Question

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?

Answer 1

n = 1,724

Step-by-step 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