Trigonometry Help: Calculating the Angle of Light Entry & Exit

AI Thread Summary
The discussion focuses on calculating the angles of light entry and exit in a triangular prism using trigonometry and Snell's law. The user initially calculates the angle of light entry as 70.6 degrees but struggles to find the exit angle, expecting it to be 20.3 degrees. Clarifications about the normal to the surface reveal that the angles should be measured relative to this normal, not the surface itself. The conversation emphasizes the importance of understanding the geometric relationships between the light ray and the prism's surfaces to correctly apply Snell's law. Ultimately, the user seeks guidance on determining the correct angles and the concept of the normal in this context.
JoeyBob
Messages
256
Reaction score
29
Homework Statement
See attached
Relevant Equations
Snell's law
So what I did first was made the face of the triangle flat and calculated the angle the light entered it. This means the light enters the triangle from the base corner angle (so (180-38.8)/2) of 70.6 degrees.

1sin(70.6)=1.47sin(angle)

angle=39.915

Now I need to find the angle it exits. But this part doesn't make sense to me because Ill just be left with my original angle.

1.47sin(39.915)=sin(angle)

angle=70.6 degrees

The answer is suppose to be 20.3. I am obviously messing up with the geometry of the problem though I am not sure how. Trigonometry I was never good at.
 

Attachments

  • question.PNG
    question.PNG
    8.1 KB · Views: 139
Physics news on Phys.org
JoeyBob said:
Homework Statement:: See attached
Relevant Equations:: Snell's law

So what I did first was made the face of the triangle flat and calculated the angle the light entered it. This means the light enters the triangle from the base corner angle (so (180-38.8)/2) of 70.6 degrees.

1sin(70.6)=1.47sin(angle)

angle=39.915

Now I need to find the angle it exits. But this part doesn't make sense to me because Ill just be left with my original angle.

1.47sin(39.915)=sin(angle)

angle=70.6 degrees

The answer is suppose to be 20.3. I am obviously messing up with the geometry of the problem though I am not sure how. Trigonometry I was never good at.
In the standard form of Snell's law, the angles are to the normal. Does it look like the ray enters at 70 degrees to the normal?
 
haruspex said:
In the standard form of Snell's law, the angles are to the normal. Does it look like the ray enters at 70 degrees to the normal?
The normal? Are you saying it entered at 0 degrees?
 
JoeyBob said:
The normal? Are you saying it entered at 0 degrees?
By normal, I mean the normal to the surface the ray is passing through. The diagram shows the normal as a dotted line.
What is the angle the ray makes to that on entry?
 
haruspex said:
By normal, I mean the normal to the surface the ray is passing through. The diagram shows the normal as a dotted line.
What is the angle the ray makes to that on entry?
Im not sure, that's what's making me not get the answer. would be 180-38.8 then divided by 2 = 70.6

This would be the angle at each bottom corner. And if I rotate the tringle so the side the light hit is flat, its a rotation of 70.6 degrees.
 
JoeyBob said:
180-38.8 then divided by 2 = 70.6
That gives you the angle the base of the prism makes to each side of the prism. Since the ray comes parallel to the base, it is also the angle between the ray and the side of the prism.
But you want the angle between the ray and the normal to the side of the prism.
 
haruspex said:
That gives you the angle the base of the prism makes to each side of the prism. Since the ray comes parallel to the base, it is also the angle between the ray and the side of the prism.
But you want the angle between the ray and the normal to the side of the prism.
How do I do that or where can I read about how to find the normal?
 
JoeyBob said:
How do I do that or where can I read about how to find the normal?
Normal to a surface means at right angles to it. If the ray is at angle θ to the surface then it is at angle π/2-θ to the normal..
 
haruspex said:
Normal to a surface means at right angles to it. If the ray is at angle θ to the surface then it is at angle π/2-θ to the normal..

Yes, the angle is between the laser and the dotted line (perpendicular to the flat side of the triangle). I understand that, but I don't know how to find such an angle.

So your saying its pi/2-angle, but what angle?
 
  • #10
JoeyBob said:
Yes, the angle is between the laser and the dotted line (perpendicular to the flat side of the triangle). I understand that, but I don't know how to find such an angle.

So your saying its pi/2-angle, but what angle?
Draw the prism.
Point A is the apex of the prism, B the lower left corner and C is the lower right corner.
What is the value of angle ABC?
Point E is where the ray hits the prism. D is where the ray has come from.
Draw line DE.
What is the geometric relationship between lines DE and BC?
What is the value of the angle DEB?
Line FE is the normal to the prism at point E.
What is angle FEB?
So what is angle FED?
 
Back
Top