Quick Question about Refraction and Light

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The discussion centers on calculating the smallest angle of incidence for a light ray entering a prism with a 62.0° apex angle and an index of refraction of 1.58. The critical angle for light exiting the prism was determined to be 39.27°. Participants clarified that the angle x2 in Snell's Law should be derived from the geometry of the prism, specifically considering the angles formed between the normals and the light ray. The solution involves understanding the relationship between the angles and applying Snell's Law correctly. Overall, the conversation emphasizes the importance of geometric relationships in solving refraction problems.
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Quick Question about Refraction and Light!

Homework Statement



http://www.webassign.net/hrw/hrw7_33-55.gif


Suppose a prism has apex angle ϕ = 62.0° and index of refraction n = 1.58. What is the smallest angle of incidence θ for which a ray can enter the left face of the prism and exit the right face?

Homework Equations


Snell's Law - - n1sin(x1) = n2sin(x2)
Critical Angle -- sin(xc) = n2/n1



The Attempt at a Solution


I figured out the critical angle at exit from the right side (39.27). I'm just confused about what the angle (x2) would become in the Snell's Law equation. Would it be 180-39.27?

Could someone let me know if what I did was incorrect so far, or how to find the angle x2 for the Snell's Law equation. Thanks!
 
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The angle between the two normals is equal to the angle of the prism.
Look at the triangle formed between the two normals and the light ray inside the prism.
This will give you the angle x2.
Good luck
 


I got the correct answer. Thanks for the help nasu!
 

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