Obtain an expression for minimum thickness of anti-reflecting film.

Anti-reflecting (AR) film: A thin transparent film of index \(\mu\) and thickness t coated on glass of index \(\mu_g\), with \(1<\mu<\mu_g\).

        R1    R2          incident (normal)
   air   ^     ^          |
 ========|=====|==========v====  air–film surface (phase change π)
   film μ, t   |
 ==============|===============  film–glass surface (phase change π)
   glass μg

Phase changes: Ray R1 is reflected at the air–film surface (rarer to denser), and ray R2 at the film–glass surface (also rarer to denser). Each undergoes a phase change of π, so the two cancel and there is no net additional path difference.

Path difference (normal incidence, \(\cos r=1\)): \(\Delta=2\mu t\)

Destructive interference (no reflection): \(2\mu t=(2n+1)\dfrac{\lambda}{2}\), n = 0, 1, 2, …

The minimum thickness corresponds to n = 0:

\[ 2\mu t=\frac{\lambda}{2}\Rightarrow\boxed{t_{min}=\frac{\lambda}{4\mu}} \]

So the film is a quarter-wave layer (optical thickness \(\mu t=\lambda/4\)).

Index condition for complete cancellation (equal amplitudes): \(\left(\dfrac{\mu-1}{\mu+1}\right)=\left(\dfrac{\mu_g-\mu}{\mu_g+\mu}\right)\Rightarrow\mu=\sqrt{\mu_g}\).

Example: MgF₂ (μ = 1.38) at λ = 5500 Å gives \(t_{min}=5500/(4\times1.38)\approx996\) Å ≈ 0.1 µm.