To see something, we need a source of light and an ob ject off which the light bounces, and then reaches the eyes of the observer.
Light travels in a straight-line path between the source of the light and the object reflecting it, then in another straight line between that object and our eyes.
The first question is illuminated by a lightbulb.
A laser pointer is useful for studying light propagation.
Most light sources do not emit light as a single beam.
The bulb sends light in different directions.
The wal s, floor, and ceiling are affected by these ideas.
There are two possible models of how extended sources emit light.
The rays are sent in different directions.
The experiments are done in a dark room.
We predict a dark shadow behind the lightbulb and a pencil behind the rays that don't reach the wall.
There is a dark shadow on the wall.
In experiment 1, we predict a light shadow with a fuzzy shaded lightbulb and place a pencil on the wall.
We predict a shadow illuminated screen with a hint of a shadow.
The fuzzy, light shadow is not as dark as in experiment 1.
The wall will be dark.
The bulb is facing the wal.
The wall will have a bulb on it.
The wall is dark.
The result remains the same if we cover the first hole and poke a hole in a different place.
Both models predicted the outcome of the experiment.
Light sources, light propagation, and shadows are sent by each point.
Each point of the laser light and the sun send one ray.
The points send rays.
Parallel rays are the only rays that reach Earth.
There are multiple rays diverging from that point.
The hole traveled in different directions.
Light from a point-like light source must be represented using multiple rays, with the exception of laser light, which can be represented using one ray.
There was a new phenomenon revealed in the experiments.
There is no light behind the object.
There was no dark shadow on the wal in the second experiment.
A semi-shadow is a region with some light but not all of it.
It is a fuzzy shadow.
The sun can be represented as a collection of parallel rays.
We can represent the laser's narrow beam of light with one ray.
On a sunny day, a streetlight pole casts a shadow on the ground.
When held vertically, the meter stick casts a 0.70 m shadow.
This can be used to determine the height of the pole.
We sketch the situation first.
The sun is represented as paral el rays hitting Earth's surface.
The stick has a shadow of 0.70 m.
The magnitude is reasonable and the unit is correct.
The shadow was cast by the stick.
The pole height is still 13.7 m, but the gles are equal.
If you hold a candle flame from a blank wall, you don't see the flame on the wal.
The wall is illumi nated by light coming from the points on the candle flame since each point emits light in all directions.
We can use a piece of cardboard with a small hole in it to make a projection of the flame on a wal in a dark room.
Conceptual Ex ercise 21.2 explains how this projection is formed.
From the bottom ing experiment, ray 1 can see.
We predict that a piece of stiff paper with a smal hole in it will show an upside-down projection of the flame on the wal because of candle.
The candle flame is upside down on the wall when you perform the ex periment.
We have a sketch of the situation.
The ray diagram can be used to predict the light source.
If you move the candle closer to the hole, the light from the candle will reach the wall.
The light from the candle flame can be seen in all directions.
They should be represented away from the wall.
Most of the rays are bigger than the others.
A lightproof box with a side world on a wall is what this camera is.
A small hole in one wall and a photographic plate or film inside the box ings are projected upside down.
Pinhole cameras were used to make photographs before the invention of modern cameras.
You would shine intense light on the person for a long time to photograph them.
The light reflected off the person would travel through the hole and form a projection of the person on the film.
A light source in front of a screen with a smal hole in it will be projected upside down on a wal behind the screen.
Light reflects along straight lines in this section.
A mirror is used with the represent to reflect a single ray.
The reflected laser light light emitted from a source and traveling beam is shown.
The ror is the angle between the incident beam AO and the normal line CO.
The table shows that the tion came out of the page.
The top view is always equal to the other.
They form atractor with the mirror surface.
This result can be used to predict the outcome of the experiment.
The table shows the relative directions of light beams from a mirror.
Two mirrors stand on a table with their faces facing the same direction.
1i mirror 2 passes over the target.
The first figure shows the situation.
Laser orientations and posi tions will work.
The beam will pass over the target if we work backward.
Start by drawing the beam backwards from the target to mirror 2, then to mirror 1 and finally to an appropriate position for the laser pointer.
The figure is below.
The process in steps is represented by mirror 1 and then mirror 2 by the incident ray.
The target is shown in the figure below because the ray that hits mirror 2 is equal to 2u.
The sum of the angles in a triangle equals 180 degrees and the orientation of the normal line in the figure and the normal lines to the mirrors makes a 90 degree angle with each equal angle of incidence and reflection.
The reflection on the other side of the mirror is shown above.
The angle between the incident and the reflected ray incident and reflected angles relative to mirror 1's nor- is 21u mal line are the same.
