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24:[["$","$L33",null,{"feature":"flashcards"}],"$L34",["$","$L35",null,{"href":"/a-level/physics/eduqas/the-nature-of-waves/flashcards","label":"The Nature of Waves","subjectCode":"a_level_physics__eduqas","surface":"flashcards"}],["$","$L36",null,{"runner":["$","div",null,{"className":"min-h-0 flex-1 overflow-hidden bg-background","children":["$","$L37",null,{"decks":[{"id":"59924628-20b4-407f-b15a-89956ad4429c","title":"The Nature of Waves Flashcards","topicId":"56b46c5c-0cf7-5d81-bd1f-d530139d1be5","topicSlug":"the-nature-of-waves","cards":[{"id":"e01ebe76-4bc8-4348-9372-0415e783b0a4","cardType":"basic","points":2,"front":"What characterizes the energy and matter transfer in a progressive wave?","back":"

Energy is transferred in the direction of wave travel.
There is no overall transfer of matter.

","mcqOptions":[{"text":"

Energy is transferred in the direction of wave travel.
There is no overall transfer of matter.

","correct":true,"explanation":"A progressive wave carries energy through the medium in its direction of travel, while the particles only oscillate about equilibrium positions and undergo no net displacement."},{"text":"

Matter is transferred in the direction of wave travel.
There is no overall transfer of energy.

","correct":false,"explanation":"This reverses the defining transfers. A progressive wave transfers energy in the direction of travel, but oscillating particles return about their equilibrium positions, so matter is not transferred overall."},{"text":"

Energy is transferred in the direction of wave travel.
There is an overall transfer of matter.

","correct":false,"explanation":"Particle oscillations can be mistaken for bulk movement of the medium. Energy travels with a progressive wave, but the particles have no net displacement, so matter is not transferred overall."},{"text":"

Energy is not transferred in the direction of wave travel.
There is no overall transfer of matter.

