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Assoli18 [71]
4 years ago
11

A light beam is traveling through an unknown substance. When it strikes a boundary between that substance and the air (nair≈1),

the angle of reflection is 29.0∘ and the angle of refraction is 39.0∘. What is the index of refraction n of the substance?
Physics
1 answer:
SashulF [63]4 years ago
6 0

Answer:

0.7707

Explanation:

From Snell's law,

n(1) * sin θ1 = n(2) * sinθ2

Where n(1) = refractive index of air = 1.0003

θ1 = angle of incidence

n(2) = refractive index of second substance

θ2 = angle of refraction

The angle of reflection through the unknown substance is the same as the angle of incidence of air. Thus this means that θ1 = 29°

=> 1.0003 * sin29 = n(2) * sin39

n(2) = (1.0003 * sin29) / sin39

n(2) = 0.7707

Explanation:

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