Optical Resonators
7.2 Modes of a Rectangular Cavity and the Open Planar Resonator
Consider a rectangular cavity of dimensions 2a×2b ×d as shown in Fig. 7.1.
Starting from Maxwell’s equations [see Eqs. (2.1), (2.2), (2.3), and (2.4)] one can show that the electric and magnetic fields satisfy a wave equation of the form given by
∇2E−n20 c2
∂2E
∂t2 =0 (7.1)
where c represents the velocity of light in free space and n0represents the refractive index of the medium filling the rectangular cavity. Equation (7.1) has been derived inChapter 2.
If the walls of the rectangular cavity are assumed to be perfectly conducting then the tangential component of the electric field must vanish at the walls. Thus ifnˆ represents the unit vector along the normal to the wall then we must have
E× ˆn=0 (7.2)
on the walls of the cavity.
Let us consider a Cartesian component (say x component) of the electric vector;
this will also satisfy the wave equation, which in the Cartesian system of coordinates will be given by
∂2Ex
∂x2 +∂2Ex
∂y2 +∂2Ex
∂z2 = n20 c2
∂2Ex
∂t2 (7.3)
In order to solve Eq. (7.3) we use the method of separation of variables and write
Ex=X(x)Y(y)Z(z)T(t) (7.4)
y 2a
2b
d x
Fig. 7.1 A rectangular cavity of dimensions 2a×2b×d
7.2 Modes of a Rectangular Cavity and the Open Planar Resonator 145 Substituting this in Eq. (7.3) and dividing by ExWe obtain
1 X
∂2X
∂x2 + 1 Y
∂2Y
∂y2 + 1 Z
∂2Z
∂z2 = n20 c2T
∂2T
∂t2 (7.5)
Thus the variables have indeed separated out and we may write 1
X
∂2X
∂x2 = −k2x (7.6)
1 Y
∂2Y
∂y2 = −ky2 (7.7)
1 Z
∂2Z
∂z2 = −k2z (7.8)
and
n20 c2T
∂2T
∂t2 = −k2 (7.9)
where
k2=kx2+k2y+k2z (7.10) Equation (7.9) tells us that the time dependence is of the form
T(t)=Ae−iωt (7.11)
whereω=c k/n0represents the angular frequency of the wave and A is a constant.
It should be mentioned that we could equally well have chosen the time dependence to be of the form eiωt. Since Exis a tangential component on the planes y=0, y= 2b, z=0, and z=d, it has to vanish on these planes and the solution of Eqs. (7.7) and (7.8) would be sin kyy and sin kzz, respectively, with
ky=nπ
2b, kz= qπ
d , n, q=0, 1, 2, 3, ... (7.12) where we have intentionally included the value 0, which in this case would lead to the trivial solution of Exvanishing everywhere (The above solutions are very similar to the ones discussed in Example 3.3). In a similar manner, the x and z dependences of Eywould be sin kxx and sin kzz, respectively, with
kx= mπ
2a, m=0, 1, 2, 3,. . . (7.13)
and kzgiven by Eq. (7.12). Finally the x and y dependences of Ez would be sin kxx and sin kyy respectively.
Now, because of the above forms of the x dependence of Eyand Ez,∂Ey/∂y, and
∂Ez/∂z would vanish on the surfaces x=0 and x=2 a. Thus on the planes x=0 and x=2a, the equation∇.E=0 leads to∂Ex/∂x =0.Hence the x dependence of Exwill be of the form cos kxx with kxgiven by Eq. (7.13). Notice that the case m=0 now corresponds to a nontrivial solution.
In a similar manner, one may obtain the solutions for Ey and Ez. The complete solution (apart from the time dependence) would therefore be given by
Ex=E0xcos kxx sin kyy sin kzz Ey=E0ysin kxx cos kyy sin kzz Ez=E0zsin kxx sin kyy cos kzz
(7.14)
where E0x, E0y, and E0zare constants. The use of Maxwell’s equation∇.E = 0, immediately gives
E0.k=0 (7.15)
wherek= ˆxkx+ ˆyky+ ˆzkz. Since the coefficients E0x, E0yand E0zhave to satisfy Eq. (7.15) it follows that for a given mode, i.e., for a given set of values of m, n, and q only two of the components of E0can be chosen independently. Thus a given mode can have two independent states of polarization.
Note that when one of the quantities m, n, or q is zero, then there is only one possible polarization state associated with the mode. Thus if we consider the use with m=0, n=0, q=0, then Ex=E0xsin kyy sin kzz, Ey=0, Ez=0. Thus the only possible case is with the electric vector oriented along the x-direction.
Using Eqs (7.10), (7.12), and (7.13), we obtain ω2= c2k2
n20 = c2 n20
kx2+k2y+k2z
= c2π2 n20
m2 4a2 + n2
4b2 +q2 d2 or
ω= cπ n0
m2 4a2+ n2
4b2 +q2 d2
1/ 2
(7.16) which gives us the allowed frequencies of oscillation of the field in the cavity. Field configurations given by Eq. (7.14) represent standing wave patterns in the cavity and are called modes of oscillation of the cavity. These are similar to the acoustic modes of vibration of an acoustic cavity (like in a musical instrument such as a guitar and veena) and represent the only possible frequencies that can exist within the cavity.
