Ask Question
7 November, 12:16

Suppose you illuminate two thin slits by monochromatic coherent light in air and find that they produce their first interference minima at / pm 35.09 degrees on either side of the central bright spot. You then immerse these slits in a transparent liquid and illuminate them with the same light. Now you find that the first minima occur at / pm 19.48 degrees instead. a) What is the index of refraction of this liquid?

+4
Answers (1)
  1. 7 November, 12:26
    0
    n = 1,724

    Explanation:

    The interference pattern for two slits for the case of constructive interference is described by the expression

    d sin θ = m λ

    Where d is the separation of the slits, lm the wavelength and m an integer

    When the slits are submerged the transparent liquid the wavelength of the light changes according to the relationship

    λₙ = λ₀ / n

    Where λ₀ is the wavelength in vacuum or air and n is the index of refraction of the material

    Let's apply these relationships to our case, write the equations for the two situations

    Air

    d sin θ = m λ₀

    Liquid

    d sin θ₂ = m λₙ

    d sin θ₂ = m λ₀ / n

    Let's pass the variables to the left

    sin θ = m λ₀ / d

    n sin θ₂ = m λ₀ / d

    Let's match the two equations

    sin θ = n syn θ₂

    n = sin θ / sin θ₂

    Let's calculate

    n = sin 35.09 / sin 19.48

    n = 1,724
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “Suppose you illuminate two thin slits by monochromatic coherent light in air and find that they produce their first interference minima at ...” in 📗 Physics if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers