Light, Reflection, & Mirrors

Light, Reflection, & Mirrors

AP Physics B

Almost everything we see is made visible by the light it reflects. Some materials, such as air, water, or window glass, allow light to pass through. Other materials, such as thin paper or frosted glass, allow the passage of light in diffused directions so that we can’t see objects through them.

1. Scientists now agree that light has a dual nature, part particle and part wave.

NOTES 13 : Lights and Optics

Early Concepts of Light and SOUND

Light has been studied for thousands of years. Some ancient Greek philosophers thought that light consists of tiny particles, which enter the eye to create the sensation of vision. Others thought that vision resulted from streamers or filaments emitted by the eye making contact with an object.

27.1 Early Concepts of Light

Up until the time of Newton and beyond, most philosophers and scientists thought that light consisted of particles. However, one Greek, Empedocles, thought that light traveled in waves. One of Newton’s contemporaries, the Dutch scientist Christian Huygens, also argued that light was a wave.


2. The particle theory was supported by the fact that light seemed to move in straight lines instead of spreading out as waves do.

3. Wave Theory -Huygens showed that under some circumstances light does spread out and other scientists found evidence to support the wave theory. The wave theory became the accepted theory in the nineteenth century.


4. In 1905, Einstein published a theory explaining the photoelectric effect. According to this theory, light consists of particles called photons, massless bundles of concentrated electromagnetic energy.Scientists now agree that light has a dual nature, part particle and part wave.


What is the nature of light?


5. Michelson’s experimental value for the speed of light was 299,920 km/s, which is usually rounded to 300,000 km/s or 3X 108 m/s

27.2 The Speed of Light

It was not known whether light travels instantaneously or with finite speed until the late 1600s. Galileo tried to measure the time a light beam takes to travel to a distant mirror, but it was so short he couldn’t begin to measure it. Others tried the experiment at longer distances with lanterns they flashed on and off between distant mountaintops. All they succeeded in doing was measuring their own reaction times.


Olaus Roemer

The first demonstration that light travels at a finite speed was supplied by the Danish astronomer Olaus Roemer about 1675. Roemer carefully measured the periods of Jupiter’s moons.

• The innermost moon, Io, revolves around Jupiter in 42.5 hours.

• The Io disappears periodically into Jupiter’s shadow, so this period could be measured with great accuracy.


• Roemer found that while Earth was moving away from Jupiter, the periods of Io were all somewhat longer than average.

• When Earth was moving toward Jupiter, the measured periods were shorter than average.

• Roemer estimated that the cumulative discrepancy amounted to about 22 minutes.


Light coming from Io takes longer to reach Earth at position D than at position A. The extra distance that the light travels divided by the extra time it takes gives the speed of light.


Christian Huygens

Christian Huygens correctly interpreted this discrepancy.• The Io passed into Jupiter’s shadow at the

predicted time.• The light did not arrive until it had traveled the extra

distance across the diameter of Earth’s orbit. • This distance is now known to be 300,000,000 km.


Using the travel time of 1000 s for light to move across Earth’s orbit makes the calculation of the speed of light quite simple:

The speed of light is 300,000 km/s.


Albert MichelsonThe most famous experiment measuring the speed of light was performed by the American physicist Albert Michelson in 1880.

• Light was directed by a lens to an octagonal mirror.• A beam of light was reflected to a stationary mirror on a mountain 35

km away and then reflected back.• The distance was known, so Michelson had to find only the time it

took to make a round trip.


• When the mirror was spun, short bursts of light reached the stationary mirror and were reflected back to the spinning octagonal mirror.

• If the rotating mirror made one-eighth rotation while the light made the trip, the mirror reflected light to the observer.

• If the mirror was rotated too slowly or too quickly, it would not be in a position to reflect light.


a. Light is reflected back to the eyepiece when the mirror is at rest.


a. Light is reflected back to the eyepiece when the mirror is at rest. b. Reflected light fails to enter the eyepiece when the mirror spins too slowly . . .


a. Light is reflected back to the eyepiece when the mirror is at rest. b. Reflected light fails to enter the eyepiece when the mirror spins too slowly . . . c. . . . or too fast.


a. Light is reflected back to the eyepiece when the mirror is at rest. b. Reflected light fails to enter the eyepiece when the mirror spins too slowly . . . c. . . . or too fast. d. When the mirror rotates at the correct speed, light reaches the eyepiece.


When the light entered the eyepiece, the time for the light to make the trip and the time for the mirror to make one eighth of a rotation were the same. Michelson divided the 70-km round trip distance by this time and found the speed of light was 299,920 km/s, which is usually rounded to 300,000 km/s.Michelson received the 1907 Nobel Prize in physics for this experiment.


Facts about LightThe speed of light, c, is constant in a vacuum.

Light can be:•REFLECTED •ABSORBED•REFRACTED

Light is an electromagnetic wave in that it has wave like properties which can be influenced by electric and magnetic fields.

The speed of light in a vacuum is a universal constant. Light is so fast that if a beam of light could travel around Earth, it would make 7.5 trips in one second. Light takes 8 minutes to travel from the sun to Earth and 4 years from the next nearest star, Alpha Centauri. 6. The distance light travels in one year is called a light-year.


think!Light entered the eyepiece when Michelson’s octagonal mirror made exactly one eighth of a rotation during the time light traveled to the distant mountain and back. Would light enter the eyepiece if the mirror turned one quarter of a rotation in this time?


think!Light entered the eyepiece when Michelson’s octagonal mirror made exactly one eighth of a rotation during the time light traveled to the distant mountain and back. Would light enter the eyepiece if the mirror turned one quarter of a rotation in this time?

Answer:

Yes, light would enter the eyepiece whenever the octagonal mirror turned in multiples of 1/8 rotation— ¼, ½, 1, etc.—in the time the light made its round trip.


What was Michelson’s experimental value for the speed of light?


Facts about Light It is a form of Electromagnetic Energy It is a part of the Electromagnetic Spectrum and the only part we

can really see

Web Group Research- 5 minutes Electromagnetic waves

Group 1- Radio waves

Group 2- Microwaves

Group 3- Infrared waves

Group 4 – Light waves

Group 5- Ultraviolet rays

Group 6- Gamma rays

7. This energy travels in a wave that is partly electric and partly magnetic. Such a wave is an electromagnetic wave.

The electromagnetic spectrum consists of radio waves, microwaves, infrared, light, ultraviolet rays, X-rays, and gamma rays.

CW : Electromagnetic Waves p 408 Define Electromagnetic waves .Define the Electromagnetic SpectrumDraw and label the electromagnetic spectrumGive examples of these waves.Define each frequency in scientific notation format and standard decimal format

Light is a portion of the family of electromagnetic waves that includes radio waves, microwaves, and X-rays. The range of electromagnetic waves is called the electromagnetic spectrum.

