Refraction and Total Internal Reflection

[blaziken's_charizard]

http://img210.imageshack.us/img210/6094/questionuz8.png

i) What is the angle of refraction of the ray RS after it strikes the glass prism of refractive index=1.5?

ii) If the critical angle is 41.8 degrees, deduce whether the ray will suffer total internal reflection.


2. Homework Equations :

Snell's Law
3.

1) Snell's Law:

n=sin i / sin r

1.5=sin45/sinR

sinR * 1.5 = sin45
sinR = sin45 / 1.5
R = sin^-1 (sin45 / 1.5)
R= 28.1 degrees

2) I have no idea what I could do to deduce whether it would suffer TIR or not... please help. Thank you :)

 

[SporadicSmile]

total internal reflection occurs for angles greater than the critical angle, that should solve that one for you =].

 

[blaziken's_charizard]

But you see the angle 28.1 isn't the incident angle made by the ray when its leaving the slow medium for the fast medium (glass => air).

I have to somehow find this incident angle, or so I assume >_> I hope to god it's not trigonometry, but I have a feeling it is as such V_V

 

[SporadicSmile]

it will invole trig =P

draw the ray on the diagram, remembering the angle is with respect to the normal of the surface, then just have to work out the angle that it is going to cross the glass-air interface at

id draw a diagram of what i have, but this computer has nothing i can do that on -.-

basically, draw your rays, draw the normals to the two surfaces, and use trig to solve, it isn't THAT difficult.

 

[blaziken's_charizard]

Ok... so it makes 28.1 degress... I think that the 90 degrees will be cut in half by the normal making a 45 degree angle in the triangle formed.

The problem now is that I can't identify where the angle is going to be formed with the normal...

 

[SporadicSmile]

http://www.omgimages.net/img/294/medium_54676.PNG

since i haven't posted enough to put urls on yet =P

you want to calculate the angle between the ray and the normal, the red and grey lines on that diagram.

 

[blaziken's_charizard]

The angle will be 16.9 degress... so therefore no TIR occurs >_>

Thanks. This was an exam question. I therefor got it wrong ;_; I drew the normal in the wrong place.