7.7 Thin film interference  (Page 2/6)

Thin film interference is most constructive or most destructive when the path length difference for the two rays is an integral or half-integral wavelength, respectively. That is, for rays incident perpendicularly, $2t={\lambda }_n,{\mathrm{2\lambda }}_n,{\mathrm{3\lambda }}_n,\dots$ or $2t={\lambda }_n/\mathrm{2,}{\mathrm{3\lambda }}_n/\mathrm{2,}{\mathrm{5\lambda }}_n/\mathrm{2,}\dots$ . To know whether interference is constructive or destructive, you must also determine if there is a phase change upon reflection. Thin film interference thus depends on film thickness, the wavelength of light, and the refractive indices. For white light incident on a film that varies in thickness, you will observe rainbow colors of constructive interference for various wavelengths as the thickness varies.

Soap bubbles: more than one thickness can be constructive

(a) What are the three smallest thicknesses of a soap bubble that produce constructive interference for red light with a wavelength of 650 nm? The index of refraction of soap is taken to be the same as that of water. (b) What three smallest thicknesses will give destructive interference?

Strategy and Concept

Use [link] 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 ${\mathrm{2\lambda }}_n$ .