Physics Notes: Sight And Light

Free Web Hosting Provider - Web Hosting - E-commerce - High Speed Internet - Free Web Page

1. A wave is a disturbance of a medium which transports energy through a medium without permanently transporting matter. In a wave, particles of the medium are temporily displaced and then return to their original position.

Amplitude (A) - maximum displacement of particle from its equilibrium.

Frequency - number of waves that pass a point in a unit of time.
Wavelength (A to F)(B to G)(D to I)(E to J) - distance between two point in phase. It is symbolised by l (lambda).

Displacement: how far a particle is from equalibrium, it is a distance with a direction since it is a vector quantity; symbol is d and SI unit is m (for meters).
Speed - distance travelled by wave in unit time: the speed with which the wave moves through a medium is the product of the wavelength and the frequency; SI unit is m/s (v = ln or v = l f)

A Transverse Wave: As a Transverse wave passes through a point, the particles vibrate at right angles to the direction in which the wave is moving, eg electromagnetic waves

A Longitudinal Wave: As a Longitudinal wave passes through a point, the particles vibrate parallel to the direction in which the wave is moving, eg sound

4. An Electromagnetic wave consist of repeating patterns of electric and magnetic forces. They exist is the form of a transverse wave. eg light waves, radio waves, microwaves, X-rays, etc. Electromagnetic waves require a vibrating energy source but do not require a medium. All Electromagnetic waves have a velocity of approximately 3 x 10^8 m/s however their frequency (f), wavelength l (lamba) and energy levels (amplitude) vary.

A Mechanical wave is a disturbance that must pass through a medium. eg sound waves, water waves, etc. These require a vibrating energy source to create the disturbance and an elastic medium to transfer the disturbance. The medium does not absorb any energy.

5. Displacement Time Graphs :
Displacement Time Graphs are drawn to represent the motion of a single particle in an elastic medium as the result of transverse or longitudinal waves passing through
- a compression is a crest and rerefraction a trough
- Displacement refers to how far a particle is from equalibrium.
- Time refers to the period(T) to complete 1 wave.
To produce a Displacement Time Graph all you need is the period. For accurate drawing you will also need an amplitude.

Displacement Distance Graph
-Distance refers to the distance between two particles is phase (wavelength)
-Displacement refers to how far a particle is from equalibrium.
To produce a Displacement Distance Graph all you need is the wavelength. For accurate drawing you will also need an amplitude.

6. The wave equation is expresses as V= ln where
V = Velocity (m/s)
l = Frequency (Hz)
n = Wavelength (m)
Period(t) = 1/Frequency

7. Light is electromagnetic radiation:
Characteristics of light as an electromagnetic wave:
-it travels through a vacuum
-its speed in a vacuum is 3 x 10^8 m/s
-visible light ranges from 700 nm for red light to 400 nm for violet (blue) light where 1 nanometer = 1 x 10^-9 m
-the electromagnetic spectrum written in order of decreasing wavelength and increasing frequency: radiowaves (Emergency and Defence, A.M, F.M and T.V, microwave and Radar), Infra Red, Visible light (red, orange, yellow, green, blue, indigo, and violet), UV, X-rays

8. (Mechanical Waves) A wave is a means of transfering energy which passes through a medium. The particles of the medium are displaced as the wave passes through them, but return to their original position after the wave has passed , eg a cork in a tub. A wave is a way of transfering energy from one place to another. It could be likened to the energy transfer of an elastic collision.

9. The relationship between waves and rays is that they both travel in straight lines.

10. A ray of light approaching and reflecting off of a flat mirror, follows the law of reflection.

The incident ray of light (I) is the ray approaching the mirror.
The reflected ray of light (R) is the ray which leaves the mirror.
The Normal is a line that can be drawn perpendicular to the surface of the mirror where the incident ray strikes the mirror. The normal line divides the angle between the incident ray and the reflected ray into two equal angles.
The angle between the incident ray and the normal is known as the angle of incidence.
The angle between the reflected ray and the normal is known as the angle of reflection.

11. Reflection involves a change in direction of the light ray. The angle of incidence is the angle between the normal and the incident ray. The angle of reflection is the angle between the normal and the reflected ray. According to the law of reflection, the angle of incidence equals the angle of reflection.
The law of reflection is stated as q_i=q_rwhereq_i is the Angle of Incidence, q_ris the Angle of Reflection

12. A real image is an image formed on the same side of the mirror as the object and light passes through the actual image location. A virtual image is an image formed behind the mirror.

13. The image location is defined as the location where reflected rays intersect.

Images formed by plane mirrors are virtual, upright, left-right reversed, the same distance from the mirror as the object's distance, and the same size as the object.

