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## 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

_{i}sin[tex]\theta[/tex]

_{i}= n

_{t}sin[tex]\theta[/tex]

_{t}

## The Attempt at a Solution

I drew a diagram like this:

air (n = 1.00)

________________ light ray[tex]\uparrow[/tex]

water (n = 1.33) light ray[tex]\uparrow[/tex]

________________ light ray[tex]\uparrow[/tex]

glass (n = 1.50) light ray[tex]\uparrow[/tex]

The ray of light travels from the glass upward.

n

_{glass}sin90 = n

_{water}sin[tex]\theta[/tex]

_{water}

n

_{water}sin[tex]\theta[/tex]

_{water}= n

_{air}sin[tex]\theta[/tex]

_{air}

Therefore, using equality of alternate angles:

n

_{glass}sin90 = n

_{air}sin[tex]\theta[/tex]

_{air}

(1.50)*(sin90) = (1.00)*(sin[tex]\theta[/tex]

_{air})

Solution: undefined

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