Answer Geometric Optics Problem: Find Reflection Angle for Index > 1.42

[leolaw]

Hey guys! I need some helps on an optics problem. It is from Giancoli Chapter 23, question 41:
Question:
A beam of light enters the end of an optic fiber (attachment B). Show that we can guarantee total internal reflection at the side surface of the material (at point a), if the index of refraction is greater than about 1.42. In other words, regardless of the angle $$\alpha$$, the light beam reflects back into the material at point a.

I start by calculating the critical angle in the material (see attachment A), which is 45.233. But then I am not sure how to do the next step cause i don't seem to know what I am looking for.

Can someone just help out? Thx

 

[marlon]

Total internal reflection occurs if the angle of the reflected beam is 90° or parallel to the refracting/reflecting plane between the two media.

The critical angle is calculated from $$\theta_c = sin^{-1}( \frac{1}{n})$$

Now, since n must be bigger then 1.42 you know that $$\frac{1}{sin( \theta_c)} > 1,42$$ and thus $$sin( \theta_c) < 0.714$$

The angle gamma MUST BE BIGGER then the critical angle theta_c which is about 45.6°

Since the gamma = 90° - beta you have that 90° - beta > 45.6 and thus 44.4 > beta.

Applying Snell's law for alpha and beta clearly shows that this upper limt for beta is indeed easy to reach by varying alpha...

marlon

 

[Doc Al]

Hint: Realize the the smallest angle of incidence at the side surface ($\gamma$) will occur when $\alpha$ equals 90 degrees. Apply Snell's law twice: once at the end surface; once at the side surface (assuming the critical angle). Combine those two equations and you can solve for n.

 

[leolaw]

Applying Snell's law for alpha and beta clearly shows that this upper limt for beta is indeed easy to reach by varying alpha...

marlon

So I have $$\gamma > 44.77$$, and $$\beta > 45.23$$, but now i don't understand what do you mean by the upper limit.

How do you find it, like what equation do i have to graph on the calculator in order to find the limit?

We have $$n_1 = 1 n_2> 1.42 , \alpha_1 = \alpha, \alpha_2 = 45.23$$

 

[Doc Al]

Much better than graphing would be to derive the minimum index of refraction that would guarantee total internal reflection at the side surface. (Follow the hints in my last post.)

 

[marlon]

Leo, as a matter of fact Doc Al is right. It is indeed better to follow the 'other way around' approach

marlon

 

[leolaw]

Using Snell's Law, I have these two equations:
The one on the side, (1)(sin 90) = (n)(sin gamma)
and the other one on the surface that ensure to have total internal reflection is,
(sin gamma) = 1/n.

when I combine these two equations, i have (sin 90) = n/n .
I am kind of lost here

 

[Doc Al]

Careful: The equation at the end surface should be $1 = n \sin\beta$. Combine that with $1 = n \sin\gamma$ at the side surface.

What's the relationship between $\beta$ and $\gamma$? Hint: use a trig identity to eliminate those angles.

 

[leolaw]

After combing these two equations, I have $$1 = \frac{sin \beta}{sin \gamma}$$, which shows me that $$\beta = \gamma$$

 

[Doc Al]

Look at the diagram. $\beta$ and $\gamma$ are part of the same right triangle.

 

[leolaw]

Sorry that I may be asking dumb question, but then how does 1.42 play in this part?
So now i have:
$$n_{air} * sin(\alpha) = n_{water} * sin (45)$$

 

[Doc Al]

Keep going until you solve for n, and then you'll see. There are several ways to go. Here's what I wanted you to deduce from my last post: Since $\beta$ and $\gamma$ add to 90 degrees, then $\sin\beta = \cos\gamma$. You take it from here.

 

[leolaw]

Got it, thx