Light in a Prism problem

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Homework Statement


A light ray traveling in the horizontal direction is incident onto a prism as shown in the figure. At what angle relative to horizontal does the light ray emerge from the second face of the shown prism if the prism has an index of refraction of 1.5 and is surrounded by air? The cross section of the prism is in the shape of an equilateral triangle.





Homework Equations


Snells Law.

The Attempt at a Solution




Okay, if the light is heading into the left side of the triangle, then the angle it makes with the normal is 30 degrees since the angle the ray to the left side of the triangle would be 60 degrees. That means the refracted ray will make 19.47 degrees with the normal inside the prism. Now, I believe I should do 90 - 19.47 to get the angle that will make a triangle to connect me to the right side of the prism. Then I can do 180-(90-19.47)-60 = 30+19.47 which gets me 49.47 degrees. Now I need the angle normal to the right side so I do 90-49.47 and I get 40.53 degrees. Snells law tells me that the refracted ray on the right side should be arcsin(1.5*sin(40.53)), which is about 77 degrees. This answer is off by 30 degrees for some reason. The answer is 47 degrees but I get 77 degrees.
 

Answers and Replies

  • #2
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Draw a rough sketch and see whether the refracted ray is meeting the opposite face of the prism or which face of the prism! Check the angle of 49.47 you have got is between refracted ray and which face of prism. Just do not play with numbers visualize what is happening.
 
  • #3
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Draw a rough sketch and see whether the refracted ray is meeting the opposite face of the prism or which face of the prism! Check the angle of 49.47 you have got is between refracted ray and which face of prism. Just do not play with numbers visualize what is happening.
There's a ray of light going right on the triangle's left face horizontally. It is then refracted downwards and hits the right face. It is then further refracted downwards. The 49.47 degrees should be the angle between the ray of light and the right side of the triangle.
 
  • #4
gneill
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Snells law tells me that the refracted ray on the right side should be arcsin(1.5*sin(40.53)), which is about 77 degrees. This answer is off by 30 degrees for some reason. The answer is 47 degrees but I get 77 degrees.
That would be with respect to the local normal to the prism. What's the normal's angle with respect to the horizontal? (You were asked for the ray's angle with respect to the horizontal).
 
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  • #5
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That would be with respect to the local normal to the prism. What's the normal's angle with respect to the horizontal? (You were asked for the ray's angle with respect to the horizontal).
Aha! Thank you! The being off by 30 degrees should of really gave it away. Thanks again for your help :)
 

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