","correct":false,"explanation":"Although matter has no overall transfer, energy is carried by a progressive wave in its direction of travel. Otherwise the wave could not transmit energy between locations."}]},{"id":"5e7b8df3-7475-4ef6-b1c2-2b2b3176f9b9","cardType":"basic","points":1,"front":"How does the direction of oscillation in a transverse wave compare to the direction of energy transfer?","back":"The oscillations are perpendicular (at right angles) to the direction of energy transfer.","mcqOptions":[{"text":"The oscillations are perpendicular (at right angles) to the direction of energy transfer.","correct":true,"explanation":"In a transverse wave, particles oscillate at right angles to the direction in which the wave transfers energy, as occurs for electromagnetic waves and waves on a stretched string."},{"text":"The oscillations are parallel (in the same line) to the direction of energy transfer.","correct":false,"explanation":"Parallel oscillations define a longitudinal wave, not a transverse wave. In a transverse wave, oscillations are perpendicular, or at right angles, to the direction of energy transfer."},{"text":"The oscillations are antiparallel (in the opposite direction) to the direction of energy transfer.","correct":false,"explanation":"Treating the two directions as opposite confuses orientation with sense. Transverse oscillations occur perpendicular to the direction of energy transfer, not along the same line in either sense."},{"text":"The oscillations are diagonal (at 45 degrees) to the direction of energy transfer.","correct":false,"explanation":"A fixed diagonal angle is not the defining geometry of a transverse wave. Its oscillations must be perpendicular, meaning exactly at right angles, to the direction of energy transfer."}]},{"id":"72deea2a-cee1-4d4e-bd7c-91a0bd9fd8b8","cardType":"basic","points":1,"front":"How does the direction of oscillation in a longitudinal wave compare to the direction of energy transfer?","back":"The oscillations are parallel to the direction of energy transfer.","mcqOptions":[{"text":"The oscillations are parallel to the direction of energy transfer.","correct":true,"explanation":"In a longitudinal wave, particles oscillate back and forth along a line parallel to the direction of energy transfer, producing alternating compressions and rarefactions."},{"text":"The oscillations are perpendicular to the direction of energy transfer.","correct":false,"explanation":"Perpendicular oscillations are characteristic of transverse waves. In a longitudinal wave, particles oscillate parallel to the direction in which the wave transfers energy."},{"text":"The oscillations are diagonal to the direction of energy transfer.","correct":false,"explanation":"A 45-degree orientation does not define a standard wave type. Longitudinal-wave particles oscillate along a line parallel to the direction of energy transfer."},{"text":"The oscillations are antiparallel to the direction of energy transfer.","correct":false,"explanation":"Antiparallel wrongly treats the back-and-forth oscillation as having one permanently opposite direction. The line of oscillation in a longitudinal wave is parallel to energy transfer."}]},{"id":"a7858f7d-c6f0-4e22-88f6-e75b70fb3c33","cardType":"cloze","points":null,"front":"In a longitudinal wave, a [ ] is a region where particles are closer together, while a rarefaction is where particles are further apart.","back":"In a longitudinal wave, a compression is a region where particles are closer together, while a rarefaction is where particles are further apart.","mcqOptions":[{"text":"compression","correct":true,"explanation":"A compression is a region of a longitudinal wave where neighbouring particles are closer together than at equilibrium, producing increased density and usually increased pressure."},{"text":"rarefaction","correct":false,"explanation":"A rarefaction is the opposite region, where particles are further apart and density is reduced. The region where particles are closer together is called a compression."},{"text":"crest","correct":false,"explanation":"A crest is the maximum positive displacement on a transverse-wave representation, not a close-packed region. Closer particles in a longitudinal wave form a compression."},{"text":"node","correct":false,"explanation":"A node is a point of zero displacement in a stationary wave, not a region of increased particle density. A close-packed region is a compression."}]},{"id":"4605a890-9180-4711-b667-dde454b299d7","cardType":"cloze","points":null,"front":"In a longitudinal wave, a compression is a region where particles are closer together, while a [ ] is where particles are further apart.","back":"In a longitudinal wave, a compression is a region where particles are closer together, while a rarefaction is where particles are further apart.","mcqOptions":[{"text":"rarefaction","correct":true,"explanation":"A rarefaction is a region of a longitudinal wave where neighbouring particles are further apart than at equilibrium, giving lower density and usually lower pressure."},{"text":"compression","correct":false,"explanation":"A compression contains particles that are closer together, so it is the opposite of the described region. Particles further apart form a rarefaction."},{"text":"trough","correct":false,"explanation":"A trough is a point of maximum negative displacement on a transverse-wave graph, not a spread-out region. In a longitudinal wave, this region is a rarefaction."},{"text":"antinode","correct":false,"explanation":"An antinode is a position of maximum oscillation amplitude in a stationary wave. A longitudinal-wave region where particles are further apart is called a rarefaction."}]},{"id":"4889444a-28c5-4189-af0c-4337e52d8e99","cardType":"basic","points":1,"front":"State the wave equation relating wave speed ($c$), frequency ($f$), and wavelength ($\\lambda$).","back":"$$$c = f\\lambda$$","mcqOptions":[{"text":"$$$c = f\\lambda$$","correct":true,"explanation":"During one cycle, the wave advances by one wavelength, so its speed equals cycles per second multiplied by distance per cycle: $c=f\\lambda$."},{"text":"$$$c = \\frac{f}{\\lambda}$$","correct":false,"explanation":"Dividing frequency by wavelength gives dimensions of $\\mathrm{s^{-1}m^{-1}}$, not speed. Multiplying frequency by wavelength gives the correct relation, $c=f\\lambda$."