# Homework Help: Optics: Total Internal Reflection with Triangular Prisms

1. Apr 6, 2013

### everlasting89

1. The problem statement, all variables and given/known data
When the striking ray is held perpendicular to the prism, there are four general configurations possible (in the attachment below). Use the figure on the next page to determine analytically which of these four configurations will result in total internal reflection of the light ray. Show analytical work and try each setup. Sketch the results.

2. Relevant equations

sin θ = (n1 / n2)(sin 90)
Angles of the prism: 90, 59, and 31 degrees

3. The attempt at a solution

Found the critical angle:
sin θ = (n1 / n2)(sin 90)
sin θ = (1.5/1)(sin 90)
sin θ = (1.5/1)(1)
θ = 41.8 degrees

And I know that for total internal reflection the angle of incidence must be larger than the critical angle. However, I am stuck on what (and how) to do next.

Any help is much appreciated!

#### Attached Files:

• ###### TIR Problem.JPG
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Views:
294
Last edited: Apr 6, 2013
2. Apr 6, 2013

### ehild

Welcome to PF!

What are the angles of the prism? They must be given.

Draw the rays inside the prism and figure out the angle of incidence when the ray strikes the other side.

ehild

3. Apr 7, 2013

### everlasting89

Thanks, ehild. The prism is 60, 59, and 30 degrees. How do I determine what the rays inside the prism look like?

everlasting89

4. Apr 7, 2013

### ehild

You mean 90 instead of 59, I guess

The ray enters at the first surface at zero angle of incidence (perpendicular to the plane) so it does not change direction in the prism. At what angle with respect to the normal does the ray strike the other surface? It is just simple geometry. I draw one of the configuration. Is the angle of incidence θ smaller of larger than the critical angle of 41.8°?

ehild

#### Attached Files:

• ###### totalrrefl.JPG
File size:
5 KB
Views:
356
Last edited: Apr 7, 2013
5. Apr 7, 2013

### everlasting89

Ok, I think I get it. Thanks! So the incidental angle would be 30 degrees which is smaller than the critical angle of 41.8. Thus, it will not result in total internal reflection.

6. Apr 7, 2013