[{"data":1,"prerenderedAt":1293},["ShallowReactive",2],{"questions:light":3},{"bank":4,"contentHash":1277,"dependencyHashes":1278,"releaseId":1292},{"schemaVersion":5,"conceptId":6,"revision":5,"title":7,"intro":8,"sections":9,"questions":46,"sourceIds":1260,"reviewStatus":1273,"authoring":1274},1,"light","Light: question bank","80 questions across sources and sinks of light, shadows and sundials, plane and curved mirrors, refraction and total internal reflection, lenses and the eye, dispersion and the rainbow, colour, the speed of light's history, and light beyond the visible. Every numeric answer is computed and checked in the generator, never guessed.",[10,14,18,22,26,30,34,38,42],{"id":11,"title":12,"description":13},"luminous-sources","Luminous sources and straight lines","Luminous vs non-luminous objects, seeing by reflected light, straight-line travel, pinhole cameras, and transparent, translucent and opaque materials.",{"id":15,"title":16,"description":17},"shadows-sundials","Shadows, umbra, penumbra and sundials","Shadow-size geometry, umbra and penumbra, sundials and Zero Shadow Day.",{"id":19,"title":20,"description":21},"reflection","Reflection: plane and curved mirrors","The law of reflection, plane mirror images, lateral inversion, diffuse vs regular reflection, multiple mirrors, and curved mirrors.",{"id":23,"title":24,"description":25},"refraction","Refraction, refractive index and total internal reflection","Bending at a boundary, Snell's law, apparent depth, the critical angle, total internal reflection and optical fibres.",{"id":27,"title":28,"description":29},"lenses-eye","Lenses and the eye","Convex and concave lenses, the lens formula, magnifiers, the eye as an optical system, and corrective lenses.",{"id":31,"title":32,"description":33},"dispersion-rainbow","Dispersion and the rainbow","Splitting white light with a prism, the visible spectrum, and the geometry of the rainbow.",{"id":35,"title":36,"description":37},"colour","Colour: mixing and scattering","Why objects have colour, additive and subtractive mixing, Rayleigh scattering, and why the sky is blue and sunsets are red.",{"id":39,"title":40,"description":41},"speed-history","The speed of light and its history","The speed of light, historical measurements by Rømer, Fizeau and Michelson, light-years and look-back time.",{"id":43,"title":44,"description":45},"beyond-wave-particle","Beyond visible light: waves and particles","The electromagnetic spectrum beyond visible light, and the evidence that light behaves as both a wave and a particle.",[47,71,89,107,124,145,159,177,197,215,233,244,254,271,289,302,312,331,349,367,385,396,406,423,441,460,469,488,506,523,540,550,560,570,588,598,606,615,634,644,662,679,687,696,715,733,753,764,782,796,814,831,841,860,878,896,916,926,943,962,970,986,1005,1015,1027,1037,1055,1073,1085,1096,1115,1124,1142,1153,1171,1190,1209,1220,1230,1249],{"id":48,"section":11,"level":49,"prompt":50,"check":51,"hints":67,"solution":68,"skills":69},"light.q001","foundation","Which of these is luminous — it makes its own light?",{"kind":52,"options":53,"correct":66},"choice",[54,57,60,63],{"id":55,"label":56},"a","The Moon",{"id":58,"label":59},"b","A firefly",{"id":61,"label":62},"c","A mirror",{"id":64,"label":65},"d","A book page",[58],[],"A firefly makes its own light with a chemical reaction. The Moon, a mirror and a book page only reflect light that comes from somewhere else.",[70],"luminous",{"id":72,"section":11,"level":49,"prompt":73,"check":74,"hints":85,"solution":86,"skills":87},"light.q002","Why can you not see anything in a completely dark, sealed room?",{"kind":52,"options":75,"correct":84},[76,78,80,82],{"id":55,"label":77},"Your eyes stop working in the dark",{"id":58,"label":79},"There is no light to enter your eye from any object",{"id":61,"label":81},"Dark rooms absorb your eyesight",{"id":64,"label":83},"Objects change colour to black in the dark",[58],[],"You see something only when light from it enters your eye. With no light source at all, nothing can reflect or emit light, so nothing reaches your eye.",[88],"seeing",{"id":90,"section":11,"level":49,"prompt":91,"check":92,"hints":103,"solution":104,"skills":105},"light.q003","A