Why is reflectance not zero at all wavelengths?

β depends on λ, so the phase slip 2β only equals π at the design condition. Away from that wavelength the two reflections no longer cancel.

Why show √(n₀n_s)?

For a single low-index film on a higher-index substrate, that geometric mean is the index that minimizes reflectance at the design wavelength when thickness is λ/(4n_f).

Do real camera lenses use one layer?

Modern systems stack many layers to widen the low-reflection band and handle oblique rays; the simulator isolates the physics of the simplest case.

Is normal incidence a big limitation?

Yes. At large incidence polarization splits s and p reflectance and the effective optical thickness changes; AR coatings for windows and camera lenses are optimized with those effects.

Anti-Reflection Coating

A single transparent film on glass can reduce reflection at a design wavelength by destructive interference between waves reflected at the air–film and film–substrate interfaces. At normal incidence the reflectance follows the two-beam formula R = |(r₀₁ + r₁₂e^{2iβ})/(1 + r₀₁r₁₂e^{2iβ})|² with β = 2π n_f d / λ and Fresnel amplitude coefficients r_ij from the indices. Choosing d = λ₀/(4n_f) gives β = π/2 at λ₀ so the reflections can cancel partially if the indices are chosen well; the ideal quarter-wave index for a single layer on glass in air is n_f ≈ √(n_air n_glass). The graph scans R(λ) across the visible band. Absorption, oblique incidence, and bandwidth limits of real coatings are omitted — this is the textbook single-layer AR model.

Who it's for: Undergraduate electromagnetism and optical coating courses introducing thin-film matrix methods later.

Key terms

Anti-reflection
Quarter-wave plate
Thin film
Fresnel coefficients
Interference
Reflectance
Refractive index matching
Spectral tuning

Live graphs

How it works

A thin dielectric film on glass uses path interference between reflections at the two interfaces; at quarter-wave optical thickness reflection can nearly vanish at one wavelength.

Key equations

β = 2π n_f d / λ , R = |(r₀₁ + r₁₂e^{2iβ})/(1 + r₀₁r₁₂e^{2iβ})|²

Frequently asked questions

Why is reflectance not zero at all wavelengths?: β depends on λ, so the phase slip 2β only equals π at the design condition. Away from that wavelength the two reflections no longer cancel.
Why show √(n₀n_s)?: For a single low-index film on a higher-index substrate, that geometric mean is the index that minimizes reflectance at the design wavelength when thickness is λ/(4n_f).
Do real camera lenses use one layer?: Modern systems stack many layers to widen the low-reflection band and handle oblique rays; the simulator isolates the physics of the simplest case.
Is normal incidence a big limitation?: Yes. At large incidence polarization splits s and p reflectance and the effective optical thickness changes; AR coatings for windows and camera lenses are optimized with those effects.

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Brewster Angle

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Fermat's Principle

OPL = n₁AP+n₂PB vs hit point; minimum = Snell path.

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Chromatic Aberration

Cauchy n(λ); thin-lens f(λ); paraxial rays R/G/B.

Launch Simulator

Anti-Reflection Coating

Live graphs

How it works

Key equations

Frequently asked questions

More from Optics & Light

Michelson Interferometer

Mach–Zehnder Interferometer

Sagnac (Ring) Interferometer

Brewster Angle

Fermat's Principle

Chromatic Aberration

Anti-Reflection Coating

Live graphs

How it works

Key equations

Frequently asked questions