Question

Working on your car you spill oil (index of refraction = 1.55) on the ground into a puddle of water (n = 1.33). You notice a rainbow pattern appear across the oil slick. You realize you can calculate where constructive and destructive interference occurs based on the thickness of the oil slick. (Assume that the average wavelength is 524 nm.)

Required:
a. What are the first three thicknesses necessary for constructive interference?
b. What are the first three thicknesses necessary for destructive interference?

Answer 1

For constructive interference

The first three thicknesses are

`t_1 =85nm`

`t_2 =254nm`

`t_3 =423nm`

For destructive interference

The first three thicknesses are

`t_D__(1)} = 169nm`

`t_D__(2)} = 338nm`

`t_D__(3)} = 507nm`

Step-by-step explanation:

The diagram for this question is shown on the first uploaded image

From the question we are told that

The refractive index of oil is
`n_o = 1.55`

The refractive index of water is
`n_w = 1.33`

The average wavelength is
`\lambda_a = 524 nm = 524*10^(-9)m`

For constructive interference the thickness is mathematically represented as

`t = (m + (1)/(2) ) (\lambda )/(2 n_w)`

Where m is the order of the interference with a value from
`m = 0,1 , -1 , 2 , -2 , ...`

For m = 0

The thickness of the oil slick would be

`t_1 = 0 + (1)/(2) * (\lambda )/(2 * n_w)`

Substituting value

`t_1 = 0 + (1)/(2) * (524 *10^(-9))/(2 * 1.55)`

`t_1 =85nm`

For m = 1

The thickness of the oil slick would be

`t_2 = 1 + (1)/(2) * (\lambda )/(2 * n_w)`

Substituting value

`t_2 = 1 + (1)/(2) * (524 *10^(-9))/(2 * 1.55)`

`t_2 =254nm`

For m = 2

The thickness of the oil slick would be

`t_3 = 2 + (1)/(2) * (\lambda )/(2 * n_w)`

Substituting value

`t_3= 2 + (1)/(2) * (524 *10^(-9))/(2 * 1.55)`

`t_3 =423nm`

For destructive interference the thickness is mathematically represented as

`t = (m \lambda )/(2 n_w)`

For m = 1

`t_D__(1)} = (\lambda )/(2 * n_w)`

Substituting value

`t_D__(1)} = (524 *10^(-9) )/(2 * 1.55)`

`t_D__(1)} = 169nm`

For m = 2

`t_D__(2)} = (\lambda )/(2 * n_w)`

Substituting value

`t_D__(2)} = ( 2 * 524 *10^(-9) )/(2 * 1.55)`

`t_D__(2)} = 338nm`

For m = 3

`t_D__(3)} = (\lambda )/(2 * n_w)`

Substituting value

`t_D__(3)} = ( 3 * 524 *10^(-9) )/(2 * 1.55)`

`t_D__(3)} = 507nm`