Question

A thin layer of oil with index of refraction no = 1.47 is floating above the water. The index of refraction of water is nw = 1.3. The index of refraction of air is na = 1. A light with wavelength λ = 775 nm goes in from the air to oil and water.

Answer 1

Answer:

A thin layer of oil with index of refraction ng = 1.47 is floating above the water. The index of refraction of water is nw = 1.3. The index of refraction of air is na= 1. A light with wavelength λ = 775 nm goes in from the air to oil and water.

 Part (a) Express the wavelength of the light in the oil,  λ₀, in terms of  λ  and n⁰ (b) Express the minimum thickness of the film that will result in destructive interference, t min, in terms of  λ o

(c) Express tmin in terms of  λ  and no.

(d) Solve for the numerical value of tmin in nm.

Explanation:

n₀ = 1.47

refraction of water = 1.3

refraction of air = 1

wavelength λ = 775 nm

(a) wavelength of light in water ⇒  λ₀ = λ / n₀

(b) minimum thickness of the film that will result in destructive interference

t(min) = λ₀ / 2

(c) the express t(min)

t =  λ /2n₀

(d) the thickness is

t = 775 / 2(1.47)

= 263.61 nm