Anti Reflection Coating Calculator

A single‑layer anti‑reflection design often targets destructive interference between reflections from the top and bottom film boundaries. A common normal‑incidence choice is:

• n₁ ≈ √(n₀·nₛ) (index matching for one layer)
• d = λ₀ / (4 n₁) (quarter‑wave thickness)

For oblique incidence and polarization, this calculator evaluates reflectance using the thin‑film characteristic matrix method. The phase thickness is:

δ = 2π n₁ d cosθ₁ / λ

The reported reflectance is R = |r|², with unpolarized results averaged from s and p components.

1. Enter the ambient index, substrate index, and design wavelength.
2. Choose incidence angle and polarization for your optical setup.
3. Select Auto for a quick quarter‑wave design, or Manual.
4. Set a reflectance threshold to estimate useful bandwidth.
5. Press Calculate to view results above the form.
6. Use Download buttons to export CSV or PDF reports.

1) Why anti‑reflection coatings matter

A bare air–glass interface reflects a noticeable fraction of light. For typical glass with refractive index near 1.52, normal‑incidence Fresnel reflectance is about 4% per surface, which can reduce transmission and increase glare. In imaging systems, that extra reflection lowers contrast, elevates stray light, and can create ghost images.

2) The single‑layer quarter‑wave concept

A classic approach uses a single thin dielectric film designed so reflections from its two boundaries interfere destructively at a target wavelength. The quarter‑wave condition sets the film’s optical thickness to one quarter of the design wavelength, making the phase shift between reflected beams approximately 180° at the target.

3) Index matching guidance

For a one‑layer design at normal incidence, the best‑known index choice is n₁ ≈ √(n₀·nₛ). With air (n₀≈1.00) and glass (nₛ≈1.52), the ideal film index is near 1.233. In practice, materials with n around 1.23 are limited, so designers often pick common low‑index films and accept a small residual reflectance.

4) Practical materials and thickness example

Magnesium fluoride (MgF₂) is a widely used low‑index coating with n≈1.38 in the visible. At λ₀=550 nm, the quarter‑wave physical thickness is roughly d ≈ 550/(4·1.38) ≈ 99.6 nm. Silicon dioxide (SiO₂) is another common film (n≈1.46), giving a quarter‑wave thickness near 94 nm at the same wavelength.

5) Angle of incidence and polarization effects

As incidence angle increases, the internal film angle changes via Snell’s law and the effective optical thickness becomes n₁ d cosθ₁. Also, s and p polarizations respond differently because their optical admittances differ. This calculator lets you compare unpolarized, s, and p cases and see how reflectance can shift at higher angles.

6) Reflectance spectra and bandwidth

A single‑layer AR typically produces a narrow reflectance minimum around λ₀, with reflectance rising away from the design wavelength. Many practical designs aim for sub‑1% reflectance over a useful band rather than a perfect zero at one wavelength. The bandwidth estimate here scans your chosen wavelength range and reports where reflectance stays below your threshold.

7) Using manual mode for optimization

Auto mode quickly proposes a quarter‑wave film using the index‑matching rule. Manual mode is useful when you have a known coating material, or when manufacturing constraints require a different thickness. By adjusting n₁ and d, you can trade a slightly higher minimum reflectance for better performance across a wider spectral band.

8) Interpreting results for real optics

Treat results as an engineering estimate for a single dielectric layer on a non‑absorbing substrate. Real coatings may have dispersion, absorption, and multi‑layer stacks that broaden bandwidth and reduce sensitivity to angle. Still, a single‑layer model is valuable for quick feasibility checks, material comparisons, and target‑wavelength planning.

1) What does “quarter‑wave” mean in this context?

It means the film optical thickness is λ₀/4, so reflections from the film’s two interfaces return roughly 180° out of phase at the design wavelength, reducing net reflection.

2) Why is the “ideal” film index √(n₀·nₛ)?

For a single non‑absorbing layer at normal incidence, that index balances impedance between ambient and substrate, enabling destructive interference to cancel reflected amplitude most effectively at the design wavelength.

3) Can a single layer give zero reflection over all wavelengths?

No. A single layer typically produces a narrow minimum near λ₀. Broad low‑reflection performance usually requires multi‑layer stacks that shape the reflectance spectrum across a wider band.

4) Why do s and p results differ at higher angles?

At oblique incidence, the boundary conditions differ for electric fields perpendicular or parallel to the plane of incidence. This changes effective admittance and shifts the reflectance minimum differently for s and p polarization.

5) What thickness should I try for MgF₂ at 550 nm?

Using n≈1.38, a quarter‑wave thickness is about 550/(4·1.38) ≈ 100 nm. Use Auto mode or enter n₁=1.38 in Manual mode to verify reflectance.

6) How is the bandwidth estimate computed?

The calculator scans reflectance over your wavelength range and finds the continuous region around λ₀ where reflectance stays below your chosen threshold (for example, 1%). More scan steps improve the estimate.

7) Does this model include absorption or dispersion?

No. It assumes real refractive indices and a single non‑absorbing layer. If your film or substrate is absorbing or strongly dispersive, measured performance may differ from this idealized prediction.