Common Question
What Portion of a Mirror is Required?
The diagram below depicts a 6-foot tall man (represented by a black arrow) standing in front of a plane mirror. To see the image of his feet, he must sight along a line towards his feet; and to see the image of the top of his head, he must sight along a line towards the top of his head. The ray diagram depicts these lines of sight and the complete path of light from his extremities of the person to the mirror and to the eyes. In order to view his image, the man must look as low as point X (to see his feet) and as high as point Y (to see the tip of his head). The man only needs the portion of mirror extending between points X and Y in order to view his entire image. All other portions of the mirror are useless to the task of this man viewing his own image.

The diagram depicts some important information about plane mirrors. If you measure the height of the man (the vertical arrow) and measure the distance between points X and Y. The man is twice as tall as the distance between points X and Y. In other words, to view an image of yourself in a plane mirror, you will need an amount of mirror equal to one-half of your height. A 6-foot tall man needs 3-feet of mirror (positioned properly) in order to view his entire image.

What if the man stood a different distance from the mirror? Would the man need a different amount of mirror than one-half his height? These questions can be explored with the help of another ray diagram. The diagram below depicts a man standing different distances from a plane mirror. Ray diagrams for each situation (standing close and standing far away) are drawn. To assist in distinguishing between the two ray diagrams, they have been color coded. Red and blue light rays have been used for the situation in which the man is standing far away; green and purple light rays have been used for the situation in which the man is standing far away.

The two ray diagrams above demonstrate that the distance which a person stands from the mirror will not effect the amount of mirror which the person needs to see their image. Indeed in the diagram, the man's line of sight crosses the mirror at the same locations. A 6-foot man needs 3-feet of mirror to view his whole image regardless of where he is standing. In fact, the man needs the exact same 3-feet of mirror.

14. Steps of Geometric Constructions:
1.

Pick a point on the top of the object and draw two incident rays traveling towards the mirror. Draw one ray so that it passes exactly through the focal point on the way to the mirror. Draw the second ray such that it travels exactly parallel to the principal axis.

Once these incident rays strike the mirror, reflect them according to the two rules of:
"Any incident ray traveling parallel to the principal axis on the way to the mirror will pass through the focal point upon reflection."
"Any incident ray passing through the focal point on the way to the mirror will travel parallel to the principal axis upon reflection."

Mark the image of the top of the object.The image point of the top of the object is the point where the two reflected rays intersect. This is the point where all light from the top of the object would intersect upon reflecting off the mirror.

Repeat the process for the bottom of the object if it is not on the principle axis.

Mirror Situations and Images

Case 1: The object is located beyond the center of curvature (C)
Case 2: The object is located at the center of curvature (C)
Case 3: The object is located between the center of curvature (C) and the focal point (F)
Case 4: The object is located at the focal point (F)
Case 5: The object is located in front of the focal point (F)

Case 1: The object is located beyond C
The image will always be located somewhere in between the center of curvature and the focal point.
The image will be inverted i.e. the image is upside down.
The magnification is less than 1 i.e. the image is diminished.
The image is a real image.

Case 2: The object is located at C
The image will also be located at the center of curvature.
The image will be inverted.
The magnification is exactly 1. The image dimensions are equal to the object dimensions.
The image is a real image.

Case 3: The object is located between C and F
The image will be located beyond the center of curvature.
The image will be inverted.
The magnification is greater than 1. The image is enlarged.
The image is a real image.

Case 4: The object is located at F
No image is formed.
The light rays are traveling parallel to each other and cannot produce an image.

Case 5: The object is located in front of F
The image is a virtual image. The image will always be located somewhere on the opposite side of the mirror.
The image will be an upright image.
The magnification is greater than 1. The image is enlarged
The image location can only be found by extending the reflected rays backwards beyond the mirror. The point of their intersection is the virtual image location.

15. All images produced by convex mirrors are:
located behind the convex mirror
a virtual image
an upright image
Diminished (reduced in size i.e., smaller than the object)

Convex mirrors always produce images which share these characteristics.
The location of the object does not effect the characteristics of the image.
The diagram below shows seven different object locations (drawn and labeled in red) and their corresponding image locations
(drawn and labeled in blue).

The diagram shows that as the object distance is decreased, the image distance is decreased and the image size is increased. So as an object approaches the mirror, its virtual image on the opposite side of the mirror approaches the mirror as well; and at the same time, the image is becoming larger.

16. An image formed on the same side of the mirror, it is termed a real image.
An image formed on the opposite side of the mirror, it is termed a virtual image.
An increase in size of the image relative to the object is termed magnification.
A decrease in size of the image relative to the object is termed diminished.
An image the is upside down is termed inverted.
An image the is the right way up is termed upright.

