Download Optical Resonators

Optical Resonators
APPH E4901/3 Applied Physics Seminar
1
Assignment from Last Week
• What
are the resonant optical frequencies
of a cube? Numbers please.
• What
are the resonant frequencies of an
“optical fiber ring”?
• What
is a Fabry-Pérot interferometer? And
other types of interferometers, … ?
2
http://www.rsoftdesign.com/aboutUs.php
3
http://www.feynmanlectures.info
http://www.youtube.com/watch?v=j3mhkYbznBk
http://www.youtube.com/watch?v=kd0xTfdt6qw&feature=relmfu
Resonators
• Richard Feynman’s “Cavity Resonators” lecture
• Next week: “The Great Seal Bug Story”
4
omagnetic
and V to
f our ideal
is that for
ossible to
containing
blem now,
feW cases.
ll produce
ce. Also,
ges on the
e changes,
me capacishown in
s Land C
h less than
glect them.
ant, and a
L
c
R
Fig. 23-1.
real resistor.
Equivalent circvit of q
5
Fig. 23-3 (b) is better. In fac
impedance of a real, physica
use in the artificial model o
Let's look a little more
of an inductance goes as wL
circuit": all we see is the res
becomes much larger than
tance. As we go'still highe
impedance is proportional t
(a)
frequencies a condenser is a
(b)
Fig. 23-3.
(c)
The equivalent circuit of
a real inductance at higher frequencies.
thing else, it ·draws no curre
into the capacitance betwee
the current in the coil jump
around and around where
intended that the current sho
path of least impedance.
If the subject had been
called "the high-frequency b
happens in all subjects. In
the speed of sound when th
It doesn't mean that there i
should be redesigned. So t
going to work as a good ind
frequencies. For high frequ
6
23-2 A capacitor at
hi
negative charge on the other; and there will be a uniform electric field between the
plates.
Now suppose that instead of DC, we put an AC of low frequency on the plates.
(We will find out later what is "low" and what is "high".) Say we connect the capacitor to a lower-frequency generator. As the voltage alternates, the positive
charge on the top plate is taken off and negative charge is put on. While that is
happening, the electric field disappears and then builds up in the 9Pposite direction.
0
-
,
CURVE f,
r-
SURFACE
S
0
0
0
0
<l9
t
E
h
1_
0
0
B
T
l'
CURVE f2
liNES OF B
LINES OF E
( 0)
. .Fig, 23-4.
(b)
The electric and magnetic fields between the plates of a capacitor.
23-2
7
the induced field tends to reduce the electric field farther out. The
E = E, + E2 is then
E
= E,
+ E2
=
(1 - w:n
S
(23.8)
Eoe'·'.
N
ric field in the capacitor is no longer uniform; it has the parabolic
by the broken line in Fig. 23-5. You see that our simple capacitor is
y complicated.
now use our results to calculate the impedance
of the capacitor
I'
/
r'
ncies. Knowing the electric field, we could compute
the charges on
I
I
d find out how the current through the capacitor
depends on the
I
but we are not interested in that problem for the moment.o We are
23-5. the
The frequency
electric field between
d in seeing what happens as we continue to go upFig.with
the capacitor plates at high frequency,
are neglected.)
happens at even higher frequencies. Aren't (Edge
we effects
already
finished?
we have corrected the electric field, which means that the magnetic
calculated is no longer right. The magnetic field of Eq. (23.5) is
right, but it is only a first approximation. So let's call it B,. We
write Eq. (23.5) as
E
T
/
clw
- iwr E i(>lt
B 1-2c
oe.
2
(23.9)
a
A
c
r
8
s
g
a
t
f
m
N
f
a
s
Th e fu
in num
Th e f
Co mp
zero a
x ap p
-0.
Fig. 23 -6.
The Bessel function Jo(xJ.
We g
more
find t
9
rs as they are, taking into account all the fields in the spaces in
tance, if we want a resonant circuit at high frequencies we will
one using a coil and a parallel-plate capacitor.
eady mentioned that the paralIel-plate capacitor we have been
e of the aspects of both a capacitor and an indnctance. With the
are charges on the surfaces of the plates, and with the magnetic
ck emf's. Is it possible that we already have a resonant circuit?
uppose we pick a frequency for which the electric field pattern
me radius inside the edge of the disc; that is, we choose wal c
. Everywhere on a circle c'oaxial with the plates the electric field
w snppose we take a tbin metal sheet and cut a strip just wide
ween the plates of the capacitor. Then we bend it into a cylinder
nd at the radius where the electric field is zero. Since there are
here, when we put this conducting cylinder in place, no currents
d there will be no changes in the electric and magnetic fields. We
put a direct short circUlt across the capacitor without changing
ok wbat we have; we have a complete cylindrical can with elecic fields inside and no connection at all to the outside world.
