Question

A thin film of ethyl alcohol (n=1.36) is spread on a glass slide (n=1.5). When illuminated with white light it shows a color pattern in reflection. If the strongest light in normal incidence is green (500 nm), how thick the film is?

Answer 1

184 nm

Step-by-step explanation:

Given that ;

the refractive index (n) of ethyl alcohol = 1.36

wavelength (
`\lambda`) = 500 nm = 500 × 10⁻⁹ m

The difference in optical path length between the adjacent light rays can be expressed by the formula:

`\lambda = 2n_fdcos \theta_f`

where;

`\theta_f` = relative phase difference

d = distance between points of reflections

`n_f` = refractive index of the film

If we replace
`m \lambda_f` for
`\lambda` in the above equation; we have;

`m \lambda_f = 2n_fdcos \theta`

where; m represents the order of the pattern of the light rays;

and θ = 0; m = 1

Making d the subject of the formula; we have:

`d = (m \lambda_f)/(2n_fcos \theta_f)`

`d = (1* 500 nm)/(2*(1.36)cos(0))`

d = 184 nm

∴ The thickness of the film = 184 nm