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GR0877 Problem 76

PROBLEM STATEMENT: This problem is still being typed.

SOLUTION: (B) According to Snell’s law \sin{\theta} = n \sin{\beta} , where \beta is the angle of refraction. The critical angle \alpha_c for the total internal reflection is determined by \sin{\alpha_c} = 1/n, where \alpha_c + \beta = \pi/2. From these three equations it is easy to obtain the critical angle of incidence: \theta = \sin^{-1}{\sqrt{n^2-1}}. Thus, we must choose between (A) and (B). Now, if we increase \theta, \beta will increase and \alpha will decrease, which prevents the total internal reflection. Therefore, choice (B) is correct.

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  1. Anne
    08.11.2011 at 23:30

    How do you get to sqrt(n^2-1)?

  2. Anne
    08.11.2011 at 23:41

    Nevermind, tricky trig identities.

  3. cizmoe
    10.11.2011 at 05:31

    Incase it’s not as ‘easy’ for you (such as in my case): [1] rearrange so that, \beta=\pi/2-\alpha [2] sub this into \sin\theta=n \sin\beta so that you have \sin\theta=n\sin(\pi/2-\alpha) [3] use the trig identity, \sin(\pi/2-a)=\cos a [4] therefore you are left with \sin\theta=n\cos(\alpha) [5] use a pythagorean’s theorem on your equation \sin\alpha=1/n to give you \cos\alpha=\sqrt{n^2-1}/n 5) [6] finally this gives you \sin\theta=n\sqrt{n^2-1}/n 5)=\sqrt{n^2-1}

  4. Steve
    10.11.2011 at 20:43

    I just muddled through those calculations, and even with your guidance, it took several tries and over 10 minutes. So here’s a cheap way for those like me: In the limiting case that n=1, there is no refraction and the angle would have to be 0 degrees – otherwise, the light would eventually exit the cable. Only A, B, and E have the n^2-1 term, so we can eliminate C and D. Next, going by intuition, you’d expect the light to stay in the cable for theta below a certain critical angle, and exit the cable if theta exceeds that angle. Therefore, I’d go with the less than sign in choice A.

    • Peter
      08.11.2012 at 00:22

      This is the way to do, Steve! Thanks!

  5. physics_dan
    01.04.2014 at 06:50

    Yeah I’d agreed with Steve, it takes too long to do these trig calculations.
    First approach must be taking limits, as n goes to 1, only way for the light to stay in the optical fiber is theta = 0. Only (B) makes sense, cause others are either theta > 90 degree or theta > 0.

  6. 20.10.2014 at 11:17

    I also had troubles with the maths solution of the problem even if i took the right one during the simulation. Although it can seems difficult the solution is quite simple:
    $sin(/alpha)= 1/n && sin(/theta)=n /cdot cos(/alpha)$
    if we try to solve the solution by inverting the first equation it becomes quite different from the solution proposed by ETS:
    $/theta = asin { n /cdot cos [ asin (1/n) }$.

    The trick is changing the second equation in
    $ sin(/theta)=n /cdot sqrt( 1- sin^2( /alpha) )
    and using the first equation right now.

    I hope to be helpful, apologize 4 my bad eng 🙂

  7. 20.10.2014 at 11:20

    No preview, no party. i repost it with correct formulas

    I also had troubles with the maths solution of the problem even if i took the right one during the simulation. Although it can seems difficult the solution is quite simple:
    \sin(\alpha)= 1/n && \sin(\theta)=n \cdot \cos(\alpha)$
    if we try to solve the solution by inverting the first equation it becomes quite different from the solution proposed by ETS:
    \theta = \asin { n \cdot \cos[\asin(1/n)}.
    The trick is changing the second equation in
    \sin(\theta)=n \cdot \sqrt(1- \sin^2(\alpha))
    and using the first equation right now.
    I hope to be helpful, apologize 4 my bad eng 🙂

  8. Moose
    14.10.2017 at 23:44

    From the equation, sin(alpha critical) = 1/n, draw the triangle, giving values to the legs, i.e. Opp./Hyp. Then name remaining leg as you know (n^2 -1)^1/2, then other equation is sin(theta) = n sin(beta), but now you have the triangle, just “read off” what sin(beta) is, sin(beta) = (n^2 -1)^1/2/n, which gives sin(theta) = (n^2 -1)^1/2, giving you theta.

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