beam of light passing through a dusty room looks like a straight line. What does this show?",{"kind":52,"options":93,"correct":102},[94,96,98,100],{"id":55,"label":95},"Light only travels through dust",{"id":58,"label":97},"Light travels in a straight line",{"id":61,"label":99},"Dust makes its own light",{"id":64,"label":101},"Light bends around dust particles",[58],[],"The dust particles simply scatter a little light sideways into your eye at every point along the beam's path, revealing its straight-line path.",[106],"straight-line",{"id":108,"section":11,"level":49,"prompt":109,"check":110,"hints":121,"solution":122,"skills":123},"light.q004","Sort: which of these is non-luminous?",{"kind":52,"options":111,"correct":120},[112,114,116,118],{"id":55,"label":113},"The Sun",{"id":58,"label":115},"A lit candle",{"id":61,"label":117},"Venus, the evening star",{"id":64,"label":119},"A lightning flash",[61],[],"Venus is a planet; it has no light of its own and is visible only because it reflects sunlight very well.",[70],{"id":125,"section":11,"level":126,"prompt":127,"check":128,"hints":139,"solution":142,"skills":143},"light.q005","core","A pinhole camera shows an image of a candle flame that is upside down. Why?",{"kind":52,"options":129,"correct":138},[130,132,134,136],{"id":55,"label":131},"The pinhole flips images on purpose",{"id":58,"label":133},"Light from the top of the flame travels in a straight line down through the pinhole to the bottom of the screen, and vice versa",{"id":61,"label":135},"The candle flame is actually upside down",{"id":64,"label":137},"Light bends as it passes through the tiny hole",[58],[140,141],"Trace one ray from the very top of the flame through the hole.","Straight lines that cross swap top and bottom.","Because light travels in straight lines, a ray from the top of the object must cross a ray from the bottom exactly at the pinhole, landing the top's ray at the bottom of the screen and the bottom's ray at the top.",[144],"pinhole",{"id":146,"section":11,"level":126,"prompt":147,"check":148,"hints":153,"solution":156,"skills":157},"light.q006","A pinhole camera box is 20 cm long. A tree 12 m tall stands 8 m from the pinhole. Using image height = object height × (box length ÷ object distance), how tall is the image, in centimetres?",{"kind":149,"answer":150,"tolerance":151,"unit":152},"number",0.3,0.5,"cm",[154,155],"Convert everything to the same units first.","image = object × (box length ÷ distance to object).","Image height = 1,200 cm × 0.20 m ÷ 8 m = 1,200 × 0.20 ÷ 8 = 30 cm. Working entirely in metres: 12 m × 0.20 m ÷ 8 m = 0.30 m = 30 cm.",[144,158],"ratio",{"id":160,"section":11,"level":126,"prompt":161,"check":162,"hints":173,"solution":174,"skills":175},"light.q007","A material lets you see a blurry, unclear shape of what is behind it, but not sharp detail. What is it?",{"kind":52,"options":163,"correct":172},[164,166,168,170],{"id":55,"label":165},"Transparent",{"id":58,"label":167},"Translucent",{"id":61,"label":169},"Opaque",{"id":64,"label":171},"Luminous",[58],[],"Translucent materials (like frosted glass or butter paper) let some light through but scatter it, so shapes appear blurred rather than sharp.",[176],"transparent-translucent-opaque",{"id":178,"section":11,"level":179,"prompt":180,"check":181,"hints":192,"solution":194,"skills":195},"light.q008","stretch","A torch is switched on inside a sealed, perfectly black box with no gaps. From outside, in a dark room, you see nothing. What is the best explanation?",{"kind":52,"options":182,"correct":191},[183,185,187,189],{"id":55,"label":184},"The torch instantly stopped working",{"id":58,"label":186},"Light travels in straight lines and none can escape the box, since it is opaque and sealed",{"id":61,"label":188},"Light needs an audience to exist",{"id":64,"label":190},"Black boxes absorb electricity, not light",[58],[193],"Light cannot bend around corners on its own to escape a sealed box.","Light needs a path to travel along; a sealed opaque box blocks every straight-line path out, so however bright the torch, no light reaches your eye