Which statement correctly describes a transverse wave?
Student Revision Notes
Term 3 Physics Quick Review
Grade 9 Advanced Physics — Term 3 Revision Notes
1. Waves
A wave transfers energy from one place to another without transferring matter permanently.
Types of waves
- Transverse wave: particles vibrate perpendicular to the direction of travel. Examples: rope waves, light waves.
- Longitudinal wave: particles vibrate parallel to the direction of travel. Example: sound in air.
- Surface wave: occurs at the boundary between two media, such as water ripples.
Important wave quantities
- Amplitude: maximum displacement from equilibrium. Larger amplitude usually means greater energy.
- Wavelength, \(\lambda \): distance between two consecutive crests or compressions.
- Period, T: time for one complete oscillation.
- Frequency, f: number of oscillations per second, measured in hertz.
- Wave speed: speed of wave travel.
Key equations:
- \(f = 1/T\)
- \(v = \lambda f\)
Reflection of pulses
- At a fixed end, a reflected pulse is inverted.
- At a free end, a reflected pulse remains upright.
Superposition
When two waves overlap, the resultant displacement is the algebraic sum of the individual displacements.
- Same direction: constructive interference.
- Opposite direction: destructive interference.
Standing waves
Standing waves form when two waves of the same frequency travel in opposite directions and interfere.
- Node: point of zero/minimum displacement.
- Antinode: point of maximum displacement.
- For a string fixed at both ends: \(\lambda_n = 2L/n\).
2. Sound
Sound is a mechanical longitudinal wave. It travels through matter as pressure oscillations.
Sound properties
- Sound travels as compressions and rarefactions.
- Sound needs a medium; it cannot travel through a perfect vacuum.
- Sound generally travels fastest in solids, slower in liquids, and slowest in gases.
- In air, sound speed increases as temperature increases.
Pitch and loudness
- Pitch depends on frequency. Higher frequency means higher pitch.
- Loudness depends mainly on amplitude/intensity. Larger amplitude means louder sound.
- Sound level is measured in decibels, dB.
Doppler effect
The Doppler effect is the change in detected frequency due to relative motion between source and observer.
- Approaching source/observer: detected frequency increases.
- Moving apart: detected frequency decreases.
Resonance
Resonance occurs when a system is driven at its natural frequency, causing a large increase in amplitude. Musical instruments use resonance in strings and air columns.
3. Light and Illumination
Light sources and media
- Luminous source: produces its own light, such as the Sun or a lamp.
- Non-luminous object: seen by reflected light, such as the Moon or a book.
- Transparent: allows clear transmission of light.
- Translucent: allows light through but scatters it.
- Opaque: blocks most light.
Light quantities
- Luminous flux, P: total visible light output, unit lumen (lm).
- Illuminance, E: light falling on unit area, unit lux (lx).
- Luminous intensity: light emitted in a direction, unit candela (cd).
Useful equations:
- \(E = P/A\)
- For a point source: \(E = P/(4\pi r^2)\)
Illuminance follows an inverse-square relationship:
- Double distance → illuminance becomes one-quarter.
- Triple distance → illuminance becomes one-ninth.
4. Diffraction and Color
Diffraction
Diffraction is the bending or spreading of waves as they pass an edge or opening. It is most noticeable when the gap size is similar to the wavelength.
Color of light
Color depends on wavelength and frequency.
- Red light: longer wavelength, lower frequency.
- Violet light: shorter wavelength, higher frequency.
Additive color mixing
Primary colors of light:
- Red
- Green
- Blue
Combinations:
- Red + green = yellow
- Green + blue = cyan
- Red + blue = magenta
- Red + green + blue = white
Object color depends on which wavelengths are present in incident light and which wavelengths are reflected or absorbed.
Pigments and dyes
Pigments use subtractive color mixing.
- Cyan absorbs red.
- Magenta absorbs green.
- Yellow absorbs blue.
Mixing pigments usually absorbs more wavelengths and gives darker colors.
5. Polarization and Malus's Law
Polarization is the restriction of light vibrations to one direction. Only transverse waves can be polarized.
Polarization by filtering
A polarizing filter allows one vibration direction to pass and blocks the perpendicular component.
Polarization by reflection
Reflected glare can be partially polarized. Polarizing sunglasses reduce glare by blocking much of this reflected polarized light.
Malus's law
For polarized light passing through an analyzer:
\(I_2 = I_1 \cos^2\theta \)
Where:
- \(I_1\) is intensity before the analyzer.
- \(I_2\) is intensity after the analyzer.
- \(\theta \) is the angle between the polarizer and analyzer axes.
Special cases:
- \(\theta \) = 0° → maximum transmitted intensity.
- \(\theta \) = 90° → zero/minimum transmitted intensity.
- For unpolarized light passing through the first ideal polarizer: \(I_1 = I_0/2\).
Exam Tips
- Always check units: Hz, m/s, m, s, dB, lm, lx, cd.
- For formula questions, write the formula first, substitute values, then calculate.
- For wave diagrams, mark the direction of travel and direction of particle vibration.
- For superposition, add displacements with signs.
- For Malus's law, decide whether the light entering the first polarizer is unpolarized or already polarized.
Waves • pages 115-116
OBJ01: Differentiate between transverse, longitudinal, and surface waves and give examples.
When particles vibrate parallel to the wave direction, the wave is:
A student shakes one end of a rope up and down while the wave travels horizontally. What type of wave is produced?
Which example is best classified as a longitudinal mechanical wave?
In a longitudinal wave, the particles of the medium vibrate ____ the direction of energy transfer.
A surface water wave is different from a pure transverse wave because the water particles near the surface usually move in:
Which statement correctly compares transverse and longitudinal waves?
Which pair correctly matches wave type and example?
A wave has crests and troughs clearly visible on a stretched string. Which property identifies it as transverse?
A slinky is pushed and pulled along its length. Regions of compression move down the spring. The wave is:
Compare transverse and longitudinal waves using direction of particle vibration and give one example of each.
States transverse vibration is perpendicular to direction of travel. States longitudinal vibration is parallel to direction of travel. Gives a valid transverse example such as rope wave or light. Gives a valid longitudinal example such as sound in air.
- 1 mark: States transverse vibration is perpendicular to direction of travel.
- 1 mark: States longitudinal vibration is parallel to direction of travel.
- 1 mark: Gives a valid transverse example such as rope wave or light.
- 1 mark: Gives a valid longitudinal example such as sound in air.
Explain why a water ripple is usually classified as a surface wave.
States it occurs at a boundary/interface. Mentions air-water or water surface. Explains motion has both up-down and back-forth components. States energy travels across the surface.
- 1 mark: States it occurs at a boundary/interface.
- 1 mark: Mentions air-water or water surface.
- 1 mark: Explains motion has both up-down and back-forth components.
- 1 mark: States energy travels across the surface.
A slinky is pushed and pulled along its length. Identify the wave type and justify.
Identifies the wave as longitudinal. Mentions compressions and rarefactions. States particle vibration is parallel to wave direction. Gives a clear link to the slinky motion.
- 1 mark: Identifies the wave as longitudinal.
- 1 mark: Mentions compressions and rarefactions.
- 1 mark: States particle vibration is parallel to wave direction.
- 1 mark: Gives a clear link to the slinky motion.
List three wave types and one example for each.
Names transverse wave with valid example. Names longitudinal wave with valid example. Names surface wave with valid example. Examples are correctly matched to wave types.
- 1 mark: Names transverse wave with valid example.
- 1 mark: Names longitudinal wave with valid example.
- 1 mark: Names surface wave with valid example.
- 1 mark: Examples are correctly matched to wave types.
Waves • pages 118-120
OBJ02: Calculate frequency from period and vice versa; apply v = λf to calculate speed, wavelength, or frequency.
