chapter 33 light-reflection and refraction
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Chapter 33Light: Reflection and
Refraction
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Introduction
We see an object in one of two ways:
1. the object may reflect light. (this chapter)
2. the object may emit light. (quantum theory)
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33-1 The Ray (Particle) Model of Light
• Evidence suggests that light travels in straight lines under a wide variety of circumstances.
• We infer the positions of objects by assuming that light moves from the objects to our eyes in straight lines.
• This is the ray model of light.• Newton used the ray model.• Ray model explains reflection, refraction, and the
formation of images by mirrors and lenses. • This subject is often referred to as geometrical
optics.
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Light rays come from each single point on an object.
Light rays travel in straight lines.
Ray Model
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What Can Happen to Light?
When it strikes a surface of an object light can be:
1. Reflected
2. Transmitted
3. Absorbed (and transformed to thermal energy)
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Reflected
Perfect Mirror
Incident lightPractically all light is reflected.
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Transmitted
Window
Incident light
Most light is transmitted.
A little light is reflected.
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Absorbed
Chunk of very very black iron
Absorbed light energy heats up iron.
Incident light
Practically no light is reflected.
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33-2 The Speed of Light and Index of Refraction
• The accepted value today for the speed of light in a vacuum is c = 2.99792458 x 108 m/s. We usually round off to c = 3.0 x 108 m/s.
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Index of Refraction
The ratio of the speed of light in a vacuum (c = 3.00 x 108 m/s) to its speed v in a material is called the index of refraction, n.
n =
where n 1
c v
speed of light in a vacuum
speed of light in the medium
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33-3 Reflection; Image Formation by a Plane Mirror
The angle of incidence, i, equals the angle of reflection, r.
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narrow beam of light
Reflection
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Diffuse reflection
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mirror
white paper
narrow beam of light
specular reflection
diffuse reflection
17image distanceobject distance
virtual image
real image
Image Formation—Plane Mirror
=
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Image Formation—Plane Mirror
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Conceptual Example 33-1
How tall must a full-length mirror be?
A women 1.60 m tall stands in front of a vertical plane mirror. What is the minimum height of the mirror, and how high must it lower edge be above the floor, if she is to see her whole body? (Assume her eyes are 10 cm below the top of her head.)
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r
i
AE/2
AE/2
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Conceptual Example 33-2
Is the photo upside down?
Close examination of the photograph on the first page of this chapter reveals that the top portion, the image of the Sun is seen clearly, whereas in the lower portion, the image of the Sun is partially blocked by the tree branches. Why isn’t this reflection in the water an exact replica of the real scene?
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33-4 Formation of Images by Spherical Mirrors
• The most common curved mirrors are spherical, which means they form a section of a sphere.
• There are two kinds of spherical mirrors, convex and concave.
• Convex: reflection takes place on the outer surface of the spherical shape.
• Concave: reflection takes place on the inner surface of the sphere. Looks like a cave.
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Gives close-up view.
Gives wide-angle view.
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These rays miss the mirror.
r >> d
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Rays do not all come to a single point.
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Rays parallel to the principal axis of the mirror come to a focus at F, called the focal point, as long as the mirror is small in diameter, d, as compared to its radius of curvature, r. In that case will be small and the rays will cross each other at very nearly the same point.
d
d << rFB FAr 2
f =
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Object Not at Infinity
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image point
Complete image (real) of object is inverted.
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Mirror Equation
+ =1 do
1 di
1 f
where f = r/2
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Lateral Magnification
m = hi ho
di do
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Example 33-3
Image in a concave mirror.
A 1.50-cm-high diamond ring is placed 20.0 cm from a concave mirror whose radius of curvature is 30.0 cm. Determine (a) the position of the image, and (b) its size.
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Conceptual Example 33-4
Reversible rays.
If the object in Example 23-3 is placed instead where the image is in Fig. 23-15, where will the new image be?
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Example 33-5
Object closer to a concave mirror.
A 1.00-cm-high object is placed 10.0 cm from a concave mirror whose radius of curvature is 30.0 cm. (a) Draw the ray diagram to locate the position of the image. (b) Determine the position of the image and the magnification analytically.
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Example 33-5b
Object farther from a concave mirror.
A 1.00-cm-high object is placed 45.0 cm from a concave mirror whose radius of curvature is 30.0 cm. (a) Draw the ray diagram to locate the position of the image. (b) Determine the position of the image and the magnification analytically.
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Example 33-6
Convex rearview mirror.
A convex rearview mirror has a radius of curvature of 40.0 cm. Determine the location of the image and its magnification for an object 10.0 m from the mirror.