outside.",[196,106],"opaque",{"id":198,"section":15,"level":49,"prompt":199,"check":200,"hints":211,"solution":212,"skills":213},"light.q009","A shadow forms because:",{"kind":52,"options":201,"correct":210},[202,204,206,208],{"id":55,"label":203},"Light bends around an opaque object",{"id":58,"label":205},"An opaque object blocks light travelling in straight lines",{"id":61,"label":207},"The object absorbs all nearby light",{"id":64,"label":209},"Shadows are a trick of the eye, not real light",[58],[],"Because light cannot bend around an opaque object on its own, the region directly behind the object, where straight-line rays cannot reach, stays dark: a shadow.",[214],"shadows",{"id":216,"section":15,"level":49,"prompt":217,"check":218,"hints":229,"solution":230,"skills":231},"light.q010","The umbra of a shadow is:",{"kind":52,"options":219,"correct":228},[220,222,224,226],{"id":55,"label":221},"The grey, partly lit fringe of a shadow",{"id":58,"label":223},"The fully dark centre of a shadow, where no part of the source is visible",{"id":61,"label":225},"Another name for the object itself",{"id":64,"label":227},"Only found during eclipses",[58],[],"The umbra is the completely dark region: standing there, you would not be able to see any part of the light source at all.",[232],"umbra-penumbra",{"id":234,"section":15,"level":126,"prompt":235,"check":236,"hints":238,"solution":241,"skills":242},"light.q011","A 10 cm tall toy is placed 40 cm from a small torch bulb. A wall stands 160 cm from the same bulb. Using shadow = object × (lamp-to-screen ÷ lamp-to-object), how wide is the shadow, in centimetres?",{"kind":149,"answer":237,"tolerance":151,"unit":152},40,[239,240],"Find lamp-to-screen ÷ lamp-to-object first.","Multiply the object's height by that ratio.","Ratio = 160 ÷ 40 = 4. Shadow = 10 cm × 4 = 40 cm.",[243],"shadow-size",{"id":245,"section":15,"level":126,"prompt":246,"check":247,"hints":249,"solution":252,"skills":253},"light.q012","Using shadow = object × D ÷ d: a 30 cm puppet is held 60 cm from a lamp, with a screen 300 cm from the same lamp. How tall is the shadow, in centimetres?",{"kind":149,"answer":248,"tolerance":5,"unit":152},150,[250,251],"D is lamp-to-screen, d is lamp-to-object.","Find the ratio first, then multiply by the object's height.","D ÷ d = 300 ÷ 60 = 5. Shadow = 30 × 5 = 150 cm.",[243],{"id":255,"section":15,"level":126,"prompt":256,"check":257,"hints":268,"solution":269,"skills":270},"light.q013","Why does a bird flying 50 m above the ground cast no visible shadow, even in bright sunshine?",{"kind":52,"options":258,"correct":267},[259,261,263,265],{"id":55,"label":260},"Birds are too light to block sunlight",{"id":58,"label":262},"The Sun's own angular width blurs the shadow's umbra completely away at that height, leaving only a faint, invisible penumbra",{"id":61,"label":264},"Birds move too fast for a shadow to form",{"id":64,"label":266},"Sunlight cannot reach 50 m above the ground",[58],[],"The Sun is not a point source; at 50 m the umbra has shrunk to nothing and only a very wide, faint penumbra remains, too dim and spread out to notice.",[232],{"id":272,"section":15,"level":179,"prompt":273,"check":274,"hints":285,"solution":286,"skills":287},"light.q014","On a Zero Shadow Day in Bengaluru, a vertical pole at local noon:",{"kind":52,"options":275,"correct":284},[276,278,280,282],{"id":55,"label":277},"Casts its longest shadow of the year",{"id":58,"label":279},"Casts no shadow at all, because the Sun is exactly overhead",{"id":61,"label":281},"Casts a shadow pointing north instead of south",{"id":64,"label":283},"This never happens in India",[58],[],"On the two days a year the Sun's declination exactly equals a location's latitude (only possible between the tropics), the Sun is exactly overhead at local noon, and a vertical pole casts no shadow at all.",[288],"zero-shadow-day",{"id":290,"section":15,"level":179,"prompt":291,"check":292,"hints":296,"solution":298,"skills":299},"light.q015","Using shadow length = height ÷ tan(Sun's altitude), what is the shadow of a 1 m stick when the Sun is 60° above the horizon? Give your answer to 2 decimal