WAVE
LIGHT WAVES:
WAVES OF LIGHT CLASS: XII BY SEMESTER 1: Nur widiyawati LIGHT WAVES OF LIGHT WAVES SMA pgri 1 Kendal SK CHART EXERCISE MOTIVATION TP KD
COMPETENCY STANDARD:
LIGHT WAVES WAVES OF LIGHT STANDARDS APPLYING THE CONCEPT OF COMPETENCE AND PRINCIPLES TROUBLESHOOTING SYMPTOM IN WAVES
BASIC COMPETENCY:
LIGHT WAVES OF LIGHT WAVES AND COMPETENCE OF SYMPTOMS describe CHARACTERISTICS OF SOUND AND LIGHT WAVES
Objectives After studying this material, you should be able to::
LIGHT WAVES OF LIGHT WAVES Objectives After studying this material, learners are expected to: Formulate incident light on the double slit Interfrensi Identify the properties of light through the experimental group to analyze the application of problem-solving nature of light in the discussion and information
How does the peacock colors formed?:
LIGHT WAVES WAVES OF LIGHT How peacock colors formed? The colors on the peacock feather is not caused by pigments in the feathers. If it is not produced by pigments, how beautiful the color is formed? P E N G A N T A R
Peacock feather:
The color of peacock feathers on a peacock feather is affected by interference. Multi-layer structure produces interference that amplifies hair for some colors such as blue and green colors will change when you look at an angle different colors on a butterfly and humming bird is also the effect of the interference
Slide 7:
LIGHT WAVES WAVES OF LIGHT WAVES OF LIGHT DIFFRACTION INTERFRENSI polarization double slit slit THIN single slit diffraction grating reflectance absorption DOUBLE SELECTIVE REFLECTION refraction scattering
A. INTERFRENSI LIGHT:
LIGHT WAVES OF LIGHT WAVES A. INTERFRENSI Interfrensi LIGHT Light is a blend of two or more coherent waves. Two waves of coherent to say, if the two waves have the same frequency and amplitude and phase differences remain. Interfrensi interfrensi light will produce a pattern that is the pattern of bright lines dark stripe pattern CHART
1. Double Gap Interfrensi Young:
LIGHT WAVES WAVES OF LIGHT 1. Young's double slit Interfrensi To get two coherent light then the light source is passed to the two slit (double slit). Light from each slit to be coherent because it comes from one light source. Each slot can serve as a light source. The light source should be monochromatic results from the superposition of light emanating from the two slits can be a maximum or minimum. Maximum Interfrensi produce a bright line, while the minimum Interfrensi produce a dark line
Slide 10:
Occurred and no interference can be described as in Figure 2.3. Figure 2.3. (A) there is no interference, (b) interference
Slide 11:
Sources: www.fisikon.com
Consider the following animation:
LIGHT WAVES OF LIGHT WAVES Consider the following animated dp d sin θ θ ℓ d = distance between two slits (m) ℓ = slit to screen distance (m) p = distance to the light pattern interfrensi maximum Interfrensi central bright line d sin θ = m λ Or dp = λ ℓ m b. Minimum Interfrensi dark line d sin θ = (2m-1) ½ λ Or dp = (2m-1) ½ λ ℓ m = 1,2,3, ... The distance between two lines of bright or 2 dark lines are formulated: Δd p = λ ℓ CHART
2. Interfrensi thin layer:
LIGHT WAVES WAVES OF LIGHT 2. Interfrensi thin layer of bright lines Interfrensi maximum 2 nd Cos r = (2 m - 1) ½ λ m = 1,2,3 ... Minimum Interfrensi dark line 2 nd m r = λ Cos. ir OABC ndn = refractive index of thin layers of thick layers of d = r = ray angle bias CHART
Diffraction:
LIGHT WAVES OF LIGHT WAVES Diffraction the single slit diffraction if the wavefront through a narrow slit, then this wave will experience bending (diffraction). Deflection of light (diffraction) can occur when light passes through a single slit and will produce a pattern of bright and dark lines. the occurrence of diffraction is shown in the following animation CHART
a. The dark line to m occurs if: d Sin θ = m λ m = 1,2,3, ... Or d p = m λ ℓ:
LIGHT WAVES OF LIGHT WAVES a. The dark line to m occurs if: d Sin θ = m λ m = 1,2,3, ... Or d p = λ ℓ m b. A bright line to m occurs if: d Sin θ = (2m + 1) ½ λ m = 1,2,3, ... Or dp = (2m + 1) ½ λ ℓ Wavefronts Gap Single Gel. CHART screen light
2. Multiple slit diffraction (lattice):
LIGHT WAVES WAVES OF LIGHT 2. Multiple slit diffraction (lattice) If the wavefront through a narrow slit, then this wave will experience bending (diffraction). If light passes through a gap compound (lattice), then the light will experience diffraction, here the white light through a diffraction grating so that the experience diffraction and decomposes according to the length galombang respectively. Consider the following animation CHART
a. The maximum diffraction d Sin θ = m λ m = 0,1,2,3, ... b. Diffraction minimum d Sin θ = (2m - 1) ½ λ m = 0,1,2,3, ... d = 1 N:
LIGHT WAVES OF LIGHT WAVES d = distance between the openings in the lattice N = number of lines per cm grating order m = a. The maximum diffraction d Sin θ = m λ m = 0,1,2,3, ... b. Diffraction minimum d Sin θ = (2m - 1) ½ λ m = 0,1,2,3, ... d = 1 N Lattice White light diffraction Order Order to zero to 1 to 2 Order Order Order to 1 to 2 FIGURE
d. Polarization:
LIGHT WAVES OF LIGHT WAVES d. Polarization Polarization Polarization reflectance because the event is partially absorb vibration direction transverse waves so that waves vibrate only have one direction only. If the light comes on a flat mirror at an angle of 57 °, then this is a beam of polarized light. Consider the following animation i = 57 ° i = 57 ° flat mirror CHART
Polarization due to refraction and reflection:
LIGHT WAVES OF LIGHT WAVES polarization due to refraction and reflection angle reflectance If (i) + angle of refraction (r) = 90 ° Or (r) = 90 ° - i, then applies: Tg i = n 2 n 1 i i i is the angle of polarization 'N r flat mirror Medium 1 (n 1) Medium 2 (n 2) FIGURE
Polarization due to double refraction (bias Twins):
LIGHT WAVES WAVES OF LIGHT If the light through the glass, the light passing with the same pace in all directions, because the glass has only one refractive index, however, certain crystalline materials (calcite and quartz) has a refractive index of more than one, so the speed of light is not the same. So light through this material will experience double refraction (twins) Polarization due to double refraction (bias Twins) FIGURE
Polarization due to selective absorption:
LIGHT WAVES OF LIGHT WAVES polarization due to selective absorption of polarized light passing Polarisator with intensity I 1 = ½ I 0. Function analyzer to analyze the light that is passed polarisator, eyes to see the brightest light, dim light later when polarisator and analyzer are perpendicular, then it was dark. The intensity of light emitted from the analyzer satisfies the equation I 2 = I 1 cos 2 θ = ½ I 0 cos 2 θ CHART
BECAUSE Polarization Scattering:
LIGHT WAVES OF LIGHT WAVES AS polarization of polarized light scattering by the Earth's atmosphere, resulting in sky blue. This happens because the most effective blue light is scattered when compared with other light EXAMPLE PROBLEM: The sun is in a state is not polarized falls on the surface of the water. The reflected linearly polarized light (n water = 4 / 3) 1. determine the angle between the incident light with surface water 2. determine the bias point of CHART
PROBLEMS:
LIGHT WAVES WAVES OF LIGHT PROBLEMS If a wave through a narrow slit, it will happen: Reflection Refraction Diffraction Polarization interfrensi At trial Young (double slit), if the distance between the two gap be two times the original, then the distance anara two successive dark lines be .......... 4 times 2 times its original originally originally ¼ ½ times its original time unchanged Fixed 1 and 2 3 and 4 5 and 6 7 and 8
To determine the wavelength of monochromatic light used experimental young that their data as follows: the distance between two slits 0.3 mm, slit to screen distance of 50 cm and the distance between the two dark lines with a dark line on the screen to 3 = 1 mm. Wavelength of monochromatic light is ....... a. 400 nm b. C. 480 nm 500 nm d. 580 nm e. 600 nm:
