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23:[["$","$L32",null,{"feature":"flashcards"}],"$L33",["$","$L34",null,{"href":"/a-level/physics/edexcel/luminosity-intensity-and-distance/flashcards","label":"Luminosity, intensity and distance","subjectCode":"a_level_physics","surface":"flashcards"}],["$","$L35",null,{"runner":["$","div",null,{"className":"min-h-0 flex-1 overflow-hidden bg-background","children":["$","$L36",null,{"decks":[{"id":"f206a0ec-5909-4a7a-9d9e-3b391e8acaff","title":"Luminosity, intensity and distance Flashcards","topicId":"487fab84-b846-52ca-8cab-7a9f9134ef9f","topicSlug":"luminosity-intensity-and-distance","cards":[{"id":"a1ee9ae5-1b9b-4bc4-8943-0f0534c1663f","cardType":"basic","points":1,"front":"What physical quantity does luminosity ($L$) represent for a star?","back":"The total power emitted as electromagnetic radiation.","mcqOptions":[{"text":"The total power emitted as electromagnetic radiation.","correct":true,"explanation":"Luminosity is the total rate at which a star emits electromagnetic radiation, so it measures the star’s total emitted power, independent of the observer’s distance."},{"text":"The total energy received by an observer each second.","correct":false,"explanation":"This describes received power or intensity at the observer, not luminosity. Luminosity is the total power emitted by the star in all directions."},{"text":"The energy per unit area received from the star.","correct":false,"explanation":"Energy received per unit area is related to intensity and depends on distance. Luminosity instead measures the star’s total electromagnetic power output."},{"text":"The surface temperature of the star’s emitting layer.","correct":false,"explanation":"Surface temperature affects the radiation emitted, but it is not luminosity. Luminosity specifically means the total electromagnetic power emitted by the star."}]},{"id":"5e4084a6-f8b9-4d66-9634-1caac23b5ae4","cardType":"cloze","points":null,"front":"Luminosity is an intrinsic property, meaning it does [ ] on the distance of the observer.","back":"Luminosity is an intrinsic property, meaning it does not depend on the distance of the observer.","mcqOptions":[{"text":"not depend","correct":true,"explanation":"Luminosity is intrinsic, so it is fixed by the star itself and does not change when the observer moves nearer to or farther from it."},{"text":"depend","correct":false,"explanation":"Distance affects the observed intensity, not the star’s luminosity. The missing words must indicate that luminosity is independent of the observer’s distance."},{"text":"only depend","correct":false,"explanation":"Luminosity is not determined by observer distance at all. The star’s intrinsic power output remains unchanged, whereas the received intensity varies with distance."},{"text":"vary","correct":false,"explanation":"An observer’s movement changes the measured intensity, but not the star’s intrinsic luminosity. The correct completion states that luminosity does not depend on distance."}]},{"id":"e5963651-72ce-4e42-bcba-e662c3ea4f8e","cardType":"basic","points":2,"front":"How is the total energy ($E$) emitted by a star calculated from its luminosity ($L$)?","back":"$$E = L \\times t$ (Luminosity multiplied by time in seconds).","mcqOptions":[{"text":"$$E = L \\times t$ (Luminosity multiplied by time in seconds).","correct":true,"explanation":"Energy transferred equals power multiplied by time. Using luminosity as the power and measuring time in seconds gives the emitted energy in joules."},{"text":"$$E = rac{L}{t}$ (Luminosity divided by time in seconds).","correct":false,"explanation":"Dividing power by time gives the wrong relationship and incorrect dimensions. Emitted energy is found by multiplying luminosity by the duration of emission."},{"text":"$$E = L + t$ (Luminosity added to time in seconds).","correct":false,"explanation":"Power and time cannot be added because they have different physical units. Energy is obtained by multiplying luminosity by time, not combining their numerical values."},{"text":"$$E = L \\times t^2$ (Luminosity multiplied by time in seconds squared).","correct":false,"explanation":"Squaring the time introduces an extra time factor and gives incorrect units. The energy emitted is proportional linearly to time, so use luminosity multiplied by time."}]},{"id":"77f1fdf2-c924-4c67-a645-795abd966d44","cardType":"basic","points":1,"front":"What is the SI unit for intensity ($I$)?","back":"$$\\mathrm{W\\,m^{-2}}$ (Watts per square metre).","mcqOptions":[{"text":"$$\\mathrm{W\\,m^{-2}}$ (Watts per square metre).","correct":true,"explanation":"Intensity is power received per unit area, so its SI unit is watts per square metre, written as $\\mathrm{W\\,m^{-2}}$."},{"text":"$$mathrm{W\\,m^{-1}}$ (Watts per metre).","correct":false,"explanation":"Intensity involves area, not just length. The area in the denominator requires square metres, so the correct unit is $\\mathrm{W\\,m^{-2}}$."},{"text":"$$mathrm{J\\,m^{-2}}$ (Joules per square metre).","correct":false,"explanation":"Joules per square metre measures energy per area, whereas intensity is power per area. The time component means its unit must be watts per square metre."},{"text":"$$mathrm{W\\,m^{2}}$ (Watts square metre).","correct":false,"explanation":"Area is in the denominator because intensity is power distributed over an area. Multiplying by square metres reverses this relationship and gives the wrong dimensions."}]},{"id":"b2284805-57e6-425b-b42a-582102849c1d","cardType":"cloze","points":null,"front":"Intensity is defined as the power received per unit area [ ] to the radiation.","back":"Intensity is defined as the power received per unit area perpendicular to the radiation.","mcqOptions":[{"text":"perpendicular","correct":true,"explanation":"Intensity is defined using the area perpendicular to the direction of radiation. This is the effective cross-sectional area receiving the radiation directly."},{"text":"parallel","correct":false,"explanation":"An area parallel to the radiation receives no direct flux through its surface. The definition uses the area perpendicular to the radiation direction."},{"text":"adjacent","correct":false,"explanation":"“Adjacent” does not specify the orientation required for defining intensity. The relevant area is perpendicular to the direction in which the radiation travels."},{"text":"concentric","correct":false,"explanation":"Concentric describes surfaces sharing a centre, not the orientation of an area relative to radiation. Intensity uses an area perpendicular to the radiation."}]},{"id":"c83b180f-17a9-4df7-8dac-5617ea984a84","cardType":"basic","points":2,"front":"Contrast luminosity and intensity regarding an observer moving away from a star.","back":"

