Download SIMG-303-20033 Solution Set #5

SIMG-303-20033
Solution Set #5
1. Describe completely the state of polarization of each of the following waves:
(a) E [z, t] = x̂E0 cos [k0 z − ω 0 t] − ŷE0 cos [k0 z − ω 0 t]
Both components are traveling down the z-axis towards +∞. The two polarizations have the same magnitude and the same phase, but note that the y-component
is multiplied by −1. This electric field could be rewritten as:
E [z, t] = x̂E0 cos [k0 z − ω 0 t] − ŷE0 cos [k0 z − ω0 t]
= x̂E0 cos [k0 z − ω 0 t] + ŷE0 cos [k0 z − ω 0 t − π]
Thus when the x-component reaches its maximum positive value, the y-component
reaches its negative extremum. The state of polarization is linear and oriented at
an angle of −45◦ measured from the x-axis.
£ ¡
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−1
(b) E [z, t] = x̂E0 sin 2π z · λ−1
−
ν
t
−
ŷE
sin
2π
z
·
λ
−
ν
t
0
0
0
0
0
This is just a disguised version of the previous question. Apply the definitions for
the wavevector k and the angular temporal frequency ω :
£ ¡
¢¤
£ ¡
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− ŷE0 sin 2π z · λ−1
E [z, t] = x̂E0 sin 2π z · λ−1
0 − ν 0t
0 − ν 0t
= x̂E0 sin [k0 z − ω 0 t] − ŷE0 sin [k0 z − ω 0 t]
h
h
πi
πi
= x̂E0 cos k0 z − ω0 t −
− ŷE0 cos k0 z − ω 0 t −
2
2
i
h
h
πi
π
+ ŷE0 cos k0 z − ω 0 t − + π
= x̂E0 cos k0 z − ω0 t −
2
2
h
h
πi
πi
+ ŷE0 cos k0 z − ω 0 t +
= x̂E0 cos k0 z − ω0 t −
2
2
The state of polarization again is linear and oriented θ = −45◦ measured from
the x-axis.
£
¤
(c) E [z, t] = x̂E0 sin [ω 0 t − k0 z] + ŷE0 sin ω0 t − k0 z − π4
Note that the component waves now travel towards −∞. I suggest rewriting in
terms of cosine functions, though this is not necessary:
h
πi
E [z, t] = x̂E0 sin [ω 0 t − k0 z] + ŷE0 sin ω0 t − k0 z −
4
h
h
πi
π πi
= x̂E0 cos ω0 t − k0 z −
+ ŷE0 cos ω 0 t − k0 z − −
2
4 ¸2
∙
h
πi
3π
+ ŷE0 cos ω 0 t − k0 z −
= x̂E0 cos ω0 t − k0 z −
2
4
¶¸
∙ µ
´i
h ³
π
3π
+ ŷE0 cos − ω 0 t − k0 z −
= x̂E0 cos − ω 0 t − k0 z −
2
4
¸
∙
i
h
π
3π
+ ŷE0 cos k0 z − ω 0 t +
= x̂E0 cos k0 z − ω 0 t +
2
4
so the phase of the y-component is smaller than that of the x-component by π4
radians. The two magnitudes are identically E0 , but the phase change is not a
1
multiple of π radians (necessary for linear polarization) or π2 radians (for circular
polarization), so the state of polarization must be elliptical. We need to determine
the handedness. If we use the “angular momentum” convention (my preference),
then the field observed at a fixed z looking back at the source rotates counterclockwise if the polarization is right-handed. Note that in the so-called “optics” or
“screwy” convention, then a field that is right-hand elliptically polarized (RHEP)
field rotates rclockwise. Look at the field for a fixed value of z (say z = 0):
∙
¸
h
πi
3π
E [z, t] = x̂E0 cos −ω 0 t +
+ ŷE0 cos −ω 0 t +
2
4
¸
∙
i
h
π
3π
+ ŷE0 cos ω 0 t −
= x̂E0 cos ω 0 t −
2
4
The Argand diagram of this field shows that it rotates counterclockwise, so the
field is RHEP in the angular momentum convention.
¤
£
(d) E [z, t] = x̂E0 sin [ω 0 t − k0 z] + ŷE0 sin ω0 t − k0 z + π2
Again, this wave travels towards −∞. Rewrite as cosines:
h
πi
E [z, t] = x̂E0 sin [ω 0 t − k0 z] + ŷE0 sin ω0 t − k0 z +
2
h
h³
πi
π´ πi
= x̂E0 cos ω0 t − k0 z −
+ ŷE0 cos ω 0 t − k0 z +
−
2
2
2
h
πi
= x̂E0 cos k0 z − ω 0 t +
+ ŷE0 cos [k0 z − ω 0 t]
2
Again, the magnitudes of the two components are equal, but the phase difference
is an odd multiple of π2 radians, so the polarization is circular. For handedness,
look back at the source from a fixed location (z=0) and determine which direction
the field rotates:
h
πi
+ ŷE0 cos [−ω0 t]
E [z, t] = x̂E0 cos −ω 0 t +
2
h
πi
+ ŷE0 cos [ω 0 t]
= x̂E0 cos ω 0 t −
2
Again, the E vector rotates counterclockwise, so the polarization is RHCP in the
angular-momentum convention.
2
2. A beam of linearly polarized light is incident on an ideal linear polarizer. Find the
angle between the axis of the directlion of oscillation of the electric field and the ideal
linear polarizer if the 25% of the “irradiance” (or “intensity”) is transmitted.
