Question

Some mirrors for infrared lasers are constructed with alternating layers of hafnia and silica. Suppose you want to produce constructive interference from a thin film of hafnia \((n=1.90)\) on \(\mathrm{BK}-7\) glass \((n=1.51)\) using infrared radiation of wavelength \(1.06 \mu \mathrm{m} .\) What is the smallest film thickness that would be appropriate, assuming that the laser beam is oriented at right angles to the film?

Short answer

Answer: The smallest film thickness of hafnia that would cause constructive interference is approximately \(0.278\, \mathrm{\mu m}\).

Step-by-step solution

Identify the relevant formula for constructive interference in thin films

For a beam of light incident at right angles to a thin film, the condition for constructive interference is given by: $$ 2nt = m \lambda $$ Where \(n\) is the refractive index of the film, \(t\) is the thickness of the film, \(m\) is the order of the interference, and \(\lambda\) is the wavelength of the light in vacuum.

Solve for the smallest thickness of the hafnia layer

In this problem, we know the refractive index of hafnia \((n = 1.90)\), the wavelength of the incident light in vacuum \((\lambda = 1.06 \, \mathrm{\mu m})\), and we seek the smallest thickness of the hafnia layer, so we will choose the smallest value of \(m\), which is \(m = 1\). Now, we solve for the thickness \(t\): $$2(1.90)t = 1(1.06\, \mathrm{\mu m})$$ Divide both sides by \(2(1.90)\): $$t = \frac{1.06\, \mathrm{\mu m}}{2(1.90)}$$ Calculate the value of \(t\): $$t = \frac{1.06\, \mathrm{\mu m}}{3.80}\approx 0.278\, \mathrm{\mu m}$$ So, the smallest film thickness of hafnia that would cause constructive interference is approximately \(0.278\, \mathrm{\mu m}\).