Question

(n = 1.33). When visible light (400nm – 700 nm) shines normally on the film an observed from above, the bright reflected light looks kind of "purplish" with red light of wavelength 650 nm mixed some blue giving it that appearance and "greenish" light of wavelength 520 nm is completely destroyed. Determine the thickness of the acetone film. (you only need to worry about the red and green wavelengths, not the blue)

Answer 1

Complete Question

A thin film in air is made by putting a thin layer of acetone (n = 1.25) on a layer of water

(n = 1.33). When visible light (400 nm – 700 nm) shines normally on the film an observed from above, the bright reflected light looks kind of “purplish” with red light of wavelength 650 nm mixed some blue giving it that appearance and “greenish” light of wavelength 520 nm is completely destroyed. Determine the thickness of the acetone film. (you only need to worry about the red and green wavelengths, not the blue)

Answer:

The thickness of acetone is
`d = 104 nm`

Step-by-step explanation:

A diagram showing this process is shown on the first uploaded image

From the question we are told that

The refractive index of acetone is
`n_a =1.25`

The refractive index of water is
`n_w = 1.33`

The wavelength of the reflected light is
`\lambda_r = 650nm = 650 *10^(-9)m`

The wavelength of the destroyed light is
`\lambda_g = 520nm = 520 *10^(-9)m`

Looking at the given data we can see that the

`n_a < n_w`

This means that the light which the acetone-water layer would reflect will have a phase shift of
`\pi`

Again this make us to understand that the light reflected at the acetone layer will also have a phase shift of
`\pi`

Since they would be having the same phase shift the two light would interfere

For interference the condition for minima is mathematically represented as

`2 n_a d = (m + (1)/(2) ) \lambda_g`

Where d is the thickness of acetone

`d = (\lambda_g)/(4 n_a)`

Substituting values

`d = (520 *10^(-9))/(4 * 1.25)`

`d = 104 *10^(-9)m`

`d = 104 nm`