Question

What does this diagram indicate about the indexes of refraction of the three mediums?

Answer 1

The index for Medium 2 is larger than the index for Medium 1.

The index for Medium 2 is equal to the index for Medium 3.

The index for Medium 2 is equal to the index for Medium 1.

The index for Medium 3 is larger than the index for Medium 2.

Answer 2

When light goes to a medium with a higher index of refraction, the light is bent toward the surface normal. When it moves into a medium with a lower index of refraction, the light bends away from the surface normal. The higher the index, the more the light is bent.

Looking at your picture, light in the medium with n=n3 is bent the most toward the surface normal. Light in the medium with n=n2 bends away from the surface normal, so we know that n1>n2.

We see the last arrow (the last red arrow on the bottom) is also bent away from the surface normal, so we know that n3>n1. Putting all that together we know that n3>n1>n2.

Answer 3

So the answer could be A or D. Right?

Answer 4

@PsiSquared

Answer 5

use snells law; remember that angles are measured with respect to the verticle.

n1*sin(angle 1) = n2*sin(angle 2)

Looking at diagram:

angle 2 is greater then angle 1 so sin(angle 2)>sin(angle 1) and therefore n2<n1 or else the equation would not hold true.

Now; sin (angle 3) < sin (angle 2) and < sin (angle 1)

therefore n3>n2 and n2<n1!!!

Answer 6

Only D is the correct answer. None of the indices of refraction are equal. If the indices of refraction of two materials were equal, light would not be bent passing from one material into the other. As for the first answer, you can see that I showed that n2 is actually the smallest index of refraction.