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Lasers

17.1 Conditions for Lasing

A laser requires three conditions:

  1. Population inversion: N2>N1N_2 > N_1 for the lasing transition (between levels 2 and 1), achieved by pumping.
  2. Stimulated emission dominance: The stimulated emission rate must exceed the absorption rate: N2>N1N_2 > N_1.
  3. Optical feedback: A resonant cavity ( two mirrors) provides positive feedback.

Threshold condition: The gain per round trip must exceed the losses:

R1R2e2gL1R_1 R_2\,e^{2gL} \geq 1

Where R1,R2R_1, R_2 are mirror reflectivities, gg is the gain coefficient, and LL is the cavity length.

The threshold gain:

gth=12Lln(R1R2)=αi+αmg_{\text{th} = -\frac{1}{2L}\ln(R_1 R_2) = \alpha_i + \alpha_m}

Where αi\alpha_i is the internal loss and αm=ln(R1R2)/(2L)\alpha_m = -\ln(R_1 R_2)/(2L) is the mirror loss.

17.2 Laser Types

TypeGain mediumWavelengthCharacteristics
He-NeGas632.8 nmCW, low power (\sim1 mW), high coherence
Ar+^+Gas488, 514 nmCW, multiline, moderate power
CO2_2Gas10.6 μ\muMHigh power (kW), efficient (\sim20%)
Nd:YAGSolid state1064 nmPulsed or CW, high power
Ti:SapphireSolid state700—1000 nmTunable, femtosecond pulses
GaAs/InPSemiconductor0.8—1.6 μ\muMCompact, efficient, diode laser
DyeLiquidTunableWide tuning range

17.3 Gaussian Beam Optics

The fundamental mode (\text{TEM_}{00}) of a laser cavity is a Gaussian beam:

E(r,z)=E0w0w(z)exp ⁣(r2w(z)2)exp ⁣(i[kz+kr22R(z)ζ(z)])E(r, z) = E_0\frac{w_0}{w(z)}\exp\!\left(-\frac{r^2}{w(z)^2}\right)\exp\!\left(-i\left[kz + \frac{kr^2}{2R(z)} - \zeta(z)\right]\right)

Beam parameters:

  • Beam waist: w0w_0 (minimum spot size)
  • Rayleigh range: zR=πw02/λz_R = \pi w_0^2/\lambda
  • Beam radius: w(z)=w01+(z/zR)2w(z) = w_0\sqrt{1 + (z/z_R)^2}
  • Radius of curvature: R(z)=z[1+(zR/z)2]R(z) = z[1 + (z_R/z)^2]
  • Divergence angle: θ=λ/(πw0)\theta = \lambda/(\pi w_0)
Worked Example 17.1: Gaussian Beam Focusing

A He-Ne laser (λ=632.8\lambda = 632.8 nm) has a beam waist w0=0.3w_0 = 0.3 mm.

(a) Rayleigh range: zR=π(0.3×103)2/(632.8×109)=π×9×108/6.328×107=0.447z_R = \pi(0.3 \times 10^{-3})^2/(632.8 \times 10^{-9}) = \pi \times 9 \times 10^{-8}/6.328 \times 10^{-7} = 0.447 m.

(b) Beam radius at z=2z = 2 m: w=0.31+(2/0.447)2=0.31+20.0=0.3×4.58=1.37w = 0.3\sqrt{1 + (2/0.447)^2} = 0.3\sqrt{1 + 20.0} = 0.3 \times 4.58 = 1.37 mm.

(c) Divergence: θ=632.8×109/(π×0.3×103)=6.71×104\theta = 632.8 \times 10^{-9}/(\pi \times 0.3 \times 10^{-3}) = 6.71 \times 10^{-4} rad =0.67= 0.67 mrad.

At a distance of 1 km, the beam radius would be wθ×1000=0.67w \approx \theta \times 1000 = 0.67 m (ignoring the waist contribution, valid for zzRz \gg z_R).

Common Pitfalls (Additional)

  1. Coherence length limits interferometer arm difference: In a Michelson interferometer, the path difference must not exceed the coherence length lc=λ2/Δλl_c = \lambda^2/\Delta\lambda for fringes to be visible. White light fringes are visible only for near-zero path difference (lc1.5μl_c \sim 1.5\,\muM), while laser fringes remain visible for path differences of many metres.

