Question

A thin film of oil with a thickness of 90 nm rests on top of a pool of water. When white light incident on the film is reflected, what color (wavelength of the first order) is seen? The refractive index of the oil is noil =1.45 and of water is nwater=1.33

Answer 1

m=0, λ₀ = 522 10⁻⁹ m

Step-by-step explanation:

In this case we can see that the first boy M The reflected light suffers an interference phenomenon, let's analyze what happens on each surface

* When light passes from a medium with a lower refractive index to a medium with a higher refractive index, it undergoes a phase change of pi rad (180º).

This occurs when passing from air to oil, but not at the oil-water interface.

* within the material the wavelength changes by the refractive index

λₙ = λ₀ / n

therefore introducing this into the constructive interference equation and assuming almost perpendicular incidence remains

2 t = (m + ½) λₙ

2t = (m + ½) λ₀ / n

λ₀ = 2t n / (m + ½)

light in the first order (m = 1)

λ₀ = 2 90 10⁻⁹ 1.45 / (1+ ½)

λ₀ = 174 10⁻⁹ m

the light for zero order (m = 0)

λ₀ = 2 90 10⁻⁹ 1.45 / (0+ ½)

λ₀ = 522 10⁻⁹ m

this radiation in the visible range in the green region