A wave has wavelength \(\lambda \) = 0.5 m and frequency \(f = 8 Hz. Calculate its speed.\)
A wave has wavelength \(\lambda \) = 2 m and frequency \(f = 3 Hz. Calculate its speed.\)
A wave has a period of 0.2 s. What is its frequency?
A wave has a period of 0.5 s. What is its frequency?
A wave has a period of 0.025 s. What is its frequency?
A wave has a period of 4 s. What is its frequency?
A wave has wavelength 2 m and frequency 8 Hz. What is its speed?
A wave has wavelength 0.75 m and frequency 20 Hz. What is its speed?
A wave has wavelength 1.5 m and frequency 4 Hz. What is its speed?
Which equation is used to find wave speed from wavelength and frequency?
A wave has period 0.20 s and wavelength 1.5 m. Calculate frequency and wave speed.
Uses \(f = 1/T\). Calculates \(f = 5 Hz. Uses\)\(v = \lambda f\). Calculates \(v = 7.5 m/s with units.\)
- 1 mark: Uses \(f = 1/T\).
- 1 mark: Calculates \(f = 5 Hz.\)
- 1 mark: Uses \(v = \lambda f\).
- 1 mark: Calculates \(v = 7.5 m/s with units.\)
Explain the difference between period and frequency and state their relationship.
Defines period as time for one cycle. Defines frequency as cycles per second. States \(f = 1/T\) or \(T = 1/f. Uses correct units\): s and Hz.
- 1 mark: Defines period as time for one cycle.
- 1 mark: Defines frequency as cycles per second.
- 1 mark: States \(f = 1/T\) or \(T = 1/f.\)
- 1 mark: Uses correct units: s and Hz.
A wave travels at 24 m/s with frequency 6 Hz. Find wavelength.
Writes \(v = \lambda f\). Rearranges \(\lambda \) = v/f. Substitutes 24/6. Gives \(\lambda \) = 4 m.
- 1 mark: Writes \(v = \lambda f\).
- 1 mark: Rearranges \(\lambda \) = v/f.
- 1 mark: Substitutes 24/6.
- 1 mark: Gives \(\lambda \) = 4 m.
Describe how wavelength changes when frequency increases while wave speed remains constant.
States \(v = \lambda f\). States v is constant in the same medium. Explains wavelength and frequency are inversely related. Concludes wavelength decreases.
- 1 mark: States \(v = \lambda f\).
- 1 mark: States v is constant in the same medium.
- 1 mark: Explains wavelength and frequency are inversely related.
- 1 mark: Concludes wavelength decreases.
Waves • pages 121-122
OBJ03: Describe that a mechanical wave is inverted if reflected from a fixed end and remains upright if reflected from a free end.
When a wave pulse reflects from a fixed end, what happens to it?
When a wave pulse reflects from a free end, what happens to it?
A pulse reaches a wall where the end of the rope is fixed. The reflected pulse is:
A pulse reaches a loose ring that can move freely on a vertical rod. The reflected pulse is:
Why does a fixed end produce an inverted reflected pulse?
A crest pulse returns from a fixed end. What shape returns?
A trough pulse returns from a free end. What shape returns?
A mechanical pulse reflects from a boundary and returns upside down. The boundary was most likely:
Which statement is true for both fixed-end and free-end reflection?
A pulse on a spring reflects upright. Which diagram label should be assigned to the boundary?
Describe how a pulse reflects from a fixed end and from a free end.
Fixed-end reflection is inverted. Free-end reflection remains upright. Both reflected pulses travel back in the opposite direction. Uses correct terms fixed and free boundary.
- 1 mark: Fixed-end reflection is inverted.
- 1 mark: Free-end reflection remains upright.
- 1 mark: Both reflected pulses travel back in the opposite direction.
- 1 mark: Uses correct terms fixed and free boundary.
A crest pulse reflects from a fixed wall. Draw/describe the reflected pulse.
Identifies reflected pulse as a trough. Mentions inversion. States direction is reversed. Sketch/description shows correct displacement sign.
- 1 mark: Identifies reflected pulse as a trough.
- 1 mark: Mentions inversion.
- 1 mark: States direction is reversed.
- 1 mark: Sketch/description shows correct displacement sign.
Explain the force reason for inversion at a fixed end.
Fixed end cannot move. Boundary exerts opposite reaction force. Displacement reverses sign. Reflected pulse is inverted.
- 1 mark: Fixed end cannot move.
- 1 mark: Boundary exerts opposite reaction force.
- 1 mark: Displacement reverses sign.
- 1 mark: Reflected pulse is inverted.
A pulse reflects upright. What can you infer about the boundary?
Boundary is free/movable. Pulse is not inverted. Reflected pulse travels back along the medium. Gives a correct physical reason or example.
- 1 mark: Boundary is free/movable.
- 1 mark: Pulse is not inverted.
- 1 mark: Reflected pulse travels back along the medium.
- 1 mark: Gives a correct physical reason or example.
Waves • pages 116-118
OBJ04: Describe wave properties: amplitude, energy, wavelength, speed, phase, period, and frequency.
What is the amplitude of a wave?
If the amplitude of a mechanical wave increases while other factors remain constant, what happens to the wave energy?
Which wave property is the distance between two consecutive crests?
Which property tells how many complete oscillations occur each second?
Two waves have the same wavelength, but wave A has twice the amplitude of wave B. Which statement is correct?
Points on two identical waves are in phase when they:
Which pair of wave properties are reciprocals?
A wave with larger amplitude appears:
What is wave speed?
If the period of a wave increases while its speed is unchanged, what happens to frequency?
Define amplitude, wavelength, period, and frequency.
Amplitude is maximum displacement. Wavelength is distance for one complete cycle. Period is time for one cycle. Frequency is cycles per second.
- 1 mark: Amplitude is maximum displacement.
- 1 mark: Wavelength is distance for one complete cycle.
- 1 mark: Period is time for one cycle.
- 1 mark: Frequency is cycles per second.
Explain how amplitude is related to wave energy.
Larger amplitude means larger displacement. Larger amplitude waves carry more energy. Comparison assumes same medium/other factors similar. Gives a valid example such as louder sound or bigger rope wave.
- 1 mark: Larger amplitude means larger displacement.
- 1 mark: Larger amplitude waves carry more energy.
- 1 mark: Comparison assumes same medium/other factors similar.
- 1 mark: Gives a valid example such as louder sound or bigger rope wave.
Describe what it means for two points on a wave to be in phase.
Points are at the same stage of oscillation. They move in the same way at the same time. Examples: crest with crest or trough with trough. Uses phase terminology correctly.
- 1 mark: Points are at the same stage of oscillation.
- 1 mark: They move in the same way at the same time.
- 1 mark: Examples: crest with crest or trough with trough.
- 1 mark: Uses phase terminology correctly.
A wave has frequency 10 Hz. What does this mean physically?
Frequency is cycles per second. 10 Hz means 10 oscillations each second. Unit Hz equals s^-1. Relates to the vibrating source/medium.
- 1 mark: Frequency is cycles per second.
- 1 mark: 10 Hz means 10 oscillations each second.
- 1 mark: Unit Hz equals s^-1.
- 1 mark: Relates to the vibrating source/medium.
Waves • pages 122-123
OBJ05: Sketch snapshots for superposition of two overlapping pulses and find resultant amplitude.
A +5 cm pulse meets a −5 cm pulse. What is the resultant amplitude at complete overlap?
Constructive interference occurs when:
Two pulses of +2 cm and +3 cm overlap at the same point. What is the resultant displacement?
A +4 cm pulse overlaps a −4 cm pulse. What is the resultant amplitude?
Which condition produces constructive interference?
Which condition produces destructive interference?
During superposition, after two pulses pass through each other, they generally:
A +6 cm pulse and a −2 cm pulse overlap completely. What is the resultant displacement?
Which principle allows the resultant displacement to be found by adding individual displacements?
Two identical downward pulses each have amplitude −3 cm. At complete overlap, the resultant is:
Two pulses +5 cm and −3 cm overlap. Find resultant displacement and name the interference type.
