798
nm
413.5
nm
—
nm
—
nm
—
nm
—
798
nm
413.5
nm
—
nm
—
nm
—
nm
—
The Thin Film Interference Calculator determines which wavelengths undergo constructive or destructive interference when light reflects from a thin transparent film. By computing the optical path difference 2nt and accounting for phase shifts at each boundary, it predicts reflected colors, anti-reflection coating thicknesses, and the conditions for maximum or minimum reflectance.
Thin film interference creates the brilliant colors seen in soap bubbles, oil slicks on water, anti-reflection coatings on lenses, and the iridescent wings of butterflies. The effect occurs because light reflected from the top and bottom surfaces of a thin film travels different path lengths and may also undergo phase shifts, leading to wavelength-dependent constructive or destructive interference in the reflected beam.
When light hits a thin film of refractive index n and thickness t, partial reflections occur at both the top and bottom surfaces. The reflected beams interfere based on two factors:
1. Optical path difference: The beam reflecting from the bottom surface travels an extra distance of 2t through the film (at normal incidence), with an effective path difference of:
$$\Delta = 2nt$$
because the wavelength inside the film is λ/n.
2. Phase shifts on reflection: When light reflects from a surface where the refractive index increases (going from low n to high n), it undergoes a phase shift of π (equivalent to half a wavelength). No phase shift occurs at a high-to-low boundary.
For a soap film in air (n₁ < n_film > n₂): both reflections involve the same type of boundary (one low-to-high, one high-to-low), so the net extra phase shift is zero. Constructive interference in reflection requires:
$$2nt = \left(m + \tfrac{1}{2}\right)\lambda, \quad m = 0, 1, 2, \ldots$$
For a coating on glass (n₁ < n_film < n₂): the top reflection (low-to-high) gives a π shift, and the bottom reflection (also low-to-high, from film to glass) gives another π shift — the two cancel. Wait — actually only the top surface gives a shift if n₁ < n_film, and the bottom gives a shift if n_film < n₂. With two shifts totaling 2π (equivalent to zero net shift), the constructive condition becomes the same as the soap film case. However, when exactly one reflection undergoes a phase shift, the conditions swap. The calculator handles both cases via the boundary condition selector.
An anti-reflection coating is designed so that destructive interference occurs for reflected light — the film thickness is chosen as t = λ/(4n) so that 2nt = λ/2, and with the appropriate phase shift the reflected beams cancel.
The optical path difference 2nt is the key quantity. Compare it with the wavelength to determine whether a given color is enhanced or suppressed in reflection. The constructive and destructive wavelengths shown correspond to the strongest reflection and strongest transmission, respectively. For anti-reflection applications, you want destructive interference at the design wavelength. For decorative or sensor applications, constructive interference determines the visible color.
Inputs
Results
A soap film (n = 1.33) with thickness 300 nm has an optical path difference of 798 nm. At m = 1 it constructively reflects about 532 nm (green) light. The bubble appears green at this thickness.
Inputs
Results
A MgF₂ coating (n = 1.38) on glass needs t ≈ 199 nm for first-order destructive interference at 550 nm. This is the standard quarter-wave anti-reflection coating thickness used on camera lenses.
Soap bubbles have varying thickness across their surface due to gravity and surface tension. Different thicknesses satisfy the constructive interference condition for different wavelengths, so each region reflects a different color. As the bubble thins, the colors shift until the film becomes too thin (< ~25 nm) and appears black because no visible wavelength satisfies constructive interference.
A quarter-wave coating has thickness t = λ/(4n), making the optical path difference 2nt = λ/2. Combined with the appropriate phase shift at the boundaries, the two reflected beams are exactly out of phase and cancel. This minimizes reflection at the design wavelength, typically chosen as 550 nm (green) for visible-light optics.
A phase shift of π at a boundary is equivalent to adding half a wavelength to the path difference. If both reflections have the same type of shift (both π or both zero), the net effect is zero and standard conditions apply. If only one reflection shifts, the constructive and destructive conditions swap.
Oil films on water have varying thickness and the refractive index of oil (≈1.5) is higher than both air above and water below. White light contains all visible wavelengths, and at each point only certain wavelengths satisfy constructive interference — the rest are transmitted. The reflected colors depend on the local film thickness.
Yes. At non-normal incidence, the path difference becomes 2nt cos θ_r, where θ_r is the refraction angle inside the film. This calculator assumes normal incidence (θ = 0). At larger angles, the effective thickness decreases and colors shift toward shorter wavelengths (blue shift).
For perfect cancellation, the coating should have n = √(n₁ × n₂), where n₁ is the surrounding medium and n₂ is the substrate. For glass (n₂ ≈ 1.52) in air (n₁ = 1), the ideal coating has n ≈ 1.23. MgF₂ (n = 1.38) is commonly used as the closest practical material.
Roboculator Team
The Roboculator Team explains calculations, planning tools, and practical formulas in clear language for real-life situations.
How helpful was this calculator?
Be the first to rate!