17. The mirror formula is stated as 1/f = 1/u + 1/v where
f : Focal Length (m)
u: Object Distance (m)
v: Image Distance (m)

The magnification equation is stated as m = |v/u| = Ho/Hi where
m: Magnification (times)
u: Object Distance (same unit as v)
v: Image Distance (same unit as u)
|...|: magnitude
Ho: Height of Object (same unit as Hi)
Hi: Height of Image (same unit as Ho)

The sign conventions for the given quantities in the mirror equation and magnification equations are
f is +ve if the mirror is a concave mirror
f is -ve if the mirror is a convex mirror
v is +ve if the image is a real image and located on the object's side of the mirror.
v is -ve if the image is a virtual image and located behind the mirror.
hi is +ve if the image is an upright image (and therefore, also virtual)
hi is -ve if the image an inverted image (and therefore, also real)

18. Parabolic reflects can be used to produce parallel beams by placing the object at the focal point. Putting the bright light source any where else will cause the rays to focus.

19. Shaving and make-up also produce magnified upright images but you have to stand close to the mirror.
Dentists use a concave mirror placed close to the back of the tooth to give a magnified upright image.
Convex mirror give a wide range of view so are often used by long vehicles, as security mirrors in corner of shops, at blind corners.

20. In a vacuum, light travels with a speed of 3 x 10^8 m/s. This value will vary depending upon the medium of which it must pass through. The value will differ depending of the optical density of the medium rather then the density refering to the mass volume ratio.

21. Refraction is the bending of light as it enters a medium of different optical density. Light is refracted only when it hits a boundary at an angle. It is not refracted if it strikes perpendicular to the boundary.

When passing from air into glass, both the speed and the wavelength decrease. Finally, and most importantly, the light is observed to change directions as it crosses the boundary separating the air and the glass. The transmitted wave experiences this refraction at the boundary. As seen in the diagram below, each individual wavefront is bent only along the boundary. Once the wavefront has passes across the boundary, it travels in a straight line. For this reason, refraction is called a boundary behavior. A ray is drawn perpendicular to the wavefronts; this ray represents the direction which the light wave is traveling. Observe that the ray is a straight line inside of each of the two media, but bends at the boundary.

It is due to the difference of speed of one side of the wave compared to the other.

22.

When light enters a more optically dense medium, its speed is reduced. The angle of refraction is less than the angle of incidence. The refracted ray is said to be bent "toward the normal."

When light enters a less optically dense medium, its speed is increased. The angle of refraction is greater than the angle of incidence. The refracted ray is said to be bent "away from the normal."

The optical density of a medium determines the speed of light in the given medium. When drawing the wave fronts of a refracted wave, the distance between wavefronts differs.

23. The Absolute Index of Refraction is a constant that is characteristic of the substance. It is the ratio of the speed of light in a vacuum to the speed of light in that substance. Its variable is n and it has no units. It is calculated by the formula n = c/v where
n is the index of refraction
c is the speed of light in a vacuum
v is the speed of light in the medium.
The index of refraction of a vacuum (approximately air) is: n = 1.00

24. Snell's law states that the ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant (or the ratio of the indices of refraction of the two mediums). It is presented as sinq₁/sinq₂ = n₂/n₁=*Nwhich can be expressed as
n₁sinq₁= n₂sinq₂ where
n₁ is the index of refraction of the incident medium
n₂ is the index of refraction of the refractive medium,
q₁ is the angle of incidence
q₂ is the angle of refraction
N is the relative refractive index

25. Example: Measure , calculate , and draw in the refracted ray with the calculated angle of refraction.

In each of these two example problems, the angle of refraction is the variable to be determined. The indices of refraction (n₁ and n₂) are given and the angle of incidence needs to be measured. With three of the four variable known, substitution into Snell's law followed by algebraic manipulation will lead to the answer.

First, use a protractor to measure the angle of incidence. An appropriate measurement would be some angle close to 45-degrees.

Second, list all known values and the unknown value for which you wish to solve:
Given:
n₁ = 1.00
n₂ = 1.33
q₁ = 45 degrees
q₂= ?

Fourth, substitute known values into the equation and algebraically manipulate the equation in order to solve for the unknown variable - "q₂"
1.00 * sine (45 degrees) = 1.33 * sine (q₂)
0.7071 = 1.33 * sine (q₂)
0.532 = sine (q₂)
sine^-1 (0.532) = sine^-1 ( sine (q₂))
32.1 degrees = q₂

Proper algebra yields the answer of 32.1 degrees for the angle of refraction. The diagram showing the refracted ray is shown below.

Example B
Begin by measuring the angle of incidence with your protractor ---> 60 degrees.
List known Values:
n₁=1.00
n₂=1.52
q₂ = 60 degrees
q₂ = ??