Luminosity remains constant.
Intensity decreases.

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

Luminosity remains constant.
Intensity decreases.

","correct":true,"explanation":"Luminosity is an intrinsic property of the star and does not depend on distance. Intensity decreases as the same power spreads over a larger spherical area."},{"text":"

Luminosity decreases.
Intensity remains constant.

","correct":false,"explanation":"Moving away does not alter the star’s total power output, so luminosity remains constant. The radiation spreads over a larger area, causing intensity to decrease."},{"text":"

Luminosity decreases.
Intensity decreases.

","correct":false,"explanation":"Only the received intensity changes with distance. The star’s total emitted power, or luminosity, remains constant while the radiation spreads out."},{"text":"

Luminosity remains constant.
Intensity remains constant.

","correct":false,"explanation":"Although luminosity remains constant, intensity is not constant because it follows the inverse-square law. Increasing distance spreads the radiation over a larger area."}]},{"id":"2cf7703c-237e-42b9-88a2-bbb814f90c82","cardType":"cloze","points":null,"front":"For radiation spreading uniformly, the inverse-square law states that intensity is proportional to $\\text{[...]}$.","back":"For radiation spreading uniformly, the inverse-square law states that intensity is proportional to $\\boxed{\\frac{1}{d^2}}$.","mcqOptions":[{"text":"$$frac{1}{d^2}$","correct":true,"explanation":"For uniform spherical spreading, the radiation is distributed over an area proportional to $d^2$. Therefore intensity is proportional to the reciprocal, $\\frac{1}{d^2}$."},{"text":"$$frac{1}{d}$","correct":false,"explanation":"Using only the first power of distance misses the fact that spherical surface area grows with $d^2$. The inverse-square law therefore requires $\\frac{1}{d^2}$."},{"text":"$$d^2$","correct":false,"explanation":"This gives intensity increasing with distance, reversing the physical relationship. As distance increases, the radiation spreads over a larger area, so intensity is proportional to $\\frac{1}{d^2}$."},{"text":"$$frac{1}{2d^2}$","correct":false,"explanation":"The extra factor of $\\frac{1}{2}$ is not part of the inverse-square dependence. For a fixed luminosity, intensity is proportional to exactly $\\frac{1}{d^2}$."}]},{"id":"a1ea4aae-2e7b-45f1-a690-f3b41bebcea9","cardType":"basic","points":2,"front":"Why is $4\\pi d^2$ used in the equation $I = \\frac{L}{4\\pi d^2}$?","back":"It is the surface area of the sphere the radiation has spread over at distance $d$.","mcqOptions":[{"text":"It is the surface area of the sphere the radiation has spread over at distance $d$.","correct":true,"explanation":"Radiation from a point-like star spreads uniformly over a spherical surface. A sphere of radius $d$ has area $4\\pi d^2$, giving the denominator in the intensity equation."},{"text":"It is the area of the circle seen by the observer at distance $d$.","correct":false,"explanation":"Radiation spreads over the entire spherical surface centred on the star, not only a circular face. The relevant area is the sphere’s surface area, $4\\pi d^2$."},{"text":"It is the volume of the sphere the radiation has spread through at distance $d$.","correct":false,"explanation":"Intensity is power per area, so a surface area is required. The volume of a sphere has units of cubic metres and cannot produce the correct intensity units."},{"text":"It is the circumference of the sphere the radiation has spread around at distance $d$.","correct":false,"explanation":"A circumference is a length and does not represent the area over which the power is distributed. The radiation covers a spherical surface with area $4\\pi d^2$."}]},{"id":"8935150b-842a-496b-bcde-9261cc7724b2","cardType":"basic","points":1,"front":"If the distance from a star triples, by what factor does the measured intensity change?","back":"It becomes $\\frac{1}{9}$ of the original intensity.","mcqOptions":[{"text":"It becomes $\\frac{1}{9}$ of the original intensity.","correct":true,"explanation":"Intensity follows the inverse-square law, so tripling the distance divides the intensity by $3^2 = 9$. The measured intensity is therefore $/9$ of its original value."},{"text":"It becomes $\\frac{1}{3}$ of the original intensity.","correct":false,"explanation":"This treats intensity as inversely proportional to distance rather than distance squared. Squaring the factor of 3 gives 9, so the correct result is $/9$ of the original intensity."