Adds displacements algebraically. Calculates +2 cm. Identifies partial destructive interference. Includes correct sign/direction.
- 1 mark: Adds displacements algebraically.
- 1 mark: Calculates +2 cm.
- 1 mark: Identifies partial destructive interference.
- 1 mark: Includes correct sign/direction.
Explain the principle of superposition for overlapping pulses.
Resultant displacement is found point by point. Individual displacements are added algebraically. Applies only while pulses overlap. Pulses continue after passing through in a linear medium.
- 1 mark: Resultant displacement is found point by point.
- 1 mark: Individual displacements are added algebraically.
- 1 mark: Applies only while pulses overlap.
- 1 mark: Pulses continue after passing through in a linear medium.
Sketch/describe complete constructive interference of two equal upward pulses.
Shows two upward pulses overlap. Resultant amplitude is sum of amplitudes. Resultant is larger than either pulse. Pulses reappear after overlap.
- 1 mark: Shows two upward pulses overlap.
- 1 mark: Resultant amplitude is sum of amplitudes.
- 1 mark: Resultant is larger than either pulse.
- 1 mark: Pulses reappear after overlap.
Sketch/describe complete destructive interference of equal opposite pulses.
Shows crest and equal trough overlap. Resultant displacement is zero at complete overlap. Identifies destructive interference. Pulses continue after crossing.
- 1 mark: Shows crest and equal trough overlap.
- 1 mark: Resultant displacement is zero at complete overlap.
- 1 mark: Identifies destructive interference.
- 1 mark: Pulses continue after crossing.
Waves • pages 123
OBJ06: Define nodes and antinodes and describe how they are formed.
What is the name of a point that has almost no motion in a standing wave?
For a string fixed at both ends, what is the relation between string length L and wavelength for harmonic n?
In a standing wave, a node is a point where:
An antinode is a point where:
How are standing waves formed?
In a string fixed at both ends, the fixed ends are:
The distance between two adjacent nodes in a standing wave is:
A point on a standing wave moves with maximum amplitude. It is called:
Which statement is true for a standing wave pattern?
Between two adjacent antinodes, there is usually:
Define node and antinode in a standing wave.
Node has zero/minimum displacement. Antinode has maximum displacement. Both belong to a standing wave pattern. Uses correct terminology.
- 1 mark: Node has zero/minimum displacement.
- 1 mark: Antinode has maximum displacement.
- 1 mark: Both belong to a standing wave pattern.
- 1 mark: Uses correct terminology.
Explain how a standing wave is formed on a string.
Incident and reflected waves overlap. Waves have same frequency. They travel in opposite directions. Superposition forms fixed nodes and antinodes.
- 1 mark: Incident and reflected waves overlap.
- 1 mark: Waves have same frequency.
- 1 mark: They travel in opposite directions.
- 1 mark: Superposition forms fixed nodes and antinodes.
Describe the standing wave pattern for the first harmonic on a string fixed at both ends.
Nodes at both ends. One antinode in the middle. Length equals half a wavelength. States \(\lambda \) = 2L for first harmonic.
- 1 mark: Nodes at both ends.
- 1 mark: One antinode in the middle.
- 1 mark: Length equals half a wavelength.
- 1 mark: States \(\lambda \) = 2L for first harmonic.
What is the distance between adjacent nodes in terms of wavelength?
States adjacent node spacing is \(\lambda \)/2. Mentions adjacent antinode spacing is also \(\lambda \)/2 if included. Explains using standing wave geometry. Uses correct symbol \(\lambda \).
- 1 mark: States adjacent node spacing is \(\lambda \)/2.
- 1 mark: Mentions adjacent antinode spacing is also \(\lambda \)/2 if included.
- 1 mark: Explains using standing wave geometry.
- 1 mark: Uses correct symbol \(\lambda \).
Waves • pages 124
OBJ07: Determine wavelengths for the first several harmonics on a string under tension: λn = 2L/n.
A string of length L = 1.2 m is fixed at both ends. What is the wavelength for harmonic n = 1?
A string of length L = 1.2 m is fixed at both ends. What is the wavelength for harmonic n = 2?
A string fixed at both ends has length L = 2 m. What wavelength is required for harmonic n = 1?
A string fixed at both ends has length L = 2 m. What wavelength is required for harmonic n = 2?
A string fixed at both ends has length L = 2 m. What wavelength is required for harmonic n = 4?
A string fixed at both ends has length L = 1.5 m. What wavelength is required for harmonic n = 3?
A string fixed at both ends has length L = 0.8 m. What wavelength is required for harmonic n = 2?
For a string fixed at both ends, which harmonic has wavelength 2L?
For a fixed string, the third harmonic wavelength is:
As harmonic number n increases on the same string, the allowed wavelength:
For a fixed string of length 1.2 m, calculate wavelengths of the first three harmonics.
Uses \(\lambda_n = 2L/n\). λ1 = 2.4 m. λ2 = 1.2 m. λ3 = 0.8 m.
- 1 mark: Uses \(\lambda_n = 2L/n\).
- 1 mark: λ1 = 2.4 m.
- 1 mark: λ2 = 1.2 m.
- 1 mark: λ3 = 0.8 m.
Derive/justify the relationship \(\lambda_n = 2L/n\) for a string fixed at both ends.
States both ends are nodes. A whole number of half-wavelengths fits on the string. Writes L = nλ/2. Rearranges to \(\lambda_n = 2L/n\).
- 1 mark: States both ends are nodes.
- 1 mark: A whole number of half-wavelengths fits on the string.
- 1 mark: Writes L = nλ/2.
- 1 mark: Rearranges to \(\lambda_n = 2L/n\).
Explain what happens to wavelength as harmonic number increases.
States \(\lambda_n = 2L/n\). Harmonic number is in denominator. Therefore wavelength decreases. Gives example n=1 vs n=2.
- 1 mark: States \(\lambda_n = 2L/n\).
- 1 mark: Harmonic number is in denominator.
- 1 mark: Therefore wavelength decreases.
- 1 mark: Gives example n=1 vs n=2.
A string length is 0.90 m. Find the wavelength of the third harmonic.
Identifies n=3. Uses λ3 = 2L/3. Substitutes 2(0.90)/3. Gives 0.60 m.
- 1 mark: Identifies n=3.
- 1 mark: Uses λ3 = 2L/3.
- 1 mark: Substitutes 2(0.90)/3.
- 1 mark: Gives 0.60 m.
Sound • pages 132
OBJ08: Define sound as pressure oscillation transmitted through matter; explain that sound is a longitudinal wave.
Which example is a longitudinal wave?
Which description of sound is most accurate?
Sound is best described as:
Why can sound not travel through a perfect vacuum?
In air, sound travels mainly as:
The particle motion in a sound wave in air is:
A compression in a sound wave is a region of:
A rarefaction in a sound wave is a region of:
Which property does sound share with other waves?
When a tuning fork vibrates, it produces sound because it:
Explain why sound is called a longitudinal pressure wave.
Sound consists of pressure oscillations. It has compressions and rarefactions. Particle motion is parallel to wave travel. It requires matter/a medium.
- 1 mark: Sound consists of pressure oscillations.
- 1 mark: It has compressions and rarefactions.
- 1 mark: Particle motion is parallel to wave travel.
- 1 mark: It requires matter/a medium.
Define compression and rarefaction in a sound wave.
Compression is high pressure/high density. Rarefaction is low pressure/low density. They alternate as sound travels. They transmit sound energy through matter.
- 1 mark: Compression is high pressure/high density.
- 1 mark: Rarefaction is low pressure/low density.
- 1 mark: They alternate as sound travels.
- 1 mark: They transmit sound energy through matter.
Why can sound not travel in a perfect vacuum?
Sound is mechanical. It needs particles/matter. Vacuum has no medium to vibrate. Therefore no pressure oscillations can be transmitted.
