Example 27.6 from College Physics, 27.7 Thin Film Interference

Calculating Non-reflective Lens Coating Using Thin Film Interference

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 that 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? The index of refraction of glass is 1.52.

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Strategy
Refer to Figure 27.33 and use $n_{1} = 1.00$ for air, $n_{2} = 1 . 38$ , and $n_{3} = 1 . 52$ . Both ray 1 and ray 2 will have a $λ / 2$ shift upon reflection. Thus, to obtain destructive interference, ray 2 will need 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,
$2 t = \frac{λ_{n_{2}}}{2},$
27.33

where $λ_{n_{2}}$ is the wavelength in the film and is given by $λ_{n_{2}} = \frac{λ}{n_{2}}$ .

Thus,

$2 t = \frac{λ / n_{2}}{2} .$
27.34

Solving for $t$ and entering known values yields

$\begin{array}{l} t & = & \frac{λ / n_{2}}{4} = \frac{(550 nm) / 1.38}{4} \\ = & 99.6 nm. \end{array}$
27.35
Discussion
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 will be reduced in intensity. These films are called non-reflective coatings; this is only an approximately correct description, though, since other wavelengths will only be partially cancelled. Non-reflective coatings are used in car windows and sunglasses.

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