Math 2.0

Optics, electromagnetic theory Definition

Physical optics

Builds on Wave equationHelmholtz equationRefractive index

Definition

Let Ω⊆ℝ3\Omega \subseteq \R^3 be a region with refractive index nn. Let cc be the speed of light in vacuum. Physical optics (scalar wave optics) describes light in Ω\Omega by a real field U(x,t)U(x, t) that solves the wave equation n(x)2c2∂2U∂t2=ΔU.\frac{n(x)^2}{c^2} \frac{\partial^2 U}{\partial t^2} = \Delta U. A monochromatic field of angular frequency ω\omega has the form U(x,t)=Re⁡(u(x)e−iωt)U(x, t) = \operatorname{Re}\left(u(x) e^{-i \omega t}\right). Its complex amplitude uu solves the Helmholtz equation Δu+k2n(x)2u=0,k=ωc=2πλ,\Delta u + k^2 n(x)^2 u = 0, \qquad k = \frac{\omega}{c} = \frac{2\pi}{\lambda}, and λ\lambda is the vacuum wavelength. The observable intensity is I(x)=∣u(x)∣2I(x) = \abs{u(x)}^2, up to a constant factor.

Remarks

Because the equations are linear, fields add, and intensities do not: for u=u1+u2u = u_1 + u_2 one gets I=∣u1∣2+∣u2∣2+2Re⁡(u1u2‾)I = \abs{u_1}^2 + \abs{u_2}^2 + 2 \operatorname{Re}(u_1 \overline{u_2}), and the last term produces interference fringes. Example: two plane waves eikxsin⁡αe^{i k x \sin\alpha} and e−ikxsin⁡αe^{-i k x \sin\alpha} in vacuum give I=4cos⁡2(kxsin⁡α)I = 4 \cos^2(k x \sin\alpha), a pattern of fringes with spacing λ/(2sin⁡α)\lambda / (2 \sin\alpha). Geometrical optics is the limit k→∞k \to \infty, where u≈AeikLu \approx A e^{i k L} and LL solves the eikonal equation. The scalar model ignores polarisation; the vector theory uses Maxwell's equations for the electric and magnetic fields.

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Further reading

  1. Web https://en.wikipedia.org/wiki/Physical_optics
About this page
TagJE101TDP4Q
URIm2://JE101TDP4Q
Kinddefinition
WritingW0 stub
MSC78A10, 78A45
DefinesPhysical optics
UsesWave equation, Helmholtz equation, Refractive index
Changed3 November 2026

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