places.",{"kind":149,"answer":293,"tolerance":294,"unit":295},0.58,0.02,"m",[297],"Shadow = height ÷ tan(altitude).","tan(60°) ≈ 1.732, so shadow = 1 ÷ 1.732 ≈ 0.58 m.",[300,301],"sundial","trigonometry",{"id":303,"section":15,"level":179,"prompt":304,"check":305,"hints":308,"solution":310,"skills":311},"light.q016","Using shadow length = height ÷ tan(Sun's altitude), what is the shadow of a 1 m stick when the Sun is only 15° above the horizon? Give your answer to 2 decimal places.",{"kind":149,"answer":306,"tolerance":307,"unit":295},3.73,0.05,[297,309],"A low Sun altitude gives a long shadow.","tan(15°) ≈ 0.268, so shadow = 1 ÷ 0.268 ≈ 3.73 m — much longer than the stick, as expected near sunrise or sunset.",[300,301],{"id":313,"section":15,"level":314,"prompt":315,"check":316,"hints":327,"solution":329,"skills":330},"light.q017","challenge","Delhi (latitude 28.6°N) never has a Zero Shadow Day. Why not?",{"kind":52,"options":317,"correct":326},[318,320,322,324],{"id":55,"label":319},"Delhi is too far north of the Tropic of Cancer (23.44°N) for the Sun to ever pass directly overhead",{"id":58,"label":321},"Delhi has too much pollution for the effect to be visible",{"id":61,"label":323},"Zero Shadow Days only happen south of the equator",{"id":64,"label":325},"Delhi's poles are always tilted",[55],[328],"Compare Delhi's latitude with the Tropic of Cancer's 23.44°.","A Zero Shadow Day needs the Sun's declination to exactly equal your latitude, which only happens between the Tropics of Cancer and Capricorn (±23.44°). Delhi, at 28.6°N, lies outside that band.",[288],{"id":332,"section":19,"level":49,"prompt":333,"check":334,"hints":345,"solution":346,"skills":347},"light.q018","The law of reflection states that the angle of incidence:",{"kind":52,"options":335,"correct":344},[336,338,340,342],{"id":55,"label":337},"Is always 90° from the mirror surface",{"id":58,"label":339},"Equals the angle of reflection, both measured from the normal",{"id":61,"label":341},"Is always twice the angle of reflection",{"id":64,"label":343},"Depends on the colour of light",[58],[],"Both angles are measured from the normal (the line at right angles to the mirror surface) and are always equal.",[348],"law-of-reflection",{"id":350,"section":19,"level":49,"prompt":351,"check":352,"hints":363,"solution":364,"skills":365},"light.q019","An image in a plane mirror is:",{"kind":52,"options":353,"correct":362},[354,356,358,360],{"id":55,"label":355},"Smaller than the object",{"id":58,"label":357},"Upside down",{"id":61,"label":359},"The same size, upright, and laterally inverted",{"id":64,"label":361},"Always blurry",[61],[],"A plane mirror always gives a same-size, upright, virtual image that is laterally inverted (left and right swapped, as seen from behind the glass).",[366],"plane-mirror",{"id":368,"section":19,"level":49,"prompt":369,"check":370,"hints":381,"solution":382,"skills":383},"light.q020","A rough surface like paper reflects light diffusely because:",{"kind":52,"options":371,"correct":380},[372,374,376,378],{"id":55,"label":373},"It absorbs all the light",{"id":58,"label":375},"Its many tiny, differently angled bumps send parallel rays off in many different directions",{"id":61,"label":377},"It has no normal at all",{"id":64,"label":379},"Paper does not obey the law of reflection",[58],[],"Every tiny bump still obeys the law of reflection individually, but because the bumps face in many directions, parallel incoming rays scatter every which way rather than staying parallel — this is what makes diffuse reflection.",[384],"diffuse-regular",{"id":386,"section":19,"level":126,"prompt":387,"check":388,"hints":392,"solution":394,"skills":395},"light.q021","A ray hits a plane mirror at 25° to the mirror's surface (not the normal). What is the angle of reflection, measured from the normal, in degrees?",{"kind":149,"answer":389,"tolerance":390,"unit":391},65,0,"°",[393],"Angles are measured from the normal, not the surface.","The normal is 90° from the surface, so the angle of incidence from the normal is 90 − 25 = 65°. Reflection equals incidence: 65°.",[348],{"id":397,"section":19,"level":126,"prompt":398,"check":399,"hints":401,"solution":403,"skills":404},"light.q022","Two