- 1 mark: Sound is mechanical.
- 1 mark: It needs particles/matter.
- 1 mark: Vacuum has no medium to vibrate.
- 1 mark: Therefore no pressure oscillations can be transmitted.
List three wave quantities that sound has and define one.
Names frequency. Names wavelength. Names speed or amplitude. Defines one correctly with unit if relevant.
- 1 mark: Names frequency.
- 1 mark: Names wavelength.
- 1 mark: Names speed or amplitude.
- 1 mark: Defines one correctly with unit if relevant.
Sound • pages 133-134
OBJ09: Explain that the speed of sound varies with different mediums and temperatures.
What usually happens to the speed of sound in air when temperature increases?
In which medium does sound usually travel fastest?
In general, sound travels fastest in:
The speed of sound in air increases when:
Which ordering is generally correct for speed of sound?
Why does sound travel faster in water than in air?
What is required for the speed of sound to be defined in a medium?
If sound enters a warmer region of air, its speed usually:
Which statement is true?
Two students clap underwater and in air. The sound reaches a nearby sensor faster through:
Explain how the speed of sound depends on medium.
Sound generally travels fastest in solids. Slower in liquids. Slowest in gases. Explains using particle closeness/elastic properties.
- 1 mark: Sound generally travels fastest in solids.
- 1 mark: Slower in liquids.
- 1 mark: Slowest in gases.
- 1 mark: Explains using particle closeness/elastic properties.
Describe how increasing air temperature affects speed of sound.
States speed of sound in air increases. Warmer particles move faster/have more kinetic energy. Vibrations transfer more quickly. Conclusion linked to temperature.
- 1 mark: States speed of sound in air increases.
- 1 mark: Warmer particles move faster/have more kinetic energy.
- 1 mark: Vibrations transfer more quickly.
- 1 mark: Conclusion linked to temperature.
Order air, water, and steel from lowest to highest speed of sound and justify.
Air is lowest. Water is intermediate. Steel/solid is highest. Justification refers to medium structure.
- 1 mark: Air is lowest.
- 1 mark: Water is intermediate.
- 1 mark: Steel/solid is highest.
- 1 mark: Justification refers to medium structure.
Why is there no sound transmission through space between planets?
Space is nearly vacuum. Sound requires matter. No pressure oscillations can propagate. Contrasts with light/electromagnetic waves if relevant.
- 1 mark: Space is nearly vacuum.
- 1 mark: Sound requires matter.
- 1 mark: No pressure oscillations can propagate.
- 1 mark: Contrasts with light/electromagnetic waves if relevant.
Sound • pages 135
OBJ10: Define sound pitch and relate it to frequency.
Sound pitch is directly related to:
A sound of 800 Hz compared with 400 Hz is:
Pitch of a sound is mainly related to:
A high-pitched sound has:
A 900 Hz tone and a 300 Hz tone are played at the same loudness. Which has higher pitch?
If a string instrument note becomes higher in pitch, its frequency has:
Which unit is used to measure the physical quantity that controls pitch?
Two sounds have equal amplitude but different frequencies. What perceptual difference is most likely?
A low-pitched sound corresponds to:
A vibrating object makes 440 oscillations each second. The sound frequency is:
Define pitch and state its relationship with frequency.
Pitch is how high or low a sound seems. Pitch depends on frequency. Higher frequency gives higher pitch. Lower frequency gives lower pitch.
- 1 mark: Pitch is how high or low a sound seems.
- 1 mark: Pitch depends on frequency.
- 1 mark: Higher frequency gives higher pitch.
- 1 mark: Lower frequency gives lower pitch.
Compare 250 Hz and 1000 Hz sounds in terms of pitch.
1000 Hz has higher frequency. Therefore 1000 Hz sound has higher pitch. 250 Hz has lower pitch. Frequency unit Hz identified.
- 1 mark: 1000 Hz has higher frequency.
- 1 mark: Therefore 1000 Hz sound has higher pitch.
- 1 mark: 250 Hz has lower pitch.
- 1 mark: Frequency unit Hz identified.
A guitar string vibrates faster after tightening. Predict the pitch change.
Faster vibration means higher frequency. Pitch increases. Mentions relationship pitch-frequency. Answer is clearly linked to the scenario.
- 1 mark: Faster vibration means higher frequency.
- 1 mark: Pitch increases.
- 1 mark: Mentions relationship pitch-frequency.
- 1 mark: Answer is clearly linked to the scenario.
Explain why amplitude alone does not determine pitch.
Amplitude relates mainly to loudness. Frequency relates to pitch. Two sounds can have same amplitude but different pitch. Uses correct distinction.
- 1 mark: Amplitude relates mainly to loudness.
- 1 mark: Frequency relates to pitch.
- 1 mark: Two sounds can have same amplitude but different pitch.
- 1 mark: Uses correct distinction.
Sound • pages 135
OBJ11: Define loudness and relate it to amplitude of a sound wave.
Sound loudness is mainly related to:
When does resonance occur?
Loudness of a sound is mainly related to:
Which sound wave is likely to be louder?
If a speaker vibrates air with larger pressure variations, the sound is perceived as:
Two tones have the same frequency but different amplitudes. They differ mainly in:
A whisper and a shout may have similar pitch but different:
What happens to the perceived loudness when sound amplitude decreases significantly?
Which graph would represent the louder of two sounds at the same frequency?
Loudness is a perception related to sound intensity, but in simple wave models it is strongly linked to:
Define loudness and relate it to wave amplitude.
Loudness is perception of how strong a sound is. It is related to amplitude/intensity. Greater amplitude means louder sound. Smaller amplitude means quieter sound.
- 1 mark: Loudness is perception of how strong a sound is.
- 1 mark: It is related to amplitude/intensity.
- 1 mark: Greater amplitude means louder sound.
- 1 mark: Smaller amplitude means quieter sound.
Compare two sounds with the same frequency but different amplitudes.
Same frequency means same pitch. Larger amplitude sound is louder. Smaller amplitude sound is quieter. Uses amplitude/loudness terms correctly.
- 1 mark: Same frequency means same pitch.
- 1 mark: Larger amplitude sound is louder.
- 1 mark: Smaller amplitude sound is quieter.
- 1 mark: Uses amplitude/loudness terms correctly.
Explain how a pressure-time graph shows a louder sound.
Louder sound has larger pressure variation. Graph has taller peaks/deeper troughs. Frequency can remain unchanged. Links larger amplitude to loudness.
- 1 mark: Louder sound has larger pressure variation.
- 1 mark: Graph has taller peaks/deeper troughs.
- 1 mark: Frequency can remain unchanged.
- 1 mark: Links larger amplitude to loudness.
Why does moving away from a sound source usually reduce loudness?
Sound energy spreads out. Less energy reaches each unit area. Amplitude/intensity at ear decreases. Perceived loudness decreases.
- 1 mark: Sound energy spreads out.
- 1 mark: Less energy reaches each unit area.
- 1 mark: Amplitude/intensity at ear decreases.
- 1 mark: Perceived loudness decreases.
Sound • pages 135
OBJ12: Describe sound level and define decibel (dB) as a unit of measuring sound level.
What is the unit used for sound level?
What does it usually mean that the decibel scale is logarithmic?
Sound level is commonly measured in:
The symbol dB represents:
A sound level meter gives a reading of 85 dB. What does this value describe?
Which quantity is measured in hertz, not decibels?
Which is the most appropriate unit for comparing noise near a construction site?
A larger dB reading usually indicates:
Why is dB useful for sound levels?
Which pair is correctly matched?
Define sound level and state its unit.
Sound level compares sound intensity/loudness level. Unit is decibel. Symbol is dB. Uses correct context such as noise measurement.
- 1 mark: Sound level compares sound intensity/loudness level.
- 1 mark: Unit is decibel.
- 1 mark: Symbol is dB.
- 1 mark: Uses correct context such as noise measurement.
Distinguish between hertz and decibel.
