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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 size 12{2t} {} farther than ray 1. When this distance is an integral or half-integral multiple of the wavelength in the medium ( λ n = λ / n size 12{λ rSub { size 8{n} } =λ/n} {} , where λ size 12{λ} {} is the wavelength in vacuum and n size 12{n} {} is the index of refraction), constructive or destructive interference occurs, depending also on whether there is a phase change in either ray.

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.

Strategy

Refer to [link] and use n 1 = 100 size 12{n rSub { size 8{1} } ="100"} {} for air, n 2 = 1 . 38 size 12{n rSub { size 8{2} } =1 "." "38"} {} , and n 3 = 1 . 52 size 12{n rSub { size 8{3} } =1 "." "52"} {} . Both ray 1 and ray 2 will have a λ / 2 size 12{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 size 12{2t} {} .

Solution

To obtain destructive interference here,

2 t = λ n 2 2 , size 12{2t= { {λ rSub { size 8{n rSub { size 6{2} } } } } over {2} } ,} {}

where λ n 2 size 12{λ rSub { size 8{n rSub { size 6{2} } } } } {} is the wavelength in the film and is given by λ n 2 = λ n 2 size 12{λ rSub { size 8{n rSub { size 6{2} } } } = { {λ} over {n rSub {2} } } } {} .

Thus,

2 t = λ / n 2 2 . size 12{2t= { { {λ} slash {n rSub { size 8{2} } } } over {2} } } {}

Solving for t size 12{t} {} and entering known values yields

t = λ / n 2 4 = ( 550 nm ) / 1.38 4 = 99.6 nm. alignl { stack { size 12{t= { { {λ} slash {n rSub { size 8{2} } } } over {4} } = { { { \( "550"`"nm" \) } slash {1 "." "38"} } over {4} } } {} #="99" "." 6`"nm" {} } } {}

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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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, 2 t = λ n , n , n , or 2 t = λ n / 2, n / 2, n / 2, . 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?

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Source:  OpenStax, College physics for ap® courses. OpenStax CNX. Nov 04, 2016 Download for free at https://legacy.cnx.org/content/col11844/1.14
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