mirrors are set at 60° to each other. Using images = (360 ÷ angle) − 1, how many images of an object between them are formed?",{"kind":149,"answer":400,"tolerance":390},5,[402],"First find 360 ÷ angle.","360 ÷ 60 = 6. Images = 6 − 1 = 5.",[405],"multiple-mirrors",{"id":407,"section":19,"level":126,"prompt":408,"check":409,"hints":420,"solution":421,"skills":422},"light.q023","You are 1.5 m in front of a plane mirror. How far are you from your own image?",{"kind":52,"options":410,"correct":419},[411,413,415,417],{"id":55,"label":412},"0.75 m",{"id":58,"label":414},"1.5 m",{"id":61,"label":416},"3 m",{"id":64,"label":418},"4.5 m",[61],[],"A plane mirror places the image exactly as far behind the mirror as the object is in front, so you and your image are 1.5 + 1.5 = 3 m apart.",[366],{"id":424,"section":19,"level":179,"prompt":425,"check":426,"hints":437,"solution":439,"skills":440},"light.q024","What is the shortest height of a plane mirror that lets a 160 cm tall person see the whole of themselves, head to toe?",{"kind":52,"options":427,"correct":436},[428,430,432,434],{"id":55,"label":429},"160 cm",{"id":58,"label":431},"120 cm",{"id":61,"label":433},"80 cm",{"id":64,"label":435},"It depends only on how far back they stand",[61],[438],"The mirror height needed is exactly half the person's height.","A plane mirror needs to be only half a person's height, positioned correctly, to show the whole body — and remarkably, this does not change with distance from the mirror.",[366],{"id":442,"section":19,"level":179,"prompt":443,"check":444,"hints":455,"solution":457,"skills":458},"light.q025","A concave mirror used as a torch reflector places the bulb:",{"kind":52,"options":445,"correct":454},[446,448,450,452],{"id":55,"label":447},"At the centre of curvature",{"id":58,"label":449},"At the principal focus, so reflected rays leave roughly parallel",{"id":61,"label":451},"Anywhere at all — position makes no difference",{"id":64,"label":453},"Behind the mirror",[58],[456],"Think about the reverse of what a mirror does to incoming parallel rays.","Placing a small bulb at the focus is exactly the reverse of a mirror focusing parallel rays to a point: light leaving the focus reflects off the mirror as a roughly parallel beam.",[459],"curved-mirrors",{"id":461,"section":19,"level":179,"prompt":462,"check":463,"hints":465,"solution":467,"skills":468},"light.q026","A concave mirror has a radius of curvature of 40 cm. What is its focal length, in centimetres?",{"kind":149,"answer":464,"tolerance":390,"unit":152},20,[466],"f = R ÷ 2.","Focal length is always half the radius of curvature: f = 40 ÷ 2 = 20 cm.",[459],{"id":470,"section":19,"level":314,"prompt":471,"check":472,"hints":483,"solution":485,"skills":486},"light.q027","A concave shaving mirror (f = 25 cm) is held 15 cm from a face, closer than its focal length. What kind of image forms?",{"kind":52,"options":473,"correct":482},[474,476,478,480],{"id":55,"label":475},"Real, inverted, diminished",{"id":58,"label":477},"Virtual, upright, magnified",{"id":61,"label":479},"Real, upright, same size",{"id":64,"label":481},"No image forms at all",[58],[484],"An object closer than the focus always gives a virtual, upright, magnified image in a concave mirror.","Using 1\u002Fv = 1\u002Ff − 1\u002Fu with f=25, u=15: 1\u002Fv = 1\u002F25 − 1\u002F15 = −2\u002F75, so v = −37.5 cm (virtual, behind the mirror), magnification = 37.5÷15 = 2.5× and upright.",[459,487],"mirror-formula",{"id":489,"section":19,"level":314,"prompt":490,"check":491,"hints":502,"solution":504,"skills":505},"light.q028","A convex mirror is used for a car's side mirror instead of a concave one because a convex mirror:",{"kind":52,"options":492,"correct":501},[493,495,497,499],{"id":55,"label":494},"Magnifies distant traffic for extra safety",{"id":58,"label":496},"Always gives an upright, wider-field, diminished image regardless of distance",{"id":61,"label":498},"Reflects only red light, which is easier to see",{"id":64,"label":500},"Is simply cheaper to manufacture",[58],[503],"Think about what convex mirrors are stamped with on real cars.","A