Example 7.1 As a specific example we consider a mode with m=0, n=1, and q=1
7.2 Modes of a Rectangular Cavity and the Open Planar Resonator 147 Thus kx=0, ky=π/ 2b, kz=π/ d and using Eq. (7.14), we have
Ex=E0xsin kyy sin kzz=E0xsinπ 2by
sinπ dz Ey=0
Ez=0
Using the time dependence of the form e–iωtand expanding the sine functions into exponentials, we may write
E= 1 (2i)2xˆ
e−i(ωt−kyy−kzz)+e−i(ωt−kyy+kzz)+e−i(ωt+kyy−kzz)+e−i(ωt+kyy+kzz)
(7.17) Thus the total field inside the cavity has been broken up into four propagating plane waves; in Eq. (7.17) the first term on the right-hand side represents a wave propagating along the (+y, +z) direction, the second along (+y, –z) direction, the third along (–y, +z) direction, and the fourth along (–y, –z) direction, respectively. These four plane waves interfere at every point inside the cavity to produce a standing wave pattern. However, since kyand kztake discrete values, the plane waves which constitute the mode make discrete angles with the axes.
Example 7.2 If we take a cavity with a=b=1 cm and d=20 cm and consider the mode with m=0, n
=1, q=106 then
kx=0, ky=π/ 2cm−1, kz=106π/ 20 cm−1 implying
k≈106π/ 20 cm−1andν=ck/2π=7.5×1014Hz which lies in the optical region. For such a case
θy=cos−1 ky
k ≈89.9994◦ θz=cos−1
kz
k ≈0.0006◦
andθx=0 because of whichθy+θz=90◦. It may be noted that the component waves are propagating almost along the z-axis. In general,
cos2θx+cos2θy+cos2θz=1
Further for m=0, n=0, q=0 the cavity mode can be thought of as a standing wave pattern formed by eight plane waves with components ofk given by
±kx,±ky,±kz .
Example 7.3 Let us now consider a few hundred nanometer-sized rectangular cavity (also referred to as a microcavity) filled with free space. Let 2a=2b=d=500 nm. We now calculate the wavelengths (λ=c/ν) of oscillation corresponding to some of the lower order modes which can be obtained from Eq. (7.16) as
m n q λ(nm)
1 0 0 1000
0 1 0 1000
0 0 1 1000
1 1 0 707.1
1 0 1 707.1
0 1 1 707.1
1 1 1 577.4
2 0 0 500
Note that since the cavity dimensions are of the order of optical wavelength, in the optical wavelength region, the wavelengths of oscillation of the modes are well separated. Also the cavity cannot support any mode at wavelengths longer than 1000 nm. If we place an atom in such a cavity and if the atom has energy levels separated by energy difference corresponding to a wavelength of say 800 nm with emission spectral width of about 10 nm, then since there are no possible modes in the cavity corresponding to this wavelength region, the atom would be inhibited from emitting radiation. Thus it is possible to inhibit spontaneous emission from atoms and increase the lifetime of the level. Microcavities of dimensions comparable to optical wavelength are now being extensively investigated for various applications includ- ing suppressing spontaneous emission or for enhancing spontaneous emission, for lowering threshold for laser oscillation, etc. (see, e.g., Vahala (2003) and Gerard (2003)).
If we had chosen even one of the dimensions to be much larger then the mode spacing would be much smaller. As an example if we assume 2a=2b=500 nm and d=10,000 nm, then the wavelength corresponding to various low-order modes would be
m n q λ(nm)
1 0 0 1000
0 1 0 1000
1 1 0 707.1
1 0 1 998.8
0 1 1 998.8
1 1 1 706.7
2 0 0 500
0 0 21 952.3
0 0 22 909.1
0 0 23 869.5
0 0 24 833.3
It can be noted that since the value of d is large compared to wavelength around 900 nm, the mode spacing is small.
Using Eq. (7.16) we can show (see Appendix E) that the number of modes per unit volume in a frequency interval fromνtoν+ dνwill be given by
p(ν)dν=8πn30
c3 ν2dν (7.18)
where n0represents the refractive index of the medium filling the cavity. For a typ- ical atomic system dν∼3×109Hz atν=3×1014Hz and the number of modes per unit volume would be (assuming n0=1)
p(ν)dν= 8πn30
c3 ν2dν= 8×π×1×
3×10142
3×1083 ×3×109≈2×108cm−3
Thus for cavities having typical volumes of 10 cm3, the number of possible oscil- lating modes within the linewidth will be 2×109which is very large. To achieve a very small number of possible oscillating modes within the linewidth of the atomic transition, the volume of the cavity has to be made very small. Thus to achieve a single mode of oscillation within the linewidth the volume of the cavity should
7.2 Modes of a Rectangular Cavity and the Open Planar Resonator 149 be of the order of 5× 10–9 cm3. This corresponds to a cube of linear dimen- sion of the order of 17μm. Optical microcavities having such small dimensions can be fabricated using various techniques and are finding applications for studying strong interactions between atoms and radiation (cavity quantum electrodynamics), for inhibiting spontaneous emission, or for enhancing spontaneous emission and as filters for optical fiber communication systems. For a nice review, readers are referred to Vahala (2003).