},{"text":"$$$c = \\frac{\\lambda}{f}$$","correct":false,"explanation":"Dividing wavelength by frequency confuses frequency with period. Since $T=1/f$ and $c=\\lambda/T$, substitution gives $c=f\\lambda$, not $\\lambda/f$."},{"text":"$$$c = \\frac{1}{f\\lambda}$$","correct":false,"explanation":"Taking the reciprocal gives units of time per distance, corresponding to inverse speed. Wave speed is distance per cycle multiplied by cycles per second: $c=f\\lambda$."}]},{"id":"804e29bf-a76f-47b4-a727-6ab0dd8243e8","cardType":"basic","points":1,"front":"What wave property can be determined from the horizontal spacing of peaks on a displacement-position graph?","back":"Wavelength ($\\lambda$).","mcqOptions":[{"text":"Wavelength ($\\lambda$).","correct":true,"explanation":"On a displacement-position graph, the horizontal axis represents distance. The distance between successive points in phase, such as adjacent peaks, is the wavelength $\\lambda$."},{"text":"Amplitude ($A$).","correct":false,"explanation":"Amplitude is read as the maximum vertical displacement from equilibrium, not the horizontal peak spacing. On a displacement-position graph, horizontal spacing between peaks gives wavelength."},{"text":"Frequency ($f$).","correct":false,"explanation":"Frequency counts oscillations per second and requires time information. Since this graph’s horizontal axis is position, the distance between adjacent peaks is wavelength $\\lambda$."},{"text":"Period ($T$).","correct":false,"explanation":"Period is the time between successive equivalent points and is obtained from a displacement-time graph. Horizontal spacing on a displacement-position graph instead gives wavelength."}]},{"id":"7eab9bb8-4eae-4635-a988-78f0e10d94d5","cardType":"basic","points":1,"front":"What wave property can be determined from the horizontal spacing of peaks on a displacement-time graph?","back":"Period ($T$).","mcqOptions":[{"text":"Period ($T$).","correct":true,"explanation":"On a displacement-time graph, the horizontal axis represents time. The time between successive points in phase, such as adjacent peaks, is the period $T$."},{"text":"Wavelength ($\\lambda$).","correct":false,"explanation":"Wavelength is a spatial separation and requires a displacement-position graph. On a displacement-time graph, the horizontal time between adjacent peaks is the period $T$."},{"text":"Frequency ($f$).","correct":false,"explanation":"Frequency is the reciprocal of the measured peak spacing, $f=1/T$, rather than the spacing itself. The horizontal interval between peaks is the period."},{"text":"Amplitude ($A$).","correct":false,"explanation":"Amplitude is measured vertically from equilibrium to a peak. The horizontal separation of adjacent peaks on a displacement-time graph instead measures the period $T$."}]},{"id":"d823f414-bafb-4f44-bfa8-f893e888cfff","cardType":"cloze","points":null,"front":"A [ ] is a line or surface joining points that oscillate in phase, while a ray shows the direction of wave propagation.","back":"A wavefront is a line or surface joining points that oscillate in phase, while a ray shows the direction of wave propagation.","mcqOptions":[{"text":"wavefront","correct":true,"explanation":"A wavefront joins locations at the same phase, meaning they are at corresponding points in their oscillations. It may be represented as a line or a surface."},{"text":"wavelength","correct":false,"explanation":"Wavelength is the distance between consecutive points in phase, rather than the line or surface joining such points. The required term is wavefront."},{"text":"ray","correct":false,"explanation":"A ray indicates the direction in which wave energy propagates; it does not join points oscillating in phase. The line or surface described is a wavefront."},{"text":"nodal line","correct":false,"explanation":"A nodal line joins points with zero oscillation in a stationary wave, not points sharing an arbitrary phase. The required line or surface is a wavefront."}]},{"id":"1cddf86e-6d8e-4dc7-8a9c-81aa25df0939","cardType":"cloze","points":null,"front":"A wavefront is a line or surface joining points that oscillate in phase, while a [ ] shows the direction of wave propagation.","back":"A wavefront is a line or surface joining points that oscillate in phase, while a ray shows the direction of wave propagation.","mcqOptions":[{"text":"ray","correct":true,"explanation":"A ray is an imaginary line indicating the direction of wave propagation and energy transfer. It is perpendicular to the associated wavefronts."},{"text":"wavefront","correct":false,"explanation":"A wavefront joins points that oscillate in phase; it does not indicate the propagation direction. A ray shows the direction in which the wave travels."},{"text":"node","correct":false,"explanation":"A node is a stationary-wave point with zero amplitude, not a representation of propagation direction. The direction of wave travel is shown by a ray."},{"text":"wavelength","correct":false,"explanation":"Wavelength measures the distance between consecutive points in phase rather than showing a direction. A ray represents the direction of wave propagation."}]},{"id":"eb3ada10-ebd1-4251-afd4-004c51133aab","cardType":"basic","points":1,"front":"What is the geometric relationship between rays and wavefronts?","back":"Rays are always at right angles (perpendicular) to wavefronts.","mcqOptions":[{"text":"Rays are always at right angles (perpendicular) to wavefronts.","correct":true,"explanation":"Rays show the direction of propagation, which is normal to surfaces of constant phase. They therefore meet wavefronts at right angles."},{"text":"Rays are always parallel to wavefronts.","correct":false,"explanation":"Treating rays as parallel to wavefronts confuses the direction of travel with a surface of constant phase. Rays are perpendicular to wavefronts."},{"text":"Rays are always at 45° to wavefronts.","correct":false,"explanation":"There is no fixed 45° relationship between these representations. Since propagation occurs normal to each constant-phase surface, rays are at right angles to wavefronts."},{"text":"Rays are always at changing angles to wavefronts.","correct":false,"explanation":"Wavefront curvature may vary, but each ray remains locally normal to the wavefront. Rays are therefore always perpendicular to wavefronts, not at arbitrary changing angles."}]},{"id":"446e49f7-ef56-4148-ab0c-64c0d3bcadcf","cardType":"basic","points":2,"front":"What two criteria must be met for two points on a wave to be in phase?","back":"

