Maximum Angle of Incidence for Glass-Water Boundary?

[Suavez]

Homework Statement

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?

Homework Equations

Snell's Law: n_isin\theta_i = n_tsin\theta_t

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.

n_glasssin90 = n_watersin\theta_water

n_watersin\theta_water = n_airsin\theta_air

Therefore, using equality of alternate angles:

n_glasssin90 = n_airsin\theta_air


(1.50)*(sin90) = (1.00)*(sin\theta_air)

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

 

[Rake-MC]

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

 

[Suavez]

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

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?

 

[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.

 

[Suavez]

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.

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

 

[Rake-MC]

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

 

[Suavez]

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

Ok, so that makes it 45 degrees?