1. crown glass has an index of refraction of 1.52. when light passes through the glass, at which angle to the normal will it be refracted along the surface of the glass-air boundary?
2.glycerine has an index of refraction of 1.473, and liquid water has an index of refraction of 1.333. what is the critical angle for light passing from glycerine to liquid water?

One

I'm assuming (perhaps incorrectly) that the ray is coming in perpendicularly to the surface of the crown glass. The question is a bit ambiguous, so the assumption may not be correct.

If I am correct.

Formula

n1*sin(θ1) = n2*sin(θ2)

Givens

n1 = 1

n2 = 1.52

θ1 = 90 degrees.

θ2 =  ?

Solution

1 * 1 = 1.52 * sin(θ2)

1/1.52 = sin(θ2)

0.6578 = sin(θ2)

θ2 = sin-1(0.6578)

θ2 = 41.14 degrees.

Two

The critical angle is the one that produces 90o as the exit point. The formula is the same for this question as it was for the first question

Givens

n1 = 1.473

n2 = 1.33

θ1 = ?

θ2 = 90o

Solution

1.473 sin(θ1) = 1.333 Sin(90)

sin(θ1)  =  1.333 / 1.473  * 1

sin(θ1) = 0.90496

θ1 = sin-1(0.90496)

θ1 = 64.82 degrees

tbh i dont know ether

your answer is, orbit around the nucleus