Same displacement from equilibrium.
Moving in the same direction.

","mcqOptions":[{"text":"

Same displacement from equilibrium.
Moving in the same direction.

","correct":true,"explanation":"Points in phase occupy the same stage of their oscillation. They must therefore have equal displacement from equilibrium and be moving in the same direction."},{"text":"

Same displacement from equilibrium.
Moving in the opposite direction.

","correct":false,"explanation":"Equal displacement alone is insufficient because the points may be at different stages of the cycle. Opposite motion indicates different phases; in-phase points move in the same direction."},{"text":"

Opposite displacement from equilibrium.
Moving in the same direction.

","correct":false,"explanation":"Opposite displacement does not represent the same stage of an oscillation, even if the velocities point the same way. In-phase points have the same displacement and direction of motion."},{"text":"

Same speed through equilibrium.
Moving in the same direction.

","correct":false,"explanation":"Equal speed and direction can occur at different displacements, so they do not uniquely establish equal phase. The points must share both displacement and direction of motion."}]},{"id":"a203c443-68d5-44fd-ba9a-99fce7f28310","cardType":"basic","points":1,"front":"Describe the separation of two points in antiphase in terms of wavelength ($\\lambda$).","back":"They are separated by an odd number of half-wavelengths (e.g. $0.5\\lambda$ or $1.5\\lambda$).","mcqOptions":[{"text":"They are separated by an odd number of half-wavelengths (e.g. $0.5\\lambda$ or $1.5\\lambda$).","correct":true,"explanation":"Antiphase points differ in phase by $[0m\\pi$ radians, corresponding to half a cycle. Adding any whole wavelength preserves this relation, giving an odd number of half-wavelengths."},{"text":"They are separated by an even number of half-wavelengths (e.g. $1.0\\lambda$ or $2.0\\lambda$).","correct":false,"explanation":"An even number of half-wavelengths is a whole number of wavelengths, so the points are in phase. Antiphase requires an odd number of half-wavelengths."},{"text":"They are separated by a whole number of wavelengths (e.g. $1.0\\lambda$ or $2.0\\lambda$).","correct":false,"explanation":"A whole-number wavelength separation produces a phase difference that is a multiple of $2\\pi$, making the points in phase. Antiphase requires odd half-wavelength separation."},{"text":"They are separated by an odd number of quarter-wavelengths (e.g. $0.25\\lambda$ or $0.75\\lambda$).","correct":false,"explanation":"Quarter-wavelength separations correspond to phase differences of $\\frac{\\pi}{2}$ or related odd quarter-cycle values, not antiphase. Antiphase points are separated by odd half-wavelengths."}]},{"id":"da551902-55ae-4f2d-a8c2-1079c39ea0d1","cardType":"cloze","points":null,"front":"In radians, a phase difference of $\\text{[...]}$ corresponds to points being in antiphase.","back":"In radians, a phase difference of $\\boxed{\\pi}$ corresponds to points being in antiphase.","mcqOptions":[{"text":"$$\\pi$","correct":true,"explanation":"Antiphase points are half an oscillation cycle apart. Since one complete cycle is $2\\pi$ radians, half a cycle gives a phase difference of $\\pi$ radians."},{"text":"$$2\\pi$","correct":false,"explanation":"A phase difference of $2\\pi$ radians is one complete cycle, so the points are in phase rather than antiphase. Antiphase corresponds to $\\pi$ radians."},{"text":"$$\\frac{\\pi}{2}$","correct":false,"explanation":"A phase difference of $\\frac{\\pi}{2}$ radians is one-quarter of a cycle, not half a cycle. Points in antiphase differ by $\\pi$ radians."},{"text":"$$0$","correct":false,"explanation":"A zero phase difference means the points are at exactly the same stage and are in phase. Antiphase requires a phase difference of $\\pi$ radians."}]},{"id":"0a536f61-9425-4af2-a0e1-f1ab0ffc6138","cardType":"basic","points":1,"front":"Define polarisation.","back":"The restriction of wave oscillations to one plane or direction.","mcqOptions":[{"text":"The restriction of wave oscillations to one plane or direction.","correct":true,"explanation":"Polarisation confines a wave’s oscillations to a single plane or direction. Only transverse waves can be polarised because their oscillations are perpendicular to propagation."},{"text":"The reflection of wave oscillations into one plane or direction.","correct":false,"explanation":"Reflection changes a wave’s direction at a boundary and is not the definition of polarisation. Polarisation is the restriction of oscillations to one plane or direction."},{"text":"The restriction of wave propagation to one plane or direction.","correct":false,"explanation":"Polarisation restricts the direction of oscillation, not the direction in which the wave propagates. The wave can continue travelling while its oscillations are confined to one plane."