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RRHS Physics Unit 4 Slide #1 Module 4.1 – Introduction to Waves Waves are caused by vibrations, such as objects undergoing simple harmonic motion. Although.

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Presentation on theme: "RRHS Physics Unit 4 Slide #1 Module 4.1 – Introduction to Waves Waves are caused by vibrations, such as objects undergoing simple harmonic motion. Although."— Presentation transcript:

1 RRHS Physics Unit 4 Slide #1 Module 4.1 – Introduction to Waves Waves are caused by vibrations, such as objects undergoing simple harmonic motion. Although water waves, sound waves, springs, and light all seem very different, they share many properties that can be explained using a wave model. This module introduces trainees to some general wave properties which will later be applied to specific types of waves.

2 RRHS Physics Unit 4 Slide #2 Period and Frequency Period – time it takes for one complete cycle Frequency – number of cycles per unit time

3 RRHS Physics Unit 4 Slide #3 Example A child swings back and forth on a swing 15 times in 30.0 s. Determine the frequency and period of the swing.

4 RRHS Physics Unit 4 Slide #4 Wave Terminology Wave – a disturbance that transfers energy

5 RRHS Physics Unit 4 Slide #5 Mechanical Waves Transverse Wave Longitudinal Wave

6 RRHS Physics Unit 4 Slide #6 Wave Terminology

7 RRHS Physics Unit 4 Slide #7 1D Wave Properties Wave Simulator Wave speed depends only on the medium Heavy medium Light Medium Upright reflected wave Light mediumHeavy Medium Inverted reflected wave

8 RRHS Physics Unit 4 Slide #8 Superposition 1.When waves collide they simply pass through one another unchanged. They continue on as if there were no interaction. 2.While the waves overlap, they temporarily produce a resultant wave due to interference. The displacement of the medium is the sum of the displacements of each component wave  Constructive Interference  Destructive Interference

9 RRHS Physics Unit 4 Slide #9 Superposition Constructive InterferenceDestructive Interference Node

10 RRHS Physics Unit 4 Slide #10 Resonance and Standing Waves Resonance – achieved when energy is added to a system at the same frequency as its natural frequency; Results in maximum amplitude. Standing Wave – example of resonance

11 RRHS Physics Unit 4 Slide #11 Check Your Learning 1.The ? of a wave depends only on the medium in which it is travelling. a.Frequency b.Period c.Speed d.Wavelength (c) speed 2.When a wave passes from one medium to another, the ? must stay the same. a.Amplitude b.Frequency c.Speed d.Wavelength (b) frequency

12 RRHS Physics Unit 4 Slide #12 Check Your Learning 3.A wave in which the medium moves parallel to the medium is called a ? wave. a.Electromagnetic b.Longitudinal c.Mechanical d.Transverse (b) longitudinal 4.The vertical distance from the top of a crest to the bottom of a trough is 34.0 cm. The amplitude of this wave is a.8.5 cm b.17.0 cm c.34.0 cm d.68.0 cm (b) 17.0 cm

13 RRHS Physics Unit 4 Slide #13 Check Your Learning 5.A pulse goes into a medium that is less dense. The reflected pulse is a.Faster b.Inverted c.Larger d.Upright (d) upright 6.Resonance occurs when one object causes a second object to vibrate. The second object must have the same natural a.Amplitude b.Frequency c.Speed d.Wavelength (b) frequency

14 RRHS Physics Unit 4 Slide #14 Check Your Learning 7.A wave source has a period of 0.20 s. What is the frequency? a.0.20 Hz b.1.0 Hz c.5.0 Hz d.20. Hz (c) 5.0 Hz

15 RRHS Physics Unit 4 Slide #15 Wave Equation Wave velocity – the velocity at which the wave crests (or any other part of the wave) move; not the same as the velocity of a particle of the medium.

16 RRHS Physics Unit 4 Slide #16 Example 1 A hiker shouts toward a vertical cliff 685 m away. The echo is heard 4.00 s later. The wavelength of the sound is m. a.What is the speed of sound in air? b.What is the frequency? c.What is the period of the wave?