Substitute and solve:
1.00*sine(60)=1.52*sine(q₂)
0.8660 = 1.52 *sine(q₂ )
0.570 = sine(q₂ )
34.7 degrees = q₂

Now draw the refracted ray at an angle of 34.7 degrees from the normal - see diagram below

Example Problem C
A ray of light in air is approaching the a layer of crown glass at an angle of 42.0 degrees. Determine the angle of refraction of the light ray upon entering the crown glass and upon leaving the crown glass. The refractive index for air is 1.00 and for crown glass 1.52 .

Refraction is taking place at two boundaries. Light refracts upon entering the layer (boundary #1: air to crown glass) and again
upon leaving the layer (boundary #2: crown glass to air).
Diagram:

Note that the angle of refraction at boundary #1 is the same as the angle of incidence at boundary #2.

Given:
Boundary #1
n₁ = 1.00 (from table)
n₂ = 1.52 (from table)
q₁= 42.0 degrees

Boundary #2
n₁ = 1.52 (from table)
n₂ = 1.00 (from table)
Find q₂ at boundary #1and q₂at boundary #2

Now list the relevant equation (Snell's Law), substitute known values into the equation, and perform the proper algebraic steps to solve for the unkown. Begin the process at boundary #1 and then repeat for boundary #2 until the final answer is found.

Boundary #1:
n₁sinq₁= n₂sinq₂
1.00 * sine (42.0 degrees) = 1.52 * sine (q₂)
0.669 = 1.52 * sine (q₂)
0.4402 = sine (q₂)
sine^-1 (0.4402) = sine^-1 ( sine (q₂))
26.1 degrees = q₂

The value of 26.1 degrees corresponds to the angle of refraction at boundary #1. Since boundary #1 is parallel to boundary #2, the angle of refraction at boundary #1 will be the same as the angle of incidence at boundary #2 (see diagram above). So now repeat the process in order to solve for the angle of refraction at boundary #2.

Boundary #2:
n₁sinq₁= n₂sinq₂
1.52 * sine (26.1 degrees) = 1.00 * sine (q₂)
1.52 * (0.4402) = 1.00 * sine (q₂)
0.6691 = sine (q₂)
sine^-1 (0.6691) = sine^-1 ( sine (q₂))
42.0 degrees = q₂

Example D
Perform the necessary calculations at each boundary in order to trace the path of the light ray through the following series of layers. Use a protractor and a ruler and show all your work.

First, draw normal and measure the angle of incidence at first boundary --> 30 degrees
Then, use the given n values and Snell's Law to calculate the q values at each boundary.
The angle of refraction at one boundary becomes the angle of incidence at the next boundary; e.g., the q₂ at the air-flint glass boundary is the q₁ at the flint glass-water boundary. Here are the calculated q₂ values:
air --> flint glass: 18 deg.
flint glass --> water: 22 deg.
water --> diamond: 12 deg.
diamond --> cub. zirconium: 13 deg.
cub. zirconium --> air: 30 deg.

The ray of light approached the top surface of the layer at 42 degrees and exitted through the bottom surface of the layer with the same angle of 42 degrees. The light ray refracted one direction upon entering and the other direction upon exiting; the two individual effects have balance each other and the ray is moving in the same direction. The important concept is this:

When light approaches a layer which has the shape of a parallelogram that is bounded on both sides by the same material, then the angle at which the light enters the material is equal to the angle at which light exits the layer.

If when sighting at an object, light from that object changes media on the way to your eye, a visual distortion
is likely to occur. If a pencil was submerged in water and viewed from the side. As you sighted at the portion of the pencil located above the water's surface, light travels directly from the pencil to your eye. Since this light does not change medium, it will not refract. As you sighted at the portion of the pencil which was submerged in the water, light traveled from water to air (or from water to glass to air). This light ray changed medium and subsequently underwent refraction. As a result, the image of the pencil appears to be broken. Furthermore, the portion of the pencil which is submerged in water appears to be wider than the portion of the pencil which is not submerged.

In this case, the only light which undergoes refraction is the light which travels from the submerged portion of the pencil, through the water, across the boundary, into the air, and ultimately to the eye. At the boundary, this ray refracts. The eye-brain interaction cannot account for the refraction of light. The brain judges the image location to be the location where light rays appear to originate from. This image location is the location where either reflected or refracted rays intersect. The eye and brain assume that light travels in a straight line and then extends all incoming rays of light backwards until they intersect. Light rays from the submerged portion of the pencil will intersect in a different location than light rays from the portion of the pencil which extends above the surface of the water. For this reason, the submerged portion of the pencil appears to be in a different location than the portion of the pencil which extends above the water. The diagram to the right shows a top view of the light path from the submerged portion of the pencil to each of your two eyes. Only the left and right extremities (edges) of the pencil are considered. The blue lines depict the path of light to your right eye and the red lines depict the path of light to your left eye. Observe that the light path has bent at the boundary. Dashed lines represent the extensions of the lines of sight backwards into the water. Observe that the these extension lines intersect at a given point; the point represents the image of the left and the right edge of the pencil. Finally, observe that the image of the pencil is wider than the actual pencil. A ray model of light which considers the refraction of light at boundaries adequately explains the broken pencil observations.