},{"text":"The restriction of wave oscillations to two planes or directions.","correct":false,"explanation":"Two perpendicular oscillation planes do not constitute plane polarisation. Polarisation restricts the oscillations to one plane or direction rather than retaining two components."}]},{"id":"a5c1eb40-88a4-4819-9767-438b45057219","cardType":"basic","points":2,"front":"Why is it impossible to polarise longitudinal waves?","back":"Oscillations are only along the direction of travel, so there is no sideways oscillation to restrict.","mcqOptions":[{"text":"Oscillations are only along the direction of travel, so there is no sideways oscillation to restrict.","correct":true,"explanation":"Longitudinal oscillations are parallel to propagation, leaving no perpendicular oscillation direction that can be selected. Consequently, a longitudinal wave cannot be polarised."},{"text":"Oscillations are only perpendicular to the direction of travel, so there is no parallel oscillation to restrict.","correct":false,"explanation":"Oscillations perpendicular to travel define a transverse wave, which can be polarised. Longitudinal waves cannot be polarised because their oscillations occur only along the direction of travel."},{"text":"Wavelengths are too large for a polarising filter, so the waves cannot pass through its gaps.","correct":false,"explanation":"Polarisation is determined by oscillation orientation, not whether a wavelength fits through a filter. Longitudinal oscillations lie along the propagation direction, leaving no sideways component to restrict."},{"text":"Oscillations occur in all sideways planes at once, so no single plane can be isolated.","correct":false,"explanation":"Oscillations in many sideways planes describe an unpolarised transverse wave, which can be polarised by selecting one plane. Longitudinal waves instead oscillate only along their direction of travel."}]},{"id":"5676e9a1-132e-4730-8ff7-d99fe506b25d","cardType":"basic","points":1,"front":"What does the ability to polarise electromagnetic waves prove about their nature?","back":"It is evidence that electromagnetic waves are transverse.","mcqOptions":[{"text":"It is evidence that electromagnetic waves are transverse.","correct":true,"explanation":"Only transverse waves can be polarised because their oscillations can be restricted to a particular direction perpendicular to the direction of travel. Electromagnetic waves are therefore transverse."},{"text":"It is evidence that electromagnetic waves are longitudinal.","correct":false,"explanation":"Longitudinal waves oscillate parallel to their direction of travel and cannot be plane-polarised. Polarisation therefore shows that electromagnetic waves are transverse."},{"text":"It is evidence that electromagnetic waves are mechanical.","correct":false,"explanation":"Polarisation concerns the direction of oscillations, not whether a wave requires matter for propagation. It demonstrates that electromagnetic waves are transverse, rather than mechanical."},{"text":"It is evidence that electromagnetic waves are stationary.","correct":false,"explanation":"Polarisation does not determine whether a wave is progressive or stationary. It is possible only for transverse oscillations, so it provides evidence that electromagnetic waves are transverse."}]},{"id":"43989360-afeb-49ee-b194-882baaae265a","cardType":"cloze","points":null,"front":"For ideal plane-polarised radiation, the intensity $I$ transmitted through an analyser at angle $\\theta$ follows the relationship $$\\text{[...]}$$.","back":"For ideal plane-polarised radiation, the intensity $I$ transmitted through an analyser at angle $\\theta$ follows the relationship $$\\boxed{I = I_0 \\cos^2 \\theta}$$.","mcqOptions":[{"text":"$$I = I_0 \\cos^2 \\theta$","correct":true,"explanation":"Malus’s law states that the transmitted intensity equals the initial plane-polarised intensity multiplied by the square of the cosine of the angle between the transmission axes."},{"text":"$$I = I_0 \\cos \\theta$","correct":false,"explanation":"Using cosine without squaring confuses the electric-field amplitude with intensity. Amplitude varies as $\\cos\\theta$, but intensity is proportional to amplitude squared, giving $I=I_0\\cos^2\\theta$."},{"text":"$$I = I_0 \\sin^2 \\theta$","correct":false,"explanation":"Sine squared reverses the predicted maximum and minimum, incorrectly giving zero transmission for parallel axes. Malus’s law uses $\\cos^2\\theta$, so $I=I_0\\cos^2\\theta$."},{"text":"$$I = \\frac{I_0}{\\cos^2 \\theta}$","correct":false,"explanation":"The initial intensity must multiply the angular factor, not be divided by it. Division would predict an unphysical infinite intensity at $90°$; Malus’s law is $I=I_0\\cos^2\\theta$."}]},{"id":"4f2e232d-4170-44af-9bae-11d5f74a6357","cardType":"basic","points":2,"front":"In a polarisation practical, at what angles $\\theta$ between the polariser and analyser will the intensity be maximum and minimum?","back":"