8. Drawing :Electromagnetic Waves

The lowest frequency of light we can see appears red. The highest visible light, violet, has nearly twice the frequency of red light.9. Electromagnetic waves of frequencies lower than the red of visible light are called infrared. Heat lamps give off infrared waves. 10. Electromagnetic waves of frequencies higher than those of violet are called ultraviolet. They are responsible for sunburns.

Electromagnetic Waves

What are the waves of the electromagnetic spectrum?

27.3 Electromagnetic Waves

11. Light doesn’t travel through a mirror, but is returned by the mirror’s surface. These waves are reflected.

12. When waves strike the surface of a medium at an angle, their direction changes. These waves are refracted. Usually waves are partly reflected and partly refracted when they fall on a transparent medium.

13.

CW : Comparison of Materials: Score - # of Information

DEFINITION:

EXAMPLE OF MATERIALS

TRANSPARENTMATERIALS P409

OPAQUEMATERIALS P 410

WHAT HAPPENS TO LIGHT WHEN IT HITSTHIS TYPE OF MATERIAL?

Light passes through materials whose atoms absorb the energy and immediately reemit it as light.

Light and Transparent Materials

Materials that transmit light are transparent. Glass and water are transparent.Materials that are springy (elastic) respond more to vibrations at some frequencies than at others. The natural vibration frequencies of an electron depend on how strongly it is attached to a nearby nucleus.

27.4 Light and Transparent Materials

The electrons of atoms in glass can be imagined to be bound to the atomic nucleus as if connected by springs.


Electrons in glass have a natural vibration frequency in the ultraviolet range.

• In ultraviolet light, resonance occurs as the wave builds a large vibration between the electron and the nucleus.

• The energy received by the atom can be either passed on to neighboring atoms by collisions or reemitted as light.

• If ultraviolet light interacts with an atom that has the same natural frequency, the vibration amplitude is unusually large.


The atom typically holds on to this energy for about 1 million vibrations or 100 millionths of a second.During this time, the atom makes many collisions with other atoms and gives up its energy in the form of heat. That’s why glass is not transparent to ultraviolet.


When the electromagnetic wave has a lower frequency than ultraviolet, as visible light does, the electrons are forced into vibration with smaller amplitudes.

• The atom holds the energy for less time, with less chance of collision with neighboring atoms.

• Less energy is transferred as heat. • The energy of the vibrating electrons is

reemitted as transmitted light.


Glass is transparent to all the frequencies of visible light. The frequency of the reemitted light is identical to that of the light that produced the vibration to begin with. The main difference is a slight time delay between absorption and reemission.This time delay results in a lower average speed of light through a transparent material.


A light wave incident upon a pane of glass sets up vibrations in the atoms. Because of the time delay between absorptions and reemissions, the average speed of light in glass is less than c.


In a vacuum, the speed of light is a constant 300,000 km/s; we call this speed of light c.

• Light travels slightly less than c in the atmosphere, but the speed is usually rounded to c.

• In water, light travels at 75% of its speed in a vacuum, 0.75c.

• In glass, light travels at about 0.67c, depending on glass type.

• In a diamond, light travels at only 0.40c. When light emerges from these materials into the air, it travels at its original speed, c.


Materials such as glass and water are not as rigid to light waves.

• When light shines perpendicularly on the surface of still water, about 2% of its energy is reflected and the rest is transmitted.

• When light strikes glass perpendicularly, about 4% of its energy is reflected.

• Except for slight losses, the rest is transmitted.

Reflection

Infrared waves, with frequencies lower than visible light, vibrate not only the electrons, but also the entire structure of the glass. This vibration of the structure increases the internal energy of the glass and makes it warmer.Glass is transparent to visible light, but not to ultraviolet and infrared light.


What kind of materials does light pass through?


In opaque materials, any coordinated vibrations given by light to the atoms and molecules are turned into random kinetic energy—that is, into internal energy.

27.5 Opaque Materials

Materials that absorb light without reemission and thus allow no light through them are opaque. Wood, stone, and people are opaque.In opaque materials, any vibrations from light are turned into random kinetic energy—that is, into internal energy.The materials become slightly warmer.


Metals are also opaque. In metals, the outer electrons of atoms are not bound to any particular atom. When light shines on metal and sets these free electrons into vibration, their energy does not “spring” from atom to atom.It is reemitted as visible light. This reemitted light is seen as a reflection. That’s why metals are shiny.


Our atmosphere is transparent to visible light and some infrared, but almost opaque to high-frequency ultraviolet waves. The ultraviolet that gets through is responsible for sunburns. Clouds are semitransparent to ultraviolet, so you can get a sunburn on a cloudy day. Ultraviolet also reflects from sand and water, so you can sometimes get a sunburn while in the shade of a

beach umbrella.


think!Why is glass transparent to visible light but opaque to ultraviolet and infrared?


think!Why is glass transparent to visible light but opaque to ultraviolet and infrared?

Answer: The natural frequency of vibration for electrons in glass matches the frequency of ultraviolet light, so resonance in the glass occurs when ultraviolet waves shine on it. These vibrations generate heat instead of wave reemission, so the glass is opaque to ultraviolet. In the range of visible light, the forced vibrations of electrons in the glass result in reemission of light, so the glass is transparent. Lower-frequency infrared causes entire atomic structures to resonate so heat is generated, and the glass is opaque.


The color of an opaque object is the color of the light it reflects.

28.2 Color by Reflection

The colors of most objects around you are due to the way the objects reflect light.The color of an opaque object is the color of the light it reflects.Light reflects from objects similar to the way sound “reflects” from a tuning fork when another sets it into vibration.


If the material is transparent, the reemitted light passes through it. If the material is opaque, the light passes back into the medium from which it came. This is reflection.Most materials absorb light of some frequencies and reflect the rest. If a material absorbs light of most visible frequencies and reflects red, for example, the material appears red.

________________________________________END of CWCount the information = Score =_________


13. When white light falls on a flower, light of some frequencies is absorbed by the cells in the flower and some light is reflected.

14. Cells that contain chlorophyll absorb light of most frequencies and reflect the green part, so they appear green. The petals of a red rose, on the other hand, reflect primarily red light, with a lesser amount of blue.


15. Petals of most yellow flowers, such as daffodils, reflect red and green as well as yellow. Yellow daffodils reflect light of a broad band of frequencies.

16. The reflected colors of most objects are not pure single-frequency colors, but a spread of frequencies. So something yellow, for example, may simply be a mixture of colors without blue and violet—or it can be red and green together.


