Applying Snell's Law Inside & Outside a Prism

[SpiraRoam]

Homework Statement

I'm having trouble / uncertainties about the third missing angle (from left to right) in the picture. The first angle of reflection was easy as it's just 34 degrees as well and I used snell's law rearranged to find the first angle of refraction as 13.36 degrees from the rearranged equation.

I'm wondering whether I should use snell's law again minus the refractive index of 1, as the angle is inside the prism, which would give me 5.48 degrees reflected for the 3rd and 4th angles. I'm not sure whether there is an extension to the snell's law rearranging I should take into account? There must be a relevance for the 17 degrees I've been shown that I'm not including in my workings?

Homework Equations

n1sinΘ1=n2sinΘ2

Θ2=sin^-1((1)sin(34) )
(2.42)

The Attempt at a Solution

Sticking with 5.48 degrees and using it for the reflected angle also.

 

[gneill]

Remember that light paths are reversible. So you could shine a ray back along the exit path and it would follow the same path back through the prism and exit where the original ray entered. You can treat the 17° angle like the angle of incidence for that "reverse" path.

 

[SpiraRoam]

Thanks a lot gneill I didn't know that - need to read more carefully in future!