convex mirror never flips character the way a concave mirror does: it is always virtual, upright and diminished, trading true size and distance judgement for a much wider field of view.",[459],{"id":507,"section":23,"level":49,"prompt":508,"check":509,"hints":520,"solution":521,"skills":522},"light.q029","A straw in a glass of water looks bent at the surface because:",{"kind":52,"options":510,"correct":519},[511,513,515,517],{"id":55,"label":512},"The straw actually bends in water",{"id":58,"label":514},"Light changes speed and direction (refracts) as it crosses from water into air",{"id":61,"label":516},"Water magnifies the straw",{"id":64,"label":518},"It's an illusion with no real cause",[58],[],"Light travelling from the straw, through water, into air bends at the surface because its speed changes there — refraction — making the submerged part appear to be in a different place.",[23],{"id":524,"section":23,"level":49,"prompt":525,"check":526,"hints":537,"solution":538,"skills":539},"light.q030","Light travelling from air into water (a denser material) bends:",{"kind":52,"options":527,"correct":536},[528,530,532,534],{"id":55,"label":529},"Away from the normal",{"id":58,"label":531},"Towards the normal",{"id":61,"label":533},"Not at all, ever",{"id":64,"label":535},"Only if it is red light",[58],[],"Entering a denser (slower) material, light always bends towards the normal, unless it arrives exactly along the normal (0°), where there is no bend at all.",[23],{"id":541,"section":23,"level":126,"prompt":542,"check":543,"hints":545,"solution":547,"skills":548},"light.q031","Using refractive index n = speed in vacuum ÷ speed in material, and water's speed of light of about 2.25 × 10⁸ m\u002Fs, what is water's refractive index (to 2 decimal places)?",{"kind":149,"answer":544,"tolerance":294},1.33,[546],"Divide the vacuum speed by the speed in water.","n = 3.00 × 10⁸ ÷ 2.25 × 10⁸ ≈ 1.33.",[549],"refractive-index",{"id":551,"section":23,"level":126,"prompt":552,"check":553,"hints":555,"solution":557,"skills":558},"light.q032","Using apparent depth = real depth ÷ refractive index, how deep does a 1.2 m pool look from directly above, in water (n = 1.33)? Give your answer to 2 decimal places.",{"kind":149,"answer":554,"tolerance":294,"unit":295},0.9,[556],"Divide the real depth by the refractive index.","Apparent depth = 1.2 ÷ 1.33 ≈ 0.90 m.",[559],"apparent-depth",{"id":561,"section":23,"level":126,"prompt":562,"check":563,"hints":566,"solution":568,"skills":569},"light.q033","A 3 m deep swimming pool is viewed from directly above. Using apparent depth = real depth ÷ 1.33, how deep does it look, to 2 decimal places?",{"kind":149,"answer":564,"tolerance":565,"unit":295},2.26,0.03,[567],"Apparent depth = real depth ÷ refractive index.","Apparent depth = 3.0 ÷ 1.33 ≈ 2.26 m.",[559],{"id":571,"section":23,"level":126,"prompt":572,"check":573,"hints":584,"solution":585,"skills":586},"light.q034","Total internal reflection can only happen when light travels:",{"kind":52,"options":574,"correct":583},[575,577,579,581],{"id":55,"label":576},"From a less dense material into a denser one",{"id":58,"label":578},"From a denser material into a less dense one, above the critical angle",{"id":61,"label":580},"Through a perfect vacuum",{"id":64,"label":582},"At exactly 0° to the normal",[58],[],"Total internal reflection needs light trying to leave a denser (slower) material for a less dense one, and only at angles greater than or equal to the critical angle.",[587],"total-internal-reflection",{"id":589,"section":23,"level":179,"prompt":590,"check":591,"hints":593,"solution":595,"skills":596},"light.q035","Using critical angle = arcsin(1 ÷ refractive index), what is water's critical angle (n = 1.33), to 1 decimal place?",{"kind":149,"answer":592,"tolerance":150,"unit":391},48.8,[594],"Critical angle = arcsin(1 ÷ n).","arcsin(1 ÷ 1.33) ≈ 48.8°.",[597],"critical-angle",{"id":599,"section":23,"level":179,"prompt":600,"check":601,"hints":603,"solution":604,"skills":605},"light.q036","What is the critical angle for glass (n = 1.50), to 1 decimal place?",{"kind":149,"answer":602,"tolerance":150,"unit":391},41.8,[594],"arcsin(1 ÷ 1.50) ≈ 41.8°.",[597],{"id":607,"section":23,"level":179,"prompt":608,"check":609,"hints":611,"solution":613,"skills":614},"light.q037","Diamond has a refractive index of 2.42. What is its critical angle, to 1 decimal place?",{"kind":149,"answer":610,"tolerance":150,"unit":391},24.4,[612],"A larger refractive index gives a smaller critical angle.","arcsin(1 ÷ 2.42) ≈ 24.4°. This very small critical angle traps light inside a cut diamond, bouncing many times before escaping — most of its sparkle.",[597],{"id":616,"section":23,"level":314,"prompt":617,"check":618,"hints":629,"solution":631,"skills":632},"light.q038","An optical fibre carries light down its length by:",{"kind":52,"options":619,"correct":628},[620,622,624,626],{"id":55,"label":621},"Total internal reflection off the boundary with the surrounding cladding, at an angle beyond the critical angle",{"id":58,"label":623},"Ordinary reflection off a silvered inner coating",{"id":61,"label":625},"Bending the glass itself around corners",{"id":64,"label":627},"Converting light to electricity and back at each bounce",[55],[630],"Think about what keeps light bouncing rather than leaking out sideways.","Light stays trapped inside the fibre's core by repeatedly striking the boundary with the cladding at an angle steeper than the critical angle, undergoing total internal reflection each time.",[633],"optical-fibre",{"id":635,"section":23,"level":314,"prompt":636,"check":637,"hints":639,"solution":641,"skills":642},"light.q039","Using Snell's law n₁sinθ₁ = n₂sinθ₂, light inside glass (n=1.50) strikes a glass-water (n=1.33) boundary at 30°. What angle does it take in the water, to 1 decimal place?",{"kind":149,"answer":638,"tolerance":151,"unit":391},34.3,[640],"Rearrange Snell's law for sinθ₂.","1.50 × sin30° = 1.33 × sinθ₂, so sinθ₂ = 0.75 ÷ 1.33 = 0.564, θ₂ ≈ 34.3°.",[643],"snells-law",{"id":645,"section":27,"level":49,"prompt":646,"check":647,"hints":658,"solution":659,"skills":660},"light.q040","A convex (converging) lens is:",{"kind":52,"options":648,"correct":657},[649,651,653,655],{"id":55,"label":650},"Thinner in the middle than at the edges",{"id":58,"label":652},"Thicker in the middle than at the edges, and bends parallel rays to a real focus",{"id":61,"label":654},"Always used only for magnifying",{"id":64,"label":656},"The same as a concave mirror",[58],[],"A convex lens bulges outward, thicker in the middle, and bends parallel light rays inward to meet at a real focal point.",[661],"lenses",{"id":663,"section":27,"level":126,"prompt":664,"check":665,"hints":675,"solution":676,"skills":677},"light.q041","A convex lens (f = 10 cm) has an object placed at 30 cm, well beyond the focus. What kind of image forms?",{"kind":52,"options":666,"correct":674},[667,668,670,672],{"id":55,"label":477},{"id":58,"label":669},"Real, inverted, smaller than the object",{"id":61,"label":671},"Real, upright, the same size",{"id":64,"label":673},"No image forms",[58],[],"Using 1\u002Fv = 1\u002Ff − 1\u002Fu: 1\u002Fv = 1\u002F10 − 1\u002F30 = 1\u002F15, v = 15 cm. Magnification = 15 ÷ 30 = 0.5, so the image is real, inverted and diminished.",[678],"lens-formula",{"id":680,"section":27,"level":126,"prompt":681,"check":682,"hints":683,"solution":685,"skills":686},"light.q042","A convex lens has a focal length of 10 cm. An object is placed exactly at 20 cm (2f). Using 1\u002Fv = 1\u002Ff − 1\u002Fu, where does the image form, in centimetres?",{"kind":149,"answer":464,"tolerance":5,"unit":152},[684],"Substitute f=10, u=20 into 1\u002Fv = 1\u002Ff − 1\u002Fu.","1\u002Fv = 1\u002F10 − 1\u002F20 = 1\u002F20, so v = 20 cm — the image forms at 2f too, the same size as the object.",[678],{"id":688,"section":27,"level":126,"prompt":689,"check":690,"hints":691,"solution":693,"skills":694},"light.q043","Using magnifying power M = 25 cm ÷ focal length, what magnification does a 5 cm focal-length magnifying glass give?",{"kind":149,"answer":400,"tolerance":390},[692],"Divide 25 cm by the focal length.","M = 25 ÷ 5 = 5×.",[695],"magnifier",{"id":697,"section":27,"level":179,"prompt":698,"check":699,"hints":710,"solution":712,"skills":713},"light.q044","The