Question

Suppose you illuminate two thin slits by monochromatic coherent light in air and find that they produce their first interference minima at ± 35.09 ∘ 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 ± 19.38 ∘ instead. What is the index of refraction of this liquid?

Answer 1

Answer:

1.7323

Explanation:

To develop this problem, it is necessary to apply the concepts related to refractive indices and Snell's law.

From the data given we have to:

$n_{air}=1$

$\theta_{liquid} = 19.38\°$

$\theta_{air}35.09\°$

Where n means the index of refraction.

We need to calculate the index of refraction of the liquid, then applying Snell's law we have:

$n_1sin\theta_1 = n_2sin\theta_2$

$n_{air}sin\theta_{air} = n_{liquid}sin\theta_{liquid}$

$n_{liquid} = \frac{n_{air}sin\theta_{air}}{sin\theta_{liquid}}$

Replacing the values we have:

$n_{liquid}=\frac{(1)sin(35.09)}{sin(19.38)}$

$n_liquid = 1.7323$

Therefore the refractive index for the liquid is 1.7323