We know how to direct the laser.
The law of reflection can now be applied to a relationship.
Everyone in the room can see the bright oriented at different angles when we shine a laser beam on a wal.
The law of reflection states that the light beam should reflect at a par- incident light.
A light beam shining on a smooth mirror can be represented by a single light ray.
The reflected laser light is not a single ray.
The light from the laser hits the parts of the surface that are oriented in different ways, and the light is reflected in many directions.
The wall was parallel to the incident.
If we had a clean mirror instead of the wall, the dust rays wouldn't stay parallel after reflection.
The parallel rays differed after reflection.
cal ed can see the reflected light when it is reflected by a "bumpy" surface.
Different parts of a light beam strike different parts of the surface at different angles with respect to the incident light.
The reflected rays go in different directions on the smooth surface.
The path of the laser beam in the ex periments is explained in Table 21.1.
Many people can see the path of light.
Sunlight coming through the church windows reflects off the dust in the air.
In the next example, we use both diffuse and specular reflection.
When light reaches the transparent sur, most of it passes into the room and off, the uncovered windows look almost black but then reflects diffusely many times inside so that little the outside walls do not.
If the windows are black, they are on the next page.
If you aren't in the one correct lo, you won't see much sunlight to your eyes.
You see very little light to explain why, our goal is to see that reflected light.
When the light shines on the rough, it reflects back at the sun in the window, which is why the walls of the house are bright.
The reflection from the window does not reach the eye.
The hole in the eye is similar to a win dow.
The pupil looks dark when the incident light enters.
The red eye effect is common at night or with low background lighting.
Light reflects from the red blood vessels in the back of the eye when the iris is open.
The reflected light makes thepupil appear red.
The sun reflects off the water's surface at the shore.
You can also see sea plants under the surface.
To see them, light must have entered the water, reflected off the rocks and plants, and then traveled from the surface to your eyes.
It is not easy to touch a rock under the surface of a pond or lake with a stick.
You missed the stick.
The baby's eyes in the photo show red circles in the dust.
The dome of a church has light from the flash.
Light rays are drawn from the red spots by shining a laser beam through the air.
We don't draw rays from the spots to our eyes for simplicity.
Light can leave through the bottom of the container.
There are red spots on the ceiling and floor of the 2 1 room and on the bottom of the container.
The path of the ray changes as it moves through the water.
When light shines at the air-water boundary at the top surface, the incident light beam is reflected back at the same angle as the angle of incidence.
Similar things happen to the light beam.
There are differences.
When ray 3 reaches the bottom water-air interface it is possible for it to partially reflect at the same angle as the angle of incidence, and for it to partially pass from the water into the air below the container.
When the incident light is represented by rays 1 and 3, it reflects back along the same line (rays 2 and 4) and passes into the second medium without Chapter 21 reflection.
The light bends and travels in a different direction if it is not in a straight line.
We could do an experiment similar to the one we did when studying the incident to measure mal line.
We could record the angles of the Normal line of incidence and refraction at the air-water and air-glass interface.
The pattern for al materials is different.
In 1621 the Dutch scientist Wil ebrord Snel found a pattern.
The number 1 to 2 is dependent on the two materials the light is traveling through.
Table 21.6 shows the ratio of the sines of the incident and refraction angles.
The glass used in Tables 21.5 and 21.6 will bend toward the normal, 1.53 if we define the index of air as 1.00.
The glass is more dense than water, and it reflects the light more toward the but light going from a higher to normal line.
As the blood's glu- is narrow, the Refractive index of blood increases.
Snel's law for this situa water is the blood.
The air of blood can help determine the concentration of blood sugar.
A small sample of blood is held by theDividing both sides of Snel's law.
The light leaves the blood and passes through the air to a row, which is higher than the normal index of refraction of tiny light detectors at the top.
The index of the blood is higher with pure blood.
The situation of the patient's blood is 1.37 instead of 1.43 as we sketch.
The incident angle is 40.0.
The angle is 61.7.
2 air is equal to 1.00.
The difficulty in touching an object under water with a stick was discussed at the beginning of the section.
Refraction can help us understand why.
You can see the example.
The air is shown below.
You can see the coin because sunlight enters your eye.
2 air is equal to 1.00.
The light rays in the water make a 42.1 incident angle relative to the normal line at the water-air interface.
The location of the coin is shown.
Suppose you light a coin with a laser light and send it to your eye.
Determine the angle of the light in the water.
When drawing diagrams, remember: a.
Most objects do not emit light.
We draw them as light-emitting objects because they are just sources of reflected light.
We draw the ones that are most convenient for describing the situation.
Light reaching our eyes is what makes us see objects.
