Question

A glass plate (n = 1.60) is covered with a thin, uniform layer of oil (n = 1.29). A light beam of variable wavelength from air is incident normally on the oil surface. Observation of the reflected beam shows constructive interference at 511 nm. Determine the minimum non-zero thickness of the oil film.

Answer 1

Answer:

The thickness of the oil film is 198 nm.

Explanation:

Given that,

Refractive index of glass plate = 1.60

Refractive index of oil = 1.29

Wavelength = 511 nm

We need to calculate the thickness of the oil film

Using formula of path difference

$2nt=k\lambda$

$t=\dfrac{k\times\lambda}{2n}$

Where, n = refractive index

t = thickness

$\lambda$ = wavelength

Put the value into the formula

$t=\dfrac{1\times511\times10^{-9}}{2\times1.29}$

$t=198\times10^{-9}\ m$

$t= 198\ nm$

Hence, The thickness of the oil film is 198 nm.