on't change even if we throwaway the edges of the plates outside
the capacitor leads. All we have left is a closed can with electric
ds inside, as shown in Fig. 23-7(a). The electric fields are osforth at the frequency w-which, don't forget, determined the
n. The amplitude of the oscillating Efield varies with the distance
he can, as shown in the graph of Fig. 23-7(b). This curve is just
he Bessel function of zero order. There is also a magnetic field
cles around the axis and oscillates in time 90° out of phase with
write out a series for the magnetic field and plot it, as shown in
23-7(c).
t we can have an electric and magnetic field inside a can with no
ons? It is because the electric and magnetic fields maintain themng E makes a B and the changing B makes an E-all according
f Maxwell. The magnetic field has an inductive aspect, and the
ii
LINES OF
/'
- - 0
0
0
®
®
it
ti
0
®
0
0
+
+
+ + +
( 0)
®
®
+
LINES OF
1.0
(b)
2.405-clw r
cBe
1.0
(c)
r
Fig. 23-7. The electric and magnetic
fields in an enclosed cylindrical can.
10
OUTPUT
INPUT
LOOP
LOOP
at the top
must be p
shown in
also, so t
of the can
We can a
what hap
magnetic
in the ma
is what gi
can.
Fig. 23-8. Coupling into and out of
a resonant cavity.
R·F
SIGNAL
GENERATOR
Fig. 23-9. A setup for observing the
cavity resonance.
>-
...crz
You
the can. W
introduce
member,
bottom p
on the ou
capacitor
insides of
Altho
gradually
if we mak
up the los
and fasten
If we now
current w
keep the o
driving so
frequency
can will b
The
the can an
The chan
motive fo
cr
:0
U
.5
>:0
>-
llw = wo/Q
Wo
Frequency
Fig. 23-10. The frequency response
curve of a resonant cavity.
circuit, th
Suppose w
as shown
whose fre
Then we c
ment that
portional
the freque
The outpu
Wo, which
11 those
like
however,
an
da
as
can
hen
y it
ent
ted
ons
use
ver,
not
the
not
3050
3300
""
'"''
,=
QJ I
Fig. 23-11.
Z1T (M"IIac)",l.. per HCOnd}
Observed resonant fre-
quencies of a cylindrical cavity.
12
tric field should meet the wa11 at right angles. We have considered the case in
ch the top and the bottom of the can are flat, but things would not be completely
h anif the top and bottom were curved. In fact, how is the can supposed to
mode, the electric and magnetic fields are as shown in Fig. 23-14. The electric
erent
a is its top and bottom, and which are its sides? It is, in fact, possible
field does not bother to go all the way across the cavity. It goes from the sides to
wnd
which
(b)
how
p, asthat there is a mode of osci11ation of the fields inside the can in which the
the ends, as shown.
Fig, 23-12. A higher-frequencyAs
mode.
fields go more or3050
less across the diameter ofthe can, as shown in Fig. 23-13.
you will probably now believe, if we go higher and higher in frequency we
striccan
3300
It is not too hard to understand
why the natural frequency of this mode
should expect to find more and more resonances. There are many different modes,
When
uld be not very different from the natural frequency of the first mode we have
each of which will have a different resonant frequency corresponding to some parary
it Suppose that instead of our cylindrical cavity we had taken a cavity
B
idered.
mode,of
thethese
electric a
ticular complicated arrangement of the electric and magnetic fields. Each
rrent
ch
was a cnbe 3 inches on a side. It is clear that this cavity would haveEthree
doesmode
not bother
field
arrangements
i.
called
a
resonant
mode. The resonance frequency field
of
each
erent
A mode with the electric field
ictedmodes, but a11 with the same frequency.
the ends, as shown.
,=
can be calculated by solving Maxwell's equations for the electric and magnetic
g
more
or
less
up
and
down
would
certainly
have
the same frequency as the
QJ I Z1T (M"IIac)",l .. per HCOnd}
As you will prob
asons
fields in the cavity.
e in which the electric field was directed right and left. If we now distort thet
should expect to find
f
cause
When we have a resonance at some particular frequency, how each
can we
knowwill ha
e into a cylinder, we will change these frequencies somewhat. We would stilI
of which
---I
J
B a little wire into the
Fig. 23-11. Observed resonant freever,
which
mode is being excited? One way is to poke
cavity
ect
them not to be changed too mnch, provided we keep the dimensions of the
ticular complicated a
E
of a So
cylindrical
cavity.
nnot
Fig.
