Ask Question
11 August, 18:29

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.

+1
Answers (1)
  1. 11 August, 19:05
    0
    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
Know the Answer?
Not Sure About the Answer?
Get an answer to your 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 ...” in 📙 Physics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers