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Sound in air oscillates parallel to its direction of travel, making it longitudinal."},{"text":"Light waves are longitudinal; sound waves are transverse.","correct":false,"explanation":"Both waves have been assigned the wrong wave type. Light is transverse because its oscillations are perpendicular to travel, while sound in air is longitudinal because its oscillations are parallel."},{"text":"Light waves are transverse; sound waves are transverse.","correct":false,"explanation":"Sound is not transverse in air: its particles vibrate parallel to the wave’s travel direction. Light is transverse, with oscillations perpendicular to its direction of travel."},{"text":"Light waves are longitudinal; sound waves are longitudinal.","correct":false,"explanation":"Light is not longitudinal: electromagnetic waves have oscillating fields perpendicular to their travel direction. Sound in air is longitudinal because it consists of parallel compressions and rarefactions."}]},{"id":"08e9547c-038d-45c2-aa42-6f5c15364739","cardType":"basic","points":1,"front":"In ray diagrams, angles of incidence, reflection and refraction are measured from which line?","back":"The normal.","mcqOptions":[{"text":"The normal.","correct":true,"explanation":"Angles of incidence, reflection and refraction are measured between the ray and the normal, an imaginary line drawn perpendicular to the boundary at the point of contact."},{"text":"The surface.","correct":false,"explanation":"The surface is the boundary itself, not the reference line for measuring these angles. Angles are measured from the normal, which is perpendicular to the surface."},{"text":"The incident ray.","correct":false,"explanation":"The incident ray is one of the lines whose angle is measured, not the reference line. The required reference is the normal at the point where the ray meets the boundary."},{"text":"The tangent.","correct":false,"explanation":"A tangent runs along the surface and is perpendicular to the normal. Reflection and refraction angles are conventionally measured from the normal, not from the tangent."}]},{"id":"1c1f71c3-5f26-491a-b40b-2dc9cd6d98cd","cardType":"basic","points":1,"front":"How does the angle of reflection compare with the angle of incidence?","back":"They are equal: $\\theta_i = \\theta_r$.","mcqOptions":[{"text":"They are equal: $\\theta_i = \\theta_r$.","correct":true,"explanation":"The law of reflection states that the angle of reflection equals the angle of incidence, with both angles measured from the normal: $\\theta_i = \\theta_r$."},{"text":"The reflection angle is twice the incidence angle: $\\theta_r = 2\\theta_i$.","correct":false,"explanation":"Reflection does not generally double the incident angle. The reflected angle equals the incident angle when both are measured from the normal."},{"text":"The reflection angle is half the incidence angle: $\\theta_r = \\frac{1}{2}\\theta_i$.","correct":false,"explanation":"The reflected ray does not have half the incident angle. The law of reflection gives equal angles measured from the normal, so $\\theta_r = \\theta_i$."},{"text":"They add to 90°: $\\theta_i + \\theta_r = 90°$.","correct":false,"explanation":"The angles are not generally supplementary when measured from the normal. They are equal, because the reflected ray leaves at the same angle on the opposite side of the normal."}]},{"id":"5034ec87-6547-45cc-a6ff-26e79f3f3f9a","cardType":"basic","points":2,"front":"A ray hits a mirror at 30° to the surface. What is the reflection angle from the normal?","back":"60°.","mcqOptions":[{"text":"60°.","correct":true,"explanation":"Angles are measured from the normal, which is perpendicular to the mirror. Therefore the reflection angle is $90° - 30° = 60°$."},{"text":"30°.","correct":false,"explanation":"30° is measured from the surface, not the normal. Since the normal is at 90° to the surface, the angle from the normal is $90° - 30° = 60°$."},{"text":"120°.","correct":false,"explanation":"120° is not the acute angle used in a ray diagram. Complementing the 30° angle to the normal gives a reflection angle of 60°."},{"text":"45°.","correct":false,"explanation":"45° would apply only if the ray made 45° with the surface or normal. Here it makes 30° with the surface, so the angle from the normal is 60°."