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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. 08.11.2011 at 23:30

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

2. 08.11.2011 at 23:41

Nevermind, tricky trig identities.

3. 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. 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.

• 08.11.2012 at 00:22

This is the way to do, Steve! Thanks!

5. 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.