Interference from Thin Films — Question 1

Question 1

A free-standing film has one reflection phase reversal

A transparent film of index $n_f=1.40$ is surrounded by air and illuminated normally at vacuum wavelength $\lambda_0=560\,\mathrm{nm}$. Consider the first two reflected rays. Reflection from lower to higher index adds phase pi; reflection from higher to lower index adds zero.

1. Find the smallest positive thickness giving a reflected maximum.

2. Find the smallest positive thickness giving a reflected minimum and explain the zero-thickness limit.

Question 1 - Solution

A free-standing film has one reflection phase reversal

(a) Method and calculation

$$\delta=\frac{4\pi n_ft}{\lambda_0}+\pi\pmod{2\pi};\quad\text{maximum}:\ 2n_ft=(m+1/2)\lambda_0.$$$$\boxed{t_{\mathrm{max,min}}=\lambda_0/(4n_f)=100\,\mathrm{nm}}.$$

(b) Interpretation and further calculation

$$\text{minimum}:\ 2n_ft=m\lambda_0\Longrightarrow\boxed{t_{\mathrm{min,positive}}=\lambda_0/(2n_f)=200\,\mathrm{nm}}.$$$$t\to0:\quad\delta\to\pi\Longrightarrow\boxed{\text{destructive reflected interference}}.$$

Physical check

For the full lossless symmetric film, zero thickness removes the layer and the exact net reflection vanishes; two-ray amplitudes alone are only an approximation to that limit.