17. White light from the sun is a composite of all the visible frequencies. The brightness of solar frequencies is uneven.

18.. The lowest frequencies of sunlight, in the red region, are not as bright as those in the middle-range yellow and green region.

19.. Humans evolved in the presence of sunlight and we are most sensitive to yellow-green. .The blue portion of sunlight is not as bright, and the violet portion is even less bright.

28.4 Sunlight

The radiation curve of sunlight is a graph of brightness versus frequency. Sunlight is brightest in the yellow-green region.

28.4 Sunlight

When red light, green light, and blue light of equal brightness are projected on a white screen, the overlapping areas appear different colors.

28.5 Mixing Colored Light

21. Light of all the visible frequencies mixed together produces white. • White also results from the combination of only red, green, and blue

light.

22. Red and green light alone overlap to form yellow. 23. Red and blue light alone produce the bluish-red color called magenta. 24. Green and blue light alone produce the greenish-blue color called

cyan.

28.5 Mixing Colored Light

When two of the three additive primary colors are combined:22. red + green = yellow23. red + blue = magenta24. blue + green = cyan

25. When we add in the third color, we get white:•yellow + blue = white•magenta + green = white•cyan + red = white

28.6 Complementary Colors

26. Mixing Colored Pigments

Of the visible frequencies, violet light is scattered the most, followed by blue, green, yellow, orange, and red, in that order.

• Violet light is scattered more than blue but our eyes are not very sensitive to violet light.

• Our eyes are more sensitive to blue, so we see a blue sky.

27. Why the Sky Is Blue

28.8 Why the Sky Is Blue

28. Water is greenish blue because water molecules absorb red.

28.10 Why Water Is Greenish Blue

Ocean water is cyan because it absorbs red. The froth in the waves is white because its droplets of many sizes scatter many colors.

28.10 Why Water Is Greenish Blue

Light doesn’t travel through a mirror, but is returned by the mirror’s surface. These waves are reflected. When waves strike the surface of a medium at an angle, their direction changes. These waves are refracted. Usually waves are partly reflected and partly refracted when they fall on a transparent medium.

29. The return of a wave back to its original medium is called reflection. Fasten a spring to a wall and send a pulse along the spring’s length.The wall is a very rigid medium compared with the spring, so all the wave energy is reflected back along the spring. Waves that travel along the spring are almost totally reflected at the wall.

29.1 Reflection

30. Incident rays and reflected rays make equal angles with a line perpendicular to the surface, called the normal.

31. The angle between the incident ray and the normal is the angle of incidence.32. The angle between the reflected ray and the normal is the angle of reflection.33. Angle of incidence = Angle of reflection

29.2 The Law of Reflection

33. The law of reflection states that the angle of incidence and the angle of reflection are equal to each other.

The Law of Reflection

34. A virtual image appears to be in a location where light does not really reach. Mirrors produce only virtual images.

Your eye cannot ordinarily tell the difference between an object and its virtual image.

• The light enters your eye in exactly the same manner as it would if there really were an object where you see the image.

• The image is the same distance behind the mirror as the object is in front of it.

• The image and object are the same size.

29.3 Mirrors

Plane MirrorSuppose we had a flat , plane mirror mounted vertically. A candle is

placed 10 cm in front of the mirror. WHERE IS THE IMAGE OF THE CANDLE LOCATED?

mirror

Object Distance, Do = 10 cm

Same side as the object?

On the surface of the mirror?

Behind the mirror?

35. Plane MirrorSuppose we had a flat , plane mirror mounted vertically. A candle is

placed 10 cm in front of the mirror. WHERE IS THE IMAGE OF THE CANDLE LOCATED?

mirror

Object Distance, Do = 10 cm Image Distance, Di = 10 cm

Do=Di, and the heights are equal as well

Virtual Image

Virtual ImagesVirtual Images are basically images which cannot be

visually projected on a screen.If this box gave off light, we could project an image of this box on to a screen provided the screen was on the SAME SIDE as the box.

You would not be able to project the image of the vase or your face in a mirror on a screen, therefore it is a virtual image.

CONCLUSION: 36. VIRTUAL IMAGES are ALWAYS on the OPPOSITE side of the mirror relative to the object.

Real ImageReal Images are ones you can project on to a screen.

For MIRRORS they always appear on the SAME SIDE of the mirror as the object.

object

image

37. The characteristics of the image, however, may be different from the original object. These characteristics are:•SIZE (reduced,enlarged,same size)•POSITION (same side, opposite side)•ORIENTATION (right side up, inverted)

What if the mirror isn’t flat?

38. Spherical Mirrors – Concave & Convex

Also called CONVERGING mirrorAlso called DIVERGING mirror

The law of reflection holds for curved mirrors. However, the sizes and distances of object and image are no longer equal.

39. The virtual image formed by a convex mirror (a mirror that curves outward) is smaller and closer to the mirror than the object is.

29.3 Mirrors

The law of reflection holds for curved mirrors. However, the sizes and distances of object and image are no longer equal.

40.The virtual image formed by a convex mirror (a mirror that curves outward) is smaller and closer to the mirror than the object is.

41. When an object is close to a concave mirror (a mirror that curves inward), the virtual image is larger and farther away than the object is.

29.3 Mirrors

Converging (Concave) Mirror42. A converging mirror is one that is spherical in

nature by which it can FOCUS parallel light rays to a point directly in front of its surface. Every spherical mirror can do this and this special point is at a “fixed” position for every mirror. We call this point the FOCAL POINT. To find this point you MUST use light from “infinity”

Light from an “infinite” distance, most likely the sun.

Images Formed by Spherical Mirrors

Spherical mirrors are made from polished sections cut from a spherical surface.

43. The center of curvature, C, is the center of the sphere, of which the mirror is a section.

Of course, you don’t really make these mirrors by cutting out part of a sphere of glass.

C

44. The radius of curvature, R, is the radius of the sphere, or the distance from V to C.

C V

R

45. The principal axis (or optical axis) is the line that passes through the center of curvature and the center of the mirror.

C V

R

Principal Axis

The center of the mirror is often called the vertex of the mirror.

42. Paraxial rays are parallel to the principal axis of the mirror (from an object infinitely far away). Reflected paraxial rays pass through a common point known as the focal point F.

C VF

42. The focal length f is the distance from P to F. Your text shows that f = R/2.

C PF

f

R

Reality check: paraxial rays don’t really pass exactly through the focal point of a spherical mirror (“spherical aberration”).

C VF

If the mirror is small compared to its radius of curvature, or the object being imaged is close to the principal axis, then the rays essentially all focus at a single point.

C VF

We will assume mirrors with large radii of curvature and objects close to the principal axis.

