7-25-00
Sections 23.6 - 23.8
The critical angle can be found from Snell's law, putting in an angle of 90° for the angle of the refracted ray. This gives:
For any angle of incidence larger than the critical angle, Snell's law will not be able to be solved for the angle of refraction, because it will show that the refracted angle has a sine larger than 1, which is not possible. In that case all the light is totally reflected off the interface, obeying the law of reflection.
Optical fibers are based entirely on this principle of total internal reflection. An optical fiber is a flexible strand of glass. A fiber optic cable is usually made up of many of these strands, each carrying a signal made up of pulses of laser light. The light travels along the optical fiber, reflecting off the walls of the fiber. With a straight or smoothly bending fiber, the light will hit the wall at an angle higher than the critical angle and will all be reflected back into the fiber. Even though the light undergoes a large number of reflections when traveling along a fiber, no light is lost.
Because light is refracted at interfaces, objects you see across an interface appear to be shifted relative to where they really are. If you look straight down at an object at the bottom of a glass of water, for example, it looks closer to you than it really is. Looking perpendicular to an interface, the apparent depth is related to the actual depth by:
A beam of light travels from water into a piece of diamond in the shape of a triangle, as shown in the diagram. Step-by-step, follow the beam until it emerges from the piece of diamond.
(a) How fast is the light traveling inside the piece of diamond?
The speed can be calculated from the index of refraction:
(b) What is 
, the angle between the normal and the beam of light inside the diamond at the water-diamond interface?
A diagram helps for this. In fact, let's look at the complete diagram of the whole path, and use this for the rest of the questions.
The angle we need can be found from Snell's law:
(c) The beam travels up to the air-diamond interface. What is 
, the angle between the normal and the beam of light inside the diamond at the air-diamond interface?
This is found using a bit of geometry. All you need to know is that the sum of the three angles inside a triangle is 180°. If
is 24.9°, this means that the third angle in that triangle must be 25.1°. So:
(d) What is the critical angle for the diamond-air interface?
(e) What happens to the light at the diamond-air interface?
Because the angle of incidence (64.9°) is larger than the critical angle, the light is totally reflected internally.
(f) The light is reflected off the interface, obeying the law of reflection. It then strikes the diamond-water interface. What happens to it here?
Again, the place to start is by determining the angle of incidence, 
. A little geometry shows that:
The critical angle at this interface is :
Because the angle of incidence is less than the critical angle, the beam will escape from the piece of diamond here. The angle of refraction can be found from Snell's law:
There are many similarities between lenses and mirrors. The mirror equation, relating focal length and the image and object distances for mirrors, is the same as the lens equation used for lenses.There are also some differences, however; the most important being that with a mirror, light is reflected, while with a lens an image is formed by light that is refracted by, and transmitted through, the lens. Also, lenses have two focal points, one on each side of the lens.
The surfaces of lenses, like spherical mirrors, can be treated as pieces cut from spheres. A lens is double sided, however, and the two sides may or may not have the same curvature. A general rule of thumb is that when the lens is thickest in the center, it is a converging lens, and can form real or virtual images. When the lens is thickest at the outside, it is a diverging lens, and it can only form virtual images.
Consider first a converging lens, the most common type being a double convex lens. As with mirrors, a ray diagram should be drawn to get an idea of where the image is and what the image characteristics are. Drawing a ray diagram for a lens is very similar to drawing one for a mirror. The parallel ray is drawn the same way, parallel to the optic axis, and through (or extended back through) the focal point. The chief difference in the ray diagram is with the chief ray. That is drawn from the tip of the object straight through the center of the lens. Wherever the two rays meet is where the image is. The third ray, which can be used as a check, is drawn from the tip of the object through the focal point that is on the same side of the lens as the object. That ray travels through the lens, and is refracted so it travels parallel to the optic axis on the other side of the lens.
The two sides of the lens are referred to as the object side (the side where the object is) and the image side. For lenses, a positive image distance means that the image is real and is on the image side. A negative image distance means the image is on the same side of the lens as the object; this must be a virtual image.
Using that sign convention gives a lens equation identical to the spherical mirror equation:
The other signs work the same way as for mirrors. The focal length, f, is positive for a converging lens, and negative for a diverging lens.
The magnification factor is also given by the same equation:
Useful devices like microscopes and telescopes rely on at least two lenses (or mirrors). A microscope, for example, is a compound lens system with two converging lenses. One important thing to note is that with two lenses (and you can extend the argument for more than two), the magnification factor m for the two lens system is the product of the two individual magnification factors.
It works this way. The first lens takes an object and creates an image at a particular point, with a certain magnification factor (say, 3 times as large). The second lens uses the image created by the first lens as the object, and creates a final image, introducing a second magnification factor (say, a factor of seven). The magnification factor of the final image compared to the original object is the product of the two magnification factors (3 x 7 = 21, in this case).