Soap Bubbles: More Than One Thickness can be Constructive

Use Figure 27.33 to visualize the bubble. Note that $n_1=n_3=1\text{.}\text{00}$ for air, and $n_2=1\text{.}\text{333}$ for soap (equivalent to water). There is a $\lambda /2$ shift for ray 1 reflected from the top surface of the bubble, and no shift for ray 2 reflected from the bottom surface. To get constructive interference, then, the path length difference ($2t$) must be a half-integral multiple of the wavelength—the first three being ${\lambda }_n/\mathrm{2,}{\mathrm{3\lambda }}_n/2$, and ${\mathrm{5\lambda }}_n/2$. To get destructive interference, the path length difference must be an integral multiple of the wavelength—the first three being $\mathrm{0,}{\lambda }_n$, and ${\mathrm{2\lambda }}_n$.

Constructive interference occurs here when

The smallest constructive thickness $t_{\text{c}}$ thus is

The next thickness that gives constructive interference is ${t\prime }_{\text{c}}={\mathrm{3\lambda }}_n/4$, so that

Finally, the third thickness producing constructive interference is ${t\prime \prime }_{\text{c}}\le {\mathrm{5\lambda }}_n/4$, so that

For destructive interference, the path length difference here is an integral multiple of the wavelength. The first occurs for zero thickness, since there is a phase change at the top surface. That is,

The first non-zero thickness producing destructive interference is

Substituting known values gives

Finally, the third destructive thickness is $2{t\prime \prime }_{\text{d}}={\mathrm{2\lambda }}_n$, so that

If the bubble was illuminated with pure red light, we would see bright and dark bands at very uniform increases in thickness. First would be a dark band at 0 thickness, then bright at 122 nm thickness, then dark at 244 nm, bright at 366 nm, dark at 488 nm, and bright at 610 nm. If the bubble varied smoothly in thickness, like a smooth wedge, then the bands would be evenly spaced.