Single-Layer Antireflection Coating: Two Wavelengths and Thickness Error
A camera lens is made of glass with refractive index n_g = 1.52. To reduce reflection at normal incidence, a single layer of magnesium fluoride with n_c = 1.38 is deposited on the glass. The layer thickness is chosen to give minimum reflection at λ0 = 550 nm in air, where n_air = 1.00. A supplier offers a different coating material with n = 1.23, but only in a thickness that is one quarter of the design wavelength in that material. The optical engineer must decide whether the existing MgF2 layer can be adjusted to suppress reflection simultaneously at 550 nm and at 400 nm, and whether the n = 1.23 coating would be better at 550 nm. The engineer also measures the actual MgF2 thickness and finds it is 10% larger than the ideal quarter-wave value. The lens is used at normal incidence, and dispersion of the indices is neglected. The team needs the minimum MgF2 thickness for destructive interference at 550 nm, an assessment of the two-wavelength requirement, and the expected effect of the 10% thickness error.
1. Determine the minimum MgF2 thickness that minimizes reflection at 550 nm. Identify the phase shifts at the two interfaces and the interference condition used.
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At the air-to-coating interface the index increases from 1.00 to 1.38, so the reflected wave undergoes a π phase shift. At the coating-to-glass interface the index increases from 1.38 to 1.52, so that reflection also undergoes a π shift. The two reflection phase shifts therefore cancel, leaving the round-trip path difference 2nt to set the condition. Destructive interference requires 2nt = (m + 1/2)λ.
The minimum nonzero thickness uses m=0: t = λ/(4n) = 550 nm/(4×1.38) = 99.6 nm. Using m=1 or larger would be thicker but not minimum.
2. Can this single MgF2 layer be made to give zero reflection at both 550 nm and 400 nm? Support your answer with the interference condition, neglecting dispersion.
Read the worked answer
For the same thickness and the same π-π phase condition, zero reflection at both wavelengths would require 2nt = (m1 + 1/2)(550 nm) = (m2 + 1/2)(400 nm), with nonnegative integers m1 and m2. Dividing gives (m1 + 1/2)/(m2 + 1/2) = 400/550 = 8/11. Cross-multiplying gives 11m1 + 5.5 = 8m2 + 4, or 11m1 - 8m2 = -1.5. Multiplying by 2 gives 22m1 - 16m2 = -3, whose left side is even and right side odd; no integer solution exists.
Therefore a single thickness satisfying this phase condition cannot null both wavelengths exactly. A practical broadband coating needs multiple layers or accepts reduced performance at one wavelength.
3. Assess the supplier's n=1.23 quarter-wave coating and the effect of making the MgF2 layer 10% thicker than the ideal value. Include whether n=1.23 is close to the ideal single-layer index.
Read the worked answer
For a single-layer antireflection coating at normal incidence, the ideal index is n_ideal = sqrt(n_air n_glass) = sqrt(1.52) = 1.233, and the ideal thickness is a quarter-wave in the coating. The n=1.23 material at quarter-wave thickness therefore nearly satisfies both conditions; its residual reflectance at 550 nm is essentially zero, much lower than that of MgF2.
For MgF2, a 10% thickness error makes 2nt = 1.10(λ/2) = 0.55λ instead of 0.50λ. The two reflected waves are then not exactly out of phase, so the reflectance no longer reaches its minimum; there is a residual reflection. The error is not catastrophic for a narrow band, but it detunes the coating away from 550 nm.