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A glass window is coated with a transparent film of refractive index n2 =1.25. This film will cause destructive interference in the reflected light when light of wavelength λ = 550nm is incident normally on the window. What is the minimum thickness of film required to produce this anti-reflection coating?

Respuesta :

Answer:

0.00000011 m

Explanation:

n = 0 for minimum thickness

t = Thickness

[tex]n_2[/tex] = Refractive index of the film = 1.25

[tex]\lambda[/tex] = Wavelength = [tex]550\times 10^{-9}\ m[/tex]

We have the formula

[tex]2n_2t=(2n+1)\dfrac{\lambda}{2}[/tex]

[tex]\\\Rightarrow t=(2n+1)\dfrac{\lambda}{4n_2}\\\Rightarrow t=(2\times 0+1)\dfrac{550\times 10^{-9}}{4\times 1.25}\\\Rightarrow t=0.00000011\ m[/tex]

The minimum thickness of the film is 0.00000011 m

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