Question

A thin film suspended in air is 0.448 μm thick and illuminated with white light that is incident perpendicularly on its surface. The index of refraction of the film is 1.37. At what wavelength will visible light reflected from the two surfaces of the film undergo fully constructive interference?

Answer 1

Answer:

λ= 2455 nm

Explanation:

Constructive interference occurs when the optical path difference is equal to an integer multiple of wavelength. Let's examine the reflection on each face

On the first surface we go from a lower medium refractive index (air) to one with a higher index (film), whereby a phase change of pi (180º) is introduced. On the second surface instead of the beginning of refraction, it is from a higher index to a smaller one, so there is no phase change, with this we can write the normal interference equation.

          2 t = (m + ½) $\lambda_{n}$

The term lann is the wavelength in the film that is related to the wavelength in the air by

          $\lambda_{n}$ = λ / n

         2t = (m + ½) λ / n

         λ = 2t n / (m + ½)

For the first constructive interference m = 0

       λ = 2 0.448 10⁻⁶ 1.37 / (0 + ½)

       λ = 2,455 10⁻⁶ m

       λ = 2,455 10⁻⁶ m (10⁹ nm / 1m)

        λ= 2455 nm

For m = 1

      λ = 2 0.448 10-6 1.37/ (1 + ½)

      λ = 0.81835 10-6 m

      λ = 818 nm

We can observe that the length and where it corresponds to ultraviolet light, so there is no constructive interference for visible light.