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Suavez
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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: nisin[tex]\theta[/tex]i = ntsin[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.
nglasssin90 = nwatersin[tex]\theta[/tex]water
nwatersin[tex]\theta[/tex]water = nairsin[tex]\theta[/tex]air
Therefore, using equality of alternate angles:
nglasssin90 = nairsin[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?