In “real life” you would minimize aberration by using a parabolic mirror.

C VF

46. Converging (Concave) Mirror Since the mirror is

spherical it technically has a CENTER OF CURVATURE, C. The focal point happens to be HALF this distance.

We also draw a line through the center of the mirror and call it the PRINCIPAL AXIS.

Ray Diagrams for Mirrors

47. We can use three “principal rays” to construct images. In this example, the object is “outside” of F.

Ray 1 is parallel to the axis and reflects through F.Ray 2 passes through F before reflecting parallel to the axis.

Ray 3 passes through C and reflects back on itself.

FC

A fourth principal ray is the one directed at the vertex V.

V

We use three “principal rays” to construct images.

C

An image is formed where the rays converge.

48. The image from a concave mirror, object outside the focal point, is real, inverted, and smaller than the object.

“Real” image: you could put a camera there and detect the image.

F

Ray Diagrams for Concave Mirrors

Two rays would be enough to show us where the image is. We include the third ray for “safety.” You don’t have to use principal rays, but they are easiest to trace.

C

49. The image from a concave mirror, object inside the focal point, is virtual, upright, and larger than the object.Ray 1: parallel to the axis then through F.

Ray 2: “through” F then parallel to the axis.

Ray 3: “through” C.

F

With this size object, there was a bit of spherical aberration present, and I had to “cheat” my C a bit to the left to make the diagram look “nice.”

What do you mean, this is a virtual image? I can see it!

C

50. You could show that if an object is placed at the focal point, reflected rays all emerge parallel, and *no image is formed.Ray 1: parallel to the axis then through F.

Ray 2: “through” F then parallel to the axis. Can’t do!

Ray 3: through C.

F

no image

*Actually, the image is formed at infinity.Worth thinking about: what if the object is placed between F and C?

Ray Diagram ReviewA ray diagram is a pictorial representation of how the

light travels to form an image and can tell you the characteristics of the image.

Principal axisfCobject

Rule One: Draw a ray, starting from the top of the object, parallel to the principal axis and then through “f” after reflection.

Ray Diagrams


Rule Two: Draw a ray, starting from the top of the object, through the focal point, then parallel to the principal axis after reflection.

Ray Diagrams


Rule Three: Draw a ray, starting from the top of the object, through C, then back upon itself.

What do you notice about the three lines? THEY INTERSECT

The intersection is the location of the image.

Ray Diagram – Image Characteristics


After getting the intersection, draw an arrow down from the principal axis to the point of intersection. Then ask yourself these questions:

1) Is the image on the SAME or OPPOSITE side of the mirror as the object?Same, therefore it is a REAL IMAGE.2) Is the image ENLARGED or REDUCED?3) Is the image INVERTED or RIGHT SIDE UP?

M= hi/ho = -si/so

With a bit of geometry, you can show that

C

52. The magnification is the ratio of the image to the object height:

F

51. The Mirror Equation

f

so

si

ho

hi

f= for a flat mirror.

53. Sign Conventions for MagnificationOrientation of image with respect to object

Sign of M Type of image

Upright + Virtual

Inverted - Real

Sign conventions for the mirror equation:

C F

f

s

s’

y

y’

54. When the object, image, or focal point is on the reflecting side of the mirror, the distance is positive.55. When the object, image, or focal point is “behind” the mirror, the distance is negative.

56. The image height is positive if the image is upright, and 57. negative if the image is inverted relative to the object.

M= hi/ho = -si/so

“When the object, image, or focal point is “behind” the mirror, the distance is negative.”

If the image distance is negative, can the image be real?

The Mirror/Lens EquationIs there any OTHER way to predict image characteristics besides

the ray diagram? YES!

51. One way is to use the MIRROR/LENS equation to CALCULATE the position of the image.

Mirror/Lens Equation58. Assume that a certain concave spherical mirror

has a focal length of 10.0 cm. Locate the image for an object distance of 25 cm and describe the image’s characteristics.

What does this tell us? First we know the image is BETWEEN “C” & “f”. Since the image distance is POSITIVE the image is a REAL IMAGE.

Real image = positive image distanceVirtual image = negative image distance

What about the size and orientation?

Mirror/Lens Equation58. Assume that a certain concave spherical mirror

has a focal length of 10.0 cm. Locate the image for an object distance of 25 cm and describe the image’s characteristics.

16.67 cm

What does this tell us? First we know the image is BETWEEN “C” & “f”. Since the image distance is POSITIVE the image is a REAL IMAGE.

Real image = positive image distanceVirtual image = negative image distance

What about the size and orientation?

59. Magnification EquationTo calculate the orientation and size of the image we

use the MAGNIFICATION EQUATION.Here is how this works:•If we get a POSITIVE magnification, the image is UPRIGHT.•If we get a NEGATIVE magnification, the image is INVERTED •If the magnification value is GREATER than 1, the image is ENLARGED.•If the magnification value is LESS than 1, the image is REDUCED.•If the magnification value is EQUAL to 1, the image is the SAME SIZE as the object.

Using our previous data we see that our image was INVERTED, and REDUCED.

60. ExampleAssume that a certain concave spherical mirror has a focal

length of 10.0 cm. Locate the image for an object distance of 5 cm and describe the image’s characteristics.

-10 cm

2x

•VIRTUAL (opposite side)•Enlarged•Upright

Characteristics?

CW: Concave Mirrors 1

A concave shaving mirror has a focal length of 33 cm. Calculate the image position of a cologne bottle placed in front of the mirror at a distance of 93 cm.

CW: Concave Mirrors 1

A concave shaving mirror has a focal length of 33 cm. Calculate the image position of a cologne bottle placed in front of the mirror at a distance of 93 cm.

1/f = 1/so + 1/si 1/.33m = 1/.93m + 1/si 1/si = 1/.33m – 1/.93m 1/si = 3.03 – 1.07 si = .51 m

CW : Concave Mirrors 1

A concave shaving mirror has a focal length of 33 cm. Calculate the magnification of the image. Is the image real or virtual? Is the image inverted or upright?

CW : Concave Mirrors 1

A concave shaving mirror has a focal length of 33 cm. Calculate the magnification of the image. Is the image real or virtual? Is the image inverted or upright?

M = - si / so = - .51m / .93 m = -.55 image is inverted , reduced

CW 2: Concave Mirror

A concave shaving mirror has a focal length of 33 cm. Calculate the image position of a cologne bottle placed in front of the mirror at a distance of 93 cm. Draw a ray diagram to confirm your results.

Draw the diagram

fC

The image is inverted and about half the height of the object.