Since refraction of light occurs when it crosses the boundary, visual distortions often occur. These distortions occur when light changes medium as it travels from the object to our eyes.

Common Question

Arthur Podd's method of fishing involves spearing the fish while standing on the shore. The actual location of a fish is shown in the diagram above. Because of the refraction of light, the observed location of the fish is different than its actual location. Indicate on the diagram the approximate location where Arthur observes the fish to be.
Must Arthur aim above or below where the fish appears to be in order to strike the fish?

Arthur must aim at a position on the water below where the fish appears to be. Since light refracts away from the normal (water to air) as Arthur sights at the fish, the refracted ray when extended backwards passes over the head of where the fish actually is.

The light refracts as it travels from the fish in the water to the eyes of the hunter. The refraction occurs at the water-air boundary. Due to this bending of the path of light, the fish appears to be in a location where it isn't. A visual distortion occurs. Subsequently, the hunter launches the spear at the location where the fish is thought to be and misses the fish.

It is sufficient to say that as the hunter with the spear sights more perpendicular to the water, the amount of refraction decreases. The most successful hunters are those who sight perpendicular to the water.

The Archer fish is unlike any other fish in that the Archer fish finds its prey living outside the water. An insect, butterfly, spider or similar creature is the target of the Archer fish's powerful spray of water. The Archer fish will search for prey that is resting upon a branch or twig above the water. The fish then positions itself underneath the prey and with pin-point accuracy knocks the prey off the branch using a powerful jet of water. The prey falls to the water, and the Archer fish swims to the surface to retrieve its meal. The feat of shooting a stream of water to knock the prey off a branch is remarkable. The fact that the Archer fish can do this time and again with pin-point accuracy is even more remarkable. But most remarkable of all is that the Archer fish can accomplish this trick despite the fact that light from the target to its eye undergoes refraction at the air-water boundary. Such refraction would cause a visual distortion, making the prey appear to be in a location where it isn't. Yet the Archer fish is hardly ever fooled. What is the secret of the Archer fish?

There is only one condition in which light can pass from one medium to another, change its speed, and still not refract. If the light is traveling in a direction which is perpendicular to the boundary, no refraction occurs. As the light wave crosses over the boundary, its speed and wavelength still change. Yet, since the light wave is appoaching the boundary in a perpendicular direction, each point on the wavefront will reach the boundary at the same time; for this reason, there is no refraction of the light. Such a ray of light is said to be approaching the boundary while traveling along the normal. (The normal is a line drawn perpendicular to the surface.)

The secret to the Archer fish's success is that it lines up its sight with the prey from a position directly underneath the prey. From this vantage point, light from the prey travels directly to the fish's eye without undergoing a change in direction. Since the light is traveling along the normal to the surface, it does not refract; the light passes straight through the water to the fish's eyes. Normally, when light from an object changes medium on the way to the eye, there is a visual distortion of the image. But as the Archer fish sights along the normal, there is no refraction and no no visual distortion of the image. From this ideal position, the Archer fish is able to hit its prey time after time. The secret of the Archer fish is to use its understanding of the physics of refraction of light. The Archer fish knows that refraction is less when sighting along the normal.

A mirage is caused by two layers of air at different temperatures. Cold air is more dense than warm air, so it creates more resistance for light, making it travel more slowly. Note how on the bottom, the edges of the "waves" are farther apart than on the top; this is because the part of the wave in the warmer air travels faster than the part in the cooler air. (Note the closeup of the "refraction zone" in the image above.) The oasis mirage occurs when the air just above the ground gets hot because the ground heats it. This effect can commonly be seen on asphalt roads during the summer. The black color of the road gets the air above it hot very quickly. At the boundary of the hot and normal air, light is refracted. The viewer no longer sees the road or desert floor, but light refracted from the blue sky which, because it is on the ground, looks like water.

When the effect appears above the water, it is often referred to as a Fata Morgana. The phrase comes from the Italian version of the name of the sorceress Morgan Le Fay from the legends of King Arthur and Camelot. In ancient times these strange effects were considered the work of witchcraft.

Fata Morganas may also be the cause of legends about phantom ships that sail the sky. Reports of the ghost ship Flying Dutchman may well have been the reflection of some distant vessel.

When several boundaries of air are involved, a mirage can become even more complex as the light is refracted multiple times. This can make natural objects, like cliffs, appear as city buildings, or castles.

