# 3.4 Interference in thin films  (Page 2/7)

 Page 2 / 7

If the film in [link] is a soap bubble (essentially water with air on both sides), then a phase shift of $\lambda \text{/}2$ occurs for ray 1 but not for ray 2. Thus, when the film is very thin and the path length difference between the two rays is negligible, they are exactly out of phase, and destructive interference occurs at all wavelengths. Thus, the soap bubble is dark here. The thickness of the film relative to the wavelength of light is the other crucial factor in thin-film interference. Ray 2 in [link] travels a greater distance than ray 1. For light incident perpendicular to the surface, ray 2 travels a distance approximately 2 t farther than ray 1. When this distance is an integral or half-integral multiple of the wavelength in the medium $\left({\lambda }_{n}=\lambda \text{/}n$ , where $\lambda$ is the wavelength in vacuum and n is the index of refraction), constructive or destructive interference occurs, depending also on whether there is a phase change in either ray.

## Calculating the thickness of a nonreflective lens coating

Sophisticated cameras use a series of several lenses. Light can reflect from the surfaces of these various lenses and degrade image clarity. To limit these reflections, lenses are coated with a thin layer of magnesium fluoride, which causes destructive thin-film interference. What is the thinnest this film can be, if its index of refraction is 1.38 and it is designed to limit the reflection of 550-nm light, normally the most intense visible wavelength? Assume the index of refraction of the glass is 1.52.

## Strategy

Refer to [link] and use ${n}_{1}=1.00$ for air, ${n}_{2}=1.38$ , and ${n}_{3}=1.52$ . Both ray 1 and ray 2 have a $\lambda \text{/}2$ shift upon reflection. Thus, to obtain destructive interference, ray 2 needs to travel a half wavelength farther than ray 1. For rays incident perpendicularly, the path length difference is 2 t .

## Solution

To obtain destructive interference here,

$2t=\frac{{\lambda }_{n2}}{2}$

where ${\lambda }_{n2}$ is the wavelength in the film and is given by ${\lambda }_{n2}=\lambda \text{/}{n}_{2}$ . Thus,

$2t=\frac{\lambda \text{/}{n}_{2}}{2}.$

Solving for t and entering known values yields

$t=\frac{\lambda \text{/}{n}_{2}}{4}=\frac{\left(500\phantom{\rule{0.2em}{0ex}}\text{nm}\right)\text{/}1.38}{4}=99.6\phantom{\rule{0.2em}{0ex}}\text{nm}.$

## Significance

Films such as the one in this example are most effective in producing destructive interference when the thinnest layer is used, since light over a broader range of incident angles is reduced in intensity. These films are called nonreflective coatings ; this is only an approximately correct description, though, since other wavelengths are only partially cancelled. Nonreflective coatings are also used in car windows and sunglasses.

## Combining path length difference with phase change

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. That is, for rays incident perpendicularly,

$2t={\lambda }_{n},2{\lambda }_{n},3{\lambda }_{n}\text{,…}\phantom{\rule{0.2em}{0ex}}\text{or}\phantom{\rule{0.2em}{0ex}}2t={\lambda }_{n}\text{/}2,3{\lambda }_{n}\text{/}2,5{\lambda }_{n}\text{/}2\text{,…}.$

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 can observe rainbow colors of constructive interference for various wavelengths as the thickness varies.

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