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A thin layer of oil of refractive index 1.30 is spread on the surface of water (n = 1.33). If the thickness of the oil is 210 nm, then what is the wavelength of light in air that will be predominantly reflected from the top surface of the oil?

Respuesta :

Answer:

546 nm

Explanation:

[tex]n_{oil}[/tex] = Index of refraction of oil = 1.30

[tex]n_{water}[/tex] = Index of refraction of water = 1.33

[tex]t[/tex] = thickness of the oil = 210 nm = 210 x 10⁻⁹ m

[tex]\lambda[/tex] = wavelength of light = ?

[tex]m[/tex] = order = 1

For reflection , the necessary condition is

[tex]2 n_{oil} t = m \lambda[/tex]

[tex]2 (1.30)(210\times 10^{-9})= (1) \lambda[/tex]

[tex]\lambda = 5.46\times 10^{-7}[/tex]

[tex]\lambda = 546[/tex] nm