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Lasers | Physics - Wyatt's Notes

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.

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

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).

  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).

flowchart TD
A[21_Lasers 17] --> B[Key Concepts]
A --> C[Core Principles]
A --> D[Practical Applications]
B --> E[Fundamental definitions]
C --> F[Design patterns]
D --> G[Real-world usage]

Lasers produce coherent light through stimulated emission, where an incoming photon triggers an excited atom to emit an identical photon. Population inversion, where more atoms are in excited than ground states, is essential for amplification. The optical cavity provides feedback, selecting specific frequencies and directions. The result is light that is monochromatic, coherent, and highly directional. Different gain media produce different wavelengths: helium-neon for red, argon-ion for blue-green, semiconductor diodes for infrared. Mode locking produces ultrashort pulses, while Q-switching produces high-energy pulses. Lasers enable precision measurements, surgery, and telecommunications.

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.