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Spherical Mirror Images
Case 1: If object is closer to the mirror than the focal point, the image will be virtual, upright, and magnified.
Case 2: If the object is farther from the mirror than the focal point, the image is real and inverted.
Case 2: Whether its magnification is greater or less than 1.0 depends on the position of the object
relative to the center of curvature, point C. d0 < C image magnification is > 1.0; d0 > C image magnification is < 1.0,
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Convex Mirrors
Focal point behind mirror.
No matter where an object is placed, the image will appear virtual and erect.
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Example 33-6
Convex rearview mirror.
A convex rearview mirror has a radius of curvature of 40.0 cm. Determine the location of the image and its magnification for an object 10.0 m from the mirror.
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Problem Solving
Sign Conventions:
(a) When the object, image, or focal point is on the reflecting side of the mirror (on the left), the corresponding distance is considered positive. If any of these point is behind the mirror (on the right) the corresponding distance is negative.
(b) The image height hi is positive if the image is upright, and negative if inverted, relative to the object (ho is always taken as positive.
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33-5 Refraction: Snell’s Law
n1sin 1 = n2sin 2
Derived experimentally in about 1621 by Willebrord Snell (1591 – 1626).
Angle of incidence
Angle of refraction
If a ray of light is incident on a surface at an angle, other than perpendicular, the ray is bent as it enters the new medium. This bending is called refraction.
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Note: Angles are measured from normal.
n2 > n1 Ray bent toward normal
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n1 > n2 Ray bent away from normal
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Example 33-7
Refraction through flat glass.
Light strikes a flat piece of glass at an incident angle of 60o, as shown. If the index of refraction of the glass is 1.50, (a) what is the angle of refraction A in the glass; (b) what is the angle B at which the ray emerges from the glass?
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Example 33-8
Apparent depth of a pool.
A swimmer has dropped her goggles in the shallow end of the pool, marked as 1.0 m deep. But the goggles don’t look that deep. Why? How deep do the goggles appear to be when you look straight down into the water?
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33-6 Visible Spectrum and Dispersion
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n Dependence on Wavelength
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Dispersion
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33-7 Total Internal Reflection; Fiber Optics
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Angle of refraction = 90o
sin C = n2
n1
Angle of incidence
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Conceptual Example 33-9
View up from under water.
Describe what a person would see who looked up at the world from beneath the perfectly smooth surface of a lake or swimming pool.
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Thin Lenses: History
• Thin lens were first used for practical purposes by a Dutch merchant, Anton van Leeuwenhoek (1632 – 1723).
• He used very small pieces of glass (it is easier to have a flawless small piece of glass than a flawless large one) and polished them so accurately that he could get magnifications of more than 200 without loss of detail.
• He was able to see blood capillaries, and tiny living animals (protozoa).
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Microscopes
• Such strong magnifying lenses are microscopes (from Greek words meaning “to see the small”).
• A microscope, like the one Leeuwenhoek used, made with one lens are called “simple” microscopes.
• If two lenses are used it is called a “compound” microscope.
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Telescopes
• The word telescope comes from the Greek “to see the distant.
• The telescope is supposed to have been invented by an apprentice-boy in the shop of the Dutch spectacle maker Hans Lipershey (ca. (1570 – 1619) in about 1608.
• Galileo Galilei (1564 – 1642), upon hearing rumors of the new device, experimented with lenses until he had built the first practical telescope in 1610.
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The Starry Messenger
(1610)• First publication dealing with telescopic observations. Saw that the milky way is composed of many
stars. Used observations to argue against the
perfection of Greek science.
• Vast expanse of the universe matched Copernicus's view.
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The Moon
• Galileo saw that the moon had mountains and craters.
• This was contrary to the Greek claim that all heavenly bodies are smooth spheres.
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Galileo's Moon
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Galileo's Moon
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Moon From Apollo 16
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Jupiter’s MoonsGalileo discovered four of Jupiter's twelve moons.Analogous to the Copernican model.Not described by Greek cosmology. Either
Greeks were wrong or the universe changed, which the Greeks claimed would not happen.
Orbiting moons would break Aristotle's crystal
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Jupiter’s Moons
EAST
JAN 7, 1610
JAN 8, 1610
JAN 10, 1610
JAN 11,1610
JAN 12, 1610
JAN 13, 1610
* * *
** *
* *
* *
* * *** **
WEST
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Jupiter’s Moons
Photographed by NASA’s Galileo spacecraft.
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Letters On Sunspots
(1613)• Correctly determined that sunspots were similar to clouds, close to the suns surface -- not planets closer to the sun than mercury.