},{"text":"It becomes 3 times the original intensity.","correct":false,"explanation":"This reverses the distance-intensity relationship: increasing distance makes the radiation spread over a larger area. Since distance triples, intensity decreases to $/9$ of its original value."},{"text":"It becomes 9 times the original intensity.","correct":false,"explanation":"Squaring the factor of 3 is appropriate, but the relationship is inverse-square, so the intensity is divided by 9 rather than multiplied by 9. The correct factor is $/9$."}]},{"id":"3b6b96bb-2451-4472-b72a-334d53422aae","cardType":"cloze","points":null,"front":"When comparing the same star at two distances, the ratio of intensities is $\\frac{I_1}{I_2} = \\text{[...]}$.","back":"When comparing the same star at two distances, the ratio of intensities is $\\frac{I_1}{I_2} = \\boxed{\\left(\\frac{d_2}{d_1}\\right)^2}$.","mcqOptions":[{"text":"$$\\left(\\frac{d_2}{d_1}\\right)^2$","correct":true,"explanation":"For the same star, luminosity is constant and $I = \\frac{L}{4\\pi d^2}$. Dividing the two equations therefore gives $\\frac{I_1}{I_2} = \\left(\\frac{d_2}{d_1}\\right)^2$."},{"text":"$$\\left(\\frac{d_1}{d_2}\\right)^2$","correct":false,"explanation":"This reverses the distance ratio. Because intensity decreases as distance squared increases, comparing $I_1$ with $I_2$ gives $\\frac{I_1}{I_2} = \\left(\\frac{d_2}{d_1}\\right)^2$."},{"text":"$$\\frac{d_2}{d_1}$","correct":false,"explanation":"This misses the square in the inverse-square law and treats intensity as inversely proportional to distance alone. The correct ratio squares the distance ratio: $\\left(\\frac{d_2}{d_1}\\right)^2$."},{"text":"$$\\frac{d_1}{d_2}$","correct":false,"explanation":"This both reverses the required distance ratio and omits the square. Applying $I \\propto \\frac{1}{d^2}$ gives $\\frac{I_1}{I_2} = \\left(\\frac{d_2}{d_1}\\right)^2$."}]},{"id":"6075e242-01e0-480c-9f80-ea0b691befb5","cardType":"basic","points":1,"front":"State the formula used to calculate distance ($d$) given luminosity ($L$) and intensity ($I$).","back":"$$d = \\sqrt{\\frac{L}{4\\pi I}}$","mcqOptions":[{"text":"$$d = \\sqrt{\\frac{L}{4\\pi I}}$","correct":true,"explanation":"Rearranging $I = \\frac{L}{4\\pi d^2}$ gives $4\\pi I d^2 = L$, so $d^2 = \\frac{L}{4\\pi I}$ and taking the positive square root gives the stated formula."},{"text":"$$d = \\sqrt{\\frac{4\\pi I}{L}}$","correct":false,"explanation":"This inverts the rearranged fraction. From $I = \\frac{L}{4\\pi d^2}$, the distance is found using $d = \\sqrt{\\frac{L}{4\\pi I}}$, with luminosity in the numerator."},{"text":"$$d = \\frac{L}{4\\pi I}$","correct":false,"explanation":"This omits the square root, so it does not correctly undo the $d^2$ in the inverse-square law. The required expression is $d = \\sqrt{\\frac{L}{4\\pi I}}$."},{"text":"$$d = \\sqrt{\\frac{L}{4\\pi I^2}}$","correct":false,"explanation":"The intensity is not squared when rearranging the equation; only the distance is squared originally. Therefore the correct result is $d = \\sqrt{\\frac{L}{4\\pi I}}$."}]},{"id":"53b1bbd6-0e23-4bd4-9b55-e34df3351ca6","cardType":"cloze","points":null,"front":"A standard candle is an astronomical object with a [ ].","back":"A standard candle is an astronomical object with a known luminosity.","mcqOptions":[{"text":"known luminosity","correct":true,"explanation":"A standard candle has a luminosity that is known or can be reliably inferred. Measuring its intensity at Earth then allows its distance to be calculated using the inverse-square law."},{"text":"known intensity","correct":false,"explanation":"Intensity depends on the object's distance from the observer, so it is measured at Earth rather than fixed in advance. A standard candle is defined by its known luminosity."},{"text":"constant distance","correct":false,"explanation":"The distance to a standard candle is what astronomers are trying to determine, so it cannot be the defining known quantity. Its known luminosity provides the reference for calculating distance."},{"text":"known radius","correct":false,"explanation":"Knowing an object's radius alone does not determine the total power it emits. Standard candles are useful because their luminosity is known, allowing distance to follow from measured intensity."}]},{"id":"e7565e01-c733-4089-a0a1-148a06a18f66","cardType":"basic","points":2,"front":"How are standard candles used to determine astronomical distances?","back":"

