Question

A thin uniform film of refractive index 1.750 is placed on a sheet of glass with a refractive index 1.50. At room temperature ( 24.0^o C), this film is just thick enough for light with a wavelength 582.9 nm reflected off the top of the film to be canceled by light reflected from the top of the glass. After the glass is placed in an oven and slowly heated to 164^o C, you find that the film cancels reflected light with a wavelength 587.0 nm.

What is the coefficient of linear expansion of the film? (Ignore any changes in the refractive index of the film due to the temperature change.) Express your answer using two significant figures.

Answer 1

The condition for destructive interference is, \(2 t=\frac{m \lambda}{n}\)

The smallest non zero thickness is, \(t=\frac{\lambda}{2 n}\)

At \(24.0^{\circ} \mathrm{C}\)

\(\begin{aligned} t_{0} &=\frac{582.9 \mathrm{~nm}}{2(1.750)} \\ &=166.54 \mathrm{~nm} \\ \text { At } & 164^{\circ} \mathrm{C} \\ t &=\frac{587.0 \mathrm{~nm}}{2(1.750)} \\ &=167.71 \mathrm{~nm} \end{aligned}\)

Now, \(t=t_{0}(1+\alpha \Delta T)\)

The coefficient of linear expansion is, \(\begin{aligned} \alpha &=\frac{t-t_{0}}{t_{0} \Delta T} \\ &=\frac{167.71 \mathrm{~nm}-166.54 \mathrm{~nm}}{166.54 \mathrm{~nm}\left(164^{\circ} \mathrm{C}-24^{\circ} \mathrm{C}\right)} \\ &=5.01 \times 10^{-5} /{ }^{\circ} \mathrm{C} \end{aligned}\)