Question

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?

Answer 1

546 nm

Step-by-step explanation:

`n_(oil)` = Index of refraction of oil = 1.30

`n_(water)` = Index of refraction of water = 1.33

`t` = thickness of the oil = 210 nm = 210 x 10⁻⁹ m

`\lambda` = wavelength of light = ?

`m` = order = 1

For reflection , the necessary condition is

`2 n_(oil) t = m \lambda`

`2 (1.30)(210* 10^(-9))= (1) \lambda`

`\lambda = 5.46* 10^(-7)`

`\lambda = 546` nm