Posted by **Andrea** on Wednesday, April 16, 2008 at 6:12pm.

A beam of light is incident on the surface of clear ice at 60 degrees with the normal. Find the angle between reflected and refracted ray.

theda 1=60

n1= 1 (air)

What would n2 be for clear ice? I know that index of refraction for water is 1.33. Would it also be 1.33?

Snell's law states that:

(sin theda 1/ sin theda 2) = (n2/n1)

sin 60 / sin theda 2 = 1.33/1(that is if n2 is 1.33)

theda 2 turns out to be 40.6 degrees. The question asks for the angle between reflected and refracted rays. Would that make 160 degrees as the final answer? Or is the final answer just 40.6 degrees?

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