}]},{"id":"b5a7b41b-cae2-4cf0-aa3d-031e2018c173","cardType":"cloze","points":null,"front":"Light entering glass from air bends [ ] the normal; light entering air from glass bends [ ] the normal.","back":"Light entering glass from air bends towards the normal; light entering air from glass bends away from the normal.","mcqOptions":[{"text":"Light entering glass from air bends towards the normal; light entering air from glass bends away from the normal.","correct":true,"explanation":"Glass is optically denser than air, so light slows down and bends towards the normal on entering glass. It speeds up and bends away when entering air."},{"text":"Light entering glass from air bends away from the normal; light entering air from glass bends towards the normal.","correct":false,"explanation":"The directions have been reversed. Entering the denser glass makes light bend towards the normal, while leaving for less dense air makes it bend away."},{"text":"Light entering glass from air bends away from the normal; light entering air from glass bends away from the normal.","correct":false,"explanation":"Light does not bend away when entering glass from air. It bends towards the normal in the denser medium, then continues away from the normal when entering air."},{"text":"Light entering glass from air bends towards the normal; light entering air from glass bends towards the normal.","correct":false,"explanation":"The first direction is correct, but the second is not. On passing from glass into air, light enters a less optically dense medium and bends away from the normal."}]},{"id":"74ce5786-b05b-4953-a704-e0e1d3def20e","cardType":"basic","points":1,"front":"What happens to a ray that crosses a boundary along the normal?","back":"It does not bend.","mcqOptions":[{"text":"It does not bend.","correct":true,"explanation":"A ray travelling along the normal has zero angle of incidence. It may change speed at the boundary, but its direction does not change, so it does not bend."},{"text":"It reflects back along the normal.","correct":false,"explanation":"A ray along the normal does not reflect back when it crosses a boundary. Its direction remains unchanged, so it passes through without bending."},{"text":"It bends towards the normal.","correct":false,"explanation":"Bending towards the normal requires the ray to meet the boundary at an angle. A ray already travelling along the normal has no sideways direction to change."},{"text":"It bends away from the normal.","correct":false,"explanation":"A ray crossing along the normal does not bend away from it. The direction stays unchanged because the angle of incidence is zero."}]},{"id":"19f1324d-db7d-4054-a738-1a4540aae0de","cardType":"basic","points":1,"front":"For light travelling from air into a material, what is the refractive index equation?","back":"$n = \\frac{\\sin i}{\\sin r}$.","mcqOptions":[{"text":"$n = \\frac{\\sin i}{\\sin r}$.","correct":true,"explanation":"For light travelling from air into a material, the refractive index is the ratio of the sine of the incidence angle to the sine of the refraction angle: $n = \\frac{\\sin i}{\\sin r}$."},{"text":"$n = \\frac{\\sin r}{\\sin i}$.","correct":false,"explanation":"This is the reciprocal of the required relationship and would normally give a value less than one for light entering glass. The correct equation is $n = \\frac{\\sin i}{\\sin r}$."},{"text":"$n = \\sin i \\sin r$.","correct":false,"explanation":"Multiplying the two sines does not represent the refractive index and gives the wrong dependence on the measured angles. The correct relationship divides $\\sin i$ by $\\sin r$."},{"text":"$n = \\sin(i-r)$.","correct":false,"explanation":"Subtracting the angles is not Snell’s law and ignores the required sine relationship. For air entering a material, use $n = \\frac{\\sin i}{\\sin r}$."}]},{"id":"03d552e2-0ea6-4981-975e-11c9997eb1c4","cardType":"basic","points":2,"front":"Light enters glass from air with $i = 35°$ and $r = 22°$. What is $n$?","back":"1.53.","mcqOptions":[{"text":"1.53.","correct":true,"explanation":"Using the refractive-index equation gives $n = \\frac{\\sin 35°}{\\sin 22°} = 1.53$ to two decimal places."},{"text":"0.65","correct":false,"explanation":"This is the reciprocal relationship, calculating $\\sin 22° / \\sin 35°$ instead of $\\sin 35° / \\sin 22°$. For light entering glass, the refractive index is 1.53."