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

Answer:

546 nm

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

$2 n_{oil} t = m \lambda$

$2 (1.30)(210\times 10^{-9})= (1) \lambda$

$\lambda = 5.46\times 10^{-7}$

$\lambda = 546$ nm