Instructor's Guide to Activity: The Critical Angle and Fiber Optics (Simulation-Based)

In today's activity, you will use simulations to measure the critical angle for a plastic prism and then investigate the effects of total internal reflection for fiber optics. The readings about these topics are found in the module on Reflection, Refraction, and Fiber Optics.

Preparatory Questions:
The following questions should have been answered before the start of the activity. Your instructor may ask you to discuss your answers with group members or as a class before continuing.

Snell's Law states that n₁ sin q₁ = n₂ sin q₂. What does each symbol in the equation represent?

n_i

represents the index of refraction of medium i; q_i represents the angle between the path of light and the normal to the surface in medium i.

Consider the situation when q₂ = 90^o. q₁ is then called the critical angle q_c. For any angles q₁ > q_c, total internal reflection occurs. Which index of refraction, n₁ or n₂ must be larger for this to occur? Explain.

For total internal reflection (TIR) to occur, light must bend away from the normal. Another way to say this is that q₂ > q₁ . Sin q₂ will then be greater than sin q₁, since sine increases with angle for angles less than 90^o. If the angle in medium 2 is larger, Snell's Law can only be satisfied if medium 2 has a smaller index of refraction. Thus n₂ < n₁ if TIR occurs as light moves from medium 1 to medium 2. Total internal reflection can only occur when light moves from a denser medium (higher n) to a rarer medium (lower n).

In the experiment you will perform, the two media will be plastic and air. You will measure the index of refraction for the plastic. Do you expect to see total internal reflection when the light travels into the plastic from the air, when it travels into the air from the plastic, or both? Explain.

In the experiment, air has the smaller index of refraction, so TIR will occur as the light travels from the plastic into the air.

What would the critical angle for the plastic be if its index were 1.54?

n₁ sin q₁ = n₂ sin q₂.
n₁ sin q_c = n₂ sin 90^o = n₂, so
q_c = sin^-¹(n₂/n₁) = sin^-¹(1.00/1.54) = 40.5^o.

If you were using total internal reflection to trap light inside a fiber, would the core (inner layer) index have to be less than or greater than the cladding (outer layer) index of refraction? Explain.

Since TIR occurs when light moves into a rarer medium, the index of the core would have to be greater than the index of the cladding.

Measuring the Critical Angle of a Plastic for a Given Wavelength

Go to the web site http://www.physics.northwestern.edu/ugrad/vpl/optics/snell.html. You will see a simulation of light moving from one medium at the top of the screen to another medium below. n₁ and n₂ represent the indices of refraction in the top and bottom medium, respectively. Notice that the different color lines do NOT represent different colors of light. The light is monochromatic laser light. The different colors identify the incident ray (blue), the reflected ray (green) and the refracted ray (red). View of Simulation
Blue incident ray and Green reflected ray in material of n_1. Red transmitted ray in material of n_2.

1. The default settings have n₁ less than n₂. Try entering different angles of incidence q₁, ranging from 1^o to 89^o. Describe what happens to the refracted beam and reflected beam as you increase the angle of incidence. Do you ever achieve total internal reflection? How do you know?
As the angle of incidence increases, the angles of reflection and refraction increase, with all three light rays moving closer to the surface. TIR will not be achieved in this situation since light is traveling from a rarer medium to a denser medium. This is verified by observation that the refracted (red) ray never reaches the surface - the angle of refraction remains well below the requisite 90 degrees.

2. Now set n₁=1.54 and n₂=1.00 to simulate light exiting plastic and entering air. Again, enter different angles of incidence q₁ ranging from 1^o to 89^o. Do you ever achieve total internal reflection with this set up? How do you know? Is this what you predicted in the Preparatory Questions?
TIR is achieved - the refracted ray approaches the surface, then vanishes. Hopefully this was predicted.

3. You will next determine the value of the critical angle for this interface by trying various values of q₁, and recording the resulting q₂, (or indicating whether total internal reflection has occurred). If you have not achieved total internal reflection for a given value of q₁, should you increase or decrease q₁in your next attempt?
Increase. The angle of incidence must be greater than (or equal to) the critical angle for TIR to occur.

4. If you have achieved total internal reflection for a given value of q₁, does that necessarily mean that q₁ is the critical angle? How will you know when the critical angle has been reached?
No. Any angle higher than the critical angle will produce TIR. The critical angle is the miminum q₁ that produces TIR.