CW 3 : Concave Mirrors

An object of height 5 cm is placed 50 cm in front of a concave mirror whose focal length is 20cm.

a. where’s the image ?

b. Is it real or virtual ?

c. Is it upright or inverted ?

d. What’s the height of the image ?

1/ so + 1/ si = 1/f

1/ 50cm + 1/si = 1/ 20cm

1/si = .03 si= 33.33 cm

Real image , Inverted,

hi /ho = - si / s o = M

hi = - si ho / so = - 33.33cm (4cm ) / 50 cm

hi = - 2.67 cm inverted

CW:

An object of height of 5 cm is placed 30 cm in front of a convex mirror whose focal length is -40cm.

A. Where’s the image ? B. is it real or virtual ? C. Is it upright or inverted ? D. What’s the height of the image ?

1/ so + 1/ si = 1/f

1/ 30cm + 1/si = 1/ -40cm

Si = - 1/.058 = -17.24cm Image is virtual Image is upright hi/ ho = - si/ so

hi= - ho si /so = - 5cm (-17.24cm)/ 30cm

hi = 2.87 cm virtual image is upright

Applications of concave mirrors.

Shaving mirrors.

Makeup mirrors.

Solar cookers.

Ant fryers.

Flashlights, headlamps, stove reflectors.

Satellite dishes (when used with electromagnetic radiation).

Kid scarers.

CF

62. Ray Diagrams for Convex Mirrors

Ray 1: parallel to the axis then “through” F.

Ray 2: “through” Vertex.

Ray 3: “through” C.

The image is virtual, upright, and smaller than the object.

CF

63. Instead of sending ray 2 “through” V, we could have sent it “through” F. The ray is reflected parallel to the principal axis.

Your text talks about all four of the “principal rays” we have used.

CF

Because they are on the “other” side of the mirror from the object, s’ and f are negative.

64. The mirror equation still works for convex mirrors.

so

f

so

ho

ho

M= hi /ho = -si /so

Convex Mirrors

65. Convex mirrors take objects in a large field of view and produce a small image Side-view mirrors on cars are convex mirrors. That’s

why they say “objects are closer than they appear”

Convex Spherical Mirrors

66. A convex spherical mirror (diverging mirror) is silvered so that light is reflected from the sphere’s outer, convex surface The image distance is always negative! The image is always a virtual image! The focal length is negative !

Ray diagrams for convex mirrors

67. The focal point and center of curvature are behind the mirror’s surface A virtual, upright image is formed behind the mirror The magnification is always less than 1

f C

68. Drawing the reference rays

Ray 1 is drawn parallel to the principal axis beginning at the top of the object. It reflects from the mirror along a line that intersects the focal point

f C

Ray 2

Ray 2 starts from the top of the object and goes as though its going to intersect the focal point but it reflects parallel to the principal axis

Ray 1

f C

Ray 2

Ray 3

Ray 3 starts at the top of the object and goes as though its going to intersect the center of curvature

Ray 1

f C

Ray 2

Ray 3

69. Convex Spherical Image Formation The image forms at the intersection of any two of

the three rays behind the mirror.

Ray 1

f C

Ray 2

Ray 3

The rays do not intersect in front of the mirror!!

The ray diagram looks like the one on the previous slide, but with the object much further away (difficult to draw).

On reflecting sidepositive.

Not on reflecting sidenegative.

70.Example: a convex rearview car mirror has a radius of curvature of 40 cm. Determine the location of the image and its magnification for an object 10 m from the mirror.

1 1 1+ =

s s' f

1 1 1 1 1

= = s' f s -0.2 m 10 m

1 1 1

= s' -0.2 m 10 m

…algebra…

s'= -0.196 m= -19.6 cm

s' -0.196 m 1m= - = - =

s 10 m 51

Remind me… what does it say on passenger side rear view mirrors?

CW : Convex Mirrors

A candle is 49 cm in front of a convex spherical mirror that has a focal length of 35 cm. What are the image distance and magnification? Is the image virtual or real? Is the image inverted or upright? Draw a ray diagram to confirm your results.

CW : Convex Mirrors

A candle is 49 cm in front of a convex spherical mirror that has a focal length of 35 cm. What are the image distance and magnification? Is the image virtual or real? Is the image inverted or upright? Draw a ray diagram to confirm your results.

1/f = 1/so + 1/si 1/ si = -2.86-2.04 -1/.35m = 1/.49m + 1/si = - 4.9 1/si = -1/.35m – 1/.49m si = - .204 m

M = - Si / S0 = - (- .204 m) / .49 m M = 0.42 Positive = virtual , upright Less than 1 = reduced

Ray diagram for the Image

f C

Remember that you only need to draw two of the three rays to find the image.

71. Applications of convex mirrors.

Passenger side rear-view mirrors.

Grocery store aisle mirrors.

Anti-shoplifting (surveillance) mirrors.

Christmas tree ornaments.

Railroad crossing mirrors.

72. Sign Conventions Introduced Today

When the object, image, or focal point is on the reflecting side of the mirror, the distance is positive.

When the object, image, or focal point is “behind” the mirror, the distance is negative.

The image height is positive if the image is upright, and negative if the image is inverted relative to the object.

Hint: download Dr. Hale’s quick reference card.

73. Summary of Sign Conventions

Object Distance. When the object is on the same side as the incoming light, the object distance is positive (otherwise is negative).

Here’s a compact way of expressing mirror and lens (coming soon) sign conventions all at once.

Image Distance. When the image is on the same side as the outgoing light, the image distance is positive (otherwise is negative). (negative image distance virtual image)

Radius of Curvature. When the center of curvature C is on the same side as the outgoing light, R is positive (otherwise is negative).

(positive image distance real image)

I really don’t understand these ray diagrams!

If, like many students, you don’t understand ray diagrams, the following supplementary (will not be covered in lecture) slides may help.

For a concave lens, the center of curvature and focal point are on the same side of the lens as the object.

C F

Concave Lens

Light coming from the object parallel to the axis will always reflect through F.

C F

Concave Lens

Light coming from the object and passing through F before it hits the mirror will always reflect parallel to the axis.

C F

Concave Lens

Light coming from the object and passing through C before it hits the mirror will always reflect back on itself.

C F

Concave Lens

Light coming from the object and striking the vertex will reflect off with an outgoing angle equal to the incoming angle. This is often more difficult to draw (unless you measure the angle).

C

Concave Lens

CF

For a convex lens, the center of curvature and focal point are on the opposite side of the lens as the object. Light from the object will never actually pass through C or F.

Convex Lens

CF

Convex Lens

Light coming from the object parallel to the axis will always reflect back as if it had come from F.

Follow the path of the light back “through” the mirror to see where it appears to have come from.

