Thin-Film Optical Coating Calculator

The optical path difference is the extra distance the light ray travelling through the film travels compared to the ray reflected at the top surface. It is calculated as OPD = 2 × n₂ × d × cos(θ₂), where n₂ is the film refractive index, d is the film thickness, and θ₂ is the refraction angle inside the film. When OPD equals a whole-number multiple of the wavelength, constructive interference occurs; when it equals a half-integer multiple, destructive interference results.

For a single-layer anti-reflective (AR) coating, the minimum thickness is λ / (4 × n₂ × cos(θ₂)), where λ is the target wavelength and n₂ is the film refractive index. This quarter-wave condition ensures the two reflected beams are half a wavelength out of phase, producing destructive interference and minimising reflection. The exact formula depends on whether phase shifts occur at both or just one of the interfaces.

Reflectivity is derived using the Fresnel equations for the two interfaces (air/film and film/substrate). For normal incidence, the amplitude reflection coefficients are r₁ = (n₁ − n₂)/(n₁ + n₂) and r₂ = (n₂ − n₃)/(n₂ + n₃). The total reflectivity R combines these coefficients along with the phase shift introduced by the film thickness and accounts for multiple internal reflections. This calculator uses the simplified two-beam approximation for typical thin-film scenarios.

A phase shift of 180° (half a wavelength) occurs when light reflects off an interface where it travels from a lower to a higher refractive index medium. If both the top and bottom interfaces cause a phase shift (n₁ < n₂ and n₂ < n₃, or n₁ > n₂ and n₂ > n₃), the two shifts cancel and the OPD alone determines the interference type. If only one interface causes a phase shift, the effective OPD is shifted by λ/2.

Constructive interference amplifies the reflected light, making the surface appear brighter at that wavelength — this is the principle behind optical filters and highly reflective coatings. Destructive interference causes the reflected beams to cancel, reducing reflection — this is exploited in anti-reflective coatings on camera lenses and eyeglasses. The type of interference depends on the OPD relative to the wavelength and the number of phase shifts at the interfaces.

The substrate refractive index determines whether a phase shift of 180° occurs at the lower film interface. If n₂ < n₃, light reflecting at the film/substrate boundary undergoes a phase inversion, which shifts the effective OPD by half a wavelength. This directly changes whether the coating acts as an anti-reflective layer or an enhancing reflector for a given film thickness, making n₃ a critical parameter in coating design.

Thin-film coatings are used widely in anti-reflective coatings on lenses and screens, high-reflectance mirrors for lasers, bandpass and notch optical filters, solar cell efficiency enhancement, semiconductor metrology, and decorative iridescent finishes. The precise control of film thickness at the nanometre scale allows engineers to tailor reflectance and transmittance at specific wavelengths for each application.

Snell's law states that n₁ × sin(θ₁) = n₂ × sin(θ₂), relating the angles of incidence and refraction at an interface between two media. In this calculator, it is used to find the refraction angle θ₂ inside the film from the incident angle θ₁ and the refractive indices n₁ and n₂. The angle θ₂ then feeds directly into the OPD formula, meaning oblique incidence reduces the effective path difference compared to normal incidence.