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The very top of the film shows no reflected light, indicating that the thickness is less than 1/4 the wavelength of visible light within the film, since the phase change on reflection for the beam from the first surface of the film is Ï radians and the phase change on reflection from the second surface of the film is 0 radians, making the interfering rays mostly out of phase from that part of the film. Given this information, what is the approximate thickness of the film at the first bright fringe counting from the top of the film, in terms of the wavelength of light in vacuum λ, and the refractive index of the film n?

Respuesta :

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To solve this problem it is necessary to apply the concepts related to interference in thin film.

By definition we know that the film thickness is given by the equation

[tex]2nt = \frac{m\lambda}{2}[/tex]

Where,

m = Any integer which represents the number of repetition of the spectrum (number of fringe)

n = Index of refraction of the film

[tex]\lambda =[/tex] Wavelength

t = Thickness

For first bright fringe we have then that m = 1, re-arrange to find t,

[tex]2nt = \frac{m\lambda}{2}[/tex]

[tex]t = \frac{1\lambda}{2*2n}[/tex]

[tex]t = \frac{\lambda}{4n}[/tex]

Therefore the thickness of film would be [tex]\lambda /4[/tex]