23-14.
the frequency
of the mode of Fig.
23-13
shouldAnother mode of a cyty
more or quencies
less the same.
through a small hole. If the electric field is along the wire, as in Fig.
field 23-15(a),
arrangements i.
lindrical
cavity.
bethe
too different from the mode of Fig. 23-8. We could make
a detailed
calcan
be
by
of
there will be relatively large currents in the wire, sapping energy from the calculated
fields,
tion
of
the
naturalfrequency
of
the
mode
shown
in
Fig.
23-13,
but
we
will
not
fields
in
the
cavity.
and the resonance will be suppressed. If tthe electric
field is as shown in Fig.
s not
f
we have
hat now. When the calculations are carried through, it is found that, for the
23-15(b), the wire will have a much smaller effect. We could find whichWhen
way the
which mode is being
ensions we have assumed, the resonant frequency comes out very close to the
Fig. 23-13. A transverse
mode
of
field points in this mode by bending
the end of the wire, as shown in Fig. 23-15(c).
cy ofresonance at 3300 megacycles.
Fig. 23-14. Another mode of a cythrough a small hole
erved
the cylindrical cavity.
Then,
as
we
rotate
the
wire,
there
will
be a big effect when the end of
thewill
wire
lindrical
cavity.
that
there
be is
relative
By
similar calculations it is possible to show that there should be still another
parallel
to
E
and
a
small
effect
when
it
is
rotated
so
as
to
be
at
90'
to
E.
E
and the resonance w
de at the other resonant frequency we (ound near 3800 megacycles. For this
nance
23-15(b), the wire w
23-9
ycles
field points in this m
might
Then, as we rotate t
ro of
parallel to E and a s
cond
s one
n out
ating
But
over
mode
nd it
have
gnetic
gnetic
ments
at the
ase in
letely
ed to
ssible
h the
3-13.
mode
have
cavity
/------
""
'"''
/------
----( 0)
E
(b)
Fig. 23-15.
( c)
A short metal wire inserted into a cavity will disturb the
Ethan when it is at right ang.'es.
(
resonance much more when it is parallel to
Fig. 23-15.
A short metal wire in
resonance much more when it is paralle
r
23-5 Cavities and resonant circuits
Although the resonant cavity we have been describing seems to 23-5
be quite
Cavities a
different from the ordinary resonant circuit consisting of an inductance and a
Although the re
capacitor, the two resonant systems are, of course, closely related. They
are both
different from the o
(b)
Fig, 23-12. A higher-frequency mode.
members of the same family; they are just two extreme cases of electromagnetic
capacitor, the two re
resonators-and there are many intermediate cases between these two
extremes.
members
of the sam
Suppose we start by considering the resonant circuit of a capacitor inresonators-and
parallel with ther
Suppose
we start by
an inductance, as shown in Fig. 23-16(a). This circuit will resonate althe
frequency
an inductance,
Wo = I/VLC. If we want to raise the resonant frequency of this circuit,
we can as sho
Wo of
= turns
I/VLC.
13
do so by lowering the inductance L. One way is to decrease the number
in If w
do so by lowering the
the coil. We can, however, go only so rar in this direction. Eventually we will get
i
r
1
8
h
L
...-'liNES
OF
8
fig. 23-16.
L
II
I
0
®
®
®
®
®
®
"e"
0
0
0
0
-
:'\
LINES OF E
-
(0)
II
I
(c)
Resonators of progressively higher resonant frequencies.
object. Our inductance is now a cylindrical hollow can attached to the edges of
the condenser plates. The electric and magnetic fields will be as shown in the
figure. Such an object is, of conrse, a resonant cavity. It is called a "loaded" cavity.
But we can still think of it as an L-C circuit in which the capacity section is the
region where we find most of the electric field and the inductance section is
that region where we find most of the magnetic field.
14
Next Week
On May 26, 1960, U.S. Ambassador to the United Nations Henry Cabot Lodge, Jr. unveiled the Great
Seal Bug before the UN Security Council to counter Soviet denunciations of American U-2 espionage.
The Soviets had presented a replica of the Great Seal of the United States as a gift to Ambassador
Averell Harriman in 1946. The gift hung in the U.S. Embassy for many years, until in 1952, during
George F. Kennan's ambassadorship, U.S. security personnel discovered the listening device embedded
inside the Great Seal. Lodge's unveiling of this Great Seal before the Security Council in 1960
provided proof that the Soviets also spied on the Americans, and undercut a Soviet resolution before
the Security Council denouncing the United States for its U-2 espionage missions
15