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exi
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Homework Statement
In this drawing:
http://img245.imageshack.us/img245/6559/physgv8.png [Broken]
Index of refraction for glass: 1.52
Index of refraction for surrounding carbon disulfide: 1.63
Incident angle at point A: 43.0°
At what angle does the ray leave the glass at point B?
Homework Equations
[tex]n_{1}sin\theta_{1} = n_{2}sin\theta_{2}[/tex] (twice)
The Attempt at a Solution
Please double-check my conceptual understanding of this.
Part 1, as light travels through the carbon disulfide into the glass:
[tex]1.63sin43 = 1.52sin\theta_{2}[/tex]
[tex]\theta_{2} = sin^{-1}\frac{1.63sin43}{1.52}[/tex]
So Θ within the glass is 47°.
Drawn out and making a triangle with the upper left hand corner of the glass, a triangle forms with point B as one of its own points, meaning that the angle with a normal line at that point would be 180 - (90 + 47) = 43°.
Part 2, as light travels through the glass back into the carbon disulfide:
[tex]1.52sin43 = 1.63sin\theta_{2}[/tex]
[tex]\theta_{2} = sin^{-1}\frac{1.52sin43}{1.63}[/tex]
Which yields 39.4923°.
Am I going about this correctly?
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