Rectangular resonator (Homework 3-323)

3. Consider again the rectangular resonator with sides of length 1 cm, 2 cm and 3 cm.

a. How many modes are there in the wavelength interval 4 / 5 cm to 8/13 cm? Hint: each mode index has two modes. Find those wavelengths.

b. How many unique resonant frequencies are there?

** c. Make a crude estimate for the number of modes in the wavelength interval 100 μm to 110 μm.


The normal modes are (problem 9.40, equation 9.204):

$ω_{l m n}=cπ*\sqrt{((l/d)^2+(m/a)^2+(n/b)^2)}$ with $b=1 cm<a=2 cm<d=3 cm$

$ω=ck$ and $k=2π/λ$ so that $2/λ=\sqrt{((l/d)^2+(m/a)^2+(n/b)^2)}$

$λ=2/\sqrt{((l/d)^2+(m/a)^2+(n/b)^2)}=2abd/\sqrt{((ab*l)^2+(b d*m)^2+(ad*n)^2 )}=$


The given condition is

$4/√5<12/√(4l^2+9m^2+36n^2 )<8/√13$ or $4/(12√5)<1/√(4l^2+9m^2+36n^2 )<8/(12√13)$


For $l m n=111$ we have $4l^2+9m^2+36n^2=49$ so $n=0$ is necessary

For $l m n=110$ we have $4l^2+9m^2+36n^2=13$

For $l m n=120$ we have $4l^2+9m^2+36n^2=40$

For $l m n=210$ we have $4l^2+9m^2+36n^2=25$

For $l m n=220$ we have $4l^2+9m^2+36n^2=52$

So the only possible mode index is $l m n=120$. If we assume there are possible TM modes with n=0 then there are possible two normal modes TE_120 and TM_120


There is just one unique resonant frequency corresponding to index $l m n=120$

$ω_120=cπ*√(1/9+4/4+0)*1/0.01=9.93*10^10 rad/sec$ (15.8 GHz)


Assume one has just the 1D case (L is the 1D length of system). The normal modes (maximum of vibration) happen when

$n*λ/2=L$ with n integer

Now go to 3D case:

$n=\sqrt{(n_1^2+n_2^2+n_3^2)}$ and the volume of “n states” is $(4πn^3)/3=4π/3*(8L^3)/λ^3$

Since we count just values of n positives wit multiply the above with 1/8 and since for every n index there are 2 modes we multiply again with 2. Therefore the volume in of “n states” is (Rayleigh-Jeans)

$N=π/3*(8L^3)/λ^3$ or if $L^3=V$ (physical volume of system)we have $|d N/dλ|=8πV/λ^4$

In our case the number of modes is ($100 μm=0.01 cm$ and $10 μm=0.001 cm$)

$ΔN=8πV/λ^4 *Δλ=(8π(1*2*3))/0.01*0.001≈1508 modes$