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Ipatiy [6.2K]
3 years ago
6

The critical angle for a certain air-liquid surface is 47.7°. What is the index of refraction of the liquid? Round to the neares

t hundredth. nair = 1.00 Answer
Physics
1 answer:
KengaRu [80]3 years ago
5 0

Answer:

The index of refraction of the liquid is 1.35.

Explanation:

It is given that,

Critical angle for a certain air-liquid surface, \theta_1=47.7^{\circ}

Let n₁ is the refractive index of liquid and n₂ is the refractive index of air, n₂ =1

Using Snell's law for air liquid interface as :

n_1\ sin\theta_1=n_2\ sin(90)

n_1\ sin(47.7)=1

n_1=\dfrac{1}{sin(47.7)}

n_1=1.35

So, the index of refraction of the liquid is 1.35. Hence, this is the required solution.

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<h3>What is a vector?</h3>

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A particle is moving with velocity v(t) = t2 _ 9t + 18 with distance, s measured in meters, left or right of zero, and t measure
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