The polarizer “passes” the portion of the electric field amplitude projected onto the axis
of the polarizer (see figure), so that the field becomes:
Eθ = E0 cos [θ]
The “irradiance” of the light is the time average of the squared magnitude of the electric
field:
Iθ = |Eθ |2 = E02 · cos2 [θ] = I0 · cos2 [θ]
So if Iθ = 12 I0 , then:
"r #
1
π
= radians = 60◦
θ = cos−1
4
3
3
3. Sketch a diagram of the windshield and dashboard of a car and use it to explain the
useful direction of polarization of polaroid sunglasses.
So the light that is reflected from the dashboard that is reflected into the eye is preferentially polarized perpendicular to the plane of the page, i.e., horizontally as seen by
the eye. We would like to have the sunglasses “block” this polarization and pass the
light that is polarized in the plane of the page, i.e., vertically as seen by the eye.
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4. Prove Malus’ law that the intensity transmitted by a pair of ideal linear polarizers
oriented with their axes at angle θ0 is proportional to cos2 [θ0 ].
This is just a variation of #2. If unpolarized light with electric field E0 passes through
one polarizer at an arbitrary angle, then the emerging electric field from the polarizer
is
E0
E0
=
E1 = Z + π
2
2
cos [θ] dθ
− π2
and the emerging irradiance is:
E1
4
The electric field is incident on a second polarizer oriented at angle θ measured relative
to the first polarizer, so the emerging electric field is:
I1 ∝
E2 = E1 · cos [θ] =
E0
cos [θ]
2
The emerging irradiance is the square:
|E0 |2
I2 ∝ |E2 | = |E1 | cos [θ] =
cos2 [θ] ∝ cos2 [θ]
4
2
2
2
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5. Consider two ideal linear polarizers that are placed with their polarizations axes orthogonal. A third ideal linear polarizer is placed betweent these two and rotated at a
constant rate ω 0 [radians per second]. Show that the emerging light “intensity” oscillates at four times the rotation frequency.
If the “middle” polarizer is oriented along the axis of either of the other polarizers,
then the output is zero because the light through the “parallel pair of polarizers” (say
that three times fast) is blocked by the orthogonal polarizer. So some light is passed
only when the middle polarizer is oriented at an angle θ that differs from the other two.
Consider the first two polarizers — the output is just that from Malus’ law:
E2 = E1 · cos [θ]
The angle between the axes of the second and third polarizer is φ = π2 − θ, so the
amplitude emerging from the third polarizer may be easily expressed in terms of the
amplitude out of the second:
i
hπ
−θ
E3 = E2 · cos [φ] = E2 · cos
2 h i
´
³ hπ i
π
cos [θ] + sin
sin [θ]
= E2 · cos
2
2
= E2 · sin [θ]
= (E1 · cos [θ]) · sin [θ]
E1
2 · sin [θ] · cos [θ]
= E1 ·
=
· sin [2θ]
2
2
BUT, θ = ω 0 t:
E1
· sin [2ω 0 t]
2
|E1 |2
2
∝ |E3 | =
· sin2 [2ω0 t]
4
E3 =
I3
BUT, sin2 [α] = 12 (1 − cos2 [2α]):
|E1 |2
4
|E1 |2
=
8
I3 ∝
¢
1¡
1 − cos2 [2 · 2ω0 t]
2
¢
¡
1 − cos2 [(4ω0 ) t]
·
.
which oscillates at an angular temporal frequency of 4ω 0 radians
sec
6
6. The wavelength of unpolarized light from a mercury source is λ = 546.072 nm. If
incident on a plate of glass at an angle θi = 58◦ 010 , the reflected light is seen to be
completely linear polarized. Find the refractive index of the glass.
If light is reflected obliquely from the sufrace of a transparent material, then both the
reflected and transparent beams are (in general) partially polarized. If incident at Brewster’s angle, the reflected beam is completely plane polarized. This angle is specified by
Brewster’s law, which evalutes this wavelength to:
n2
= tan [θB ]
n1
n2
= tan [58◦ 010 ] ' 1.601
1
7. A source of left-hand circularly polarized light at λ0 = 656 nm should be converted to
right-hand circularly polarized light by passing it through a thickness of quartz (SiO2 ),
which has ns = 1.551 and nf = 1.542.
(a) Compute the minimum thickness of a plate that will accomplish the task.
We want to create a phase delay of 12 cycle, or π radians between the two polarizations via the difference in optical path lengths. For a plate with physical thickness
d, the optical path length is nd, so the optical path lengths in the two directions
are n s d and n f d. The difference in optical path length is
The difference in the index of refraction in the two directions is:
(ns − nf ) d = (1.551 − 1.542) d = 0.009d
The difference in the number of wavelengths is the optical path difference divided
by the vacuum wavelength λ0 :
d
(ns − nf ) d
= 0.009 = difference in the number of wavelengths
λ0
λ0
π
To get a phase delay of π radians, this difference has to be 2π
= 12 wavelength:
µ
¶
d
5
1 656 nm
1
= 0.009 =⇒ d =
= 36, 444 nm = 55 λ0 ' 0.036 mm
2
λ0
2 0.009
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which is pretty thin
(b) (EXTRA CREDIT) From the result of (a), you can see that such a plate is not
very practical. Modify the design of the plate to create a practical device.
This plate is very thin. We can create a practical phase plate by constructing a
sandwich of two such plates oriented at 90 ◦ to each other so that one generates a
phase difference of 2πN + π radians in one direction, where N is a large number,
while the other generates 2πN radians of phase difference in the other direction.
The sandwich can be quite thick, but the resulting phase difference is π radians.
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