LIGHT WAVES WAVES OF LIGHT To determine the wavelength of monochromatic light used experimental young that their data as follows: the distance between two slits 0.3 mm, slit to screen distance of 50 cm and the distance between the two dark lines with a dark line on the screen to 3 = 1 mm. Wavelength of monochromatic light is ....... a. 400 nm b. C. 480 nm 500 nm d. 580 nm e. The basic principle of two 600 nm coherent light source is ........ Both are very close together Simpangannya always the same amplitude as the phase difference the two is still emitting light that passed both 1 and 2 3 and 4 5 and 6 7 and 8
When the sun light on a thin layer of oil on the top surface of the water, then the colors are seen arising due ..... a. B. Diffraction C. Dispersion Interfrensi d. Polarization e. refraction:
LIGHT WAVES WAVES OF LIGHT When light of the sun on a thin layer of oil on the top surface of the water, then the colors are seen arising due ..... a. B. Diffraction C. Dispersion Interfrensi d. Polarization e. a sky blue color refraction occurs because the sun had ......... Refraction Diffraction Dispersion Polarization Interfrensi 1 and 2 3 and 4 5 and 6 7 and 8
A beam of light passed through a diffraction grating with 5000 lines / cm will produce a bright line second with a 30 degree angle deviation of the normal line. Determine the wavelength of light used:
LIGHT WAVES OF LIGHT WAVES 1 and 2 3 and 4 5 and 6 7 and 8 A beam of light passed through a diffraction grating with 5000 lines / cm will produce a bright line second with a 30 degree angle deviation of the normal line. Determine the wavelength of light used a single slit width of 0.2 mm lurusberkas erect illuminated light wavelength 8000 nm. Diffraction pattern was captured on a screen a distance of 1 meter from the slit. Determine: the distance between the center of a bright line of the third dark line distance between the two dark lines against the dark to the 4.
WAVES OF LIGHT CLASS: XII BY SEMESTER 1: Nur widiyawati LIGHT WAVES OF LIGHT WAVES SMA pgri 1 Kendal SK CHART EXERCISE MOTIVATION TP KD
COMPETENCY STANDARD:
LIGHT WAVES WAVES OF LIGHT STANDARDS APPLYING THE CONCEPT OF COMPETENCE AND PRINCIPLES TROUBLESHOOTING SYMPTOM IN WAVES
BASIC COMPETENCY:
LIGHT WAVES OF LIGHT WAVES AND COMPETENCE OF SYMPTOMS describe CHARACTERISTICS OF SOUND AND LIGHT WAVES
Objectives After studying this material, you should be able to::
LIGHT WAVES OF LIGHT WAVES Objectives After studying this material, learners are expected to: Formulate incident light on the double slit Interfrensi Identify the properties of light through the experimental group to analyze the application of problem-solving nature of light in the discussion and information
How does the peacock colors formed?:
LIGHT WAVES WAVES OF LIGHT How peacock colors formed? The colors on the peacock feather is not caused by pigments in the feathers. If it is not produced by pigments, how beautiful the color is formed? P E N G A N T A R
Peacock feather:
The color of peacock feathers on a peacock feather is affected by interference. Multi-layer structure produces interference that amplifies hair for some colors such as blue and green colors will change when you look at an angle different colors on a butterfly and humming bird is also the effect of the interference
Slide 7:
LIGHT WAVES WAVES OF LIGHT WAVES OF LIGHT DIFFRACTION INTERFRENSI polarization double slit slit THIN single slit diffraction grating reflectance absorption DOUBLE SELECTIVE REFLECTION refraction scattering
A. INTERFRENSI LIGHT:
LIGHT WAVES OF LIGHT WAVES A. INTERFRENSI Interfrensi LIGHT Light is a blend of two or more coherent waves. Two waves of coherent to say, if the two waves have the same frequency and amplitude and phase differences remain. Interfrensi interfrensi light will produce a pattern that is the pattern of bright lines dark stripe pattern CHART
1. Double Gap Interfrensi Young:
LIGHT WAVES WAVES OF LIGHT 1. Young's double slit Interfrensi To get two coherent light then the light source is passed to the two slit (double slit). Light from each slit to be coherent because it comes from one light source. Each slot can serve as a light source. The light source should be monochromatic results from the superposition of light emanating from the two slits can be a maximum or minimum. Maximum Interfrensi produce a bright line, while the minimum Interfrensi produce a dark line
Slide 10:
Occurred and no interference can be described as in Figure 2.3. Figure 2.3. (A) there is no interference, (b) interference
Slide 11:
Sources: www.fisikon.com
Consider the following animation:
LIGHT WAVES OF LIGHT WAVES Consider the following animated dp d sin θ θ ℓ d = distance between two slits (m) ℓ = slit to screen distance (m) p = distance to the light pattern interfrensi maximum Interfrensi central bright line d sin θ = m λ Or dp = λ ℓ m b. Minimum Interfrensi dark line d sin θ = (2m-1) ½ λ Or dp = (2m-1) ½ λ ℓ m = 1,2,3, ... The distance between two lines of bright or 2 dark lines are formulated: Δd p = λ ℓ CHART
2. Interfrensi thin layer:
LIGHT WAVES WAVES OF LIGHT 2. Interfrensi thin layer of bright lines Interfrensi maximum 2 nd Cos r = (2 m - 1) ½ λ m = 1,2,3 ... Minimum Interfrensi dark line 2 nd m r = λ Cos. ir OABC ndn = refractive index of thin layers of thick layers of d = r = ray angle bias CHART
Diffraction:
LIGHT WAVES OF LIGHT WAVES Diffraction the single slit diffraction if the wavefront through a narrow slit, then this wave will experience bending (diffraction). Deflection of light (diffraction) can occur when light passes through a single slit and will produce a pattern of bright and dark lines. the occurrence of diffraction is shown in the following animation CHART
a. The dark line to m occurs if: d Sin θ = m λ m = 1,2,3, ... Or d p = m λ ℓ:
LIGHT WAVES OF LIGHT WAVES a. The dark line to m occurs if: d Sin θ = m λ m = 1,2,3, ... Or d p = λ ℓ m b. A bright line to m occurs if: d Sin θ = (2m + 1) ½ λ m = 1,2,3, ... Or dp = (2m + 1) ½ λ ℓ Wavefronts Gap Single Gel. CHART screen light
2. Multiple slit diffraction (lattice):
LIGHT WAVES WAVES OF LIGHT 2. Multiple slit diffraction (lattice) If the wavefront through a narrow slit, then this wave will experience bending (diffraction). If light passes through a gap compound (lattice), then the light will experience diffraction, here the white light through a diffraction grating so that the experience diffraction and decomposes according to the length galombang respectively. Consider the following animation CHART
a. The maximum diffraction d Sin θ = m λ m = 0,1,2,3, ... b. Diffraction minimum d Sin θ = (2m - 1) ½ λ m = 0,1,2,3, ... d = 1 N:
LIGHT WAVES OF LIGHT WAVES d = distance between the openings in the lattice N = number of lines per cm grating order m = a. The maximum diffraction d Sin θ = m λ m = 0,1,2,3, ... b. Diffraction minimum d Sin θ = (2m - 1) ½ λ m = 0,1,2,3, ... d = 1 N Lattice White light diffraction Order Order to zero to 1 to 2 Order Order Order to 1 to 2 FIGURE
d. Polarization:
LIGHT WAVES OF LIGHT WAVES d. Polarization Polarization Polarization reflectance because the event is partially absorb vibration direction transverse waves so that waves vibrate only have one direction only. If the light comes on a flat mirror at an angle of 57 °, then this is a beam of polarized light. Consider the following animation i = 57 ° i = 57 ° flat mirror CHART
Polarization due to refraction and reflection:
LIGHT WAVES OF LIGHT WAVES polarization due to refraction and reflection angle reflectance If (i) + angle of refraction (r) = 90 ° Or (r) = 90 ° - i, then applies: Tg i = n 2 n 1 i i i is the angle of polarization 'N r flat mirror Medium 1 (n 1) Medium 2 (n 2) FIGURE
Polarization due to double refraction (bias Twins):