Think of the rays that will reach the eyes of the observer.
Imagine placing a piece of glass in the water.
If the light reflects off the glass and reaches your eye, you will see it.
Light hitting the angle when traveling in water.
A good example is vegetable oil.
Part is reflected.
The magic trick described in the opening story is explained here.
In order for us to see things, they have to either reflect or emit light.
The reflecting object is different from the material around it.
Light traveled from water to air in two examples in the last section.
This behavior can be used to transmit light by optical fibers.
You perform a series of experiments in which an incident ray under water hits a water-air interface at an increasingly larger angle relative to the normal No refracted light.
The angle in the air between the reacted ray and the normal line gets bigger as the incident angle gets bigger.
The light is reflected back into the water at incident angles larger than the critical angle.
Remember that sin is 90.
If the incident angle is greater than uc, there is no solution to Snell's law, as sin u2 would be greater than 1.00.
The light is reflected back into the water.
There isn't a refracted ray.
Light travels from a me dium to a medium with a lowerRefractive index.
The Refractive index is a fundamental physical property of a substance and can be used to identify an unknown substance, confirm its purity, or measure its concentration.
Medical and industrial applications exist for refractometers.
In addition to detecting drug tampering in racehorses, veterinarians use portable refrac tometers to measure the total cholesterol in blood and urine.
As you move the light source apparatus, we have a sketch of the cal block.
The light ray 1 is what we assume.
The source of Ray 3 is always oriented in the same way as the curved 90deg in the blood.
The incident angle is the surface on the bottom and the point cal angle.
The glass-blood interface has a greater incident angle than Ray 4.
The critical angle is reflected back into the blood layer.
We drew a diagram for four lower blocks.
When the incident angle is larger than the critical angle, the detectors on the top surface stop detecting light.
We can use it.
The angle is equal to 0.850 and it is partially reflected.
When the blood was drawn from the incident.
The detectors on the top hemi blood bend back toward the normal line when the light reaches this angle.
As long as the incident into the hemispherical glass block above is small, the detectors on the top of reflected at the second interface are not shown.
Light will be detected by the net.
The critical angle is parallel to ray 1 in the top block and thus the apparatus is zero--ray 1 in the lower lows.
The first glass-blood inter- concentration has a greater incident angle than the index of refraction of the blood.
You don't see light at angles of and partial y reflected for a different sample of blood.
ray diagrams can be used to help solve light problems.
You can use the diagrams to evaluate the final answer.
If you want to understand how to interpret the mathematical description of light phenomena, be sure to use a ruler.
An observer at a lower elevation than she wants to see.
The first mirror reflects downward.
If you want to see around the second mirror, you have to travel through the tube ner of the Pentagon.
If the mirror is outside.
One mirror is pointing down.
The side of the building with the lower mirror will do.
A leaf has blown onto the lake.
The situation is sketched below.
All the known and unknown quantities are indicated.
We want to know what size leaf is needed so that any light reflected from the fish and reaching the water surface does not leave the water.
The leaf is hit by a light incident at a smaler angle.
The fish is a shining point particle and the leaf is circular.
Light rays from the object reach the observer.
Use the sketch and diagram to help.
We find that sin uc is 1.00 and it is unknown.
The angle with the greatest sin is quantities.
Evaluate the results to see if they are true.
The fish is not safe.
The equation could be used to solve the word problem.
The problem-solving procedure is changed.
A physical process is described in the equation.
The equation would provide a thick glass bottom of an aquarium and hit an inter solution if a narrow beam of light moved up through the vent.
The aquarium is surrounded by air.
The amount of the glass and water is unknown.
Determine the angle of the light in the water.
The equation appears to be based on a law.
2 sin 90 is the incident angle.
There is an incident on a different medium.
The light is not straight.
The critical angle is 48.
In this section we look at several applications of reflection, including fiber optics, mirages, and the color of the sky.
In telecommunications, fibers are used to transmit high-speed data and in medicine to see inside the human body during surgery.
Understanding the physics behind fiber optics will be helped by the example.
Imagine that you have a long glass block with a glass-air interface.
If the incident angle of light in the glass is greater than the critical angle, the light is reflected at the glass-air in terface.
It hits the bottom of the block at the same angle when it reaches the opposite side.
The situation is sketched.
Light travels from glass into the air.
The critical angle is greater than the incident angle for total internal total internal reflection.
The light is reflection and total.
The tom surfaces of the block are parallel to the top and bot.
The light is hitting the top of the glass.
The light leaves the block, the light travels the length of the block, or both.
If the light hits the top light, it will reflect back into the block.
For 45 incidence, total internal reflection occurs.