17 RRHS Physics Unit 4 Slide #17 Solution a. b. c.

18 RRHS Physics Unit 4 Slide #18 Example 2 Water waves have a wavelength of 3.2 m and a frequency of 0.78 Hz. a.At what rate does a stationary boat bob up and down? b.If the boat starts moving into the waves (in the opposite direction to) at a speed of 5.0 m/s, at what rate will the boat bob up and down now?

19 RRHS Physics Unit 4 Slide #19 Solution a.Since the boat is not moving, it will bob up and down at the same frequency as the waves. b.

20 RRHS Physics Unit 4 Slide #20 Check Your Learning Water waves in a lake travel 4.4 m in 1.8 s. The period of oscillation is 1.2 s. What is the speed and wavelength of the water waves?

21 RRHS Physics Unit 4 Slide #21 Reflection Law of Reflection

22 RRHS Physics Unit 4 Slide #22 Refraction Refraction – change in speed while going from one medium to another results in a change of direction of the wave

23 RRHS Physics Unit 4 Slide #23 Refraction

24 RRHS Physics Unit 4 Slide #24 Diffraction Diffraction – waves bend around the edges of the barrier

25 RRHS Physics Unit 4 Slide #25 Diffraction

26 RRHS Physics Unit 4 Slide #26 Interference

27 RRHS Physics Unit 4 Slide #27 Check Your Learning 1.The direction a wave moves is a.Parallel to the wavefronts. b.Perpendicular to the wavefronts. c.In the direction of increasing density. d.In the direction of increasing wavelength. (b) Perpendicular to the wavefronts. 2.The process by which a wave bounces off an obstacle in its path is called a.Diffraction b.Reflection c.Refraction d.Superposition (b) Reflection

28 RRHS Physics Unit 4 Slide #28 Check Your Learning 3.The bending of waves as they go through a small opening is called a.Diffraction b.Reflection c.Refraction d.Superposition (a) Diffraction 4.The bending of waves as they go from one medium to a new medium is called a.Diffraction b.Reflection c.Refraction d.Superposition (c) Refraction

29 RRHS Physics Unit 4 Slide #29 Check Your Learning 5.When two waves interfere with one another, the word interfere means a.One wave prevents the other wave from finishing its cycle. b.One wave stops moving while the other passes. c.The motion we observe is the sum of the motions of the two individual waves. d.The wave with the larger amplitude grows and the wave with the smaller amplitude shrinks. (c) The motion we observe is the sum of the motions of the two individual waves.

30 RRHS Physics Unit 4 Slide #30 Module Summary In this module you have learned that Mechanical waves need a medium while electromagnetic ones do not. Mechanical waves can be transverse or longitudinal. The correct terminology to use when describing waves, such as: period, frequency, crest, trough, amplitude, wavelength The speed of a wave depends only upon the medium in which it is travelling. Waves will be both reflected and transmitted at a boundary.

31 RRHS Physics Unit 4 Slide #31 Module Summary The frequency of a wave does not change when going from one medium to another one. When waves interfere with one another, they can interfere constructively or destructively before passing through one another unchanged. A standing wave is an example of resonance in a medium. All waves are governed by the wave equation

32 RRHS Physics Unit 4 Slide #32 Module Summary All two-dimensional waves obey the law of reflection, which states that the angle of incidence is equal to the angle of reflection All two-dimensional waves undergo refraction, diffraction, and interference.

33 RRHS Physics Unit 4 Slide #33 Module 4.2 – Sound Waves In this module, the wave properties studied in module 7.2 will be looked at in greater depth as they apply to sound waves. Although these properties can be observed in general with all waves, they are often easily observable and can be demonstrated using these specific types of waves. Sound waves are used in a variety of techniques in exploring for oil and minerals.

34 RRHS Physics Unit 4 Slide #34 Sound Waves Mechanical Wave (longitudinal) Series of compressions and rarefactions

35 RRHS Physics Unit 4 Slide #35 Sound Properties Sound PropertyPhysical Wave Characteristic LoudnessAmplitude PitchFrequency QualityWave Form (multiple resonant frequencies)

36 RRHS Physics Unit 4 Slide #36 Range of Hearing Human Hearing 20 Hz20000 Hz infrasonicultrasonic Range decreases as we age Many animals can hear above our range of hearing

37 RRHS Physics Unit 4 Slide #37 Speed of Sound Mechanical waves need a medium Medium determines speed of sound MaterialSpeed of Sound (m/s) Air (at 0 o C and 101 kPa)331 Helium (at 0 o C and 101 kPa) 965 Fresh water (at 20 o C)1482 Copper5010 Steel5960

38 RRHS Physics Unit 4 Slide #38 Speed of Sound in Air Mach Number Supersonic – Mach Number is greater than one

39 RRHS Physics Unit 4 Slide #39 Example A plane is flying at a speed of 855 m/s. If the air temperature is 12 o C, a.What is the speed of sound? b.What is the Mach number for the plane?

40 RRHS Physics Unit 4 Slide #40 Solution a. b.

41 RRHS Physics Unit 4 Slide #41 Doppler Effect Observer A hears a higher frequency

42 RRHS Physics Unit 4 Slide #42 Sonic Boom

43 RRHS Physics Unit 4 Slide #43 Check Your Learning 1.Which of the following is NOT a property of sound? a.Amplitude b.Frequency c.Mass d.Wavelength (c) Mass 2.The average human ear cannot hear frequencies above a.20 Hz b.2000 Hz c Hz d Hz (c) Hz

44 RRHS Physics Unit 4 Slide #44 Check Your Learning 3.When we describe something as supersonic we mean it is a.Faster than the speed of sound b.Higher in frequency than Hz c.Lower in frequency than 20 Hz d.Slower than the speed of sound (a) Faster than the speed of sound 4.When the amplitude of a sound wave increases, a.The wavelength of the sound decreases b.The sound gets louder c.The pitch increases d.The speed of sound increases (b) The sound gets louder

45 RRHS Physics Unit 4 Slide #45 Check Your Learning 5.Sound is a longitudinal wave because a.The oscillations in pressure are in the same direction as the wave moves. b.The oscillations in pressure are perpendicular to the direction that the wave moves. c.The wavelength is long compared to light waves. d.The wavelength is always longer than the amplitude. (a) The oscillations in pressure are in the same direction as the wave moves. 6.The wavelength of a sound wave can be calculated by a.Multiplying the amplitude by the frequency b.Dividing the amplitude by the frequency c.Multiplying the speed by the frequency d.Dividing the speed by the frequency (d) Dividing the speed by the frequency (the wave equation)

46 RRHS Physics Unit 4 Slide #46 Check Your Learning 7.The speed of sound in air at 7.0 o C is a.331 m/s b.332 m/s c.335 m/s d.338 m/s (c) 335 m/s (using v= T) 8.A person is behind an ambulance as it moves away from her. The pitch of the sound that she hears is a.Lower than if the ambulance was stationary. b.The same as if the ambulance was stationary. c.Higher than if the ambulance was stationary. (a) Lower than if the ambulance was stationary, since the wavelength will be larger behind the ambulance.

47 RRHS Physics Unit 4 Slide #47 Closed Air Columns Standing wave in a closed air column requires a node at the closed end and an antinode at the open end of the air column Full Standing Wave

48 RRHS Physics Unit 4 Slide #48 Resonant Lengths

49 RRHS Physics Unit 4 Slide #49 Example 1 Calculate the first 3 resonant lengths for a 512 Hz tuning fork, assuming that the air temperature is 20.0 o C.

50 RRHS Physics Unit 4 Slide #50 Solution

51 RRHS Physics Unit 4 Slide #51 Fixed Length Closed Air Column Multiple frequencies produced

52 RRHS Physics Unit 4 Slide #52 Example 2 A 15.0 cm test tube is blown across so that it resonates. If the air temperature is 20.0 o C, calculate the fundamental frequency and the first two overtones.

53 RRHS Physics Unit 4 Slide #53 Solution

54 RRHS Physics Unit 4 Slide #54 Overtones and Harmonics Harmonics – Whole number multiples of the fundamental frequency For closed air columns, only the odd number harmonics are present.

