Let\'s apply Snell\'s law to the refraction of light across a water–air interfac

Question

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

a)

You are spearfishing from a boat and eye a large bass swimming below. It is apparently at an angle of 32.4 from the normal. At what angle should you aim your spear?

Express your answer in degrees to three significant figures.

Explanation / Answer

n1sin(theta1) = n2sin(theta2)

where n1 = R.I for water = 1.33, n2 = R.I for air = approx 1

theta1 = 90-60 = 30 , theta2 = to be found

so, 1.33sin30 = 1sin(theta2) so, theta2 = 41.682 degrees

now, n1 = 1, n2 = 1.33 , theta1 = to be found, theta2 = 32.4

so, 1sin(theta1) = 1.33sin32.4

so, theta1 = 45.45 degrees from the normal

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