Measure the intensity ($I$) at Earth.
Use the known luminosity ($L$) in the inverse-square law to solve for $d$.

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

Measure the intensity ($I$) at Earth.
Use the known luminosity ($L$) in the inverse-square law to solve for $d$.

","correct":true,"explanation":"A standard candle supplies a known luminosity. Comparing it with the intensity measured at Earth in $I = \\frac{L}{4\\pi d^2}$ allows the distance $d$ to be calculated."},{"text":"

Measure the luminosity ($L$) at Earth.
Use the known intensity ($I$) in the inverse-square law to solve for $d$.

","correct":false,"explanation":"Luminosity is the object's total power output and is not measured directly at Earth as the received quantity. Astronomers measure intensity and use the standard candle's known luminosity to find $d$."},{"text":"

Measure the intensity ($I$) at Earth.
Use the known distance ($d$) in the inverse-square law to solve for $L$.

","correct":false,"explanation":"The distance is the unknown being determined, not a known input. The method uses measured intensity together with known luminosity in the inverse-square law to calculate $d$."},{"text":"

Measure the luminosity ($L$) at Earth.
Use the known luminosity ($L$) in the inverse-square law to solve for $d$.

","correct":false,"explanation":"The received measurement at Earth is intensity, not luminosity. Using luminosity twice provides no distance information; the correct method combines measured $I$ with known $L$."}]},{"id":"f36078c5-c4ab-405d-89d0-0a5830b179b9","cardType":"basic","points":2,"front":"State two assumptions required for the equation $I = \\frac{L}{4\\pi d^2}$ to be valid.","back":"

Radiation is emitted uniformly in all directions.
No absorption or scattering by interstellar dust/gas.

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

Radiation is emitted uniformly in all directions.
No absorption or scattering by interstellar dust/gas.

","correct":true,"explanation":"The equation assumes the luminosity spreads uniformly over a spherical area $4\\pi d^2$ and that no radiation is removed before reaching the observer by absorption or scattering."},{"text":"

Radiation is emitted uniformly in one direction.
No absorption or scattering by interstellar dust/gas.

","correct":false,"explanation":"Emission in one direction would not spread over the full spherical area $4\\pi d^2$, so the formula would not apply. It requires uniform emission in all directions and no absorption or scattering."},{"text":"

Radiation is emitted unevenly in all directions.
No absorption or scattering by interstellar dust/gas.

