Q32E - University Physics with Modern Physics

Question

What is the thinnest soap film (excluding the case of zero thickness) that appears black when illuminated with light with wavelength 480 nm? The index of refraction of the film is 1.33, and there is air on both sides of the film.

Step-by-Step Solution

Answer

The thickness of soap film is 180nm.

Step 1: Snell’s Law

We are given ${\lambda }_{iar}=480nm\&n=1.33$ . The first reflected ray experiences a phase change. The red circle indicates the phase change.

But the second reflected ray, as you see below, experiences no phase change at all.

Hence, we have two reflected rays with one phase change and we need them to interact destructively when the soap illuminated by a 480 nm-light beam.

So, the thickness of the soap bubble is given by

$$\begin{gathered} 2t=m{\lambda }_{soap} \\ t=\frac{m{\lambda }_{soap}}{2} \end{gathered}$$                                                       (1)

                              

Step 2: About Snell’s Law

We know, from Snell’s Law

$$\begin{gathered} n_1{\lambda }_1=n_2{\lambda }_2 \\ {n}_{air}{\lambda }_{air}=n_{soap}{\lambda }_{soap} \end{gathered}$$

 

Solving for ${\lambda }_{soap}$ and that $n_{air}=1$

${\lambda }_{soap}=\frac{{\lambda }_{air}}{n_{soap}}$

Substitute into above equation

$t=\frac{m{\lambda }_{air}}{2n_{soap}}$

Since the question asked to exclude the case the case of zero thickness, we will not use m = 1.0.