• Observed the curious shape of Saturn, but could offer no explanation.
• Reports on the phases of Venus -- like the moon. (This is the one piece of hard evidence against the Ptolomiac model that Galileo had.)
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Galileo’s Drawings of Sunspots
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Saturn
Galileo saw that there was something odd about the shape of Saturn, but he could not explain what is was.
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Saturn
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Saturn
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Venus
• Discovered the phases of Venus.
Saw a nearly full face.
In the Greek system Venus should not have a full face.
• Saw that the angular width of the stars did not increase over naked eye observations. This was evidence for the great distance of the stars.
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The Phases of Venus
Strong scientific argument against the Polemic system.
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Venus
Photo by Mariner spacecraft
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23-7 Thin Lenses; Ray Tracing
A lens can produce an image of an object only because it can bend light rays; but it can bend light rays only if its index of refraction differs from that of the surrounding medium.
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23-7 Thin Lenses; Ray Tracing
A thin lens is defined as a lens whose diameter is very small compared to its radius of curvature. That is:
d << r
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Converging Lenses
Any lens that is thicker in the center than at the edges will make parallel rays converge to a point, is called a converging lens.
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Focal Point for Thin Converging Lenses
• The axis of a thin lens is a straight line passing through the very center of the lens and perpendicular to the two surfaces.
• From Snell’s law: each ray is bent toward the axis at both lens surfaces.
• If rays parallel to the axis fall on a thin lens, they will be focused to a focal point, F.
• The focal point for a thin lens is defined as the image point on the principal axis for an object at infinity.
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Focal Length for Thin Converging Lenses
• The focal point for a thin lens can be found by locating the point where the Sun’s rays are brought into sharp image.
• The distance from the focal point to the center of the lens is called the focal length, f.
• If parallel rays fall on a lens from an angle to the axis, they focus at a point Fa.
• The plain in which all points such as F and Fa is called the focal plane of the lens.
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Converging Lens
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Converging Lens
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Converging Lens
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Converging Lens
Image is real.
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Diverging Lenses
• Lenses that are thinner at the center than at the edges are called diverging lenses.
• Diverging lens make parallel light diverge.
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Focal Point/Length for Thin Diverging Lenses
• The focal point, F, of a diverging lens is defined as that point from which refracted rays, originating from parallel incident rays, seem to emerge.
• The distance from the focal point, F, of a diverging lens is called the focal length.
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Diverging Lens
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Diverging Lens
Image is virtual.
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Real or Virtual
Real images form on the side of a lens that is opposite the object, and virtual images from on the same side of the object.
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Images
(a)A real, inverted image is formed by a converging lens when the object is outside the focal point.
(b)A virtual, non-inverted image is formed by a converging lens when the object is inside the focal point.
(c)A diverging lens forms a virtual, non-inverted image whether the object is inside or outside the focal point.
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Power of Lens
P 1f
The unit of lens power is the diopter (D).
1 D = 1 m-1
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23-8 The Lens Equation
The lens equation relates the image distance to the object distance and the focal length of the lens.
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Lens Equation Converging Lens
+ =1 do
1 di
1 f
where f = r/2
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Lens Equation Diverging Lens
=1 do
1 di
1 f
where f = r/2
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Sign Conventions
• The focal length (f) is positive for converging lenses and negative for diverging lenses.
• The object distance (do) is positive if it is on the side of the lens from which the light is coming
• The image distance (di) is positive if it is on the opposite side of the lens from which the light is coming; if it on the same side it is negative.
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Sign Conventions
• Equivalently, the image distance (di) is positive for a real image, negative for a virtual image.
• The height of the image (hi) is positive if the image is upright, and negative if the image is inverted relative to the object. (ho is always taken as positive.)
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Lateral Magnification
m = hi ho
di do
Applies to both converging and diverging lenses. For an upright image the magnification (m) is positive, and for an inverted image it is negative.
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23-9 Problem Solving for Lenses
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Example 23-11
Image formed by a converging lens.
What is (a) the position, and (b) the size, of the image of a large 7.6-cm-high flower placed 1.00 m from a +50.0-mm-focal-length camera lens?
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Example 23-12
Object close to converging lens.
An object is placed 10 cm from a 15-cm-focal-length converging lens. Determine the image position and size (a) analytically, and (b) using a ray diagram.
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Example 23-13
Diverging lens.
Where must a small insect be placed if a 25-cm-focal-length diverging lens is to form a virtual image 20 cm in front of the lens?
