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

b.

explanation:

10.0

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