},{"text":"1.59","correct":false,"explanation":"Dividing the angles directly, $35/22$, instead of dividing their sines gives about 1.59. Refractive index requires $\\sin i / \\sin r$, giving 1.53."},{"text":"1.50","correct":false,"explanation":"Rounding the sine values too aggressively, or using inaccurate trigonometric values, can produce this near-miss result. Using the calculator accurately gives $\\sin35°/\\sin22° = 1.53$."}]},{"id":"7200cc40-4fb6-484d-b30a-767f5ad06e87","cardType":"basic","points":1,"front":"On a graph of $\\sin i$ against $\\sin r$, what does the gradient give?","back":"The refractive index.","mcqOptions":[{"text":"The refractive index.","correct":true,"explanation":"Using Snell’s law, $n = \\frac{\\sin i}{\\sin r}$, so the gradient of a graph of $\\sin i$ against $\\sin r$ gives the refractive index."},{"text":"The critical angle.","correct":false,"explanation":"The critical angle is found from the angle at which refraction is 90°, not from this graph’s gradient. The gradient represents $\\frac{\\sin i}{\\sin r}$, which is the refractive index."},{"text":"The speed of light.","correct":false,"explanation":"The graph does not directly give a speed because its gradient is a ratio of sines. Snell’s law shows that this ratio is the refractive index, not the speed of light."},{"text":"The wavelength of light.","correct":false,"explanation":"Wavelength is not represented by the ratio of $\\sin i$ to $\\sin r$. The gradient follows Snell’s law and gives the refractive index of the material."}]},{"id":"a15f7b0b-3c22-443d-8f84-1bb0650960bc","cardType":"basic","points":2,"front":"Why aim at the centre of the curved face of a semi-circular block?","back":"The ray enters without bending.","mcqOptions":[{"text":"The ray enters without bending.","correct":true,"explanation":"Aiming at the centre means the ray travels along a radius of the curved surface, so it meets the surface normally and enters without bending."},{"text":"The ray enters with maximum bending.","correct":false,"explanation":"Maximum bending would occur at a larger angle of incidence, not along a radius. Aiming at the centre makes the ray meet the curved surface normally, so it enters without bending."},{"text":"The ray is totally internally reflected.","correct":false,"explanation":"Total internal reflection requires travel from higher to lower refractive index with an incidence angle above the critical angle. At the centre, the ray enters normally and does not reflect."},{"text":"The ray enters parallel to the flat face.","correct":false,"explanation":"The important condition is that the ray follows a radius and meets the curved face normally. This prevents bending at entry; being parallel to the flat face is not the reason."}]},{"id":"61b4ddef-ee39-44a2-892f-f8c22fa98867","cardType":"cloze","points":null,"front":"In a rectangular block, the emergent ray is usually [ ] to the incident ray but [ ].","back":"In a rectangular block, the emergent ray is usually parallel to the incident ray but shifted sideways.","mcqOptions":[{"text":"In a rectangular block, the emergent ray is usually parallel to the incident ray but shifted sideways.","correct":true,"explanation":"The two opposite parallel faces produce equal and opposite changes in direction, so the emergent ray is parallel to the incident ray but displaced sideways."},{"text":"In a rectangular block, the emergent ray is usually perpendicular to the incident ray but shifted sideways.","correct":false,"explanation":"The emergent ray is not generally at 90° to the incident ray. Refraction at parallel opposite faces restores the original direction, leaving the ray parallel but shifted sideways."},{"text":"In a rectangular block, the emergent ray is usually parallel to the incident ray but unchanged in position.","correct":false,"explanation":"Parallel faces restore the ray’s direction but do not restore its path. The ray leaves parallel to the incident ray with a lateral displacement, called a sideways shift."},{"text":"In a rectangular block, the emergent ray is usually angled to the incident ray but shifted sideways.","correct":false,"explanation":"An angled emergent ray would result if the entry and exit faces were not parallel. In a rectangular block, the faces cancel the directional changes, so the rays remain parallel."}]},{"id":"2858422d-54e8-4840-bcea-cb0c90374b1c","cardType":"cloze","points":null,"front":"At the critical angle, the refracted ray is at [ ] and travels along the [ ].","back":"At the critical angle, the refracted ray is at 90° and travels along the boundary.","mcqOptions":[{"text":"At the critical angle, the refracted ray is at 90° and travels along the boundary.","correct":true,"explanation":"At the critical angle, the refracted angle reaches 90° to the normal. The refracted ray therefore travels along the boundary between the two materials."},{"text":"At the critical angle, the refracted ray is at 45° and travels along the normal.","correct":false,"explanation":"The critical angle is defined by a refracted angle of 90°, not 45°. At this point the ray travels along the boundary, whereas a ray along the normal has zero refracted angle."},{"text":"At the critical angle, the refracted ray is at 0° and travels along the boundary.","correct":false,"explanation":"An angle of 0° means the ray travels along the normal, not the boundary. At the critical angle, the refracted ray is at 90° to the normal and follows the boundary."},{"text":"At the critical angle, the refracted ray is at 90° and travels along the normal.","correct":false,"explanation":"Although 90° is the correct angle to the normal, a ray at 90° cannot travel along the normal. It travels along the boundary, perpendicular to the normal."}]},{"id":"21f81260-be9a-44a9-a9b0-46ba2937867c","cardType":"basic","points":1,"front":"For total internal reflection, what refractive-index change must light make?","back":"From higher refractive index to lower refractive index.","mcqOptions":[{"text":"From higher refractive index to lower refractive index.","correct":true,"explanation":"Total internal reflection can occur only when light travels from a material with higher refractive index to one with lower refractive index, such as glass to air."},{"text":"From lower refractive index to higher refractive index.","correct":false,"explanation":"Light entering a higher refractive index material bends towards the normal and cannot undergo total internal reflection. TIR requires travel from higher refractive index to lower refractive index."},{"text":"From equal refractive index to lower refractive index.","correct":false,"explanation":"Equal refractive indices do not create the refraction needed for a critical angle. Total internal reflection requires the light to travel from a higher refractive index to a lower one."},{"text":"From lower refractive index to equal refractive index.","correct":false,"explanation":"A change from lower to equal refractive index does not provide the conditions for TIR. The light must travel from a higher refractive index into a lower refractive index."}]},{"id":"ae0183c2-2a4c-4e24-b638-0aa490878370","cardType":"basic","points":1,"front":"For total internal reflection, how must the angle of incidence compare with the critical angle?","back":"It must be greater than the critical angle.","mcqOptions":[{"text":"It must be greater than the critical angle.","correct":true,"explanation":"Total internal reflection occurs when the angle of incidence is greater than the critical angle, provided the light is travelling from higher to lower refractive index."},{"text":"It must be less than the critical angle.","correct":false,"explanation":"Below the critical angle, the ray refracts out of the material. TIR occurs only when the angle of incidence is greater than the critical angle."},{"text":"It must be equal to the critical angle.","correct":false,"explanation":"At exactly the critical angle, the refracted ray travels along the boundary. Total internal reflection begins only when the angle of incidence becomes greater than the critical angle."},{"text":"It must be twice the critical angle.","correct":false,"explanation":"There is no requirement for the angle to be twice the critical angle. The condition is simply that it is greater than the critical angle, while travelling into a lower-index material."}]},{"id":"5a6aa796-e4e8-4808-a1ad-dddd119b8073","cardType":"basic","points":1,"front":"What equation links critical angle and refractive index?","back":"$\\sin c = \\frac{1}{n}$.","mcqOptions":[{"text":"$\\sin c = \\frac{1}{n}$.","correct":true,"explanation":"For light passing from a material into air, the critical-angle relationship is $\\sin c = \\frac{1}{n}$, where $c$ is the critical angle and $n$ is the refractive index."