","correct":false,"explanation":"Uneven emission means the intensity varies with direction, contradicting the spherical spreading assumed by the equation. The required assumption is uniform emission in all directions, with no absorption or scattering."},{"text":"

Radiation is emitted uniformly in all directions.
Strong absorption or scattering by interstellar dust/gas.

","correct":false,"explanation":"Absorption or scattering reduces the intensity before it reaches Earth, so the inverse-square prediction alone becomes inaccurate. The equation assumes there is no absorption or scattering by intervening material."}]},{"id":"c0aa2717-1d92-4cbe-aeff-a1051f3d9acc","cardType":"cloze","points":null,"front":"Interstellar dust can cause a star to appear [ ] than predicted by the inverse-square law because it absorbs or scatters radiation.","back":"Interstellar dust can cause a star to appear dimmer than predicted by the inverse-square law because it absorbs or scatters radiation.","mcqOptions":[{"text":"dimmer","correct":true,"explanation":"Dust absorbs or scatters some of the radiation travelling from the star to Earth. Less radiation is detected than the inverse-square law predicts, so the star appears dimmer."},{"text":"brighter","correct":false,"explanation":"Dust removes radiation from the direct path by absorption or scattering, so it cannot make the star appear brighter in this context. The correct description is that the star appears dimmer."},{"text":"larger","correct":false,"explanation":"Dust affects the detected radiation and therefore apparent brightness, not the star's physical or apparent size in this statement. The star appears dimmer than the inverse-square prediction."},{"text":"hotter","correct":false,"explanation":"Dust can reduce the observed intensity, but this does not mean the star's temperature increases. The relevant observational effect is that the star appears dimmer than predicted."}]},{"id":"7d0f5293-efb2-440e-b63e-e9d91d2de9ca","cardType":"cloze","points":null,"front":"Interstellar dust can cause a star to appear dimmer than predicted by the inverse-square law because it [ ] radiation.","back":"Interstellar dust can cause a star to appear dimmer than predicted by the inverse-square law because it absorbs or scatters radiation.","mcqOptions":[{"text":"absorbs or scatters","correct":true,"explanation":"Interstellar dust reduces the radiation reaching Earth in two ways: it can absorb photons or scatter them out of the direct line of sight. Thus it absorbs or scatters radiation."},{"text":"emits or produces","correct":false,"explanation":"Dust may emit some radiation, but that is not the process causing the star to appear dimmer in this statement. The relevant processes are absorption or scattering of the star's radiation."},{"text":"reflects or amplifies","correct":false,"explanation":"Reflection does not describe the main loss of direct starlight here, and dust does not amplify the radiation. The star appears dimmer because dust absorbs or scatters radiation."},{"text":"transmits or focuses","correct":false,"explanation":"Transmission allows radiation through and focusing would concentrate it, neither of which explains the reduced observed intensity. Dust causes dimming because it absorbs or scatters radiation."}]},{"id":"0daa38ca-f2c3-4dfa-a987-54462d372238","cardType":"basic","points":1,"front":"Calculate the power collected by a detector of area $A$ in an environment with intensity $I$.","back":"$$P = I \\times A$","mcqOptions":[{"text":"$$P = I \\times A$","correct":true,"explanation":"Intensity is power received per unit area, so multiplying $I$ by the detector area $A$ gives the total collected power $P$."},{"text":"$$P = \\frac{I}{A}$","correct":false,"explanation":"Dividing intensity by area gives units of power per area squared. The collected power is found by multiplying intensity by detector area: $P = I \\times A$."},{"text":"$$P = \\frac{A}{I}$","correct":false,"explanation":"Dividing area by intensity does not produce units of power and reverses the required relationship. Since intensity is power per unit area, $P = I \\times A$."},{"text":"$$P = I \\times A^2$","correct":false,"explanation":"Squaring the area mistakenly applies an extra area factor. Collected power is directly proportional to detector area, so the correct relationship is $P = I \\times A$."}]},{"id":"c10ed211-3923-4c2a-9974-e7e86509bf83","cardType":"basic","points":1,"front":"If a star is 4 times further away than another star of identical luminosity, what is the ratio of their intensities?","back":"The further star has $1/16$ the intensity of the closer star.","mcqOptions":[{"text":"The further star has $1/16$ the intensity of the closer star.","correct":true,"explanation":"Intensity follows an inverse-square relationship with distance. Increasing the distance by a factor of 4 reduces the intensity by $4^2 = 16$, giving $1/16$."