  2. The Abbe limit is not a fundamental limit: Techniques such as STED (stimulated emission depletion), PALM (photoactivated localisation microscopy), and SIM (structured illumination microscopy) can achieve resolutions well below the Abbe limit of λ/(2NA)\lambda/(2\text{NA}). The 2014 Nobel Prize in Chemistry was awarded for super-resolution microscopy.

  3. Gaussian beams do not have sharp edges: Unlike geometrical optics rays, Gaussian beams have no well-defined edge. The beam radius ww is defined as the 1/e21/e^2 intensity radius (86.5%\sim 86.5\% of the peak). The power contained within ww is 1e286.5%1 - e^{-2} \approx 86.5\% of the total, not 100%.

  4. Spatial filtering with a pinhole: A pinhole of diameter dd in the focal plane of a lens acts as a low-pass spatial filter with cutoff frequency fc=d/(λf)f_c = d/(\lambda f). The transmitted beam approaches a Gaussian profile (Airy pattern central maximum), which is why spatial filtering is used to “clean up” laser beams.

  5. Polarisation and Brewster”s angle: At Brewster’s angle, the reflected beam is purely ss-polarised, not the transmitted beam. The transmitted beam has reduced ss-component and becomes partially pp-polarised. Complete polarisation of the transmitted beam requires many interfaces (pile-of-plates polariser).

Problems (Additional)

Problem 19: Resolution of a Telescope

The Hubble Space Telescope has a primary mirror diameter of 2.4 m and operates at λ=550\lambda = 550 nm.

(a) Calculate the angular resolution (Rayleigh criterion).

(b) What is the minimum distance on the Moon’s surface (d=384400d = 384\,400 km) that can be resolved?

(c) How does atmospheric seeing (0.5\sim 0.5 arcsec) compare with the diffraction limit?

Solution:

(a) θmin=1.22λ/D=1.22×550×109/2.4=2.80×107\theta_{\min} = 1.22\lambda/D = 1.22 \times 550 \times 10^{-9}/2.4 = 2.80 \times 10^{-7} rad =0.058= 0.058 arcsec.

(b) s=θmin×d=2.80×107×3.844×108=107.6s = \theta_{\min} \times d = 2.80 \times 10^{-7} \times 3.844 \times 10^8 = 107.6 m 108\approx 108 m.

(c) Atmospheric seeing 0.5\sim 0.5 arcsec is about 8.6 times worse than Hubble’s diffraction limit. This is why Hubble was placed in space --- ground-based telescopes are limited by seeing, not diffraction, unless adaptive optics is used.

Problem 20: Fabry--Perot Etalon

A Fabry—Perot etalon consists of two parallel reflecting surfaces with reflectance R=0.8R = 0.8 and separation d=1d = 1 mm, used at normal incidence with λ=500\lambda = 500 nm.

(a) Calculate the free spectral range (FSR) in frequency and wavelength.

(b) Calculate the finesse F\mathcal{F}.

(c) What is the minimum resolvable wavelength difference?

Solution:

(a) FSR in frequency: ΔνFSR=c/(2d)=3×108/(2×103)=1.5×1011\Delta\nu_{\text{FSR} = c/(2d) = 3 \times 10^8/(2 \times 10^{-3}) = 1.5 \times 10^{11}} Hz =150= 150 GHz.

FSR in wavelength: ΔλFSR=λ2/(2d)=(500×109)2/(2×103)=1.25×1013\Delta\lambda_{\text{FSR} = \lambda^2/(2d) = (500 \times 10^{-9})^2/(2 \times 10^{-3}) = 1.25 \times 10^{-13}} m =0.125= 0.125 nm.

(b) Finesse: F=πR/(1R)=π0.8/(10.8)=π×0.894/0.2=14.1\mathcal{F} = \pi\sqrt{R}/(1 - R) = \pi\sqrt{0.8}/(1 - 0.8) = \pi \times 0.894/0.2 = 14.1.

(c) Minimum resolvable wavelength difference (resolution):

\delta\lambda = \frac{\Delta\lambda_{\text{FSR}}{\mathcal{F}} = \frac{0.125}{14.1}\ \text{nm} = 0.0089\ \text{nm} = 8.9\ \text{pm}}

This corresponds to a resolving power R=λ/δλ=500/0.008956000\mathcal{R} = \lambda/\delta\lambda = 500/0.0089 \approx 56\,000.