CF

Light coming from the object and directed at F will always reflect back parallel to the axis.

Convex Lens


CF

Convex Lens

Light coming from the object and directed at C will always reflect back on itself.


CF

Convex Lens

Light coming from the object and striking the vertex will reflect back with an outgoing angle equal to the incoming angle. This is often more difficult to draw (unless you measure the angle).


CW:

AN OBJECT PLACED 70CM IN FRONT OF A SPHERICAL MIRROR FORMS A REAL IMAGE AT A DISTANCE OF 40 CM FROM THE MIRROR.

.a. Is the mirror concave or convex ?

b. What’s the mirror’s focal length ?

c. Is the image taller or shorter than the object ?

The fact that the image is real tells us that the mirror is not convex, since convex mirrors form only virtual images. The mirror is Concave

so = 70cm si = 40 cm

1/ so + 1/ si = 1/f 1/70cm + 1/ 40cm = 1/f 1/f = .039 f = 25.64 cm

Diffuse reflection is the reflection of light from a rough surface. Each ray obeys the law of reflection.The many different angles that incident light rays encounter at the surface cause reflection in many directions.

74. Diffuse Reflection

Diffuse reflection allows us to see most things around us. 75. Light is diffusely reflected from paper in many directions. 76. Light incident on a smooth mirror is only reflected in one direction.

29.4 Diffuse Reflection

When a wave that is traveling at an angle changes its speed upon crossing a boundary between two media, it bends.

29.6 77. Refraction

When a wave that is traveling at an angle changes its speed upon crossing a boundary between two media, it bends.Refraction is the bending of a wave as it crosses the boundary between two media at an angle.

29.6 Refraction

Water waves travel faster in deep water than in shallow water.

a. The wave refracts at the boundary where the depth changes.

29.6 Refraction

Water waves travel faster in deep water than in shallow water. a. The wave refracts at the boundary where the depth

changes. b. The sample ray is perpendicular to the wave front it

intersects.

29.6 Refraction

In drawing a diagram of a wave, it is convenient to draw lines, called wave fronts, that represent the positions of different crests.

• At each point along a wave front, the wave is moving perpendicular to the wave front.

• The direction of motion of the wave is represented by rays that are perpendicular to the wave fronts.

• Sometimes we analyze waves in terms of wave fronts, and at other times in terms of rays.

29.6 Refraction

What causes a wave to bend?

29.6 Refraction

Sound waves are refracted when parts of a wave front travel at different speeds.

29.7 Refraction of Sound

Sound refraction occurs in uneven winds or when sound is traveling through air of uneven temperature.

• On a warm day the air near the ground may be appreciably warmer than the air above.

• Sound travels faster in warmer air, so the speed of sound near the ground is increased.

• The refraction is not abrupt but gradual.

• Sound waves tend to bend away from warm ground, making it appear that the sound does not carry well.


When the layer of air near the ground is colder than the air above, the speed of sound near the ground is reduced. The higher speed of the wave fronts above causes a bending of the sound toward Earth. Sound can then be heard over considerably longer distances.


Changes in the speed of light as it passes from one medium to another, or variations in the temperatures and densities of the same medium, cause refraction.

29.8 Refraction of Light

Due to the refraction of light:• swimming pools appear shallower, • a pencil in a glass of water appears bent, • the air above a hot stove seems to shimmer, and• stars twinkle.

The directions of the light rays change because of refraction.


As a light wave passes from air into water, its speed decreases.


When light rays enter a medium in which their speed decreases, as when passing from air into water, the rays bend toward the normal. When light rays enter a medium in which their speed increases, such as from water into air, the rays bend away from the normal.The light paths are reversible for both reflection and refraction. If you can see somebody in a reflective or refractive device, such as a mirror or a prism, then that person can see you by looking through the device also.


Refraction

Example

Less Rigid

Medium

More Rigid

Medium

Refracted ray bends

towards

the normal.

Incident Ray

Refracted Ray

Index of Refraction, n

n = c / v

c : the speed of light in a vacuum, 3 x 108 m/sec v : speed of light in the medium. n : medium's index of refraction

Indices of Refraction

Vacuum 1.00Air 1.0003Water 1.33Ethanol 1.36Crown glass 1.52Quartz 1.52Diamond 2.42

Note

The speed of light has a lower speed in a more optically dense medium.

Problem

What is the speed of light in quartz?

V = c/n = 3X108 m/s / 1.52 V = Answer: 1.97 x 10 8 m/s

Snell’s Law

n1 sin 1 = n2 sin

Problem

What is the angle of refraction when a ray from air with an angle of incidence of 25 o is incident to water?

Draw the ray diagram. n1 sin θ1 = n2 sin θ2

1.0003 sin 25o = 1.33 sin θ2

Answer: 18.5 o

Total Internal Reflection

Can occur when ray goes from higher n to lower n.

Above a Critical angle (of incidence) the ray is reflected, not refracted

http://www.walter-fendt.de/ph14e/refraction.htm

For problems, set the angle of refraction to 90, and solve for critical angle

Problem

To solve for critical angle use the angle of refraction = 900

Find the critical angle for a light ray that is incident from water to air.

n1 sin θ1 = n2 sin θ2

1.33 sin θ1 = 1.0003 sin 90

θ1 = 48.80

Dispersion The index of refraction of glass is different

for the colors that make up white light because the speed of light is slightly different in glass for each frequency of light. (In vacuum all colors have speed c=3x108m/s.)

CW : Problem

The index of refraction for crown glass for red light is 1.514.What is the speed of red light in crown glass?

CW : Problem

The index of refraction for crown glass for red light is 1.514.What is the speed of red light in crown glass?

Answer: 1.98 x 10 8 m/s

CW: Refraction

A beam of light in air is incident upon a piece of glass , striking the surface at an angle of 300. If the index of refraction of the glass is 1.5 , what are the angles of reflection and refraction ?

CW: Refraction

A beam of light in air is incident upon a piece of glass , striking the surface at an angle of 300. If the index of refraction of the glass is 1.5 , what are the angles of reflection and refraction ?

Law of Reflection : 600

Snell’s Law : 350

n1Sin60o = n2sin θ2

1.0003 (.866) = 1.5 sin θ2

Angle of refraction = 19.2o Angle of reflection = 30o

Diffraction

AP Physics B

Superposition ..AKA….InterferenceOne of the characteristics of a WAVE is the ability to

undergo INTERFERENCE. There are TWO types.

We call these waves IN PHASE.

We call these waves OUT OF PHASE.

DiffractionWhen light OR sound is produced by TWO sources a

pattern results as a result of interference.