LIGHT WAVES WAVES OF LIGHT If the light through the glass, the light passing with the same pace in all directions, because the glass has only one refractive index, however, certain crystalline materials (calcite and quartz) has a refractive index of more than one, so the speed of light is not the same. So light through this material will experience double refraction (twins) Polarization due to double refraction (bias Twins) FIGURE
Polarization due to selective absorption:
LIGHT WAVES OF LIGHT WAVES polarization due to selective absorption of polarized light passing Polarisator with intensity I 1 = ½ I 0. Function analyzer to analyze the light that is passed polarisator, eyes to see the brightest light, dim light later when polarisator and analyzer are perpendicular, then it was dark. The intensity of light emitted from the analyzer satisfies the equation I 2 = I 1 cos 2 θ = ½ I 0 cos 2 θ CHART
BECAUSE Polarization Scattering:
LIGHT WAVES OF LIGHT WAVES AS polarization of polarized light scattering by the Earth's atmosphere, resulting in sky blue. This happens because the most effective blue light is scattered when compared with other light EXAMPLE PROBLEM: The sun is in a state is not polarized falls on the surface of the water. The reflected linearly polarized light (n water = 4 / 3) 1. determine the angle between the incident light with surface water 2. determine the bias point of CHART
PROBLEMS:
LIGHT WAVES WAVES OF LIGHT PROBLEMS If a wave through a narrow slit, it will happen: Reflection Refraction Diffraction Polarization interfrensi At trial Young (double slit), if the distance between the two gap be two times the original, then the distance anara two successive dark lines be .......... 4 times 2 times its original originally originally ¼ ½ times its original time unchanged Fixed 1 and 2 3 and 4 5 and 6 7 and 8
To determine the wavelength of monochromatic light used experimental young that their data as follows: the distance between two slits 0.3 mm, slit to screen distance of 50 cm and the distance between the two dark lines with a dark line on the screen to 3 = 1 mm. Wavelength of monochromatic light is ....... a. 400 nm b. C. 480 nm 500 nm d. 580 nm e. 600 nm:
LIGHT WAVES WAVES OF LIGHT To determine the wavelength of monochromatic light used experimental young that their data as follows: the distance between two slits 0.3 mm, slit to screen distance of 50 cm and the distance between the two dark lines with a dark line on the screen to 3 = 1 mm. Wavelength of monochromatic light is ....... a. 400 nm b. C. 480 nm 500 nm d. 580 nm e. The basic principle of two 600 nm coherent light source is ........ Both are very close together Simpangannya always the same amplitude as the phase difference the two is still emitting light that passed both 1 and 2 3 and 4 5 and 6 7 and 8
When the sun light on a thin layer of oil on the top surface of the water, then the colors are seen arising due ..... a. B. Diffraction C. Dispersion Interfrensi d. Polarization e. refraction:
LIGHT WAVES WAVES OF LIGHT When light of the sun on a thin layer of oil on the top surface of the water, then the colors are seen arising due ..... a. B. Diffraction C. Dispersion Interfrensi d. Polarization e. a sky blue color refraction occurs because the sun had ......... Refraction Diffraction Dispersion Polarization Interfrensi 1 and 2 3 and 4 5 and 6 7 and 8
A beam of light passed through a diffraction grating with 5000 lines / cm will produce a bright line second with a 30 degree angle deviation of the normal line. Determine the wavelength of light used:
LIGHT WAVES OF LIGHT WAVES 1 and 2 3 and 4 5 and 6 7 and 8 A beam of light passed through a diffraction grating with 5000 lines / cm will produce a bright line second with a 30 degree angle deviation of the normal line. Determine the wavelength of light used a single slit width of 0.2 mm lurusberkas erect illuminated light wavelength 8000 nm. Diffraction pattern was captured on a screen a distance of 1 meter from the slit. Determine: the distance between the center of a bright line of the third dark line distance between the two dark lines against the dark to the 4.