5. Write down at least 10 values of q₁ which you tried as you looked for the critical angle, along with the resulting sin q₂ given by the simulation. Use the given sin q₂ to calculate q₂. Your values of q₁ should show a gradual approach to the critical angle.
Here is a sample set of student data:

q₁ 10 30 50 40 45 41 40.5 40.1 40.3 40.4

sin q₂ 0.267 0.769 n/a 0.989 n/a n/a n/a 0.991 0.996 0.998

q₂ 15.5^o 50.3^o 81.5^o 82.3^o 84.9^o 86.4^o

6. Record your value for the critical angle to the nearest 0.1 degree as found from the simulation. How does this critical angle compare to the one you predicted for plastic in the Preparatory Questions?
Based on the above data, the critical angle is 40.5^o. This is in exact agreement with the calculation in the Preparatory Questions. It should be within 0.1 degree of the student's predicted value.

Applications to Optical Fibers

Go to this simulation of optical fibers by the Universitat Oldenburg, and answer the following questions. A few notes about the simulation:

The simulation is located at the bottom of the web page.
After you change parameters, you must click on the Trace button to see the effect of the new settings.
The angle denoted by i in this simulation (the angle of entry) is denoted by q₀ in the on-line readings for this course.
Assume that the fiber is in air, so n₀ is 1.0 throughout this activity.
Predict the answers to questions 8-9 before checking with the simulation.

8.	Will the angle of entry i affect whether or not the light is totally internally reflected as it enters the fiber? If so, how? If not, why not? Yes. the angle at which light enters the fiber determines its angle of incidence on the fiber wall. If the angle of entry is too big, the angle of incidence on the fiber wall will be less than the critical angle, causing light to escape.
9.	Based on what you know about total internal reflection, what range of values of n₂ (relative to the core index n₁) will keep the light trapped in the fiber? (You may assume the angle of incidence on the edge of the fiber is larger than the critical angle) Try out ns from your range and check. The cladding index n₂ must be less than the core index n₁ if TIR is to occur.
10.	Some of the n₂ values you expected to trap light in the fiber might have let light escape in the simulation due to the angle at which light strikes the edge of the fiber. In these situations would you need to increase or decrease the angle at which light strikes the top edge of the fiber (as measured to the normal) to trap the light? If light escapes, the angle of incidence at the top edge is less than the critical angle, so you need to increase the angle of incidence on the top edge.
11.	Does this desired increase or decrease in the angle at the edge of the fiber correspond to an increase or decrease of the entry angle i at which the light enters the fiber? Use the simulation to verify your answer. To increase the angle of incidence at the top edge, one must decrease the angle of entry i.
12.	Set the indices to n₁=1.54 and n₂ = 1.50, and find the largest value of i that results in total internal reflection by trying different values of i. Compare your result to the cut-off angle given by the simulation. To three decimal places, the maximum angle of entry is 20.4 degrees, in agreement with the cut-off angle calculated by the simulation.
13.	If you were designing a fiber, do you think you would prefer for your light to enter at an angle close to (but still less than) the cut-off angle, or at a much smaller angle close to zero degrees? What potential benefits do you envision this having? The primary advantage of light entering at an entry angle of close to zero is if all the light did so. If all of the light entering a fiber enters in a narrow cone, it travels essentially the same path and does not spread out, or disperse. (Such "modal dispersion" is the topic for another module.) A more basic way to look at it is that light entering along the axis of the fiber takes a more direct route and spends less time bouncing around than light that enters near the cut-off angle. Or, that light entering near the cut-off is more likely to exceed the critical angle at an edge should the fiber bend. Students might argue that light entering along the axis travels a shorter distance and so arrives more quickly. You could point out that such an advantage could be negated at the first bend. Additionally, the time it takes a single information "packet" to travel a fiber is much less important than the delay between information packets, or bits. The delay is dictated by dispersion (spreading), not by the transit time. This is an "extension" question and should probably not be graded for "correctness." It can lead to a discussion of modes and dispersion, it can be discussed in terms of path length traveled by the light, it can serve as a springboard to a discussion of acceptable bend in fibers, or it could be ignored.