Chapter III: Optical Resonators and Laser Beam Propagation
Source: Gerrard and Burch, Introduction to Matrix Methods in Optics (1975), Chapter III.
Illustrative problems
Problem 3.1 — Recovering an unknown system matrix from focal-power tests
Brief solution
1. Method.
Let the sealed system be \(M_0=(A_0,B_0;C_0,D_0)\) and let weak lenses of powers \(P_1\) and \(P_2\) be attachable at its input and output. Because equivalent focal length measures \(C=-1/f\), the four tests give
2. Key step.
3. Answer.
The independent check is \(A_0D_0-B_0C_0=1\); disagreement measures experimental or paraxial-model error.
Show detailed steps
Let the sealed system be \(M_0=(A_0,B_0;C_0,D_0)\) and let weak lenses of powers \(P_1\) and \(P_2\) be attachable at its input and output. Because equivalent focal length measures \(C=-1/f\), the four tests give
Solving sequentially produces
The independent check is \(A_0D_0-B_0C_0=1\); disagreement measures experimental or paraxial-model error.
Problem 3.2 — Sagittal separation and the optical invariant
Brief solution
1. Method.
Propagate both marginal ray vectors by the same translation matrix and solve for the plane where their heights agree. Use the paraxial sag relation \(s\simeq y^2/(2R)\) together with \(V=ny/R\). After substituting the intersection distance, all translation-dependent terms cancel and the sagittal difference is one-half of the corresponding two-ray determinant (with the book’s reduced-coordinate normalization).
2. Answer.
That determinant is the Lagrange optical invariant, so its survival is not an accident: multiplication by any unimodular transfer matrix preserves it. Replacing one marginal ray by its diametrically opposite ray reverses the determinant and hence the sagittal ordering. Physically, the wavefront that led at one equal-area crossing lags at the other.
Show detailed steps
Propagate both marginal ray vectors by the same translation matrix and solve for the plane where their heights agree. Use the paraxial sag relation \(s\simeq y^2/(2R)\) together with \(V=ny/R\). After substituting the intersection distance, all translation-dependent terms cancel and the sagittal difference is one-half of the corresponding two-ray determinant (with the book’s reduced-coordinate normalization).
That determinant is the Lagrange optical invariant, so its survival is not an accident: multiplication by any unimodular transfer matrix preserves it. Replacing one marginal ray by its diametrically opposite ray reverses the determinant and hence the sagittal ordering. Physically, the wavefront that led at one equal-area crossing lags at the other.
Problem 3.3 — General round-trip matrix for a compound dye-laser cavity
Brief solution
1. Method.
Let \(M_0\) transfer rays from the left mirror to the right mirror, and represent each mirror by \(R(P_i)\). Write the outward chain first, then the reciprocal return chain with the refractive-index sign reversed as prescribed in Chapter II. Multiplying the two translations through the intervening system and both mirror matrices yields the round-trip matrix at the output mirror.
2. Key step.
every factor is unimodular, hence \(\det M_{rt}=1\);
a round trip referenced at one physical plane must have equal diagonal elements before the final output-mirror factor is applied.
3. Answer.
Keeping the intermediate matrix in symbolic \(A_0,B_0,C_0,D_0\) form is the shortest route; expanding numerical lens trains first obscures both checks.
Show detailed steps
Let \(M_0\) transfer rays from the left mirror to the right mirror, and represent each mirror by \(R(P_i)\). Write the outward chain first, then the reciprocal return chain with the refractive-index sign reversed as prescribed in Chapter II. Multiplying the two translations through the intervening system and both mirror matrices yields the round-trip matrix at the output mirror.
Two checks simplify the algebra:
every factor is unimodular, hence \(\det M_{rt}=1\);
a round trip referenced at one physical plane must have equal diagonal elements before the final output-mirror factor is applied.
Keeping the intermediate matrix in symbolic \(A_0,B_0,C_0,D_0\) form is the shortest route; expanding numerical lens trains first obscures both checks.