Hertz measures frequency. Frequency relates to pitch. Decibel measures sound level. Sound level relates to loudness/noise.
- 1 mark: Hertz measures frequency.
- 1 mark: Frequency relates to pitch.
- 1 mark: Decibel measures sound level.
- 1 mark: Sound level relates to loudness/noise.
Explain why the decibel scale is useful.
Sound intensities cover a very wide range. dB gives manageable numbers. Used for comparing noise/sound levels. Higher dB generally means greater sound level.
- 1 mark: Sound intensities cover a very wide range.
- 1 mark: dB gives manageable numbers.
- 1 mark: Used for comparing noise/sound levels.
- 1 mark: Higher dB generally means greater sound level.
Give two examples where measuring sound level in dB is important.
Gives a valid example such as construction noise. Gives a second valid example such as headphones/traffic/aircraft. Explains health/safety or comfort relevance. Uses dB correctly.
- 1 mark: Gives a valid example such as construction noise.
- 1 mark: Gives a second valid example such as headphones/traffic/aircraft.
- 1 mark: Explains health/safety or comfort relevance.
- 1 mark: Uses dB correctly.
Sound • pages 138
OBJ13: Apply the Doppler effect equation to calculate detected frequencies and velocities.
When an ambulance approaches a stationary observer, how does the observed frequency change?
A sound source of frequency 500 Hz is moving toward the observer at 20 m/s. Assuming sound speed 343 m/s, what is the approximate observed frequency?
A stationary observer hears a siren from a car moving toward them. Compared with the source frequency, the detected frequency is:
A siren source moves away from a stationary observer. The detected frequency is:
Using \(v = 340 m/s\), a 500 Hz source moves toward a stationary detector at 20 m/s. Which expression is correct?
A 600 Hz source moves away from a stationary observer at 40 m/s. Take \(v = 340 m/s. What is the approximate observed frequency\)?
A detector moves toward a stationary 400 Hz source at 34 m/s. Take \(v = 340 m/s. What is the detected frequency\)?
A detector moves away from a stationary 800 Hz source at 17 m/s. Take \(v = 340 m/s. What is the detected frequency\)?
In the Doppler effect for sound, v usually represents:
Which situation gives the highest detected frequency?
Explain the Doppler effect using an approaching ambulance.
Relative motion between source and observer occurs. Wavefronts are compressed in front. Observed frequency/pitch increases while approaching. Observed frequency/pitch decreases after moving away.
- 1 mark: Relative motion between source and observer occurs.
- 1 mark: Wavefronts are compressed in front.
- 1 mark: Observed frequency/pitch increases while approaching.
- 1 mark: Observed frequency/pitch decreases after moving away.
A 500 Hz source moves away from a stationary observer at 20 m/s. Write the correct Doppler expression using \(v = 340 m/s.\)
Identifies source moving away. Uses denominator v + vs. Writes fd = 500 × 340/(340 + 20). States observed frequency is lower than 500 Hz.
- 1 mark: Identifies source moving away.
- 1 mark: Uses denominator v + vs.
- 1 mark: Writes fd = 500 × 340/(340 + 20).
- 1 mark: States observed frequency is lower than 500 Hz.
Calculate detected frequency for a stationary source of 600 Hz when observer moves toward it at 34 m/s. Use \(v = 340 m/s.\)
Uses fd = fs(v + vd)/v. Substitutes 600(340 + 34)/340. Calculates 660 Hz. Includes higher frequency due to approach.
- 1 mark: Uses fd = fs(v + vd)/v.
- 1 mark: Substitutes 600(340 + 34)/340.
- 1 mark: Calculates 660 Hz.
- 1 mark: Includes higher frequency due to approach.
State two factors that determine whether detected frequency increases or decreases.
Direction of relative motion: approaching or moving apart. Speed of source/observer affects size of shift. Speed of sound in medium is relevant. Correctly links approaching to higher and receding to lower frequency.
- 1 mark: Direction of relative motion: approaching or moving apart.
- 1 mark: Speed of source/observer affects size of shift.
- 1 mark: Speed of sound in medium is relevant.
- 1 mark: Correctly links approaching to higher and receding to lower frequency.
Sound • pages 141
OBJ14: Recall resonance as amplitude increase at natural frequency; explain resonance in air columns and instruments.
Which example is related to resonance in air columns?
In a string instrument, changing the effective string length changes:
Resonance occurs when a system is driven at:
The main effect of resonance is:
Which instrument uses resonance in an air column?
A tuning fork makes a nearby air column sound loudly when the length is adjusted. This is due to:
Why do musical instruments have specific shapes and lengths?
A child pumps a swing at the right time and the amplitude grows. This is an example of:
In a resonating air column, the standing wave pattern contains:
Which statement about resonance is correct?
Define resonance and give one everyday example.
Resonance is large amplitude vibration. Occurs when driving frequency matches natural frequency. Gives valid example such as swing, tuning fork, air column, instrument. Explains amplitude increase.
- 1 mark: Resonance is large amplitude vibration.
- 1 mark: Occurs when driving frequency matches natural frequency.
- 1 mark: Gives valid example such as swing, tuning fork, air column, instrument.
- 1 mark: Explains amplitude increase.
Explain resonance in an air column of a wind instrument.
Air column has natural frequencies. Sound waves reflect and form standing waves. Resonance occurs at allowed frequencies. This produces strong notes/pitches.
- 1 mark: Air column has natural frequencies.
- 1 mark: Sound waves reflect and form standing waves.
- 1 mark: Resonance occurs at allowed frequencies.
- 1 mark: This produces strong notes/pitches.
Why does changing the length of a flute or pipe change the note?
Length changes allowed wavelengths. Allowed wavelengths change resonant frequencies. Frequency determines pitch. Therefore note changes.
- 1 mark: Length changes allowed wavelengths.
- 1 mark: Allowed wavelengths change resonant frequencies.
- 1 mark: Frequency determines pitch.
- 1 mark: Therefore note changes.
Describe a laboratory demonstration of resonance using a tuning fork and tube.
Tuning fork provides periodic driving sound. Air column length is adjusted. Loud sound occurs at resonance. Conclusion: air column natural frequency matches tuning fork.
- 1 mark: Tuning fork provides periodic driving sound.
- 1 mark: Air column length is adjusted.
- 1 mark: Loud sound occurs at resonance.
- 1 mark: Conclusion: air column natural frequency matches tuning fork.
Light • pages 157-158
OBJ15: Differentiate luminous and non-luminous sources; differentiate opaque, translucent, and transparent media.
Which of the following is a luminous source?
Which description matches a transparent object?
Which object is luminous?
A non-luminous object is seen because it:
A clear glass window is best described as:
Frosted glass allows light through but does not form a clear image. It is:
A wooden door is:
Which pair is correct?
Which material allows the clearest view of an object behind it?
Which statement describes an opaque medium?
Differentiate luminous and non-luminous sources with examples.
Luminous source emits its own light. Non-luminous source does not emit its own light. Gives valid luminous example such as Sun/lamp. Gives valid non-luminous example such as Moon/book.
- 1 mark: Luminous source emits its own light.
- 1 mark: Non-luminous source does not emit its own light.
- 1 mark: Gives valid luminous example such as Sun/lamp.
- 1 mark: Gives valid non-luminous example such as Moon/book.
Compare transparent, translucent, and opaque materials.
Transparent transmits light clearly. Translucent transmits light but scatters/blurred image. Opaque blocks most light. Gives at least one correct example.
- 1 mark: Transparent transmits light clearly.
- 1 mark: Translucent transmits light but scatters/blurred image.
- 1 mark: Opaque blocks most light.
- 1 mark: Gives at least one correct example.
Why can we see a non-luminous object?
It is illuminated by a luminous source. It reflects/scatters light into our eyes. It does not produce its own light. Example provided.
- 1 mark: It is illuminated by a luminous source.
- 1 mark: It reflects/scatters light into our eyes.