28. Total internal reflection occurs when light falls on a surface of a less optically dense medium at an angle of incidence equal to or greater than the critical angle of the substance. There is no refracted ray; the angle of refraction is 90° or greater.
The Critical Angle is the angle of incidence in the more optically dense medium at which total internal reflection occurs. At this angle of incidence, the angle of refraction in the less optically dense medium is exactly 90°.
Total internal reflection only occurs when a light ray passes from a more optically dense substance into a less optically dense substance.

29. Total internal reflection occurs when light falls on a surface of a less optically dense medium at an angle of incidence equal to or greater than the critical angle of the substance.

30. If light falls on a surface of a less optically dense medium at an angle of incidence equal to or greater than the critical angle of the substance total internal reflection will occur. When using Snells Law, you need to calculate the critical angle and if the problem fits the criteria, it will occur.

31. Dispersion is the scattering of light or the separating of light into its component colors. In this means, a spectrum can be seen. A spectrum is an array of colors.

32. Dispersion occurs because light of different wavelengths travel at different speeds through a medium. Red light has the longest wavelength and the highest velocity. Violet light has the shortest wavelength and the lowest velocity.

The dispersion of colors in a prism occurs because of something called the "refractive index" of the glass. Every material has a different refractive index. When light enters a different material (for example, when light traveling through the air enters the glass of a prism), the difference in the refractive index of air and glass causes the light to bend. The angle of bending is different for different wavelengths of light. As the white light moves through the two faces of the prism, the different colors bend their different amounts and spread out into a The visible light spectrum is shown in the diagram below.

33. The traditional rainbow is sunlight spread out into its spectrum of colors and diverted to the eye of the observer by water droplets. The "bow" part of the word describes the fact that the rainbow is a group of nearly circular arcs of color all having a common center. The sun is always behind you when you face a rainbow, and that the center of the circular arc of the rainbow is in the direction opposite to that of the sun. The rain is in the direction of the rainbow.

In a rainbow, raindrops in the air act as tiny prisms. Light enters the drop, reflects off of the side of the drop and exits, and in the process is broken into a spectum just like it is in a triangular glass prism. Like this:

The angle between the ray of light coming in and the red ray coming out of the drops is 42 degrees for red, and 40 degrees for violet. You can see in this diagram that the angles cause different colors from different drops to reach your eye, forming a circular rim of color in the sky - a rainbow! In a double rainbow, the second bow is produced because droplets can have two reflections internally and get the same effect. The droplets have to be the right size to get two reflections to work.

The traditional description of the rainbow is that it is made up of seven colors - red, orange, yellow, green, blue, indigo, and violet. Actually, the rainbow is a whole continuum of colors from red to violet and even beyond the colors that the eye can see. The colors of the rainbow arise from two basic facts:
Sunlight is made up of the whole range of colors that the eye can detect. The range of sunlight colors, when combined, looks white to the eye. This property of sunlight was first demonstrated by Sir Isaac Newton in 1666.
Light of different colors is refracted by different amounts when it passes from one medium (air, for example) into another (water or glass, for example).

34. Visible light waves consists of a continuous range of wavelengths or frequencies. When a light wave with a single frequency strikes an object, two things could happen. The light wave could be absorbed by the object, in which case its energy is converted to heat; the light wave could be reflected by the object; and the light wave could be transmitted by the object.

The color of the objects which we see are largely due to the way those objects interact with light and ultimately reflect or transmit it to our eyes. The color of an object is not actually within the object itself; rather, the color is in the light which shines upon it that ultimately becomes reflected or transmitted to our eyes. Any visible light which strikes the object and becomes reflected or transmitted to our eyes will contribute to the color appearance of that object. So the color is not in the object itself, but in the light which strikes the object. The only role that the object plays is that it might contain filters capable of absorbing one or more frequencies of the visible light which shine upon it. So if an object absorbs none of the white light it appears black. A substance only reflects its own colour. When blue light is reflected off a red surface, the light is absorbed and it appears black.

35.
1. A Convex lens (converging lens)
A convex lens is always thicker in the center than at the edges.
Light traveling through the lens goes slower through the thick center and faster through the thin ends causing the rays to focus or converge.
The focal length of a convex lens is always positive.
Real images are produced when the object is outside of the focus.
No image is produced when the object is at the focus.
Virtual images are produced when the object is within the focus.

2. A Concave lens (diverging lens)
A concave lens is always thinner in the middle than at the edges.
Light traveling through a concave lens goes faster through the center and slower through the ends. This causes the rays to diverge or not to focus.
The focal length of a concave lens is always negative.
Only virtual images are produced by a concave lens.

Principal axis: If a symmetrical lens is thought of as being a slice of a sphere, then there would be a line passing
through the center of the sphere and attaching to the mirror in the exact center of the lens. This
imaginary line is known as the principal axis.

Centre of lens: A lens also has an imaginary vertical axis which bisects the symmetrical lens in two.

