How Do You Find the Angle of Incidence from a Prism to Air?

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SUMMARY

The discussion focuses on calculating the angle of incidence from a prism to air, specifically using a prism with a 60° apex angle. Participants clarify that the angle of incidence can be determined by subtracting the angle of refraction from 60°. Additionally, they explain the relevance of the 120° angle formed by the intersection of the two normals within the prism, which aids in visualizing the geometry involved in the problem. The conversation emphasizes the importance of understanding triangle properties and angle relationships in solving the problem.

PREREQUISITES
  • Understanding of basic geometry, including triangle properties and angle relationships.
  • Familiarity with the concept of refraction and Snell's Law.
  • Knowledge of prism geometry, specifically the properties of angles within a prism.
  • Ability to visualize geometric constructions and reference lines.
NEXT STEPS
  • Study the principles of Snell's Law and its application in optics.
  • Learn about the properties of prisms and how they affect light behavior.
  • Explore geometric constructions involving triangles and angle relationships.
  • Practice solving problems related to angles of incidence and refraction in various optical scenarios.
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Students studying optics, physics educators, and anyone interested in understanding the behavior of light as it passes through prisms and other optical devices.

totomyl
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Homework Statement


it is the third question:
IMG_20161002_172724.jpg


Attempt

IMG_20161002_173241.jpg


Here is my attempt at the question, i got up to the point where i need to find the angle of incidence from the prism to the air, but i don't know how to do that. The solution page says to use 60o minus the angle of refraction i just calculated, but i don't know why we need to do that.

if you could please tell me why we use that or how we get the angle from the prism to the air, that would be very helpful. thank you.
 
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The statement of the problem does not match the numbers that you are using. Setting that aside, you can find the angle of incidence by bringing into the picture the upper triangle formed by the top apex of the prism, by the point where the ray enters the prism and the point where it exits the prism. You already know that one angle is 60o. You can relate the other two angles to the angle of refraction, which you already know, and the exit of angle that is the unknown.
 
An alternative method is to note that the two normals meet in the interior of the prism at an angle of 120°. Use the triangle Δabc formed by the normals and the light ray.

upload_2016-10-2_18-35-4.png
 
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gneill said:
An alternative method is to note that the two normals meet in the interior of the prism at an angle of 120°. Use the triangle Δabc formed by the normals and the light ray.

View attachment 106850
could i ask how you got the 120 when both of the normals meet? this is where my geometry skills come in i assume.
 
totomyl said:
could i ask how you got the 120 when both of the normals meet? this is where my geometry skills come in i assume.
Yes. It's just a matter of transferring angles around various reference lines and intersections. Erect a few verticals to create triangles and reference lines. Use the fact that triangle angles sum to 180 and complimentary angles to 90. See if you can't "transfer" the prism's base angles to where b is in the diagram (erect a vertical through b to divide the angle there into two parts). Give it a try.
 
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From geometry what is the sum of the interior angles of a 4-sided figure?
 
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gneill said:
Yes. It's just a matter of transferring angles around various reference lines and intersections. Erect a few verticals to create triangles and reference lines. Use the fact that triangle angles sum to 180 and complimentary angles to 90. See if you can't "transfer" the prism's base angles to where b is in the diagram (erect a vertical through b to divide the angle there into two parts). Give it a try.
i see, thank you for the help i think i can work out the rest on my own now. this was very helpful and will be useful for future reference.
 

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