Problem 3.4 — Stability range and output-wave curvature
Brief solution
1. Method.
Write the round-trip matrix immediately before the output mirror as \((A_r,B_r;C_r,A_r)\) and multiply by \(R(P_2)\). The trace becomes \(2A_r-B_rP_2\). Stable operation requires its half-trace to lie strictly between \(-1\) and \(1\), so \(P_2\) lies between the two endpoint values
2. Key step.
with their numerical order selected from the sign of \(B_r\). For any allowed value, the fixed-point solution for \(q\) gives
3. Answer.
so the oscillating wavefront matches the output mirror. At either stability endpoint the Gaussian solution becomes marginal.
Show detailed steps
Write the round-trip matrix immediately before the output mirror as \((A_r,B_r;C_r,A_r)\) and multiply by \(R(P_2)\). The trace becomes \(2A_r-B_rP_2\). Stable operation requires its half-trace to lie strictly between \(-1\) and \(1\), so \(P_2\) lies between the two endpoint values
with their numerical order selected from the sign of \(B_r\). For any allowed value, the fixed-point solution for \(q\) gives
so the oscillating wavefront matches the output mirror. At either stability endpoint the Gaussian solution becomes marginal.
Problem 3.5 — Short single-frequency carbon-dioxide laser
Brief solution
1. Method.
A parallel output requires a plane output mirror. With cavity length \(L=0.5\ \mathrm m\), the round-trip trace condition \((A+D)/2=\cos1\) gives an input-mirror power
2. Key step.
3. Answer.
At the carbon-dioxide laser wavelength, the Gaussian formulas give \(w_0=1.76\ \mathrm{mm}\) and \(z_0=0.915\ \mathrm m\). The spot radius at the curved mirror is \(1.14w_0\), so the maximum \(1/e^2\) diameter is about 4 mm. Allowing diffraction, alignment, wall-reflection, and discharge margins leads to the practical recommendation \(\boxed{10\ \mathrm{mm}}\) for the tube or limiting diaphragm diameter.
Show detailed steps
A parallel output requires a plane output mirror. With cavity length \(L=0.5\ \mathrm m\), the round-trip trace condition \((A+D)/2=\cos1\) gives an input-mirror power
The resulting matrix is approximately
At the carbon-dioxide laser wavelength, the Gaussian formulas give \(w_0=1.76\ \mathrm{mm}\) and \(z_0=0.915\ \mathrm m\). The spot radius at the curved mirror is \(1.14w_0\), so the maximum \(1/e^2\) diameter is about 4 mm. Allowing diffraction, alignment, wall-reflection, and discharge margins leads to the practical recommendation \(\boxed{10\ \mathrm{mm}}\) for the tube or limiting diaphragm diameter.
Problem 3.6 — Internal-reflection focus inside a plano-convex lens
Brief solution
1. Method.
Use refraction matrices \(R_1,R_2\), internal-reflection matrices \(R'_1,R'_2\), and the 1-cm glass translation \(T\). Since the second surface is plane, its refraction and reflection matrices are identities. The ray after successive internal reflections is therefore generated by repeated application of \(TR'_1T\).
2. Key step.
about 1 mm inside the curved front surface. Each uncoated glass-air reflection retains approximately 0.04 of the incident power, but the Airy spot area is much smaller than the original 1-cm-radius beam. For a He–Ne wavelength, the average focal irradiance is therefore approximately
3. Answer.
This large value explains the internal-damage warning; aberration and a longer wavelength reduce it, but do not remove the hazard.
Show detailed steps
Use refraction matrices \(R_1,R_2\), internal-reflection matrices \(R'_1,R'_2\), and the 1-cm glass translation \(T\). Since the second surface is plane, its refraction and reflection matrices are identities. The ray after successive internal reflections is therefore generated by repeated application of \(TR'_1T\).
Progressively tracing the marginal ray shows that the first internal focus after two reflections is
about 1 mm inside the curved front surface. Each uncoated glass-air reflection retains approximately 0.04 of the incident power, but the Airy spot area is much smaller than the original 1-cm-radius beam. For a He–Ne wavelength, the average focal irradiance is therefore approximately
This large value explains the internal-damage warning; aberration and a longer wavelength reduce it, but do not remove the hazard.