55 RRHS Physics Unit 4 Slide #55 Check Your Learning An organ pipe is 80.0 cm long. If the temperature is 23 ºC, what are the fundamental frequency and first three audible overtones if the pipe is closed at one end?

56 RRHS Physics Unit 4 Slide #56 Check Your Learning

57 RRHS Physics Unit 4 Slide #57 Open Air Columns Open at both ends Antinode required at both ends

58 RRHS Physics Unit 4 Slide #58 Fixed Length Open Air Columns

59 RRHS Physics Unit 4 Slide #59 Example 3 Assuming an air temperature of 20.0 o C, calculate the fundamental frequency and the first two overtones for a 30.0 cm long open air column.

60 RRHS Physics Unit 4 Slide #60 Solution

61 RRHS Physics Unit 4 Slide #61 Overtones and Harmonics For open air columns, all of the harmonics are present.

62 RRHS Physics Unit 4 Slide #62 Check Your Learning An organ pipe is 80.0 cm long. If the temperature is 23°C, what are the fundamental frequency and first three audible overtones if the pipe is open at both ends?

63 RRHS Physics Unit 4 Slide #63 Check Your Learning

64 RRHS Physics Unit 4 Slide #64 Beat Frequency

65 RRHS Physics Unit 4 Slide #65 Beat Frequency Beat frequency is the absolute value of the difference between the two sources:

66 RRHS Physics Unit 4 Slide #66 Example A Hz tuning fork is sounded at the same time as a key on a piano. You count 23 beats over 8.0 s. What are the possible frequencies of the piano key?

67 RRHS Physics Unit 4 Slide #67 Check Your Learning A guitar string produces a beat frequency of 4 Hz when sounded with a 350 Hz tuning fork and a beat frequency of 9 Hz when sounded with a 355 Hz tuning fork. What is the frequency of the string? Using the first beat frequency, the possible frequencies of the string are 346 Hz or 354 Hz. Using the second beat frequency, the possible frequencies of the string are 346 Hz or 364 Hz. Since the only frequency in common is 346 Hz, this must be the frequency of the string.

68 RRHS Physics Unit 4 Slide #68 Module Summary In this module you learned that Sound waves are longitudinal mechanical waves. Sounds can be distinguished by loudness, pitch, and quality. Sound travels through air with a speed given by The mach number of an object can be calculated using

69 RRHS Physics Unit 4 Slide #69 Module Summary The Doppler Effect and sonic booms can be explained using wave theory. Air columns resonate at their natural frequencies. Closed air columns resonate at their fundamental frequency when Open air columns resonate at their fundamental frequency when

70 RRHS Physics Unit 4 Slide #70 Module Summary All of the harmonics are present in open air columns. Only the odd harmonics are present in closed air columns. When two frequencies are close but not the exact same, beats will be heard with a frequency of

71 RRHS Physics Unit 4 Slide #71 Module 4.3 – Electromagnetic Waves The wave model of light will be applied to electromagnetic waves to further study wave properties such as reflection, refraction, and diffraction. A brief introduction to geometric optics is also included. An understanding of light is important as it applies to one of the principle means through which we obtain information, using both instruments and our sense of sight.

72 RRHS Physics Unit 4 Slide #72 Light as a Wave Two basic methods of transferring energy: –Particles – for example, a baseball travelling through the air has kinetic energy which can be transferred to another object in a collision. –Waves – water waves transfer energy to the shore and cause erosion. Newton proposed a particle model Christian Huygens proposed a wave model Newton’s model initially accepted

73 RRHS Physics Unit 4 Slide #73 Light as a Wave Huygens model began to gain more acceptance for the following reasons. –double slit experiment to show that light passing through two slits demonstrated the same interference pattern as two sources of water waves; –speed of light was shown to be lower in water than in air; this supported Huygen's theory of refraction and contradicted Newton's theory of refraction. Huygens wave model replaced by Electromagnetic Wave Model

74 RRHS Physics Unit 4 Slide #74 Electromagnetic Spectrum Current model of light incorporates both waves and particles

75 RRHS Physics Unit 4 Slide #75 Reflection of Light Regular Reflection Smooth Surface Rough Surface Diffuse reflection

76 RRHS Physics Unit 4 Slide #76 Plane Mirror Virtual Image

77 RRHS Physics Unit 4 Slide #77 Speed of Light Speed accurately determined around 1900 by Michelson

78 RRHS Physics Unit 4 Slide #78 Example Using the accepted value for the speed of light, calculate the minimum frequency that would have been needed for light to be reflected into the eye of the observer in Michelson’s apparatus.

