# For which incidence angle the ray won't come out of the prism's surface...

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1. May 19, 2016

1. The problem statement, all variables and given/known data
A ray of light falls on the surface AC of a prism with a straight angle and with two sides equal. For which incidence angle the ray won't come out of the surface AB. The index of refraction for the prism is n=2.

2. Relevant equations
sina/sinb=n2/n1

3. The attempt at a solution
sinb=1 so sina=n2/n1=1/2 the angle is pi/6 rad. Am I right?

2. May 19, 2016

### blue_leaf77

Give us the related picture.

3. May 19, 2016

http://s32.postimg.org/6fsfdz69x/image.png
I am adding: pi/6 rad is the angle of incidence. The first refracted angle is pi/3 rad. sin pi/3/sina=1/2
sin a=1.7. Where am I wrong?

Last edited: May 19, 2016
4. May 19, 2016

### blue_leaf77

What is the prism apex angle? In the problem statement you said "straight angle", but what does straight mean in this case? Googling this term resulted in a straight angle being equal to 180 degrees.

5. May 19, 2016

It is pi/2 rad or 90 degrees.

6. May 19, 2016

### blue_leaf77

Take a look at the picture below,

since you know that exit angle at the last surface is 90 degrees, you should be able to calculate $\delta$, then $\epsilon$, afterwards $\gamma$ follows naturally from that upper triangle and so on till you find $\alpha$.

7. May 19, 2016

sin sigma/sin 90=1/2=> sin sigma= pi/6 rad. beta=180-90-30=60=pi/3 rad. (as the angle of two extensions is pi/2.
sin alpha/ sin 60=2/1=> sin alpha/sqrt(3)/2=2=> sin aplha=sqrt (3)=1.7. I don't know where I am wrong.

Last edited: May 19, 2016
8. May 19, 2016

### blue_leaf77

9. May 19, 2016

Just did it.

10. May 19, 2016

### blue_leaf77

This potentially means that any incident angle on a right angle prism with $n=2$ will result in a total internal reflection at the second surface and hence no light can be transmitted.

Last edited: May 19, 2016
11. May 19, 2016

What about the two equal sides of the prism. i didn't use this fact anywhere.

12. May 19, 2016

### blue_leaf77

Probably that means the prism is an isosceles triangle but that doesn't influence the propagation of the ray. Anyway you can actually prove the statement I made in post #10. Denote the exit angle out of the second side as $\eta$. In the second surface, Snell's law gives $n \sin \delta = \sin \eta$. Then $\beta = 90^o-\delta$ and Snell's law in the first side gives
\begin{aligned} \sin \alpha &= n \sin\beta = n \sin(90^o-\delta) \\ &= n\cos \delta = n \sqrt{1-\sin^2 \delta} \end{aligned}
Then by substituting Snell's law from the second side for $\sin \delta$,
$$\sin \alpha = n \sqrt{1-\frac{\sin^2 \eta}{n^2}} \\ \sin \alpha = \sqrt{n^2 - \sin^2 \eta}$$
The maximum value for the left side is unity, while the minimum value for the right side with $n=2$ is $\sqrt{2^2-1} = \sqrt{3}$. This means, for any incident angle $0\geq \alpha \geq 90^o$, there is no real solution for the exit angle $\eta$. Physically, this means that the rays, irrespective of the incident angle, will undergo total internal reflection at the second surface.