WAVES AND SOUND
Wave
• In everyday life, a wave phenomenon is familiar to us all.
• Sound, light, ocean waves, earthquakes, and radio-television transmission, is a wave phenomenon in our daily lives.
• Waves can occur if a system is disturbed from the equilibrium position and the disturbance propagates from one region to another system
This type of wave propagation based on whether or not the medium
• wave mechanics.
• waves that travel in a material called medium.
• electromagnetic waves.
• waves that do not require a medium in creepage
Types of medium wave based on the direction of the shift
• transverse wave
• shift in the medium wave direction perpendicular to the direction of travel of the wave along the medium
• longitudinal wave
• medium wave direction shifts in the direction of traveling wave along the medium
Wave length
• The distance from one peak to peak next or from one valley to the next valley, or from any point to point in question on the next repetition of the wave
• Coat wavelength (λ)
• Fre number of waves within a certain time interval
• The number of waves per second
• Symbol frequency (f)
Quency
rapid propagation of
• The relationship between the wavelength (λ) and frequency (f) with a quick wave propagation: v = f λ
• The rate of the wave is affected by the mechanical properties of the medium perambatannya
Periodic waves
• Waves that every moment is a changing pattern
• Periodic wave runs at constant speed
• The addition of frequency (f) will cause a reduction of the wavelength (λ)
NOISE
• The sound is a longitudinal wave.
• Phenomena associated with the sense of hearing the sound of us, namely our ears and our brain.
• Functioning ear receives sound waves, while the brain function translates information from the ear.
ASPECTS OF NOISE
• Three aspects of sound: sound sources, sound energy, and detector noise.
• Sound sources are all things that vibrate.
• Sound energy is transferred from the sound source in the form of sound waves, namely the longitudinal wave.
• Sound detector is a tool to capture / receive sound waves, can be our ears or other tools.
characteristics of sound
• Sound waves require a medium vine.
• Sound propagation medium can include air, soil, rock, metal, and others.
• The sound can not propagate in a vacuum
speed of sound
• Speed of sound in each medium is different.
• Speed of sound in various mediums influenced by the elastic moduli B and medium density
• Speed of sound in air is also influenced also by temperature,
where T is temperature
where T is temperature
• The sound loudness and Altitude Loudness of sound is influenced by the energy in sound waves, which is influenced by the sound wavelength
• Altitude oeh frekeunsi influenced sound waves of sound.
Beep altitude
• Altitude oeh frekeunsi influenced sound waves of sound.
• Based on the height, the sound daat tigkatan divided into three, namely: infrasound, audiosonik, and ultra sonic.
• Infrasonic sound, its frequency (¦) < 20 Hz.
• Audiosonik sound, its frequency (¦)
20 Hz £ ¦ £ 20.000 Hz.
• Ultrasonic sound, the frequency
(¦) > 20.000 Hz
(¦) > 20.000 Hz
The sound intensity
• Sound intensity is the energy carried by a sound wave through a unit area per unit time.
• Unit of sound intensity is
• The human ear can detect sounds with intensity (I)
• To produce a sound that kenyarinyannya 2-fold, takes a sound wave with intensity 10-fold.
Sound intensity level
• Sound intensity level (in dB) is defined as:
where Io is the intensity of reference, usually taken from the minimum intensity dpat heard by the human ear is
:.