Focal point: As mentioned above, light rays incident towards either face of the lens and traveling parallel to the principal axis will either converge or diverge. If the light rays converge (as in a converging lens), then they will converge to a point. This point is known as the focal point of the converging lens. If the light rays diverge(as in a diverging lens), then the diverging rays can be traced backwards until they intersect at a point. This point is known as the focal point of a diverging lens. The focal point is denoted by the letter F on the diagrams below. Note that each lens has two focal points - one on each side of the lens.

Focal length: Unlike mirrors, lenses can allow light to pass through either face, depending on where the incident rays are coming from. Subsequently, every lens has two possible focal points. The distance from the mirror to the focal point is known as the focal length (abbreviated by "f").

A lens has an imaginary point which we refer to as the 2F point. This is the point on the principal axis which is twice as far from the vertical axis as the focal point is.

37. We see an object because light from the object travels to our eyes as we sight along a line at the object. Similarly, we see an image of an object because light from the object reflects off a mirror or refracts through a transparent material and travel to our eyes as we sight at the image location of the object. From these two basic premises, we have defined the image location as the location in space where light appears to diverge from. Because light emanating from the object converges or appears to diverge from this location, a replica or reproduction of the object is created at this location.

38. Instructions for creating ray diagrams for lenses:
Rules
1. A ray passing through the optical center of the lens passes through unrefracted.
2. A ray incident upon the lens parallel to the principal axis passes through the focus; convex lenses use the focus on the opposite side of the lens

Steps
1.

Pick a point on the top of the object and draw three incident rays traveling towards the lens. Using a straight edge, accurately draw one ray so that it passes exactly through the focal point on the way to the lens. Draw the second ray such that it travels exactly parallel to the principal axis. Place arrowheads upon the rays to indicate their direction of travel. Draw the third incident ray such that it travels directly to the exact center of the lens.

Once these incident rays strike the lens, refract them according to the three rules of refraction for converging lenses. The ray that passes through the focal point on the way to the lens will refract and travel parallel to the principal axis. Use a straight edge to accurately draw its path. The ray which traveled parallel to the principal axis on the way to the lens will refract and travel through the focal point. And the ray which traveled to the exact center of the lens will continue in the same direction. Place arrowheads upon the rays to indicate their direction of travel. Extend the rays past their point of intersection.

Mark the image of the top of the object. The image point of the top of the object is the point where the two refracted rays intersect. All three rays should intersect at exactly the same point. This point is merely the point where all light from the top of the object would intersect upon refracting through the lens. Of course, the rest of the object has an image as well and it can be found by applying the same three steps to another chosen point.

Repeat the process for the bottom of the object. To determine the location, size, orientation, and type of image which is formed by the double convex lens requires determining where the image of the upper and lower extreme of the object is located and then tracing the entire image. After completing the first three steps, only the image location of the top extreme of the object has been found. Thus, the process must be repeated for the point on the bottom of the object. If the bottom of the object lies upon the principal axis (as it does in this example), then the image of this point will also lie upon the principal axis and be the same distance from the mirror as the image of the top of the object. At this point the entire image can be filled in.

It should be noted that the process of constructing a ray diagram is the same regardless of where the object is located. While the result of the ray diagram (image location, size, orientation, and type) is different, the same three rays are always drawn. The three rules of refraction are applied in order to determine the location where all refracted rays appear to diverge from (which for real images, is also the location where the refracted rays intersect).

Lens and Image Cases:

Oject is beyond the 2F point the image will be
located between the 2F point and the focal point on the opposite side of the lens.
Inverted
Reduced in size (smaller than the object)
Real

Object in located in front of the focal point, its image will be
Upright
Enlarged
Virtual

39. Instructions for creating ray diagrams for lenses:
1.A ray passing through the optical center of the lens passes through unrefracted.
2.A ray incident upon the lens parallel to the principal axis passes through the focus; concave lenses use the focus on the same side of the lens.

The method of drawing ray diagrams for a double concave lens is described below.
1.

Pick a point on the top of the object and draw three incident rays traveling towards the lens. Using a straight edge, accurately draw one ray so that it travels towards the focal point on the opposite side of the lens; this ray will strike the lens before reaching the focal point; stop the ray at the point of incidence with the lens. Draw the second ray such that it travels exactly parallel to the principal axis. Draw the third ray to the exact center of the lens. Place arrowheads upon the rays to indicate their direction of travel.

Once these incident rays strike the lens, refract them according to the three rules of refraction for double concave lenses.
The ray that travels towards the focal point will refract through the lens and travel parallel to the principal axis. Use a straight edge to accurately draw its path. The ray which traveled parallel to the principal axis on the way to the lens will refract and travel in a direction such that its extension passes through the focal point. Align a straight edge with the point of incidence and the focal point, and draw the second refracted ray. The ray which traveled to the exact center of the lens will continue to travel in the same direction. Place arrowheads upon the rays to indicate their direction of travel. The three rays should be diverging upon refraction.

