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Picture is a schematic drawing of the light undergoing interference by a thin film. Wave reflected from the top of the film is inverted; wave reflected from the bottom of the film is not inverted; refracted waves are not inverted.
Reflection at an interface for light traveling from a medium with index of refraction n 1 to a medium with index of refraction n 2 , n 1 < n 2 , causes the phase of the wave to change by π radians.

If the film in [link] is a soap bubble (essentially water with air on both sides), then a phase shift of λ / 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 ( λ n = λ / n , where λ 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.


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 λ / 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 .


To obtain destructive interference here,

2 t = λ n 2 2

where λ n 2 is the wavelength in the film and is given by λ n 2 = λ / n 2 . Thus,

2 t = λ / n 2 2 .

Solving for t and entering known values yields

t = λ / n 2 4 = ( 500 nm ) / 1.38 4 = 99.6 nm .


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.

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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,

2 t = λ n , 2 λ n , 3 λ n ,… or 2 t = λ n / 2 , 3 λ n / 2 , 5 λ 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 can observe rainbow colors of constructive interference for various wavelengths as the thickness varies.

Questions & Answers

A Pb wire wound in a tight solenoid of diameter of 4.0 mm is cooled to a temperature of 5.0 K. The wire is connected in series with a 50-Ωresistor and a variable source of emf. As the emf is increased, what value does it have when the superconductivity of the wire is destroyed?
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how does colour appear in thin films
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derivation of lateral shieft
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Refraction does not occur in concave mirror. If refraction occurs then I don't know about this.
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normal distribution of errors
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give any fix value to wave length
40 cm into change mm
Arhaan Reply
40cm=40.0×10^-2m =400.0×10^-3m =400mm. that cap(^) I have used above is to the power.
i.e. 10to the power -2 in the first line and 10 to the power -3 in the the second line.
there is mistake in my first msg correction is 40cm=40.0×10^-2m =400.0×10^-3m =400mm. sorry for the mistake friends.
40cm=40.0×10^-2m =400.0×10^-3m =400mm.
this msg is out of mistake. sorry friends​.
what is physics?
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why we have physics
Anil Reply
because is the study of mater and natural world
because physics is nature. it explains the laws of nature. some laws already discovered. some laws yet to be discovered.
physics is the study of non living things if we added it with biology it becomes biophysics and bio is the study of living things tell me please what is this?
physics is the study of matter,energy and their interactions
all living things are matter
why rolling friction is less than sliding friction
thanks buvanas
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Source:  OpenStax, University physics volume 3. OpenStax CNX. Nov 04, 2016 Download for free at http://cnx.org/content/col12067/1.4
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