},{"text":"$\\sin c = n$.","correct":false,"explanation":"The refractive index is the reciprocal of $\\sin c$ for a material-air boundary. The correct relationship is $\\sin c = \\frac{1}{n}$, not $\\sin c = n$."},{"text":"$\\cos c = \\frac{1}{n}$.","correct":false,"explanation":"The critical-angle derivation uses the refracted angle of 90° and produces a sine relationship. Therefore the correct equation is $\\sin c = \\frac{1}{n}$, not a cosine equation."},{"text":"$\\sin c = \\frac{n}{1}$.","correct":false,"explanation":"Writing $\\frac{n}{1}$ is just $n$, which would make the refractive index equal to $\\sin c$. The correct equation has the reciprocal: $\\sin c = \\frac{1}{n}$."}]},{"id":"cc5dc49d-b964-4a40-a6f2-6c5725ad34cb","cardType":"basic","points":2,"front":"A glass has $n = 1.50$. What is its critical angle?","back":"41.8°, about 42°.","mcqOptions":[{"text":"41.8°, about 42°.","correct":true,"explanation":"Using $\\sin c = \\frac{1}{n}$ gives $\\sin c = \\frac{1}{1.50} = 0.667$. Therefore $c = \\sin^{-1}(0.667) = 41.8°$, approximately 42°."},{"text":"48.2°, about 48°.","correct":false,"explanation":"This is a common calculator or complementary-angle error. Using $\\sin^{-1}(1/1.50)$ gives 41.8°, so the critical angle is about 42°, not 48°."},{"text":"33.3°, about 33°.","correct":false,"explanation":"This uses $1/n$ as though it were the angle in degrees, rather than taking the inverse sine. The calculation requires $c = \\sin^{-1}(0.667) = 41.8°$, about 42°."},{"text":"56.3°, about 56°.","correct":false,"explanation":"This value results from using the wrong trigonometric relationship or a complementary angle. The correct calculation is $\\sin^{-1}(1/1.50)$, giving 41.8°, approximately 42°."}]},{"id":"bf02553f-cd3b-44cf-8f50-529878e6d61c","cardType":"basic","points":2,"front":"How do optical fibres carry light signals long distances with little loss?","back":"Light pulses repeatedly reflect inside the core.","mcqOptions":[{"text":"Light pulses repeatedly reflect inside the core.","correct":true,"explanation":"Repeated total internal reflections keep each light pulse within the fibre’s core, so it can travel a long distance with very little energy lost through the sides."},{"text":"Light pulses are absorbed inside the core.","correct":false,"explanation":"Absorption would weaken the signal rather than guide it along the fibre. The correct mechanism is repeated reflection inside the core, caused by total internal reflection at the boundary."},{"text":"Light pulses repeatedly refract through the cladding.","correct":false,"explanation":"Refraction alone would change the direction at a boundary but would not keep the signal confined over a long distance. Repeated reflection inside the core provides the guidance."},{"text":"Light pulses are amplified throughout the core.","correct":false,"explanation":"The core does not generate new light to replace the signal. Optical fibres guide existing pulses by repeated reflection inside the core, minimising loss during transmission."}]},{"id":"2aaff10f-ef72-42c4-ba45-91bfad2877d2","cardType":"basic","points":2,"front":"Why does lower-index cladding help an optical fibre?","back":"It keeps light trapped by total internal reflection.","mcqOptions":[{"text":"It keeps light trapped by total internal reflection.","correct":true,"explanation":"The cladding has a lower refractive index than the core, allowing total internal reflection at the boundary and preventing light from escaping into the surrounding material."},{"text":"It keeps light trapped by ordinary refraction.","correct":false,"explanation":"A higher-index cladding would prevent the required total internal reflection at the core boundary. The cladding must have a lower index to keep light confined."},{"text":"It keeps light trapped by absorbing the signal.","correct":false,"explanation":"The cladding is not intended to absorb the signal, since absorption would reduce its intensity. A lower index enables total internal reflection and preserves guidance."