Light of wavelength 500 nm is incident on a thin sheet of plastic film (n= 1.50) in air. What is the thinnest the sheet can be so that the reflected light from the surfaces undergoes destructive interference?

t = 1.6667 10⁻⁷ m

Explanation:

When light falls on a thin film on the surface we have two phenomena

- a change in phase when the beam passes to a medium with a higher index

- on the other hand, in the wavelength, due to the refractive index of the medium

       λ = λ₀ / n

The second ray reflected from the other surface has no phase change

So the path difference between the two rays for destructive interference is

           2t = m   λ₀ / n

           t = m   λ₀ / 2n

Let's calculate

          t = 1  500 10⁻⁹ / 2 1.50

          t = 166.67 10⁻⁹ m

          t = 1.6667 10⁻⁷ m

answer: good job it's all correct !

explanation:

you'd use the equation that establishes a relationship between  energy  and  wavelength, which looks like this

e=h⋅cλ, where

λ  - the wavelength of the photon; c  - the speed of light, usually given as  3⋅108m/s; h  -  planck's constant, equal to  6.626⋅10−34j s

plug the value you were given for the energy of the photon in the above equation to get the wavelength

e=h⋅cλ⇒λ=h⋅ce

λ=6.626⋅10−34js⋅3⋅108ms−1.257⋅10−24j=1.581⋅10−1m