## GR0877 Problem 76

**PROBLEM STATEMENT**: *This problem is still being typed*.

**SOLUTION**: (**B**) According to Snell’s law , where is the angle of refraction. The critical angle for the total internal reflection is determined by , where . From these three equations it is easy to obtain the critical angle of incidence: . Thus, we must choose between (A) and (B). Now, if we increase , will increase and will decrease, which prevents the total internal reflection. Therefore, choice (B) is correct.

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

Nevermind, tricky trig identities.

Incase it’s not as ‘easy’ for you (such as in my case): [1] rearrange so that, [2] sub this into so that you have [3] use the trig identity, [4] therefore you are left with [5] use a pythagorean’s theorem on your equation to give you 5) [6] finally this gives you 5)=

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.

This is the way to do, Steve! Thanks!

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.

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 🙂

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 🙂

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.