},{"text":"It keeps light trapped by scattering at the boundary.","correct":false,"explanation":"Scattering would send light in random directions and increase signal loss. The lower-index cladding instead supports total internal reflection, which confines the light."}]},{"id":"a76cb5a6-47a6-4509-ae37-b8117c740d5e","cardType":"cloze","points":null,"front":"In air, compressions have particles [ ]; rarefactions have particles [ ].","back":"In air, compressions have particles closer together; rarefactions have particles more spread out.","mcqOptions":[{"text":"In air, compressions have particles closer together; rarefactions have particles more spread out.","correct":true,"explanation":"A compression is a region where particles are close together, while a rarefaction is a region where particles are separated more widely in the longitudinal sound wave."},{"text":"In air, compressions have particles more spread out; rarefactions have particles closer together.","correct":false,"explanation":"The particle spacing has been reversed: compressions are crowded regions and rarefactions are less crowded regions. This is the common mistake of swapping the two wave features."},{"text":"In air, compressions have particles closer together; rarefactions have particles equally spaced.","correct":false,"explanation":"Both regions cannot have particles closer together. Compressions are crowded, but rarefactions are regions of reduced particle density with particles more spread out."},{"text":"In air, compressions have particles more spread out; rarefactions have particles equally spaced.","correct":false,"explanation":"Both regions cannot have particles more spread out. Compressions contain increased density, whereas rarefactions contain decreased density and more widely separated particles."}]},{"id":"0c6ff7e6-3765-42dd-bcc8-e8582196790b","cardType":"basic","points":1,"front":"What does a higher sound frequency produce?","back":"A higher pitch.","mcqOptions":[{"text":"A higher pitch.","correct":true,"explanation":"Frequency measures the number of vibrations per second. A higher frequency means more vibrations each second, which the ear perceives as a higher pitch."},{"text":"A lower pitch.","correct":false,"explanation":"Pitch depends on frequency, not amplitude. Increasing amplitude makes a sound louder, while a higher frequency produces the higher pitch required here."},{"text":"A louder sound.","correct":false,"explanation":"A higher frequency does not make the sound louder. Loudness depends mainly on amplitude; frequency determines whether the pitch is higher or lower."},{"text":"A longer sound.","correct":false,"explanation":"Frequency describes vibration rate, not the duration of a sound. A higher frequency produces a higher pitch rather than a longer sound."}]},{"id":"15913f5b-57e6-4928-86ec-63f333a6123f","cardType":"basic","points":1,"front":"What does a larger sound amplitude produce?","back":"A louder sound.","mcqOptions":[{"text":"A louder sound.","correct":true,"explanation":"Amplitude is the maximum displacement of particles from their rest positions. A larger amplitude transfers more energy and is heard as a louder sound."},{"text":"A higher pitch.","correct":false,"explanation":"Pitch is determined by frequency, not amplitude. Increasing amplitude changes the sound’s loudness, whereas increasing frequency would produce a higher pitch."},{"text":"A higher frequency.","correct":false,"explanation":"A larger amplitude does not necessarily alter the wave’s frequency or period. It increases the energy carried by the wave and therefore makes it louder."},{"text":"A longer wavelength.","correct":false,"explanation":"Wavelength is related to wave speed and frequency, not directly to amplitude. A larger amplitude changes the intensity and perceived loudness of the sound."}]},{"id":"1a1fc4ce-0594-49b4-87d9-35a1789e8fb4","cardType":"basic","points":1,"front":"For one echo from a wall, how far does the sound travel?","back":"Twice the distance to the wall.","mcqOptions":[{"text":"Twice the distance to the wall.","correct":true,"explanation":"The sound travels from the source to the wall and then back to the source, so the echo path is two times the one-way distance to the wall."},{"text":"The distance to the wall.","correct":false,"explanation":"The distance to the wall is only the outgoing part of the journey. An echo also includes the return journey, making the total distance twice the wall distance."