- 1 mark: It does not produce its own light.
- 1 mark: Example provided.
Classify clear glass, frosted glass, and wood using light transmission.
Clear glass is transparent. Frosted glass is translucent. Wood is opaque. Classification linked to transmission/scattering/blocking.
- 1 mark: Clear glass is transparent.
- 1 mark: Frosted glass is translucent.
- 1 mark: Wood is opaque.
- 1 mark: Classification linked to transmission/scattering/blocking.
Light • pages 158-161
OBJ16: Define luminous flux, illuminance, and luminous intensity with SI units.
What is the SI unit of illuminance?
A point source has luminous flux \(P = 1000 lm. Calculate the illuminance on a surface at r\)= 1 m approximately.
Luminous flux is measured in:
Illuminance is measured in:
Luminous intensity is measured in:
Which quantity describes luminous flux per unit area on a surface?
Which light-quantity pair is correctly matched?
A lamp emits a total visible light output. This total output is best described as:
Which unit belongs to luminous intensity rather than illuminance?
If the same luminous flux spreads over a larger area, illuminance:
Define luminous flux, illuminance, and luminous intensity with units.
Luminous flux is total visible light output in lumen (lm). Illuminance is light per unit area in lux (lx). Luminous intensity is light in a direction in candela (cd). All units correctly stated.
- 1 mark: Luminous flux is total visible light output in lumen (lm).
- 1 mark: Illuminance is light per unit area in lux (lx).
- 1 mark: Luminous intensity is light in a direction in candela (cd).
- 1 mark: All units correctly stated.
Explain why illuminance changes when the same light spreads over a larger area.
Illuminance is luminous flux per unit area. Same flux over larger area gives smaller E. Uses \(E = P/A\). Conclusion: illuminance decreases.
- 1 mark: Illuminance is luminous flux per unit area.
- 1 mark: Same flux over larger area gives smaller E.
- 1 mark: Uses \(E = P/A\).
- 1 mark: Conclusion: illuminance decreases.
A source emits 600 lm uniformly onto 3 m². Calculate average illuminance.
Uses \(E = P/A\). Substitutes 600/3. Calculates 200. Uses unit lux.
- 1 mark: Uses \(E = P/A\).
- 1 mark: Substitutes 600/3.
- 1 mark: Calculates 200.
- 1 mark: Uses unit lux.
Match units lm, lx, and cd to the correct light quantities.
lm matched to luminous flux. lx matched to illuminance. cd matched to luminous intensity. No incorrect sound units included.
- 1 mark: lm matched to luminous flux.
- 1 mark: lx matched to illuminance.
- 1 mark: cd matched to luminous intensity.
- 1 mark: No incorrect sound units included.
Light • pages 162
OBJ17: Apply the illuminance equation for a point source to numerical problems.
A point source has luminous flux \(P = 1000 lm. Calculate the illuminance on a surface at r\)= 2 m approximately.
A point source has luminous flux \(P = 500 lm. Calculate the illuminance on a surface at r\)= 1 m approximately.
A point source emits luminous flux \(P = 800 lm. What is the illuminance at distance r\)= 1 m? Use \(E = P/(4\pi r^2)\) and \(\pi \) ≈ 3.14.
A point source emits luminous flux \(P = 800 lm. What is the illuminance at distance r\)= 2 m? Use \(E = P/(4\pi r^2)\) and \(\pi \) ≈ 3.14.
A point source emits luminous flux \(P = 1200 lm. What is the illuminance at distance r\)= 2 m? Use \(E = P/(4\pi r^2)\) and \(\pi \) ≈ 3.14.
A point source emits luminous flux \(P = 600 lm. What is the illuminance at distance r\)= 3 m? Use \(E = P/(4\pi r^2)\) and \(\pi \) ≈ 3.14.
A point source emits luminous flux \(P = 1600 lm. What is the illuminance at distance r\)= 4 m? Use \(E = P/(4\pi r^2)\) and \(\pi \) ≈ 3.14.
Which formula is suitable for illuminance from a point source spreading uniformly in all directions?
A lamp is moved from 1 m to 2 m away. If it was 80 lx at 1 m, what is it at 2 m?
A surface receives 1000 lm over 5 m². What is the average illuminance?
Calculate illuminance from a point source of 1000 lm at 2.0 m using \(E = P/(4\pi r^2)\).
Substitutes \(P = 1000 lm and r\)= 2.0 m. Calculates denominator 4πr² = 16π. Calculates E ≈ 19.9 lx. Gives correct unit lux.
- 1 mark: Substitutes \(P = 1000 lm and r\)= 2.0 m.
- 1 mark: Calculates denominator 4πr² = 16π.
- 1 mark: Calculates E ≈ 19.9 lx.
- 1 mark: Gives correct unit lux.
A lamp gives 90 lx at 1.0 m. Predict illuminance at 3.0 m.
Uses inverse-square relationship. Distance triples so illuminance divided by 9. Calculates 90/9 = 10 lx. Includes unit.
- 1 mark: Uses inverse-square relationship.
- 1 mark: Distance triples so illuminance divided by 9.
- 1 mark: Calculates 90/9 = 10 lx.
- 1 mark: Includes unit.
A surface receives 1500 lm over 6 m². Find average illuminance.
Uses \(E = P/A\). Substitutes 1500/6. Calculates 250. Unit lx.
- 1 mark: Uses \(E = P/A\).
- 1 mark: Substitutes 1500/6.
- 1 mark: Calculates 250.
- 1 mark: Unit lx.
Explain why the point-source formula contains 4πr².
Light spreads in all directions. It forms a spherical wavefront/surface. Area of sphere is 4πr². Illuminance is flux divided by area.
- 1 mark: Light spreads in all directions.
- 1 mark: It forms a spherical wavefront/surface.
- 1 mark: Area of sphere is 4πr².
- 1 mark: Illuminance is flux divided by area.
Light • pages 158-161
OBJ18: Define quantities of light such as luminous flux and illuminance with SI units.
A point source has luminous flux \(P = 2000 lm. Calculate the illuminance on a surface at r\)= 2 m approximately.
A point source has luminous flux \(P = 1500 lm. Calculate the illuminance on a surface at r\)= 3 m approximately.
Which quantity has unit lumen?
Which quantity has unit lux?
Which quantity has unit candela?
Which definition best matches illuminance?
Which symbol-unit pair is correct?
A lamp has a high luminous flux. This means it:
Which unit is not a light quantity unit?
If 400 lm falls uniformly on 2 m², the illuminance is:
State the SI units for luminous flux, illuminance, and luminous intensity.
Luminous flux: lumen (lm). Illuminance: lux (lx). Luminous intensity: candela (cd). Correct spelling/symbols.
- 1 mark: Luminous flux: lumen (lm).
- 1 mark: Illuminance: lux (lx).
- 1 mark: Luminous intensity: candela (cd).
- 1 mark: Correct spelling/symbols.
A student writes that illuminance is measured in lumens. Correct the mistake.
States illuminance is measured in lux. Lumen is unit of luminous flux. Explains illuminance is flux per unit area. Gives \(E = P/A\) if appropriate.
- 1 mark: States illuminance is measured in lux.
- 1 mark: Lumen is unit of luminous flux.
- 1 mark: Explains illuminance is flux per unit area.
- 1 mark: Gives \(E = P/A\) if appropriate.
Describe a practical situation where illuminance is more useful than luminous flux.
Gives valid situation such as desk/classroom lighting. Explains illuminance tells light falling on surface. Mentions lux measurement. Contrasts with source's total flux.
- 1 mark: Gives valid situation such as desk/classroom lighting.
- 1 mark: Explains illuminance tells light falling on surface.
- 1 mark: Mentions lux measurement.
- 1 mark: Contrasts with source's total flux.
Calculate average illuminance when 900 lm falls uniformly on 9 m².
Uses \(E = P/A\). Substitutes 900/9. Calculates 100. Uses unit lx.