Maximum: 0° (transmission axes are parallel).
Minimum: 90° (transmission axes are perpendicular).

","mcqOptions":[{"text":"

Maximum: 0° (transmission axes are parallel).
Minimum: 90° (transmission axes are perpendicular).

","correct":true,"explanation":"Malus’s law gives $I=I_0\\cos^2\\theta$. Parallel axes at $0°$ give $\\cos^2\\theta=1$, while perpendicular axes at $90°$ give $\\cos^2\\theta=0$."},{"text":"

Maximum: 90° (transmission axes are perpendicular).
Minimum: 0° (transmission axes are parallel).

","correct":false,"explanation":"This reverses the extrema. Parallel transmission axes at $0°$ pass maximum intensity, whereas perpendicular axes at $90°$ ideally transmit no radiation and therefore give minimum intensity."},{"text":"

Maximum: 45° (transmission axes are diagonal).
Minimum: 90° (transmission axes are perpendicular).

","correct":false,"explanation":"At $45°$, Malus’s law gives half the maximum intensity rather than the maximum. The maximum occurs at $0°$, while the minimum occurs at $90°$."},{"text":"

Maximum: 0° (transmission axes are parallel).
Minimum: 45° (transmission axes are diagonal).

","correct":false,"explanation":"Although parallel axes correctly produce maximum transmission, $45°$ gives $I=I_0/2$, not minimum intensity. The ideal minimum occurs when the axes are perpendicular at $90°$."}]},{"id":"7606a9b6-f870-4b5b-a608-9c3b4edb0b54","cardType":"basic","points":2,"front":"List two reasons why a polarisation practical might yield a non-zero intensity at 90°.","back":"

Background radiation.
Misalignment of the polariser and analyser.
Reflections from nearby surfaces.

