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

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  1. May 19, 2016 #1
    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. jcsd
  3. May 19, 2016 #2

    blue_leaf77

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    Give us the related picture.
     
  4. May 19, 2016 #3
    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
  5. May 19, 2016 #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.
     
  6. May 19, 2016 #5
    It is pi/2 rad or 90 degrees.
     
  7. May 19, 2016 #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##.
     
  8. May 19, 2016 #7
    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
  9. May 19, 2016 #8

    blue_leaf77

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    Please fix your comment above.
     
  10. May 19, 2016 #9
    Just did it.
     
  11. May 19, 2016 #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.
     
    Last edited: May 19, 2016
  12. May 19, 2016 #11
    What about the two equal sides of the prism. i didn't use this fact anywhere.
     
  13. May 19, 2016 #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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