},{"text":"The further star has $1/4$ the intensity of the closer star.","correct":false,"explanation":"This applies an inverse relationship but forgets to square the distance factor. Intensity varies as $1/r^2$, so four times the distance gives $1/16$ the intensity."},{"text":"The further star has $1/64$ the intensity of the closer star.","correct":false,"explanation":"This incorrectly cubes the distance factor, since $1/4^3 = 1/64$. Intensity obeys an inverse-square law, so the correct ratio is $1/4^2 = 1/16$."},{"text":"The further star has $1/2$ the intensity of the closer star.","correct":false,"explanation":"This treats the intensity as decreasing by only half despite a fourfold distance increase. The inverse-square law instead gives an intensity ratio of $1/4^2 = 1/16$."}]},{"id":"74382d01-c680-47a0-97dd-3e4afb5096a6","cardType":"cloze","points":null,"front":"In inverse-square law calculations, you must square the [ ] value, including the power of ten.","back":"In inverse-square law calculations, you must square the entire distance value, including the power of ten.","mcqOptions":[{"text":"entire distance","correct":true,"explanation":"Distance appears as $r^2$ in the inverse-square law, so the complete distance in standard form must be squared, including both its coefficient and power of ten."},{"text":"distance coefficient","correct":false,"explanation":"Squaring only the coefficient omits the power-of-ten contribution. The entire distance must be squared, so $(a \\times 10^n)^2 = a^2 \\times 10^{2n}$."},{"text":"distance exponent","correct":false,"explanation":"Squaring only the exponent is not how powers in standard form are handled. The entire distance must be squared, including its coefficient and its power of ten."},{"text":"distance unit","correct":false,"explanation":"The unit is not the only part affected by squaring. The entire numerical distance, including its coefficient and power of ten, must be squared in the calculation."}]},{"id":"9ef8ecdb-2212-4e59-834b-0ab3fa95fdd1","cardType":"basic","points":1,"front":"Why might two stars with the same luminosity appear to have different brightnesses?","back":"They are at different distances from the observer.","mcqOptions":[{"text":"They are at different distances from the observer.","correct":true,"explanation":"For stars with equal luminosity, observed intensity follows the inverse-square law. A more distant star delivers less power per unit area and therefore appears dimmer."},{"text":"They have different radii from each other.","correct":false,"explanation":"Different radii can accompany equal luminosities, but radius alone does not determine received intensity. For equal luminosity, different apparent brightnesses arise because the stars are at different distances."},{"text":"They have different temperatures from each other.","correct":false,"explanation":"Stars of different temperatures can still have the same total luminosity, but bolometric intensity depends on luminosity and distance. Their differing apparent brightnesses are caused by different distances."},{"text":"They have different masses from each other.","correct":false,"explanation":"Equal luminosity means equal total power output, regardless of stellar mass. Their received intensities can differ because inverse-square spreading makes intensity depend on distance from the observer."}]},{"id":"23c07b93-5fec-422e-8506-cc4b68a2fdad","cardType":"basic","points":1,"front":"What is the relationship between power ($P$), energy ($E$), and time ($t$)?","back":"$$P = \\frac{E}{t}$","mcqOptions":[{"text":"$$P = \\frac{E}{t}$","correct":true,"explanation":"Power is the rate at which energy is transferred, so dividing energy transferred by the time taken gives $P = E/t$, measured in watts."},{"text":"$$P = E \\times t$","correct":false,"explanation":"Multiplying energy by time gives units of joule-seconds, not watts. Power is energy transferred per unit time, so the correct relationship is $P = E/t$."},{"text":"$$P = \\frac{t}{E}$","correct":false,"explanation":"This reverses the required ratio and gives units of seconds per joule. Power is the rate of energy transfer, so energy must be divided by time: $P = E/t$."},{"text":"$$P = \\frac{E}{t^2}$","correct":false,"explanation":"Squaring time introduces an unsupported extra time factor and produces the wrong units. Power is simply energy transferred divided by elapsed time: $P = E/t$."}]}]}],"examCountdown":{"examDate":"2027-05-21","paperLabel":"Paper 1"},"heading":"Luminosity, intensity and distance","title":"Luminosity, intensity and distance Flashcards","topbar":{"backHref":"/a-level/physics/edexcel/flashcards"},"topicSidebar":"$L37"}]}],"topicId":"487fab84-b846-52ca-8cab-7a9f9134ef9f","walled":"$L38"}],"$L39","$L3a"]