","mcqOptions":[{"text":"

Background radiation.
Misalignment of the polariser and analyser.
Reflections from nearby surfaces.

","correct":true,"explanation":"Ambient radiation can reach the detector, imperfect alignment can prevent the axes being exactly perpendicular, and reflected radiation can bypass the intended path. Each can produce a residual reading."},{"text":"

No background radiation.
Perfect alignment of the polariser and analyser.
Absorption by nearby surfaces.

","correct":false,"explanation":"These conditions would reduce unwanted detector readings rather than cause them. A non-zero result can instead arise from background radiation, misalignment, or reflections from nearby surfaces."},{"text":"

Reduced source intensity.
Perpendicular alignment of the polariser and analyser.
Removal of reflections from nearby surfaces.

","correct":false,"explanation":"Lower source intensity, accurate perpendicular alignment, and removing reflections all reduce measured intensity. Residual intensity is instead explained by background radiation, misalignment, or nearby reflections."},{"text":"

Shielding from radiation.
Careful alignment of the polariser and analyser.
Blackening of nearby surfaces.

","correct":false,"explanation":"These changes suppress rather than create residual transmission at $90°$. Non-zero intensity is more plausibly produced by ambient background, imperfect alignment of the axes, or reflected radiation."}]},{"id":"463b87f5-1317-494d-ae03-b6a848717f84","cardType":"cloze","points":null,"front":"To test the polarisation model using a straight-line graph, intensity $I$ should be plotted against $\\text{[...]}$.","back":"To test the polarisation model using a straight-line graph, intensity $I$ should be plotted against $\\boxed{\\cos^2 \\theta}$.","mcqOptions":[{"text":"$$\\cos^2 \\theta$","correct":true,"explanation":"Malus’s law has the form $I=I_0\\cos^2\\theta$, so plotting $I$ against $\\cos^2\\theta$ should produce a straight line with gradient $I_0$."},{"text":"$$\\cos \\theta$","correct":false,"explanation":"Plotting against $\\cos\\theta$ overlooks that intensity depends on the square of the amplitude factor. A linear test of Malus’s law requires $I$ against $\\cos^2\\theta$."},{"text":"$$\\sin^2 \\theta$","correct":false,"explanation":"Using $\\sin^2\\theta$ reverses the angular dependence, making the horizontal variable largest when ideal transmission is smallest. The appropriate variable is $\\cos^2\\theta$."},{"text":"$$\\theta$","correct":false,"explanation":"Intensity is not directly proportional to the angle, so this graph should be curved rather than linear. Malus’s law is tested by plotting $I$ against $\\cos^2\\theta$."}]},{"id":"f2bc1065-f447-4251-adb2-a3edc288197f","cardType":"basic","points":1,"front":"How do you calculate frequency ($f$) from the wave period ($T$)?","back":"$$$f = \\frac{1}{T}$$","mcqOptions":[{"text":"$$$f = \\frac{1}{T}$$","correct":true,"explanation":"Frequency counts cycles per second, while period is the time taken for one cycle. They are reciprocals, so dividing one by the period gives the frequency."},{"text":"$$$f = T$$","correct":false,"explanation":"Equating frequency and period confuses reciprocal quantities with different units. Since one cycle takes time $T$, the number of cycles per second is $f=1/T$."},{"text":"$$$f = \\frac{2\\pi}{T}$$","correct":false,"explanation":"The factor $2\\pi$ converts ordinary frequency to angular frequency, $\\omega=2\\pi f$. Ordinary frequency is simply the reciprocal of period, so $f=1/T$."},{"text":"$$$f = \\frac{1}{T^2}$$","correct":false,"explanation":"Squaring the period gives the wrong dimensions and does not represent cycles per second. Frequency is the reciprocal of the time for one cycle, so $f=1/T$."}]}]}],"examCountdown":{"examDate":"2027-05-21","paperLabel":"Component 1"},"heading":"The Nature of Waves","title":"The Nature of Waves Flashcards","topbar":{"backHref":"/a-level/physics/eduqas/flashcards"},"topicSidebar":"$L38"}]}],"topicId":"56b46c5c-0cf7-5d81-bd1f-d530139d1be5","walled":"$L39"}],"$L3a","$L3b"]