In the case of conventional lasers the volume of the cavity is large and thus the number of oscillating modes within the linewidth of the atomic transition is very large. Thus all these oscillating modes can draw energy from the atomic system and the resulting emission would be far from monochromatic. In order to have very few oscillating modes within the cavity, if the dimensions of the cavity are chosen to be of the order of the wavelength, then the volume of the atomic system available for lasing becomes quite small and the power would be quite small.
The problem of the extremely large number of oscillating modes can be over- come by using open cavities (as against closed cavities) which consist of a pair of plane or curved mirrors facing each other. As we have seen earlier, a mode can be considered to be a standing wave pattern formed between plane waves propa- gating within the cavity withk given by
±kx,±ky,±kz
. Thus the angles made by the component plane waves with the x-, y- and z-directions will, respectively, be, cos−1(mλ/ 2a), and cos−1(nλ/ 2b), cos−1(qλ/ d). Since in open resonators, the side walls of the cavity have been removed, those modes which are propagating almost along the z-direction (i.e., with large value of q and small values of m and n) will have a loss which is much smaller than the loss of modes which make large angles with the z-axis (i.e., modes with large values of m and n). Thus on removing the side walls of the cavity, only modes having small values of m and n (∼0, 1, 2..,) will have a small loss, and thus as the amplifying medium placed inside the cavity is pumped, only these modes will be able to oscillate. Modes with larger values of m and n will have a large loss and thus will be unable to oscillate.
It should be noted here that since the resonator cavity is now open, all modes would be lossy. Thus even the modes that have plane wave components travel- ing almost along the z-direction will suffer losses. Since m and n specify the field patterns along the transverse directions x and y and q that along the longitudinal direction z, modes having different values of (m, n) are referred to as various trans- verse modes while modes differing in q-values are referred to as various longitudinal modes.
The oscillation frequencies of the various modes of the closed cavity are given by Eq. (7.16). In order to obtain an approximate value for the oscillation frequencies of the modes of an open cavity, we may again use Eq. (7.16) with the condition m, n << q. Thus making a binomial expansion in Eq. (7.16) we obtain
νmnq= c 2n0
q d +
m2 a2 +n2
b2 d 8q
1/ 2
(7.19) The difference in frequency between two adjacent modes having same values of m and n and differing in q value by unity would be very nearly given by
νq≈ c
2n0d (7.20)
which corresponds to the longitudinal mode spacing. In addition if we completely neglect the terms containing m and n in Eq. (7.19) we will obtain
νq≈q c
2n0d (7.21)
The above equation is similar to the frequencies of oscillation of a stretched string of length d.
Example 7.4 For a typical laser resonator d∼100 cm and assuming free space filling the cavity, the longitudinal mode spacing comes out to be∼150 MHz which corresponds to a wavelength spacing of approximately 0.18 pm (=0.18×10–12m) at a wavelength of 600 nm.
Problem 7.1 Show that the separation between two adjacent transverse modes is much smaller thanνq. Solution The frequency separation between two modes differing in m values by unity would be
νm≈ c 2n0
d 8a2q
m2−(m−1)2
≈νqλd 8a2
m−1
2
where we have used q ≈ 2d /λ[see Eq. (7.21)]. For typical values ofλ=600 nm, d=100 cm, a=1 cm,λd
8a2 =7.5×10−4. Thus for m∼1,νm<< νq.
It is of interest to mention that an open resonator consisting of two plane mir- rors facing each other is, in principle, the same as a Fabry–Perot interferometer or an etalon (see Section 2.9). The essential difference in respect of the geometrical dimensions is that in a Fabry–Perot interferometer the spacing between the mirrors is very small compared to the transverse dimensions of the mirrors while in an opti- cal resonator the converse is true. In addition, in the former case the radiation is incident from outside while in the latter the radiation is generated within the cavity.
Earlier we showed that the modes in closed cavities are essentially superpositions of propagating plane waves. Because of diffraction effects, plane waves cannot rep- resent the modes in open cavities. Indeed if we start with a plane wave traveling parallel to the axis from one of the mirrors, it will undergo diffraction as it reaches the second mirror and since the mirror is of finite transverse dimension the energy in the diffracted wave that lies outside the mirror would be lost. The wave reflected from the mirror will again undergo diffraction losses when it is reflected from the first mirror. Fox and Li (1961) performed numerical calculations of such a planar resonator. The analysis consisted of assuming a certain field distribution at one of the mirrors of the resonator and calculating the Fresnel diffracted field at the second mirror. The field reflected at the second mirror is used to calculate back the field distribution at the first mirror. It was shown that after many traversals between the mirrors, the field distribution settles down to a steady pattern, i.e., it does not change between successive reflections but only the amplitude of the field decays exponen- tially in time due to diffraction losses. Such a field distribution represents a normal mode of the resonator and by changing the initial field distribution on the first mirror other modes can also be obtained.