Question

In your research lab, a very thin, flat piece of glass with refractive index 2.30 and uniform thickness covers the opening of a chamber that holds a gas sample. The refractive indexes of the gases on either side of the glass are very close to unity. To determine the thickness of the glass, you shine coherent light of wavelength λ0 in vacuum at normal incidence onto the surface of the glass. When λ0= 496 nm, constructive interference occurs for light that is reflected at the two surfaces of the glass. You find that the next shorter wavelength in vacuum for which there is constructive interference is 386 nm.

Answer 1

λ₀= 495.88 nm

Step-by-step explanation:

To analyze this constructive interference interference experiment by reflection, let's look at two important aspects:

* when a ray of light passes from a medium with a lower index, they refact to another medium with a higher index, the reflected ray has a phase difference of pyres

* When a beam penetrates a material medium, the wavelength of the radiation changes according to the refractive index of the material.

λₙ = λ₀ / n

when we introduce these aspects in the expression of contributory interference, it remains

2 d sin θ = (m + ½) λ₀ / n

In general, reflection phenomena are measured at an almost normal angle, whereby θ = π/2 and sin θ = 1

2 d = (m +1/2) λ₀/ n

2n d = (m + ½) λ₀

Let's apply this expression to our case

d = (m + ½) λ₀ / 2n

Suppose we measure on the first interference, this is m = 0

d = ½ λ₀ / 2n

let's calculate

d = ½ 496 10⁻⁹ / (2 2.30)

d = 53.9 10-9 m

This is the thickness of the glass, the next wavelength that gives constructive interference is

λ₀ = 2 n d / (m + ½)

let's calculate

λ₀ = 2 2.3 5.39 10-8 / (1 + ½)

λ₀= 4.9588 10-7 m

λ₀= 495.88 nm