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Homework Problem 4
A person whose eyes are 1.62 m above the floor stands 2.10 m in front of a vertical plane mirror whose bottom edge is 43 cm above the floor. What is the horizontal distance x to the base of the wall supporting the mirror of the nearest point on the floor that can be seen reflected in the mirror?
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Homework Problem 10
How far from a concave mirror (radius 27.0 cm) must an object be place if its image is to be at infinity?
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Homework Problem 12
If you look at yourself in a shiny Christmas tree ball with a diameter of 9.0 cm when your face is 30.0 cm away from it, where is your image? Is it real or virtual? Is it upright or inverted?
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Homework Problem 15
Some rearview mirrors produce images of cars to your rear that are a bit smaller than they would be if the mirror were flat. The image is upright. Are the mirrors concave or convex? What type and height of image would such a mirror produce of a car that was 1.3 m high and was 15.0 m behind you, assuming the mirror’s radius of curvature is 3.2 m?
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Homework Problem 28
The speed of light in a certain substance is 85 percent of its value in water. What is the index of refraction of this substance?
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Homework Problem 33
Rays of the Sun are seen to make a 21.0o angle to the vertical beneath the water. At what angle above the horizon is the Sun?
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Homework Problem 42
A beam of light is emitted in a pool of water from a depth of 62.0 cm. Where must is strike the air-water interface, relative to the spot directly above it, in order that the light not exit from the water?
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Homework Problem 43
A ray of light enters a light fiber at an angle of 15o with the long axis of the fiber, as shown. Calculate the distance the light ray travels between successive reflections off the sides of the fiber. Assume that the fiber has an index of refraction of 1.6 and is 10-4 m in diameter.
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Homework Problem 51
(a) What is the power of a 29.5-cm-focal-length lens? (b) What is the focal length of a –6.25-diopter lens? Are these lenses converging or diverging?
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Homework Problem 52
A stamp collector uses a converging lens with a focal length 24 cm to view a stamp 18 cm in front of the lens. (a) Where is the image located? (b) what is the magnification?
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Homework Problem 54
An 80-mm-focal-length lens is used to focus an image on the film of a camera. The maximum distance allowed between the lens and the film plane is 120 mm. (a) How far ahead of the film should the lens be located if the object to be photographed is 10.0 m away? (b) 3.0 m away? (c) 1.0 m away? (d) What is the closest object this lens could photograph sharply?
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Newton’s Analysis of Light
2
n2
n1
=
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Chapter 33Light: Reflection and
Refraction
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Introduction
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33-1 The Ray Model of Light
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33-3 Reflection; Image Formation by a Plane Mirror
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Conceptual Example 32-1
How tall must a full-length mirror be?
A women 1.60 m tall stands in front of a vertical plane mirror. What is the minimum height of the mirror, and how high must its lower edge be from the floor, if she is to see her whole body? Assume her eyes are 10 cm below the top of her head.
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Conceptual Example 32-2
Is the photo upside down?
Close examination of the photograph on the first page of this chapter reveals that in the top portion, the image of the Sun is seen clearly, whereas in the lower portion, the image of the Sun is partially blocked by the tree branches. Why isn’t the reflection in the water the exact replica of the real scene?
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33-4 Formation of Images by Spherical Mirrors
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Example 33-3
Image in a concave mirror.
A 1.50-cm-high diamond ring is placed 20.0 cm from a concave mirror whose radius of curvature is 30.0 cm. Determine (a) the position of the image, and (b) its size.
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Conceptual Example 33-4
Reversible rays.
If the object in Example 33-3 is placed instead where the image is, where will the new image be?
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Example 33-5
Object closer to a concave mirror.
A 1.00-cm-high object is placed 10.0 cm from a concave mirror whose radius of curvature is 30.0 cm. (a) Draw a ray diagram to locate (approximately) the position of the image. (b) Determine the position of the image and the magnification analytically.
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Example 33-6
Convex rearview mirror.
A convex rearview car mirror has a radius of curvature of 16 m. Determine the location of the image and its magnification for an object 10.0 m from the mirror
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Problem Solving
Spherical Mirrors
1. Always draw a ray diagram.
2. Use Eqns. 33-3 and 33-4; it is critically important to follow the sign conventions.
3. Sign Conventions
(a) When
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33-5 Refraction: Snell’s Law
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Example 33-7
Refraction through a flat glass.
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Example 33-8
Apparent depth of a pond.
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33-6 Visible Spectrum and Dispersion
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33-7 Total Internal Reflection; Fiber Optics
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Conceptual Example 33-9
View up from under water.
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*33-8 Refraction from a Spherical Mirror
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Example 33-10
Apparent depth II.
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Example 33-11
A spherical “lens.”
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