Please help with physics problem!?

Link to picture: http://i608.photobucket.com/albums/tt165/trojans_cheerchick/prismangle_tir.png

The refracted angle is arcsin(sin(59.0)/1.47) = 35.67o. Thus the angle at the left interface in the upper triangle is 90 – 35.67 = 54.33o

Therefore on the right side the angle in the upper triangle is 180 – 90- 54.33 = 35.67o..S o the incident angle on the right face is 54.33o….

BUT the critical angle for this prism is arcsin(1/1.47) = 42.86o

So this ray is reflected internally. so the reflected angle is 54.33o….

So the angle in the right hand triangle is 35.67o…So this reflected ray strikes the lower surface at an angle with respect to the bottom of 180 – 45- 35.67 = 99.33o

This means the incident ray on the bottom surface is 9.33o

A) So the original ray will exist the bottom of the prism and the angle it makes is

B) arcsin(1.47*sin(9.33)) = 13.8o Which based on your system is a positive angle.

C)For the ray to exist the right hand side we have from the left hand side
sin(59.0) = n*sin(θ) and at the rt hand side the incident angle will be (90 – θ)

so n*sin(90 – θ) = 1 but sin(90 – θ) = cos(θ) so n*cos(θ) = 1

So if we square both eqns we get
sin^2(59.) = n^2*sin^2(θ)
and 1 = n^2*cos^2(θ)…

Now adding these we get
1+sin^2(59.0) = n^2*(sin^2(θ) + cos^2(θ)) = n^2

Therefore n = sqrt((1+sin^2(59.0)) = sqrt(1+ .734) = 1.317 = 1.32

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