Interference from Thin Films — Question 2

Question 2

Two reflection phase reversals exchange the bright and dark thickness conditions

A film of index $n_f=1.50$ lies between air ($n_0=1.00$) and a substrate of index $n_s=2.00$. Light of vacuum wavelength $600\,\mathrm{nm}$ is incident normally. Treat reflected maxima and minima by the phase difference of the first two rays.

1. Identify both reflection phase shifts and find the least positive thickness for a reflected minimum.

2. Find the least positive thickness for a reflected maximum and state whether a phase minimum necessarily means zero reflectance.

Question 2 - Solution

Two reflection phase reversals exchange the bright and dark thickness conditions

(a) Method and calculation

$$\phi_{01}=\pi,\quad\phi_{fs}=\pi\Longrightarrow\Delta\phi_r=0;\quad\delta=4\pi n_ft/\lambda_0.$$$$\delta=\pi\Longrightarrow\boxed{t_{\mathrm{min}}=\lambda_0/(4n_f)=100\,\mathrm{nm}}.$$

(b) Interpretation and further calculation

$$\delta=2\pi\Longrightarrow\boxed{t_{\mathrm{max,positive}}=\lambda_0/(2n_f)=200\,\mathrm{nm}}.$$$$I_{\min}=(\sqrt{I_1}-\sqrt{I_2})^2\Longrightarrow\boxed{I_{\min}=0\ \text{only for equal interfering amplitudes}}.$$

Physical check

The relevant reflection contribution is the difference of the two phase shifts; two pi reversals cancel in the relative phase.