79 RRHS Physics Unit 4 Slide #79 Check Your Learning 1.Why is it better when the pages of a book are rough rather than smooth and glossy? Rough pages allow light to undergo diffuse reflection, meaning the light is not all reflected in the same direction. This reduces glare from the page. 2.A particular nearsighted person can only see clearly 0.50 m from their face. How far from a plane mirror should they be to see their image clearly? They should be 0.25 m (or less) from the mirror. Because their image is the same distance behind the mirror as they are in front of it, the total distance from the person to their image will be 0.50 m.

80 RRHS Physics Unit 4 Slide #80 Check Your Learning 3.What is the angle of incidence if the angle between a reflected ray and the mirror is 34 o ? If the angle between the reflected ray and the mirror is 34 o, the angle of reflection (the angle with the normal) is 56 o (90-34). The angle of incidence must therefore be 56 o. 4.The moon is 3.85×10 8 m away from the earth. How long does it take light reflected from the moon to reach the earth?

81 RRHS Physics Unit 4 Slide #81 Coin in a Cup Demo Can See CoinCannot See CoinCan See Coin because of refraction

82 RRHS Physics Unit 4 Slide #82 Index of Refraction Index of refraction (n) defined as SubstanceIndex of Refraction Vacuum1.00 Air1.00 Water1.33 Ethyl alcohol1.36 Quartz1.46 Plexiglass1.51 Crown Glass1.52 Flint Glass1.65 Diamond2.42

83 RRHS Physics Unit 4 Slide #83 Example 1 Calculate the speed of light in water.

84 RRHS Physics Unit 4 Slide #84 Snell’s Law

85 RRHS Physics Unit 4 Slide #85 Example 2 A ray of light (travelling in air) has an angle of incidence of 30.0 o on a block of quartz and an angle of refraction of 20.0 o. What is the index of refraction for this block of quartz?

86 RRHS Physics Unit 4 Slide #86 Check Your Learning The speed of light in a clear plastic is 1.90×10 8 m/s. A ray of light travelling through air enters the plastic with an angle of incidence of 22°. At what angle is the ray refracted?

87 RRHS Physics Unit 4 Slide #87 Total Internal Reflection Total internal reflection can only occur going from a high index of refraction to a lower index of refraction Critical Angle

88 RRHS Physics Unit 4 Slide #88 Total Internal Reflection

89 RRHS Physics Unit 4 Slide #89 Total Internal Reflection Two conditions required for total internal reflection to occur: 1.The light must be travelling from a higher index of refraction to a lower index of refraction. 2.The angle of incidence must be greater than the critical angle, θ c, associated with the two materials.

90 RRHS Physics Unit 4 Slide #90 Example 3 What is the critical angle for the interface between air and water?

91 RRHS Physics Unit 4 Slide #91 Fibre Optics

92 RRHS Physics Unit 4 Slide #92 Check Your Learning The critical angle for a certain liquid-air surface is 42.9 o. What is the index of refraction for the liquid?

93 RRHS Physics Unit 4 Slide #93 Double-Slit Diffraction Central Maximum

94 RRHS Physics Unit 4 Slide #94 Double Slit Diffraction Dark Spot – Destructive Interference Bright Spot – Constructive Interference

95 RRHS Physics Unit 4 Slide #95 Small Scale Path Difference = nλBright spot

96 RRHS Physics Unit 4 Slide #96 Large Scale

97 RRHS Physics Unit 4 Slide #97 Example 1 Red light with a wavelength of 685 nm is shone through two small slits. An interference pattern is observed on a screen that is 4.2 m away. The distance between the central maximum and the second order bright spot is 3.2 cm. What was the distance between the two slits?

