Thin Film Interference

Thin film interference explains the rainbow colors in soap bubbles, oil slicks, and anti-reflective coatings. It arises because light reflects from both the top and bottom surfaces of a thin transparent film.

Phase Shifts on Reflection

A key subtlety: light undergoes a phase shift of $\pi$ (half a wavelength) when it reflects off a medium with a higher refractive index. No phase shift occurs when reflecting off a lower-index medium.

Condition for Constructive Interference

For a film of thickness $t$ and refractive index $n$ surrounded by air ($n_{\text{air}} = 1$), light reflects from the top surface (with phase shift) and the bottom surface (no phase shift). The one net phase shift from reflection means the condition for constructive interference is:

$2nt = \left(m - \frac{1}{2}\right)\lambda \quad m = 1, 2, 3, \ldots$

This simplifies for $m = 1$ (minimum thickness):

$t_{\min} = \frac{\lambda}{4n}$

General Thickness for Constructive Interference

For order $m \geq 1$:

$t = \frac{m \lambda}{2n} - \frac{\lambda}{4n} = \frac{(2m - 1)\lambda}{4n}$

However, a cleaner convention used in many textbooks counts the minimum as order $m = 1$ giving:

$t_{\min} = \frac{\lambda}{4n}$

And subsequent maxima at:

$t_m = \frac{m \lambda}{2n} \quad m = 1, 2, 3, \ldots$

(where $m=1$ gives the second constructive maximum).

Example: For $\lambda_0 = 550\,\text{nm}$ in glass ($n = 1.5$):

$t_{\min} = \frac{550}{4 \times 1.5} \approx 91.7\,\text{nm}$

Your Task

Implement the two formulas. Wavelength inputs are in nm; return thickness in nm.