
2
.
The constant
μ
o
c
(impedance of free space)
turns out to be 120
π
Ω.
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homework 18 – KIM, JI – Due: May 5 2008, 4:00 am
2
You should verify that (V
/
m)
2
Ω = W
/
m
2
.
With this we can find the intensity, which
multiplied by the area of a circle of radius
R
must gives us the source’s power
P
, therefore
4
π R
2
I
=
P
= 80 W
.
Solving for the radius we get,
R
=
radicalbigg
P
4
π I
=
radicalBigg
80 W
4
π
(0
.
224143 W
/
m
2
)
=
5
.
32939 m
.
Question 3, chap 33, sect 5.
part 1 of 1
10 points
A 30 kW beam of radiation is incident nor
mally on a surface that reflects half of the
radiation.
The speed of light is 3
×
10
8
m
/
s.
What is the force on this surface?
Correct answer: 0
.
15 mN (tolerance
±
1 %).
Explanation:
Let :
P
= 30 kW = 30000 W
and
c
= 3
×
10
8
m
/
s
.
The total force on the surface is the sum of
the force due to the reflected radiation and
the force due to the absorbed radiation, so
F
tot
=
F
r
+
F
a
= 2
P
2
c
+
P
2
c
=
3
P
2
c
=
3 (30000 W)
2 (3
×
10
8
m
/
s)
×
10
6
mN
N
=
0
.
15 mN
.
Question 4, chap 33, sect 5.
part 1 of 1
10 points
Consider a monochromatic electromagnetic
plane wave propagating in the
z
direction. At
a particular point in space, the magnitude of
the electric field is pointing in the positive
x
direction.
The permeability of free space is 1
.
25664
×
10
−
6
N
/
A
2
, the permittivity of free space is
8
.
85419
×
10
−
12
C
2
/
N
/
m
2
and speed of light
is 2
.
99792
×
10
8
m
/
s.
z
x
y
E
wave propagation
What are the directions of the instanta
neous magnetic field and the instantaneous
Poynting vector respectively?
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 Spring '08
 Turner
 Physics, Work, Light, Wavelength, Doubleslit experiment, Correct Answer, Thomas Young

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