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

In summary, the beam of light beams reflects back into the material at point a no matter the angle, as long as the index of refraction is greater than about 1.42.
  • #1
leolaw
85
1
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 [tex]\alpha[/tex], 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
 

Attachments

  • B.jpg
    B.jpg
    22.9 KB · Views: 1,201
  • A.jpg
    A.jpg
    33.8 KB · Views: 1,223
Physics news on Phys.org
  • #2
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 [tex]\theta_c = sin^{-1}( \frac{1}{n})[/tex]

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

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
 
  • #3
Hint: Realize the the smallest angle of incidence at the side surface ([itex]\gamma[/itex]) will occur when [itex]\alpha[/itex] 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.
 
  • #4
marlon said:
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 [tex] \gamma > 44.77 [/tex], and [tex] \beta > 45.23 [/tex], 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 [tex] n_1 = 1 n_2> 1.42 , \alpha_1 = \alpha, \alpha_2 = 45.23 [/tex]
 
Last edited:
  • #5
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.)
 
  • #6
Leo, as a matter of fact Doc Al is right. It is indeed better to follow the 'other way around' approach

marlon
 
  • #7
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
 
  • #8
Careful: The equation at the end surface should be [itex]1 = n \sin\beta[/itex]. Combine that with [itex]1 = n \sin\gamma[/itex] at the side surface.

What's the relationship between [itex]\beta[/itex] and [itex]\gamma[/itex]? Hint: use a trig identity to eliminate those angles.
 
  • #9
After combing these two equations, I have [tex] 1 = \frac{sin \beta}{sin \gamma}[/tex], which shows me that [tex] \beta = \gamma[/tex]
 
  • #10
Look at the diagram. [itex]\beta[/itex] and [itex]\gamma[/itex] are part of the same right triangle.
 
  • #11
Sorry that I may be asking dumb question, but then how does 1.42 play in this part?
So now i have:
[tex]n_{air} * sin(\alpha) = n_{water} * sin (45) [/tex]
 
Last edited:
  • #12
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 [itex]\beta[/itex] and [itex]\gamma[/itex] add to 90 degrees, then [itex]\sin\beta = \cos\gamma[/itex]. You take it from here.
 
  • #13
Got it, thx
 

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

What is geometric optics?

Geometric optics is the study of light as rays that travel in straight lines and interact with surfaces and objects. It is used to understand how light behaves and how it is affected by different materials and structures.

Why do we need to find the reflection angle for an index greater than 1.42?

When light travels from one medium to another, it changes direction and speed. This change is determined by the index of refraction, which is a measure of how much the speed of light is reduced in a particular medium. When the index is greater than 1.42, the angle of reflection will be different than the angle of incidence.

How is the reflection angle calculated?

The reflection angle can be calculated using the law of reflection, which states that the angle of incidence is equal to the angle of reflection. This means that the reflection angle can be found by drawing a line perpendicular to the surface at the point of incidence and measuring the angle between this line and the incident ray.

What factors affect the reflection angle?

The reflection angle is affected by the index of refraction of the medium, the angle of incidence, and the surface properties of the material. The type of material and its surface texture can also play a role in determining the reflection angle.

How is the reflection angle used in practical applications?

The reflection angle is important in many practical applications such as designing mirrors, lenses, and other optical devices. It is also used in fields such as architecture, where understanding the reflection angles of light can help in optimizing lighting and reducing glare in buildings. Additionally, the reflection angle is used in photography and cinematography to create certain effects and enhance image quality.

Similar threads

  • Introductory Physics Homework Help
Replies
4
Views
809
  • Introductory Physics Homework Help
Replies
16
Views
374
  • Introductory Physics Homework Help
Replies
1
Views
233
  • Introductory Physics Homework Help
Replies
3
Views
483
  • Introductory Physics Homework Help
2
Replies
35
Views
1K
Replies
3
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
15
Views
1K
  • Introductory Physics Homework Help
Replies
12
Views
2K
Back
Top