98 RRHS Physics Unit 4 Slide #98 Diffraction Gratings

99 RRHS Physics Unit 4 Slide #99 Diffraction Gratings Double Slit Diffraction Grating

100 RRHS Physics Unit 4 Slide #100 Example 2 Calculate the angle between the central maximum and the first order bright spot for a diffraction grating that has 3800 lines per centimetre on it if monochromatic light with a wavelength of 420 nm is shone on it.

101 RRHS Physics Unit 4 Slide #101 Check Your Learning Light with a wavelength of 542 nm is shone through a diffraction grating. The third order bright spot is observed to be 74.0 cm away from the central maximum on a screen 8.20 m away. How many lines per cm does the grating have?

102 RRHS Physics Unit 4 Slide #102 Check Your Learning

103 RRHS Physics Unit 4 Slide #103 Concave Mirrors

104 RRHS Physics Unit 4 Slide #104 Ray Diagrams Consider an object in front of a concave mirror.

105 RRHS Physics Unit 4 Slide #105 Rule 1 Any ray drawn parallel to the principal axis will reflect through the focal point.

106 RRHS Physics Unit 4 Slide #106 Rule 2 Because of the law of reflection, the opposite must also be true. Any ray drawn through the focal point must reflect parallel to the principal axis.

107 RRHS Physics Unit 4 Slide #107 Rule 3 Any ray that goes through the center of curvature hits the mirror at a 90 o angle, and so reflects back on itself. Image is real, inverted, larger

108 RRHS Physics Unit 4 Slide #108 Check Your Learning Locate and describe the images of the object in each of the following diagrams: a. The image is inverted, real, and smaller.

109 RRHS Physics Unit 4 Slide #109 Check Your Learning b. Notice that in this case, the reflected rays are spreading apart and will not cross. It is necessary to extend the rays behind the mirror until they cross. This image is larger, upright, and virtual.

110 RRHS Physics Unit 4 Slide #110 Check Your Learning c. Image is inverted, real, and the same size.

111 RRHS Physics Unit 4 Slide #111 Mirror Equation

112 RRHS Physics Unit 4 Slide #112 Mirror Equation Image height h i is positive if upright, negative if inverted (relative to the object) The image distance d i and the object distance d o positive if on the reflecting side of the mirror (real) and negative if behind the mirror (virtual) The focal length f is positive if on the reflecting side of the mirror, which will always be true for concave mirrors.

113 RRHS Physics Unit 4 Slide #113 Example 1 A concave mirror has a radius of curvature of 12.0 cm. A 1.2 cm tall object is placed a distance of 8.2 cm away from the mirror. a.Locate the image. b.Calculate the height of the image. c.Describe the image.

114 RRHS Physics Unit 4 Slide #114 Solution a.

115 RRHS Physics Unit 4 Slide #115 Solution b. c.The image is inverted (because h i is negative), larger (because h i is bigger) and real (because d i is positive).

116 RRHS Physics Unit 4 Slide #116 Convex Mirrors Rays of light diverge as if coming from the focal point

117 RRHS Physics Unit 4 Slide #117 Convex Mirrors Rules for drawing rays diagrams that we learned before are very similar for convex mirrors; instead of our incoming rays going through the focal point or the centre of curvature, they simply go toward them (since they are on the other side of the mirror). Image is always upright, smaller, virtual

118 RRHS Physics Unit 4 Slide #118 Example 2 A convex mirror has a radius of curvature of 12.0 cm. A 1.2 cm tall object is placed a distance of 8.2 cm away from the mirror. a.Locate the image. b.Calculate the height of the image. c.Describe the image.

119 RRHS Physics Unit 4 Slide #119 Solution a.Remember, since the mirror is convex, the radius of curvature and the focal length must be negative.

120 RRHS Physics Unit 4 Slide #120 Solution b. c. The image is upright (because h i is positive), smaller (because h i is smaller) and virtual (because d i is negative).