- 1 mark: Uses \(E = P/A\).
- 1 mark: Substitutes 900/9.
- 1 mark: Calculates 100.
- 1 mark: Uses unit lx.
Light • pages 159
OBJ19: Show inverse-square relation of illuminance with distance and direct relation with luminous flux.
If the distance between a lamp and a surface doubles, what happens to illuminance?
The distance becomes 2 times the original distance. What is the new illuminance relative to the original?
Illuminance from a point source varies with distance according to:
If distance from a point source triples, illuminance becomes:
If distance is halved, illuminance becomes:
If luminous flux doubles and distance is unchanged, illuminance:
A lamp gives 120 lx at 1 m. What is the illuminance at 2 m?
Which change gives the greatest illuminance?
The graph of illuminance E against 1/r² should be:
A lamp emits twice the luminous flux but is placed twice as far away. Compared with the original illuminance, the new illuminance is:
Show mathematically why illuminance from a point source follows an inverse-square law.
Starts with \(E = P/(4\pi r^2)\). For constant P, E ∝ 1/r². Explains distance is squared. Concludes doubling distance gives one-quarter illuminance.
- 1 mark: Starts with \(E = P/(4\pi r^2)\).
- 1 mark: For constant P, E ∝ 1/r².
- 1 mark: Explains distance is squared.
- 1 mark: Concludes doubling distance gives one-quarter illuminance.
A lamp gives 240 lx at 1 m. Find illuminance at 2 m and 4 m.
At 2 m, divide by 4 = 60 lx. At 4 m, divide by 16 = 15 lx. Uses inverse-square relationship. Correct units.
- 1 mark: At 2 m, divide by 4 = 60 lx.
- 1 mark: At 4 m, divide by 16 = 15 lx.
- 1 mark: Uses inverse-square relationship.
- 1 mark: Correct units.
Explain the direct relationship between luminous flux and illuminance.
Uses \(E = P/(4\pi r^2)\). At constant distance, E ∝ P. Doubling P doubles E. Halving P halves E.
- 1 mark: Uses \(E = P/(4\pi r^2)\).
- 1 mark: At constant distance, E ∝ P.
- 1 mark: Doubling P doubles E.
- 1 mark: Halving P halves E.
A lamp has twice the luminous flux but is moved three times farther away. Find the factor change in illuminance.
Flux factor is ×2. Distance factor is ÷9. Net factor is 2/9. States illuminance becomes 2/9 of original.
- 1 mark: Flux factor is ×2.
- 1 mark: Distance factor is ÷9.
- 1 mark: Net factor is 2/9.
- 1 mark: States illuminance becomes 2/9 of original.
Light • pages 165
OBJ20: Define diffraction as bending of a wave as it passes the edge of a barrier.
What is diffraction?
Diffraction is more noticeable when the opening width is:
Diffraction is:
Diffraction is strongest when the gap size is:
Which situation shows diffraction?
Diffraction can occur with:
At the edge of a barrier, a wave may bend into the shadow region. This is called:
Which observation supports the wave model of light?
If a slit becomes much wider compared with wavelength, diffraction generally:
A sound can sometimes be heard around a corner because sound waves:
Define diffraction and give one example.
Diffraction is bending/spreading of a wave. Occurs at an edge/opening/barrier. Gives valid example such as water waves through gap or sound around corner. Identifies it as wave behavior.
- 1 mark: Diffraction is bending/spreading of a wave.
- 1 mark: Occurs at an edge/opening/barrier.
- 1 mark: Gives valid example such as water waves through gap or sound around corner.
- 1 mark: Identifies it as wave behavior.
Explain why diffraction is more noticeable through a narrow gap.
Diffraction depends on gap size relative to wavelength. More noticeable when gap is similar to wavelength. Wave spreads after passing the gap. Correctly uses wavelength concept.
- 1 mark: Diffraction depends on gap size relative to wavelength.
- 1 mark: More noticeable when gap is similar to wavelength.
- 1 mark: Wave spreads after passing the gap.
- 1 mark: Correctly uses wavelength concept.
Describe how diffraction supports the wave model of light.
Light can spread/bend through a small opening. Diffraction is a property of waves. Therefore light shows wave behavior. Gives a slit/edge example.
- 1 mark: Light can spread/bend through a small opening.
- 1 mark: Diffraction is a property of waves.
- 1 mark: Therefore light shows wave behavior.
- 1 mark: Gives a slit/edge example.
Compare diffraction at a wide opening and a narrow opening.
Wide opening gives little spreading. Narrow opening gives greater spreading. Comparison uses wavelength size. Mentions wavefront bending/spreading.
- 1 mark: Wide opening gives little spreading.
- 1 mark: Narrow opening gives greater spreading.
- 1 mark: Comparison uses wavelength size.
- 1 mark: Mentions wavefront bending/spreading.
Light • pages 166
OBJ21: Describe that color of light is related to wavelength and frequency.
Which visible light color generally has the longer wavelength?
Why does an object appear red under white light?
In visible light, red light has:
Violet light has:
For light waves in vacuum, frequency and wavelength are related by:
If wavelength decreases for light travelling in the same medium, frequency:
Which visible color generally has the shortest wavelength?
Which visible color generally has the lowest frequency?
Color of monochromatic light is determined mainly by its:
Green light has a different color from blue light because it has a different:
Explain how color is related to wavelength and frequency.
Different colors have different wavelengths. Different colors have different frequencies. In the same medium, c = λf. Longer wavelength corresponds to lower frequency.
- 1 mark: Different colors have different wavelengths.
- 1 mark: Different colors have different frequencies.
- 1 mark: In the same medium, c = λf.
- 1 mark: Longer wavelength corresponds to lower frequency.
Compare red and violet light in terms of wavelength and frequency.
Red has longer wavelength. Red has lower frequency. Violet has shorter wavelength. Violet has higher frequency.
- 1 mark: Red has longer wavelength.
- 1 mark: Red has lower frequency.
- 1 mark: Violet has shorter wavelength.
- 1 mark: Violet has higher frequency.
If a visible light wave has shorter wavelength than another in the same medium, what happens to frequency?
Uses c = λf or constant speed. Frequency increases when wavelength decreases. States inverse relationship. Applies to color difference.
- 1 mark: Uses c = λf or constant speed.
- 1 mark: Frequency increases when wavelength decreases.
- 1 mark: States inverse relationship.
- 1 mark: Applies to color difference.
Arrange red, green, and violet from highest to lowest frequency.
Violet highest. Green middle. Red lowest. Order linked to wavelength/frequency relationship.
- 1 mark: Violet highest.
- 1 mark: Green middle.
- 1 mark: Red lowest.
- 1 mark: Order linked to wavelength/frequency relationship.
Light • pages 167
OBJ22: Describe primary, secondary, and complementary colors of light and explain object color by absorption/reflection.
Primary colors of light in additive mixing are:
Glare reflected from a road can be reduced using:
The primary colors of light are:
Red light plus green light produces:
Green light plus blue light produces:
Red light plus blue light produces:
Combining red, green, and blue light in suitable intensities produces:
An object appears blue under white light because it mainly:
A red apple under pure blue light may appear dark because:
Complementary colors of light combine to produce:
State the primary colors of light and describe what happens when all three combine.
Primary light colors are red, green, blue. They combine additively. Suitable intensities produce white light. Uses additive color terminology.
- 1 mark: Primary light colors are red, green, blue.
- 1 mark: They combine additively.
- 1 mark: Suitable intensities produce white light.
- 1 mark: Uses additive color terminology.
Explain why a blue object appears blue under white light.
White light contains blue and other wavelengths. Object reflects blue wavelengths. Object absorbs many other wavelengths. Reflected blue reaches eye.
- 1 mark: White light contains blue and other wavelengths.
- 1 mark: Object reflects blue wavelengths.
- 1 mark: Object absorbs many other wavelengths.
