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

  • #1
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Homework Statement


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.

Homework Equations


sina/sinb=n2/n1

The Attempt at a Solution


sinb=1 so sina=n2/n1=1/2 the angle is pi/6 rad. Am I right?
 

Answers and Replies

  • #2
blue_leaf77
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Give us the related picture.
 
  • #4
blue_leaf77
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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
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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.
It is pi/2 rad or 90 degrees.
 
  • #6
blue_leaf77
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Take a look at the picture below,
Prism.png


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
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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.
 
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  • #8
blue_leaf77
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Please fix your comment above.
 
  • #10
blue_leaf77
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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.
 
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  • #11
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This potentially means that any incident angle on a right angle prism will result in a total internal reflection at the second surface and hence no light can be transmitted.
What about the two equal sides of the prism. i didn't use this fact anywhere.
 
  • #12
blue_leaf77
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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.
 
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