At what angle does the light leave the glass

AI Thread Summary
The discussion focuses on solving a physics problem involving light refraction through a glass block surrounded by a liquid. The key equation used is Snell's law, which relates the angles of incidence and refraction based on the refractive indices of the materials involved. The initial attempt at a solution included an incorrect value of 1.6, leading to confusion about its origin. Participants emphasize the need to apply Snell's law twice—once at the interface between the liquid and glass, and again at the glass and air interface. Proper calculations and understanding of the angles are crucial for arriving at the correct answer.
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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