Locate and mark the image of the top of the object. The image point of the top of the object is the point where the three refracted rays intersect. Since the three refracted rays are diverging, they must be extended behind the lens in order to intersect. Using a straight edge, extend each of the rays using dashed lines. Draw the extensions until they intersect. All three extensions should intersect in the same location. The point of intersection is the image point of the top of the object. The three refracted rays would appear to diverge from this point. This is merely the point where all light from the top of the object would appears to diverge from after refracting through the double concave lens. Of course, the rest of the object has an image as well and it can be found by applying the same three steps to another chosen point.

Repeat the process for the bottom of the object. The goal of a ray diagram is to determine the location, size, orientation, and type of image which is formed by the double concave lens. Typically, this requires determining where the image of the upper and lower extreme of the object is located and then tracing the entire image. After completing the first three steps, only the image location of the top extreme of the object has been found. Thus, the process must be repeated for the point on the bottom of the object. If the bottom of the object lies upon the principal axis (as it does in this example), then the image of this point will also lie upon the principal axis and be the same distance from the lens as the image of the top of the object. At this point the complete image can be filled in.

The ray diagram above illustrates that the image of an object in front of a double concave lens will be upright, reduced in size (smaller than the object), and virtual.

The diagrams above shows that in each case, the image is
a virtual image
an upright image
reduced in size (i.e., smaller than the object)

Unlike converging lenses, diverging lenses always produce images which share these characteristics. The location of the object does not effect the characteristics of the image. As such, the characteristics of the images formed by diverging lenses are easily predictable.

40. Conditions of Convex Lenses : Uses
The object is located in front of the focal point (F) : Spectacles for near sightedness
The object is located just inside the focal point (F) : Magnifying Glass
The object is located beyond the center of curvature (C) : Movie Projector
The object is located at 2F (2F) : Photocopier
The object is located beyond 2F (2F) : Eye, Camera

41. To obtain this type of numerical information, it is necessary to use the lens equation and the Magnification equation. The lens equation expresses the quantitative relationship between the object distance (u), the image distance (v), and the focal length (f). The equation is stated as follows:

The Magnification equation relates the ratio of the image distance and object distance to the ratio of the image height (hi) and object height (ho). The magnification equation is stated as follows:

These two equations can be combined to yield information about the image distance and image height if the object distance, object height, and focal length are known.

The sign conventions for the given quantities in the lens equation and magnification equations are
f is -ve if the lens is a concave lens
f is +ve if the lens is a convex lens
v is +ve if the image is a real image and located behind the lens.
v is -ve if the image is a virtual image and located on the object's side of the lens.
hi is +ve if the image is an upright image.
hi is -ve if the image an inverted image.

Example A
A 4.0-cm tall light bulb is placed a distance of 45.7 cm from a double convex lens having a focal length of 15.2 cm. Determine the image distance and the image size.

Like all problems in physics, begin by the identification of the unknown information.
ho = 4.0 cm
v = 45.7 cm
f = 15.2 cm

Next identify the unknown quantities which you wish to solve for.
u = ???
hi = ???

To determine the image distance, the lens equation will have to be used. The following lines represent the solution to the image distance; substitutions and algebraic steps are shown.
1/f = 1/v + 1/u
1/(15.2 cm) = 1/(45.7 cm) + 1/u
1/0.0658cm = 1/0.0219cm + 1/u
1/0.0439 = 1/u
u = 22.8 cm

42. Farsightedness or hyperopia is the inability of the eye to focus on nearby objects. The cure for the farsighted eye centers around assisting the lens in refracting the light. Since the lens can no longer assume the convex and highly curved shape which is required to view nearby objects, it needs some help. Thus, the farsighted eye is assisted by the use of a converging lens.
This converging lens will refract light before it enters the eye and subsequently decreases the image distance. By improving the refracting ability of the eye, the image of nearby objects is once again focused upon the retinal surface.

Nearsightedness or myopia is the inability of the eye to focus on distant objects. The nearsighted eye has no difficulty viewing nearby objects. But the ability to view distant objects requires that the light be refracted less. Nearsightedness will result if the light from distant objects is refracted more than is necessary. This tends to cause the images of distant objects to form at locations in front of the retina. On the retinal surface, where the light-detecting nerve cells are located, the image is not focused. These nerve cells thus detect a blurry image of distant objects.

The cure for the nearsighted eye is to equip it with a diverging lens. Since the nature of the problem of nearsightedness is that the light is focused in front of the retina, a diverging lens will serve to reduce the total refracting power of the eye. The diverging lens acts to diverge light before it reaches the eye; this light will then be converged by the cornea and lens and produce an image on the retina.