BBC - GCSE Bitesize Science - Other applications of light : Revision, Page 4
When light travels from a medium of lower refractive index to a medium of higher refractive index it is refracted towards the normal and vice Relationship between Critical Angle and Refractive Index . Tip when carrying out the experiment. A transparent material, such as glass or water, can actually reflect light better than any mirror. Hold your laser or flashlight up to the side of the aquarium so the light is shining into the water. At this angle, called the critical angle, all the light is reflected back into the. If the light travels from a medium of high refractive index (slow speed) into one a relationship between the angles of incidence, refraction and the refractive indices. For all light incident at angles larger than this critical angle, the light will not.
So this is the passenger compartment, and it has four wheels on the car.
We're looking at it from above. And let's say it's traveling on a road.
It's traveling on a road. On a road, the tires can get good traction. The car can move pretty efficiently, and it's about to reach an interface it's about to reach an interface where the road ends and it will have to travel on mud. It will have to travel on mud. Now on mud, obviously, the tires' traction will not be as good.
Instructor's Guide to Critical Angle Activity (Equipment-Based)
The car will not be able to travel as fast. So what's going to happen? Assuming that the car, the steering wheel isn't telling it to turn or anything, the car would just go straight in this direction. But what happens right when--which wheels are going to reach the mud first? This wheel is going to reach the mud first.
Physics Study Guide
There's going to be some point in time where the car is right over here. Where it's right over here. Where these wheels are still on the road, this wheel is in the mud, and that wheel is about to reach the mud. Now in this situation, what would the car do?
What would the car do? And assuming the engine is revving and the wheels are turning, at the exact same speed the entire time of the simulation.
Well all of a sudden, as soon as this wheel hits the medium, it's going to slow down. This is going to slow down. But these guys are still on the road. So they're still going to be faster. So the right side of the car is going to move faster than the left side of the car. You see this all the time. If the right side of you is moving faster than the left side of you, you're going to turn, and that's exactly what's going to happen to the car.
The car is going to turn. It's going to turn in that direction. And so once it gets to the medium, it will now travel, it will now turn-- from the point of the view from the car it's turning to the right. But it will now travel in this direction.Total internal reflection - Geometric optics - Physics - Khan Academy
It will be turned when it gets to that interface. Now obviously light doesn't have wheels, and it doesn't deal with mud. But it's the same general idea. When I'm traveling from a faster medium to a slower medium, you can kind of imagine the wheels on that light on this side of it, closer to the vertical, hit the medium first, slow down, so light turns to the right. If you were going the other way, if I had light coming out of the slow medium, so let's imagine it this way.
Let's have light coming out of the slow medium. And if we use the car analogy, in this situation, the left side of the car is going to-- so if the car is right over here, the left side of the car is going to come out first so it's going to move faster now.
So the car is going to turn to the right, just like that. So hopefully, hopefully this gives you a gut sense of just how to figure out which direction the light's going to bend if you just wanted an intuitive sense. And to get to the next level, there's actually something called Snell's Law. And all this is saying is that this angle-- so let me write it down here--so let's say that this velocity right here is velocity 2 this velocity up here was velocity 1, going back to the original.
Actually, let me draw another diagram, just to clean it up. And also that vacuum-water interface example, I'm not enjoying it, just because it's a very unnatural interface to actually have in nature. So maybe it's vacuum and glass.
That's something that actually would exist. So let's say we're doing that. So this isn't water, this is glass. Let me redraw it. And I'll draw the angles bigger.
So let me draw a perpendicular. And so I have our incident ray, so in the vacuum it's traveling at vand in the case of a vacuum, it's actually going at the speed of light, or the speed of light in a vacuum, which is c, orkilometers per second, or million meters per second--let me write that-- so c is the speed of light in a vacuum, and that is equal to it's not exactlyI'm not going into significant digits-- this is true to three significant digits million meters per second.
The Critical Angle and Fiber Optics Equipment-Based In today's activity, you will measure the critical angle and index of refraction for a plastic prism and then use a simulation to 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. Each group needs a laser, a plastic semi-circular prism, a device to vertically spread the laser beam mounted at an appropriate height such as a cylindrical lens attached to a washer and mounted on a magnetic optical carriertwo pieces of blank paper, and a protractor.
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. What does each symbol in the equation represent? Which index of refraction, n1 or n2 must be larger for this to occur? For total internal reflection TIR to occur, light must bend away from the normal.
Sin q2 will then be greater than sin q1, since sine increases with angle for angles less than 90o. If the angle in medium 2 is larger, Snell's Law can only be satisfied if medium 2 has a smaller index of refraction.
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?
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. 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? 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.
Set up the equipment on top of a blank page of paper, as shown to the right. The laser light should shine through a device that spreads the light so you can see the path of the laser on the paper. The light should then strike the curved side of the plastic "prism", travel through the prism, and exit near the middle of the flat side of the prism.
Some light will also be reflected off the flat side and exit the curved side. View from Top not to scale 1.
Rotate the prism, observing the behavior of the light beam exiting the flat face refracted beam and the behavior of the beam reflected from the flat face. Start with the flat side perpendicular to but facing away from the laser beam, and rotate it counterclockwise, as depicted in the diagram above.
Continue rotating it until the flat side is parallel to the laser beam. Describe what happens to the refracted beam and reflected beam as you rotate the prism. How do you know when the critical angle has been reached? The derivation is shown below.
Since TIR only occurs if the refractive medium is less dense than the incident medium, the value of ni must be greater than the value of nr. If at any time the values for the numerator and denominator become accidentally switched, the critical angle value cannot be calculated.
Mathematically, this would involve finding the inverse-sine of a number greater than 1. Physically, this would involve finding the critical angle for a situation in which the light is traveling from the less dense medium into the more dense medium - which again, is not possible. This equation for the critical angle can be used to predict the critical angle for any boundary, provided that the indices of refraction of the two materials on each side of the boundary are known.
Examples of its use are shown below: Example A Calculate the critical angle for the crown glass-air boundary.
- Refraction of light
- The Critical Angle
- Refraction and Snell's law
Refer to the table of indices of refraction if necessary. The solution to the problem involves the use of the above equation for the critical angle.
Of all the possible combinations of materials that could interface to form a boundary, the combination of diamond and air provides one of the largest differences in the index of refraction values. This peculiarity about the diamond-air boundary plays an important role in the brilliance of a diamond gemstone.
Having a small critical angle, light has the tendency to become "trapped" inside of a diamond once it enters. A light ray will typically undergo TIR several times before finally refracting out of the diamond. Because the diamond-air boundary has such a small critical angle due to diamond's large index of refractionmost rays approach the diamond at angles of incidence greater than the critical angle.
This gives diamond a tendency to sparkle. The effect can be enhanced by the cutting of a diamond gemstone with a strategically planned shape.