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?
To find the angle between the reflected and refracted rays, we first need to determine the angle of reflection.
The angle of reflection is equal to the angle of incidence, which in this case is 60 degrees. So, the angle of reflection is also 60 degrees.
Now, to find the angle between the reflected and refracted rays, we can use the fact that the sum of these two angles is equal to 180 degrees since they form a straight line. Let's call this angle x.
Therefore, x + 60 = 180.
Solving for x, we find that x = 120 degrees.
So, the final answer is that the angle between the reflected and refracted rays is 120 degrees.