# Undefined Angle of Incidence?

1. Oct 13, 2008

### Suavez

1. The problem statement, all variables and given/known data
A 1.0-cm-thick layer of water stands on a horizontal slab of glass. Light from a source within the glass is incident on the glass-water boundary.
What is the maximum angle of incidence for which the light ray can emerge into the air above the water?

2. Relevant equations
Snell's Law: nisin$$\theta$$i = ntsin$$\theta$$t

3. The attempt at a solution
I drew a diagram like this:

air (n = 1.00)
________________ light ray$$\uparrow$$
water (n = 1.33) light ray$$\uparrow$$
________________ light ray$$\uparrow$$
glass (n = 1.50) light ray$$\uparrow$$

The ray of light travels from the glass upward.

nglasssin90 = nwatersin$$\theta$$water

nwatersin$$\theta$$water = nairsin$$\theta$$air

Therefore, using equality of alternate angles:

nglasssin90 = nairsin$$\theta$$air

(1.50)*(sin90) = (1.00)*(sin$$\theta$$air)

Solution: undefined
What is the maximum angle of incidence if it is undefined?

2. Oct 13, 2008

### Rake-MC

Think about where Snell's law measures $$\theta$$ from.

3. Oct 13, 2008

### Suavez

Sorry buddy, that was no help. After spending about an hour on that problem, I think I deserve a little more than that. I don't expect anyone to DO the work for me but come on now, let's be serious.

My question is: What is the maximum angle of incidence for which the light ray can emerge into the air above the water?

4. Oct 13, 2008

### Rake-MC

Yes I am aware of that, I'm showing you what you've done wrong.
For a ray of light perpendicular to the surface, you wrote: $$n_1sin(90)$$
Why did you write sin(90)?
Think of where snell's law measures $$\theta$$ from.

5. Oct 13, 2008

### Suavez

If it is incident it's perpendicular and if it's perpendicular it's 90 degrees.

6. Oct 13, 2008

### Rake-MC

90 degrees from the tangent.
doesn't Snell's law measure $$\theta$$ from the normal?

7. Oct 13, 2008

### Suavez

Ok, so that makes it 45 degrees?