- 1 mark: Reflected blue reaches eye.
Predict the appearance of a red object under pure green light and explain.
Red object mainly reflects red. Green light contains little/no red. Green is mostly absorbed. Object appears dark/blackish.
- 1 mark: Red object mainly reflects red.
- 1 mark: Green light contains little/no red.
- 1 mark: Green is mostly absorbed.
- 1 mark: Object appears dark/blackish.
Give examples of secondary colors produced by adding primary light colors.
Red + green = yellow. Green + blue = cyan. Red + blue = magenta. Any two correct combinations accepted for relevant marks.
- 1 mark: Red + green = yellow.
- 1 mark: Green + blue = cyan.
- 1 mark: Red + blue = magenta.
- 1 mark: Any two correct combinations accepted for relevant marks.
Light • pages 168
OBJ23: Describe primary and secondary pigments and effects of mixing pigments or dyes.
Mixing pigments differs from mixing light because pigments usually:
Primary pigment colors are usually:
Pigments mix by:
The primary pigments are commonly:
A cyan pigment appears cyan because it absorbs mainly:
A magenta pigment absorbs mainly:
A yellow pigment absorbs mainly:
Mixing many pigments often produces a darker color because:
Which statement best describes a dye?
What happens when cyan and yellow pigments are mixed ideally?
Explain subtractive color mixing using pigments.
Pigments absorb selected wavelengths. They reflect/transmit remaining wavelengths. Mixing pigments removes more wavelengths. Result often becomes darker.
- 1 mark: Pigments absorb selected wavelengths.
- 1 mark: They reflect/transmit remaining wavelengths.
- 1 mark: Mixing pigments removes more wavelengths.
- 1 mark: Result often becomes darker.
State the primary pigments and one color they absorb.
Primary pigments are cyan, magenta, yellow. Cyan absorbs red. Magenta absorbs green. Yellow absorbs blue.
- 1 mark: Primary pigments are cyan, magenta, yellow.
- 1 mark: Cyan absorbs red.
- 1 mark: Magenta absorbs green.
- 1 mark: Yellow absorbs blue.
Explain why mixing cyan and yellow pigments ideally gives green.
Cyan absorbs red. Yellow absorbs blue. Green is not absorbed by either ideally. Green is reflected/transmitted.
- 1 mark: Cyan absorbs red.
- 1 mark: Yellow absorbs blue.
- 1 mark: Green is not absorbed by either ideally.
- 1 mark: Green is reflected/transmitted.
Compare mixing colored light with mixing pigments.
Light mixing is additive. Pigment mixing is subtractive. RGB primary lights differ from CMY primary pigments. Explains absorption/reflection in pigments.
- 1 mark: Light mixing is additive.
- 1 mark: Pigment mixing is subtractive.
- 1 mark: RGB primary lights differ from CMY primary pigments.
- 1 mark: Explains absorption/reflection in pigments.
Light • pages 169
OBJ24: Explain polarization of light by filtering and by reflection.
What does a polarizing filter do?
If two polarizing filters are crossed at 90°, why is transmitted light nearly zero?
A polarizing filter mainly allows light vibrations in:
Only which type of wave can be polarized?
Unpolarized light contains vibrations:
After unpolarized light passes through an ideal polarizer, it becomes:
Polarizing sunglasses reduce glare mainly because reflected glare is often:
If two polarizing filters have perpendicular axes, transmitted light is:
Which example demonstrates polarization by filtering?
Why is sound in air not polarized in the usual way?
Explain polarization by filtering.
Unpolarized light has many vibration directions. Polarizing filter transmits one vibration direction. Emerging light is polarized. Blocked components reduce intensity.
- 1 mark: Unpolarized light has many vibration directions.
- 1 mark: Polarizing filter transmits one vibration direction.
- 1 mark: Emerging light is polarized.
- 1 mark: Blocked components reduce intensity.
Describe polarization by reflection and give an application.
Reflected light can become partially polarized. Glare from horizontal surfaces is often polarized. Polarizing sunglasses reduce glare. Application correctly linked to reflection.
- 1 mark: Reflected light can become partially polarized.
- 1 mark: Glare from horizontal surfaces is often polarized.
- 1 mark: Polarizing sunglasses reduce glare.
- 1 mark: Application correctly linked to reflection.
Why can light be polarized but sound in air cannot be polarized in the same way?
Light is transverse. Polarization selects transverse vibration direction. Sound in air is longitudinal. Longitudinal vibration is along travel direction.
- 1 mark: Light is transverse.
- 1 mark: Polarization selects transverse vibration direction.
- 1 mark: Sound in air is longitudinal.
- 1 mark: Longitudinal vibration is along travel direction.
Two polarizing filters are crossed at 90°. Describe the transmitted light.
First filter polarizes the light. Second filter axis is perpendicular. Very little/no light is transmitted. Explanation refers to vibration direction.
- 1 mark: First filter polarizes the light.
- 1 mark: Second filter axis is perpendicular.
- 1 mark: Very little/no light is transmitted.
- 1 mark: Explanation refers to vibration direction.
Light • pages 170
OBJ25: Apply Malus’s law to light filtered by polarizer and analyzer filters.
According to Malus’s law, when the angle between two polarizers is 90°, the transmitted intensity is approximately:
Polarized light with initial intensity I₀ = 100 units passes through an analyzer at 0°. Calculate transmitted intensity.
Malus's law relates transmitted intensity through an analyzer to:
If polarized light of intensity I₁ passes through an analyzer at \(\theta \) = 0°, I₂ is:
If polarized light passes through an analyzer at \(\theta \) = 90°, the transmitted intensity is:
If I₁ = 80 W/m² and \(\theta \) = 60°, what is I₂?
If I₁ = 100 W/m² and \(\theta \) = 45°, what is I₂?
Unpolarized light of intensity I₀ passes through the first ideal polarizer. The intensity after the first polarizer is:
Unpolarized light has I₀ = 120 W/m². After the first polarizer, then an analyzer at 60°, what is the final intensity?
In Malus's law, \(\theta \) is measured between:
State Malus's law and define the angle used in it.
Writes I2 = I1 cos²θ. I1 is intensity after first polarizer/before analyzer. I2 is intensity after analyzer. \(\theta \) is angle between polarizing axes.
- 1 mark: Writes I2 = I1 cos²θ.
- 1 mark: I1 is intensity after first polarizer/before analyzer.
- 1 mark: I2 is intensity after analyzer.
- 1 mark: \(\theta \) is angle between polarizing axes.
Polarized light of intensity 80 W/m² passes through an analyzer at 60°. Calculate final intensity.
Uses I2 = I1 cos²θ. cos60° = 0.5. cos²60° = 0.25. I2 = 20 W/m².
- 1 mark: Uses I2 = I1 cos²θ.
- 1 mark: cos60° = 0.5.
- 1 mark: cos²60° = 0.25.
- 1 mark: I2 = 20 W/m².
Unpolarized light of intensity 100 W/m² passes through a polarizer and then an analyzer at 45°. Find final intensity.
After first polarizer \(I_1 = I_0/2\) = 50 W/m². Uses I2 = I1 cos²45°. cos²45° = 0.5. I2 = 25 W/m².
- 1 mark: After first polarizer \(I_1 = I_0/2\) = 50 W/m².
- 1 mark: Uses I2 = I1 cos²45°.
- 1 mark: cos²45° = 0.5.
- 1 mark: I2 = 25 W/m².
Describe how transmitted intensity changes as analyzer angle changes from 0° to 90°.
At 0°, intensity is maximum I1. As \(\theta \) increases, intensity follows cos²θ. At 90°, intensity is zero/minimum. Explanation linked to alignment of polarizing axes.
- 1 mark: At 0°, intensity is maximum I1.
- 1 mark: As \(\theta \) increases, intensity follows cos²θ.
- 1 mark: At 90°, intensity is zero/minimum.
- 1 mark: Explanation linked to alignment of polarizing axes.