What is an antireflection coating? State the conditions on its refractive index and thickness. Find the minimum thickness of a magnesium fluoride (\(\mu=1.38\)) antireflection coating on glass (\(\mu=1.52\)) for light of wavelength 5500 Å.

Answer: A thin transparent film that cancels reflection by destructive interference; it needs \(\mu_f=\sqrt{\mu_g}\) and \(t=\lambda/4\mu_f\). For MgF₂, \(t_{min}=996\) Å \(\approx0.1\ \mu\)m.

An antireflection (AR) coating is a thin transparent film deposited on a lens or glass surface so that the rays reflected from its top and bottom surfaces cancel each other, reducing reflection and increasing transmission.

Conditions:

1. Amplitude condition: the two reflected rays must have equal amplitudes: \(\mu_f=\sqrt{\mu_{air}\,\mu_g}=\sqrt{\mu_g}\). (For \(\mu_g=1.52\) the ideal value is 1.23; MgF₂ with 1.38 is the nearest practical material.)
2. Phase condition: both reflections occur at a denser medium, so both rays get a \(\lambda/2\) phase change, which cancels. For destructive interference at normal incidence the path difference must be \(\lambda/2\): \(2\mu_ft=\lambda/2\Rightarrow t=\dfrac{\lambda}{4\mu_f}\) (a quarter-wave film).

Calculation: \[t=\frac{\lambda}{4\mu_f}=\frac{5500\times10^{-10}}{4\times1.38}=9.96\times10^{-8}\ \text{m}=996\ \text{Å}\]