At what angle does the light leave the glass

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SUMMARY

The discussion focuses on the application of Snell's Law to determine the angle at which light exits a glass block. Given a glass refractive index of 1.52 and a surrounding liquid with a refractive index of 1.79, a ray of light enters the glass at a 30.0° angle of incidence. The correct calculation involves applying Snell's Law, n1*sin(theta1) = n2*sin(theta2), twice to find the angle of refraction at both interfaces. The final angle of refraction at point B is calculated to be 53.13 degrees.

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phys62
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Homework Statement


The drawing shows a rectangular block of glass (n = 1.52) surrounded by a liquid with n = 1.79. A ray of light is incident on the glass at point A with a 30.0° angle of incidence. At what angle does the ray leave the glass at point B?

http://edugen.wiley.com/edugen/courses/crs1507/art/qb/qu/c26/ch26p_18.gif


Homework Equations


n1*sin(theta1)=n2*sin(theta2)


The Attempt at a Solution


I solved this and got the right answer by doing the steps below, however I have no idea why I put 1 in the denominater, and this is not the correct way to solve it because my friend's problem is not working out. Please help me figure out the proper way to solve this, as I've tried everything I can think of!

sin(theta2) = (1.6 sin30)/1 = 53.13 degrees
 
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Hi phys62! :wink:
phys62 said:
sin(theta2) = (1.6 sin30)/1 = 53.13 degrees

Where did 1.6 come from? :confused:

Show us your full calculations, and then we'll know how to help. :smile:
 
You need to apply Snell's law twice: Once at A; once at B. And you need to relate the angle of refraction at A to the angle of incidence at B.
 

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