- #1

JoeyBob

- 256

- 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.

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.