If a diver below the water's surface shines a light up at the bottom of the oil film, at what wavelength (as measured in water) would there be constructive interference in the light that reflects back downward

see that the entire spectrum of the visible is between the integers from 3 to 5 so only three wavelengths are reflected with constructive interference

Explanation:

This is an interference problem in thin films, the refractive index of water is 1.33 and the refractive index of oil is 1.5

Let's analyze the light beam path emitted by the diver.

* when the beam passes from the water to the oil with the highest refractive index, it has a phase change of 180º

* also the wavelength of light in a material medium changes

λ_n = λ / n

where λ_n is the wavelength in the material and λ the wavelength in the vacuum air and n the refractive index.

If we include these aspects, the constructive interference equation is

2t = (m + ½) λ_n

2nt = (m + ½) λ

let's apply this equation to our case

λ = 2nt / (m + ½)

The incidence of replacement of the oil with respect to water is

n = n_oil / n_water = 1.5 / 1.33

n = 1,128

let's calculate

λ = 2 1,128 t / (m + ½)

λ = 2,256 t / (m + ½)

In your statement you do not include the value of the oil layer that is the thin film, suppose a value to finish the calculation

t = 0.001 mm = 1 10⁻⁶ m

the formula remains

λ = 2,256 10⁻⁶ / (m + ½)

Let's find what values of m we have to cut light in the visible range (400 to 700) 10⁻⁹ m

m + ½ = 2,256 10⁻⁶ / λ

m = 2,256 10⁻⁶ / λ - ½

light purple lan = 400 10⁻⁹m

m = 2,256 10-6 / 400 10⁻⁹ - ½

m = 5.64 - 0.5

m = 5.14

m = 5

red light λ = 700 10⁻⁹m

m = 2,256 1-6 / 700 10⁻⁹ - ½

m = 3.22 - 0.5

m = 2.72

m = 3

we see that the entire spectrum of the visible is between the integers from 3 to 5 so only three wavelengths are reflected with constructive interference