How does an anti-reflecting coating work and what thickness and refractive index should it have?

In one line

An anti-reflecting coating is a thin transparent film of quarter-wave optical thickness whose two reflected beams cancel by destructive interference, reducing reflection from a lens or solar cell.

Explained simply

Glass reflects about 4% of light at each surface, causing glare and loss of brightness. A very thin coating creates a second reflection that is exactly half a wave out of step with the first, so the two cancel. Less light is reflected, so more passes through. The purplish tint on spectacle and camera lenses is this coating at work.

How it works

1. A film of index \(\mu_f\) is coated on glass of index \(\mu_g\) with \(1 < \mu_f < \mu_g\).
2. Ray 1 reflects at the air-film surface; ray 2 at the film-glass surface. Both reflect from a denser medium, so both get a \(\lambda/2\) shift, which cancel each other.
3. Path difference is just \(2\mu_f t\) (normal incidence).
4. For cancellation: \(2\mu_f t = \lambda/2\), so \(t = \lambda/(4\mu_f)\): a quarter-wave layer.
5. For complete cancellation the two reflected beams must have equal amplitudes, which needs \(\mu_f = \sqrt{\mu_g}\).
6. Coatings are designed for the middle of the visible band (about 550 nm), where the eye is most sensitive. Red and blue are reflected a little more, giving the purplish tint.

Key formula

\(t = \dfrac{\lambda}{4\mu_f}\) and ideally \(\mu_f = \sqrt{\mu_g}\). \(t\): coating thickness (m), \(\lambda\): design wavelength in air (m), \(\mu_f\): coating index, \(\mu_g\): glass index.

Quick example

Magnesium fluoride (\(\mu_f = 1.38\)) on glass for \(\lambda = 550\) nm: \(t = 550/(4\times1.38) \approx 99.6\) nm, about 100 nm. Ideal \(\mu_f\) for glass of 1.5 is \(\sqrt{1.5} \approx 1.22\); MgF₂ is used because it is close, durable and easy to deposit. Uses: spectacles, camera and microscope lenses, solar cells (silicon nitride on silicon) and display screens.

Common mistakes

• Applying the usual \(\lambda/2\) term; here both reflections have the shift, so it cancels out.
• Writing \(t = \lambda/4\) without dividing by \(\mu_f\).
• Thinking the coating destroys light energy; it redistributes it, so more is transmitted.

How MU asks it

Asked as "What is an anti-reflection coating? Derive the condition for its thickness and refractive index" (5 marks) or a numerical for minimum thickness (3 marks). Draw the two reflected rays, state that both suffer phase change, derive \(t = \lambda/4\mu_f\), add \(\mu_f = \sqrt{\mu_g}\), and list applications.