Interference Patterns

Diffraction is normally taken to refer to various phenomena which occur when a wave encounters an obstacle. It is described as the apparent bending of waves around small obstacles and the spreading out of waves past small openings

Diffraction – The Central Maximum

Suppose you had TWO sources each being allowed to emit a wave through a small opening or slit.

The distance between the slits denoted by, d.

The distance from the slit spacing to the screen is denoted by the letter, L.

If two waves go through the slit and then proceed straight ahead to the screen, they both cover the SAME DISTANCE and thus will have constructive interference. Their amplitudes will build and leave a very bright intense spot on the screen. We call this the CENTRAL MAXIMUM.

Diffraction – The Central Maximum

Here is the pattern you will see. Notice in figure 2 that there are several bright spots and dark areas in between.

The spot in the middle is the BRIGHTEST and thus the CENTRAL MAXIMUM. We call these spots FRINGES.

Notice we have additional bright spots, yet the intensity is a bit less. We denote these additional bright spots as ORDERS. So the first bright spot on either side of the central maximum is called the FIRST ORDER BRIGHT FRINGE. Figure 1 represents the intensity of the orders as we move farther from the bright central maximum.

Figure 1 Figure 2

Diffraction – Orders, mWe use the letter, m, to represent the ORDER of the fringe from the bright central.

CentralMaximum

First OrderBright Fringem = 1

First OrderBright Fringem = 1

Second OrderBright Fringem = 2

Second OrderBright Fringem = 2

It is important to understand that we see these bright fringes as a result of CONSTRUCTIVE INTERFERENCE.

Diffraction – Bright FringesThe reason you see additional bright fringes is because the waves CONSTRUCTIVELY build.

There is a difference however in the intensity as you saw in the previous slide.

As you can see in the picture, the BLUE WAVE has to travel farther than the RED WAVE to reach the screen at the position shown. For the BLUE WAVE and the RED WAVE to build constructively they MUST be IN PHASE.

Here is the question: HOW MUCH FARTHER DID THE BLUE WAVE HAVE TO TRAVEL SO THAT THEY BOTH HIT THE SCREEN IN PHASE?

Diffraction – Path differenceNotice that these 2 waves are IN PHASE. When they hit they screen they both hit at the same relative position, at the bottom of a crest.

How much farther did the red wave have to travel?

Exactly – ONE WAVELENGTH

The call this extra distance the PATH DIFFERENCE. The path difference and the ORDER of a fringe help to form a pattern.

Diffraction – Path Difference

11

2 2The bright fringes you see on either side of the central maximum are multiple wavelengths from the bright central. And it just so happens that the multiple is the ORDER.

Therefore, the PATH DIFFERENCE is equal to the ORDER times the WAVELENGTH

P.D. = mConstructive)

Diffraction – Dark Fringes

We see a definite DECREASE in intensity between the bright fringes. In the pattern we visible notice the DARK REGION. These are areas where DESTRUCTIVE INTERFERENCE has occurred. We call these areas DARK FRINGES or MINIMUMS.

Diffraction – Dark Fringes

CentralMaximum

ZERO OrderDark Fringem = 0

ZERO OrderDark Fringem = 0

First OrderDark Fringem = 1

First OrderDark Fringem = 1

It is important to understand that we see these dark fringes as a result of DESTRUCTIVE INTERFERENCE.

Diffraction – Dark FringesOn either side of the bright central maximum we see areas that are dark or minimum intensity.

Once again we notice that the BLUE WAVE had to travel farther than the RED WAVE to reach the screen. At this point , however, they are said to destructively build or that they are OUT OF PHASE.

Here is the question: HOW MUCH FARTHER DID THE BLUE WAVE HAVE TO TRAVEL SO THAT THEY BOTH HIT THE SCREEN OUT OF PHASE?

Diffraction – Path differenceNotice that these 2 waves are OUT OF PHASE. When they hit they screen they both hit at the different relative positions, one at the bottom of a crest and the other coming out of a trough. Thus their amplitudes SUBTRACT.

How much farther did the red wave have to travel?

Exactly – ONE HALF OF A WAVELENGTH

The call this extra distance the PATH DIFFERENCE. The path difference and the ORDER of a fringe help to form another pattern.

Diffraction – Path Difference

0.50.5

1.5 1.5The dark fringes you see on either side of the central maximum are multiple wavelengths from the bright central. And it just so happens that the multiple is the ORDER.

Therefore, the PATH DIFFERENCE is equal to the ORDER plus a HALF, times A WAVELENGTH

P.D. = (m+1/2)Destructive)

Path Difference - Summary

For CONSTRUCTIVE INTERFERENCE or MAXIMUMS use:

For DESTRUCTIVE INTERFERENCE or MINIMUMS use:

P.D. = m

P.D. = (m+1/2)Where “m”, is the ORDER. Refer to slides 7 and 12.

Young’s ExperimentIn 1801, Thomas Young successfully showed that light does produce an interference pattern. This famous experiment PROVES that light has WAVE PROPERTIES.

Suppose we have 2 slits separated by a distance, d and a distance L from a screen. Let point P be a bright fringe.

d

L

P

B.C.

y

We see in the figure that we can make a right triangle using, L, and y, which is the distance a fringe is from the bright central. We will use an angle, , from the point in the middle of the two slits.

We can find this angle using tangent!

Diffraction – Another way to look at “path difference”

d

P

B.C. d

P.D.

P.D.

P.D. = dsin

Notice the blue wave travels farther. The difference in distance is the path difference.

This right triangle is SIMILAR to the one made by y & L.

Similar Triangles lead to a path difference

L

y

d

dsin

These anglesAre EQUAL.

Diffraction – Putting it all togetherPath difference is equal to the following: m (m+1/2) dsin

Therefore, we can say:

Will be used to find theangle!

ExampleA viewing screen is separated from a double slit source by 1.2

m. The distance between the two slits is 0.030 mm. The second -order bright fringe ( m=2) is 4.5 cm from the central maximum. Determine the wavelength of light.

?

veConstructi bright""

045.02

100.32.1 5

mym

mxdmL )2.1

045.0(tan)(tan 11

L

y 2.15 degrees

2)15.2sin()103(

sin5x

md

5.62x10-7 m

Example

?

?0021.0

80.110450 9

d

my

mLmx

)8.1

0021.0(tan 1

d

xd

md

910450)210()067.0sin(

)21(sin

A light with wavelength, 450 nm, falls on a diffraction grating (multiple slits). On a screen 1.80 m away the distance between dark fringes on either side of the bright central is 4.20 mm. a) What is the separation between a set of slits? b) How many lines per meter are on the grating?

