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Ksenya-84 [330]
3 years ago
8

Let's apply Snell's law to the refraction of light across a water–air interface. Suppose you kneel beside the fishpond in your b

ackyard and look at one of the fish. You see it by sunlight that reflects off the fish and refracts at the water–air interface. If the light from the fish to your eye strikes the water–air interface at an angle of 60.0∘∘ to the interface, what is the angle of refraction of the ray in the air?

Physics
1 answer:
Nataliya [291]3 years ago
4 0

Answer:

The angle of refraction is 41.68°.

Explanation:

The refractive index for water is n_2 = 1.333, and for air n_1 = 1.00: the angle of light with the normal is 90^o-60^o = 30^o; therefore Snell's law gives

n_1sin(\theta_1)= n_2sin(\theta_2)

1.00*sin(\theta_1) = 1.33 sin(30^o)

sin (\theta_1) = \dfrac{1.33sin(30^o)}{1.00}

sin (\theta_1) = 0.665

\theta _1 = sin^{-1}(0.665)

\boxed{\theta_1 = 41.68^o}

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