Question

An oil slick on water is 103 nm thick and illuminated by white light incident perpendicular to its surface. What color does the oil appear (what is the most constructively reflected wavelength, in nanometers), given its index of refraction is 1.41

Answer 1

The wavelength of the reflected light is 581 nm, which is between yellow and orange, Therefore, the color that will appear in the oil will be between yellow and orange.

Step-by-step explanation:

The relation between wavelength and refractive medium is equal to:

λn = λ/n (eq. 1)

Where λ is the wavelength of light in vacuum and n is the refactive index of medium. One phase change is:

2t = λn/2

Replacing in equation 1:

λ = 4nt

n = 1.41

t = 103 nm

λ = 4 * 1.41 * 103 =580.92 nm = 581 nm

The wavelength of the reflected light is 581 nm, which is between yellow and orange, Therefore, the color that will appear in the oil will be between yellow and orange.

Answer 2

The most constructive reflected wavelength would be 576.8 nm

Step-by-step explanation:

Given the thickness of oil t = 103 nm

refractive index n = 1.41

The refractive index is related to wavelength in a particular medium with the relationship below;

λ
`_(m)` =λ / n ...............1

here λ
`_(m)` is the wavelength in the medium (oil)

λ is the wavelength of light in vacuum

n is the refractive index

most constructively reflected wavelength will occur at the point of Most constructive interference, which is at;

2 t = λ
`_(m)` / 2 .........1

Substituting the value that we have for λ
`_(m)` in equation 1 into equation 2

2 t = (λ / n)/2

4t = λ / n

λ = 4nt

λ = 4 x 1.41 x 103 nm

λ = 576.8 nm

Therefore the most constructive reflected wavelength would be

576.8 nm