121 RRHS Physics Unit 4 Slide #121 Check Your Learning A 5.3 cm tall object is placed 6.4 cm away from a spherical mirror. A virtual image is formed 4.2 cm from the mirror. a.What is the focal length of the mirror? Since the image is virtual, d i must be negative

122 RRHS Physics Unit 4 Slide #122 Check Your Learning b.What kind of mirror is it? Because the focal length was calculated to be negative, the mirror is convex. c.What is the height of the image?

123 RRHS Physics Unit 4 Slide #123 Convex (Converging) Lens Lens is thicker in the middle

124 RRHS Physics Unit 4 Slide #124 Concave (Diverging) Lens Lens is thinner in the middle

125 RRHS Physics Unit 4 Slide #125 Ray Diagrams – Rule 1 A ray drawn parallel to the axis is refracted by the lens so that it passes along a line through the focal point.

126 RRHS Physics Unit 4 Slide #126 Rule 2 A ray drawn on a line passing through the other focal point F’ emerges from the lens parallel to the axis.

127 RRHS Physics Unit 4 Slide #127 Rule 3 A ray directed to the center of the lens continues in a straight line. image is real, inverted, and smaller

128 RRHS Physics Unit 4 Slide #128 Check Your Learning Locate and describe the image in each of the following diagrams: a. Image is upright, larger, and virtual

129 RRHS Physics Unit 4 Slide #129 Check Your Learning b. Image is inverted, larger, and real.

130 RRHS Physics Unit 4 Slide #130 Check Your Learning c. Image is upright, smaller and virtual

131 RRHS Physics Unit 4 Slide #131 Lens Equation Power of a lens defined as If f is in metres, P is in diopters (D)

132 RRHS Physics Unit 4 Slide #132 Lens Equation The object distance d o is positive if it is on the same side of the lens from which the light is coming (in other words, if it is real) The image distance d i is positive if it is on the opposite side of the lens from which the light is coming (in other words, if it is real); it is negative if it is on the same side of the lens from which the light is coming (in other words, if it is virtual) The height of the image h i is positive if upright and negative if inverted relative to the object. The focal length is positive for a convex (converging) lens and negative for a concave (diverging) lens.

133 RRHS Physics Unit 4 Slide #133 Example 3 A certain lens focuses an object 22.5 cm away as an image 33.0 cm on the other side of the lens. a.Is the image real or virtual? b.What type of lens is it and what is its focal length? c.What is the power of the lens?

134 RRHS Physics Unit 4 Slide #134 Solution a.Because the image is on the other side of the lens, it must be real. b.Because the image is real, the lens must be convex, or diverging (since concave lenses will always give virtual images). A positive focal length will confirm this.

135 RRHS Physics Unit 4 Slide #135 Solution c.To calculate the power, the focal length must be in metres.

136 RRHS Physics Unit 4 Slide #136 Check Your Learning A concave lens has a focal point 20.0 cm away from the lens. A 2.1 m tall object is placed 3.0 m away from it. a.Where is the image? The image is located 0.19 m from the lens on the same side as the object.

137 RRHS Physics Unit 4 Slide #137 Check Your Learning b.How big is the image? c.Describe the image. The image is upright (because h i is positive), smaller (because h i is smaller than h o ), and virtual (because d i is negative).

138 RRHS Physics Unit 4 Slide #138 Module Summary In this module you learned that Light exhibits many wave properties and can be modeled in many situations as a wave. Light can undergo both regular reflection (mirrors) or diffuse reflection (rough surfaces) Light is just a small part of the electromagnetic spectrum. The speed of light in a vacuum is

139 RRHS Physics Unit 4 Slide #139 Module Summary The index of refraction for a medium can be calculated using The angle of incidence and angle of refraction for a refracting ray of light are related by Snell’s Law When going from a high index of refraction to a low index of refraction, there is a critical angle beyond which light cannot refract. For all angles of incidence greater than this critical angle, total internal reflection occurs.

140 RRHS Physics Unit 4 Slide #140 Module Summary Diffraction and interference of light for double slits and diffraction gratings can be modeled using the equations Ray diagrams can be used to locate images in spherical mirrors and lenses. The following equations can be applied to problems involving spherical mirrors and lenses:

141 RRHS Physics Unit 4 Slide #141 Module Summary The power of a lens can be calculated from the focal length


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