},{"text":"Three times the distance to the wall.","correct":false,"explanation":"The return path is not an additional full round trip. Sound travels to the wall and back once, giving twice the distance rather than three times it."},{"text":"Four times the distance to the wall.","correct":false,"explanation":"The echo does not travel four one-way distances. It makes one outward journey and one return journey, so the total path is twice the distance to the wall."}]},{"id":"5ffd7e0c-2c11-4bc4-a241-55a62f09af55","cardType":"basic","points":2,"front":"A student stands 85 m from a wall and times 5 echoes in 2.50 s. What is the speed of sound?","back":"340 m/s.","mcqOptions":[{"text":"340 m/s.","correct":true,"explanation":"Each echo travels 85 m to the wall and 85 m back, so five echoes cover 5 × 170 = 850 m. Dividing by 2.50 s gives 340 m/s."},{"text":"170 m/s.","correct":false,"explanation":"This uses only the one-way distance of 85 m for each echo and ignores the return journey. The correct total distance is 850 m, giving 340 m/s."},{"text":"272 m/s.","correct":false,"explanation":"This value results from using four echo journeys instead of the five measured echoes. All five round trips cover 850 m, so the speed is 340 m/s."},{"text":"425 m/s.","correct":false,"explanation":"This calculation effectively treats the total travel distance as 1,062.5 m, rather than five round trips of 170 m. The measured speed is therefore 340 m/s."}]},{"id":"37a296a2-3c61-46f4-9fdd-941380c5a853","cardType":"cloze","points":null,"front":"On an oscilloscope, vertical height shows [ ]; one cycle’s horizontal spacing gives the [ ].","back":"On an oscilloscope, vertical height shows amplitude; one cycle’s horizontal spacing gives the time period.","mcqOptions":[{"text":"On an oscilloscope, vertical height shows amplitude; one cycle’s horizontal spacing gives the time period.","correct":true,"explanation":"The vertical height of an oscilloscope trace represents amplitude, while the horizontal spacing for one complete cycle represents the time taken for one cycle, called the time period."},{"text":"On an oscilloscope, vertical height shows amplitude; one cycle’s horizontal spacing gives the frequency.","correct":false,"explanation":"Frequency is related to the horizontal spacing, but the spacing of one cycle directly measures time period. Vertical height represents amplitude, not frequency."},{"text":"On an oscilloscope, vertical height shows wavelength; one cycle’s horizontal spacing gives the time period.","correct":false,"explanation":"Wavelength is a distance measured along a wave, whereas an oscilloscope’s horizontal axis represents time. One cycle’s spacing therefore gives the time period."},{"text":"On an oscilloscope, vertical height shows frequency; one cycle’s horizontal spacing gives the wavelength.","correct":false,"explanation":"The vertical height measures amplitude, not frequency, and the horizontal spacing of a cycle measures its duration. Frequency is the reciprocal of the time period."}]},{"id":"1f453578-5519-4304-8f49-b0824335471a","cardType":"basic","points":2,"front":"One oscilloscope cycle covers 4.0 divisions at 0.00050 s per division. What is the frequency?","back":"500 Hz.","mcqOptions":[{"text":"500 Hz.","correct":true,"explanation":"The period is $4.0 \\times 0.00050 = 0.0020$ s. Using $f = \\frac{1}{T}$ gives $f = \\frac{1}{0.0020} = 500$ Hz."},{"text":"2000 Hz.","correct":false,"explanation":"This uses the time for one division as the period, ignoring that the complete cycle spans four divisions. The period is 0.0020 s, so the frequency is 500 Hz."},{"text":"8000 Hz.","correct":false,"explanation":"This results from dividing the time-per-division by four rather than multiplying by four. A cycle lasts 0.0020 s, and its frequency is therefore 500 Hz."},{"text":"50 Hz.","correct":false,"explanation":"This comes from a factor-of-ten conversion error when calculating the period. Four divisions represent 0.0020 s, not 0.020 s, so the frequency is 500 Hz."}]}]}],"examCountdown":{"examDate":"2027-05-25","paperLabel":"Paper 1"},"heading":"Light and sound","title":"Light and sound Flashcards","topbar":{"backHref":"/igcse/physics/edexcel/flashcards"},"topicSidebar":"$L36"}]}],"topicId":"dfbabace-e95b-5dc6-9720-9285b5941483","walled":"$L37"}],"$L38","$L39"]