0.067 degrees

dord

meter

linesline

metersd

110.0001924 m 5197.2 lines/m

Mirage

http://www.warren-wilson.edu/~physics/PhysPhotOfWeek/2007PPOW/20070921RoadMirage/IMG_1444MirageCrop500.jpg

MirageTotal internal reflection occurs becausehot air has a lower n, than cold air.

http://www.warren-wilson.edu/~physics/PhysPhotOfWeek/2007PPOW/20070921RoadMirage/index.html

The laser beam bends toward the normal when it enters the water, and away from the normal when it leaves.


a. The apparent depth of the glass block is less than the real depth.


a. The apparent depth of the glass block is less than the real depth. b. The fish appears to be nearer than it actually is.


a. The apparent depth of the glass block is less than the real depth. b. The fish appears to be nearer than it actually is. c. The full glass mug appears to hold more root beer than it actually does.


These effects are due to the refraction of light whenever it crosses a boundary between air and another transparent medium.

What causes the refraction of light?


A mirage is caused by the refraction of light in Earth’s atmosphere.

29.9 Atmospheric Refraction

The speed of light in air is only 0.03% less than c, but in some situations, atmospheric refraction is quite noticeable. A distorted image, called a mirage, is caused by refraction of light in Earth’s atmosphere.

• A layer of very hot air is in contact with the ground on very hot days. • Light travels faster through it than through the cooler air above. • The speeding up of the part of the wave nearest the ground

produces a gradual bending of the light rays. • Light is refracted.



Wave fronts of light travel faster in the hot air near the ground, thereby bending the rays of light upward.


A motorist experiences a similar situation when driving along a hot road that appears to be wet ahead. The sky appears to be reflected from a wet surface, but, in fact, light from the sky is being refracted through a layer of hot air. A mirage is not a “trick of the mind.” A mirage is formed by real light and can be photographed.


When you watch the sun set, you see the sun for several minutes after it has really sunk below the horizon. Since the density of the atmosphere changes gradually, refracted rays bend gradually to produce a curved path. The same thing occurs at sunrise, so our daytimes are about 5 minutes longer because of atmospheric refraction.


When the sun is near the horizon, the rays from the lower edge are bent more than the rays from the upper edge. This produces a shortening of the vertical diameter and makes the sun look elliptical instead of round.Atmospheric refraction produces a “pumpkin” sun.


think!If the speed of light were the same for the various temperatures and densities of air, would there still be mirages?


think!If the speed of light were the same for the various temperatures and densities of air, would there still be mirages?

Answer:

No! There would be no refraction if light traveled at the same speed in air of different temperatures and densities.


What causes the appearance of a mirage?


Light’s Nature

Wave nature (electromagnetic wave)

Particle nature (bundles of energy called photons)

Past- Separate Theories of Either Wave or Particle Nature Corpuscular theory of Newton (1670) Light corpuscles have mass and travel at

extremely high speeds in straight lines

Huygens (1680) Wavelets-each point on a wavefront acts as a

source for the next wavefront

Why was it difficult to prove the wave part of the nature of light?

Proofs of Wave Nature

Thomas Young's Double Slit Experiment (1807) bright (constructive) and dark (destructive)

fringes seen on screen Thin Film Interference Patterns

Poisson/Arago Spot (1820) Diffraction fringes seen within and around a

small obstacle or through a narrow opening

Proof of Particle Nature:The Photoelectric Effect

Albert Einstein 1905 Light energy is quantized Photon is a quantum or packet of energy

The Photoelectric Effect

Heinrich Hertz first observed the photoelectric effect in 1887

Einstein explained it in 1905 and won the Nobel prize for this.

Thomas Young’s Double Slit Interference Experiment Showed an interference

pattern Measured the wavelength

of the light

Two Waves Interfering

Young’s Double SlitInterference Pattern

http://galileo.phys.virginia.edu/classes/USEM/SciImg/home_files/introduction_files/doubleslit.jpg

Interference of Waves FromTwo Sources Simulation “Ripple Tank”

http://www3.interscience.wiley.com:8100/legacy/college/halliday/0471320005/simulations6e/index.htm?newwindow=true

Interference

Young’s Double Slit Interference http://galileo.phys.virginia.edu/classes

/109N/more_stuff/flashlets/youngexpt4.htm

For Constructive Interference:

The waves must arrive to the point of study in phase.

So their path difference must be integral multiples of the wavelength:

L= m

m=0,1,2,3,………

For destructive interference:

, the waves must arrive to the point of study out of phase.

So the path difference must be an odd multiple of /2:

L= m m=1/2,3/2,5/2,….

Typical Question

Where is the first location of constructive or destructive interference?

Fo Constructive Interference of Waves from Two Sourcesx=Ltan

sinL/d

L=n

For small angles:Lsin~Ltan

dsinn

ndx L

d

L

x

n=0,1,2,3,…

Double Slit Interference

dsinn

ndx L

Constructive (brights) n=0,1,2,3,…..Destructive (darks) n=1/2, 3/2, 5/2,…..

Note:To find maximum # of fringes set to 90o for n.

Question

How does x change with wavelength?

How does x change with slit distance?

ProblemTwo slits are 0.05 m apart. A laser of

wavelength 633nm is incident to the slits.

A screen is placed 2m from the slits.

a) Calculate the position of the first and second bright fringe.

b) What is the maximum number of destructive interference spots there can be on either side of the central maximum?

Diffraction Grating

http://des.memphis.edu/lurbano/vpython/matter_interactions/spectrum/spectrum_02.jpg

Diffraction Grating Large number of equally spaced parallel slits. Equations are same as for double slit interference

but first calculate the d (slit separation) from the grating density, N.

d=1/N , N slits per unit length

dsinnndx

L

Constructive (brights) n=0,1,2,3,…..Destructive (darks) n=1/2, 3/2, 5/2,…..

Problem

A neon laser of wavelength 633nm is pointed

at a diffraction grating of 3000lines/cm. Find the angle where the first bright occurs.

(Hint: slit separation d is inverse of grating density)

Diffraction

Wave bends as it passes an obstacle.

Diffraction through a Narrow Slit

Each part of the slit acts as a point source that interferes with the others.

(Based on Huygens Principle)

Pattern of Diffraction of Light through a Narrow Slit

L

wx

Intensity of the Diffraction Interference Patterns Simulation “Interference of Light”

http://www3.interscience.wiley.com:8100/legacy/college/halliday/0471320005/simulations6e/index.htm?newwindow=true

Diffraction from Narrow Slit

wsinn nw y

L w: is the width of the slit

Destructive (dark fringes): m=0,1,2,3,….

Questions

How does x change with the width?

How does x change with the wavelength

Diffraction around a Penny and Poison Spot

Example of Diffraction

Documents

Light, Reflection, & Mirrors