# Thread: Totally internally reflected ray

1. ## Totally internally reflected ray

1. The problem statement, all variables and given/known data
The diagram shows a cube of glass. A ray of light, incident at the centre of a face of the cube, at an
angle of incidence θ, goes on to meet another face at an angle of incidence of 50°, as shown
in Figure 3.
critical angle at the glass-air boundary = 45°
Figure 3
(a) Draw on the diagram the continuation of the path of the ray, showing it passing through the glass
and out into the air.

2. Relevant equations

3. The attempt at a solution
I know that the wave will be internally reflected. But I am not sure of the angle at which it will emerge from the glass to air boundary. Please advise. Is it 40 degrees at the TIR and does it emerge with an incident angle of 50?

2. Originally Posted by gbenguse78
1. The problem statement, all variables and given/known data
The diagram shows a cube of glass. A ray of light, incident at the centre of a face of the cube, at an
angle of incidence θ, goes on to meet another face at an angle of incidence of 50°, as shown
in Figure 3.
critical angle at the glass-air boundary = 45°
Figure 3
(a) Draw on the diagram the continuation of the path of the ray, showing it passing through the glass
and out into the air.

2. Relevant equations

3. The attempt at a solution
I know that the wave will be internally reflected. But I am not sure of the angle at which it will emerge from the glass to air boundary. Please advise. Is it 40 degrees at the TIR and does it emerge with an incident angle of 50?
Probably this answer comes too late. But nevertheless ...:

1. Since the critical angle is $\displaystyle \beta = 45^\circ$ (which is a little bit low in my opinion) you use a refractive index of $\displaystyle n= \sqrt{2}\approx 1.414$.

2. Draw a sketch. (Attention: My sketch is not to scale!)

3. If the ray enters the glas under an angle of $\displaystyle \theta$ then the ray includes an angle of 40° with the perpendicular. Using Snell's law (have a look here: Snell's law - Wikipedia, the free encyclopedia) you'll get:

$\displaystyle \dfrac{\sin(\theta))}{\sin(40^\circ)} = \sqrt{2}~\implies~\theta \approx 65.4^\circ$