How to Design an AR Coating

Every air–glass surface reflects about 4% of the light that hits it. In a system with several lenses, those reflections reduce transmission and produce ghost images and glare. The simplest useful antireflection (AR) coating is a single thin layer.

This guide derives the quarter-wave design conditions, works through the numbers, and then verifies the result in TFStudio.

Why a bare surface reflects 4%

When light crosses a boundary between two media, some of it reflects. At normal incidence the fraction reflected from a single surface is

R = \left(\frac{n_s - n_0}{n_s + n_0}\right)^2

At 550 nm, TFStudio's material library uses $n_0 = 1.0000$ for air and $n_s = 1.5185$ for N-BK7:

R = \left(\frac{1.5185 - 1.0000}{1.5185 + 1.0000}\right)^2 \approx 0.042385 = 4.24

That is per surface. A lens has two, so roughly 8% of your light is gone.

The idea: cancel reflection with interference

A single-layer AR coating adds one transparent film on top of the glass. Now there are two reflecting surfaces: air–film and film–glass. If we choose the film's thickness so the two reflected waves come back exactly out of step, they cancel. If we also choose the film's index so the two reflections are equal in strength, the cancellation is complete.

Condition 1 — thickness (a quarter wave).
The film must be one quarter of a wavelength thick optically, so the round trip through it delays the second reflection by half a wave:

n_1 d_1 = \frac{\lambda_0}{4}

Condition 2 — index (the geometric mean).
For the two reflections to be equal in amplitude, the film index must be the geometric mean of the media on either side:

n_1 = \sqrt{n_0 n_s}

This is the classic quarter-wave result: a quarter-wave film of admittance $\eta$ transforms the substrate so the front surface sees the incident medium exactly.

Calculations

Using the same N-BK7 value, the ideal film index is

n_1 = \sqrt{1.0000 \times 1.5185} \approx 1.2323

Here is the catch that makes AR design a real engineering problem: no durable solid material has an index that low. The closest practical, hard, stable coating material is magnesium fluoride, MgF₂. TFStudio's library gives $n_1 = 1.3785$ at 550 nm.

So we accept the compromise. With MgF₂ the reflection at the design wavelength does not reach zero, but it drops dramatically:

R = \left(\frac{n_0 n_s - n_1^2}{n_0 n_s + n_1^2}\right)^2 = \left(\frac{1.5185 - 1.3785^2}{1.5185 + 1.3785^2}\right)^2 \approx 0.012469 = 1.25

From 4.24% down to roughly 1.25% per surface, with one layer. This is why MgF₂ has been a workhorse material for consumer optics: it is inexpensive, durable, and effective.

Design at the center of the visible band, $\lambda_0 = 550\ \text{nm}$ . The physical thickness is

d_1 = \frac{\lambda_0}{4 n_1} = \frac{550}{4 \times 1.3785} \approx 99.75\ \text{nm}

That's the whole design: 99.75 nm of MgF₂ on N-BK7. Now let's verify it.

Building it in TFStudio

Open the web demo or install Windows release from github repository.

1. Set up

Substrate: N-BK7 ( $n = 1.5185$ at 550 nm)
Incident medium: Air
Add one layer: MgF₂ ( $n = 1.3785$ at 550 nm).
Set its optical thickness to 1 quarter-wave at 550 nm. TFStudio converts that to 99.75 nm.

A bare air–N-BK7 surface reflects roughly 4.24% at 550 nm.
A single quarter-wave film cancels reflection by interference; ideal index is $\sqrt{n_0 n_s}$ and thickness is $\lambda_0/(4n_1)$ .
For N-BK7, the ideal index is 1.2323; MgF₂ ( $n = 1.3785$ ) gets you to
approximately 1.25%.

How to Design an Antireflection (AR) Coating — A Step-by-Step Guide

Why a bare surface reflects 4%

The idea: cancel reflection with interference

Calculations

Building it in TFStudio

1. Set up

2. Plot the reflectance

3. Compare against bare glass

Cons

Recap