Geometric optics - thickness of acrylic ?

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Homework Help Overview

The problem involves determining the thickness of an acrylic piece that deflects a ray by 2.37 cm at an incident angle of 50.5 degrees. The context is geometric optics, specifically focusing on refraction and the application of Snell's law.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the use of triangles to analyze the path of the ray through the acrylic. There are questions about the angles involved and how to relate them to the geometry of the problem. Some suggest breaking the problem into two triangles and using trigonometric relationships to find the necessary dimensions.

Discussion Status

The discussion is ongoing, with participants exploring various geometric interpretations and relationships. Some guidance has been offered regarding the use of the law of sines and cosine to find the hypotenuse and thickness, but no consensus has been reached on the final approach.

Contextual Notes

Participants are working under the constraints of the problem statement and are questioning the assumptions related to angles and triangle configurations. There is a focus on ensuring that the angles are correctly identified and related to the geometry of the situation.

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Homework Statement


A ray is deflected by 2.37cm by a piece of acrylic. Find the thickness t of the acrylic if the incident angle is 50.5 degrees.
http://imgur.com/kx2VT5c

Homework Equations



n1sinΘ1 = n2sinΘ2

The Attempt at a Solution



n of acrylic is 1.5. Therefore, the refracted angle is 30.958 degrees. ( 1(sin(50.5)) = 1.5(sin(θ)) )

I'm not sure if I can make a triangle within the acrylic that is above the ray line where one angle is 50.042, one is 90, and the other is 30.958 and use 2.37cm as on side of the triangle.
 

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carnivalcougar said:

Homework Statement


A ray is deflected by 2.37cm by a piece of acrylic. Find the thickness t of the acrylic if the incident angle is 50.5 degrees.
http://imgur.com/kx2VT5c

Homework Equations



n1sinΘ1 = n2sinΘ2

The Attempt at a Solution



n of acrylic is 1.5. Therefore, the refracted angle is 30.958 degrees. ( 1(sin(50.5)) = 1.5(sin(θ)) )

I'm not sure if I can make a triangle within the acrylic that is above the ray line where one angle is 50.042, one is 90, and the other is 30.958 and use 2.37cm as on side of the triangle.
Break this up into two triangles.

What is the distance that the ray travels through the acrylic ?
...
 
The distance that the ray travels through the acrylic is the hypotenuse of a triangle I can make using 2.37cm as one side. However, I do not know how to find the angles of this triangle
 
carnivalcougar said:
The distance that the ray travels through the acrylic is the hypotenuse of a triangle I can make using 2.37cm as one side. However, I do not know how to find the angles of this triangle
What angle does that hypotenuse make with the normal?

What angle does the exit ray make with the normal?
 
The exit ray makes an angle of 50.5 with the normal while the refracted ray makes an angle of 30.958 with the normal.

I drew which triangle I am talking about on the diagram.
 

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carnivalcougar said:
The exit ray makes an angle of 50.5 with the normal while the refracted ray makes an angle of 30.958 with the normal.

I drew which triangle I am talking about on the diagram.
What is the complimentary angle to the 50.5° angle you have drawn ?
 
That would be 39.5 degrees
 
carnivalcougar said:
That would be 39.5 degrees
Rotate that angle 90° counter-clockwise.

Where does it now line up ?
 
I think it would be south of the X axis. Which would give me the angle I need to find the third angle of the triangle?
 

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  • #10
carnivalcougar said:
I think it would be south of the X axis. Which would give me the angle I need to find the third angle of the triangle?
Yes.
 
  • #11
Would I then use the law of sines to find the hypotenuse of this triangle?

Thanks!
 
  • #12
carnivalcougar said:
Would I then use the law of sines to find the hypotenuse of this triangle?

Thanks!
You have to add the 39.5° angle to the angle of refraction.

Then use the definition of the cosine to find the hypotenuse. Right?
 
  • #13
Oh yeah that would have been easier. Either way I got 7.0852 for the hypotenuse. Then I used the law of sines on the triangle that has the same hypotenuse but the opposite side is the normal line, and therefore, the thickness of the acrylic to find what the thickness of the acrylic is. This came out to be 6.07587cm.
 
  • #14
carnivalcougar said:
Oh yeah that would have been easier. Either way I got 7.0852 for the hypotenuse. Then I used the law of sines on the triangle that has the same hypotenuse but the opposite side is the normal line, and therefore, the thickness of the acrylic to find what the thickness of the acrylic is. This came out to be 6.07587cm.

Good !
 
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