human eye's cornea and lens together focus light onto the:",{"kind":52,"options":700,"correct":709},[701,703,705,707],{"id":55,"label":702},"Pupil",{"id":58,"label":704},"Iris",{"id":61,"label":706},"Retina",{"id":64,"label":708},"Eyelid",[61],[711],"This is the light-sensitive layer at the back of the eyeball.","Together they form a real, upside-down image on the light-sensitive retina at the back of the eye; the brain then interprets this signal correctly.",[714],"the-eye",{"id":716,"section":27,"level":179,"prompt":717,"check":718,"hints":729,"solution":731,"skills":732},"light.q045","Of the eye's roughly 59 dioptres of total focusing power, most is supplied by:",{"kind":52,"options":719,"correct":728},[720,722,724,726],{"id":55,"label":721},"The lens alone",{"id":58,"label":723},"The cornea, the fixed curved front surface, because light bends most where it first meets a denser material",{"id":61,"label":725},"The retina",{"id":64,"label":727},"The eyelashes",[58],[730],"Light bends most sharply at the very first boundary it crosses, from air into a denser material.","The cornea supplies about two-thirds of the eye's total focusing power on its own; the flexible internal lens supplies the rest and handles fine adjustment (accommodation).",[714],{"id":734,"section":27,"level":314,"prompt":735,"check":736,"hints":747,"solution":750,"skills":751},"light.q046","A student's far point is only 2 m away (they cannot focus on anything beyond that). What kind of lens do they need, and roughly what power?",{"kind":52,"options":737,"correct":746},[738,740,742,744],{"id":55,"label":739},"Converging, +2.0 D",{"id":58,"label":741},"Diverging, about −0.5 D",{"id":61,"label":743},"Diverging, about −2.0 D",{"id":64,"label":745},"No lens is needed for this condition",[58],[748,749],"Power = −1 ÷ far point in metres.","A short sight problem always needs a diverging lens.","Power = −1 ÷ far point (m) = −1 ÷ 2 = -0.5 D. Myopia (short sight) is corrected with a diverging lens.",[752],"corrective-lenses",{"id":754,"section":27,"level":314,"prompt":755,"check":756,"hints":760,"solution":762,"skills":763},"light.q047","A person's near point has drifted to 1.0 m instead of the usual 25 cm. Using power = 1\u002F0.25 − 1\u002Fnear point (m), what lens power do they need, in dioptres?",{"kind":149,"answer":757,"tolerance":758,"unit":759},3,0.2,"D",[761],"Power = 1\u002F0.25 − 1\u002Fnear point (m).","Power = 1\u002F0.25 − 1\u002F1.0 = 4 − 1 = 3 D, a converging lens for hyperopia.",[752],{"id":765,"section":31,"level":49,"prompt":766,"check":767,"hints":778,"solution":779,"skills":780},"light.q048","White light passing through a prism splits into a spectrum because:",{"kind":52,"options":768,"correct":777},[769,771,773,775],{"id":55,"label":770},"The prism creates new colours out of nothing",{"id":58,"label":772},"Glass bends different colours (wavelengths) by slightly different amounts",{"id":61,"label":774},"The prism paints the light",{"id":64,"label":776},"Only red and blue light exist inside white light",[58],[],"White light is already a mixture of colours; a prism merely separates them because its refractive index is slightly different for each wavelength, bending violet more than red.",[781],"dispersion",{"id":783,"section":31,"level":49,"prompt":784,"check":785,"hints":791,"solution":793,"skills":794},"light.q049","What is the order of colours in the visible spectrum, from longest to shortest wavelength? Name the first and last colours only.",{"kind":786,"accept":787},"text",[788,789,790],"red and violet","red, violet","red to violet",[792],"Think of a rainbow's order from the outside band to the inside band.","The spectrum runs red, orange, yellow, green, blue, indigo, violet — from the longest wavelength (red) to the shortest (violet).",[795],"spectrum",{"id":797,"section":31,"level":126,"prompt":798,"check":799,"hints":810,"solution":811,"skills":812},"light.q050","A rainbow's primary bow appears at about 42° from:",{"kind":52,"options":800,"correct":809},[801,803,805,807],{"id":55,"label":802},"The Sun itself",{"id":58,"label":804},"The point in the sky directly opposite the Sun (the antisolar point)",{"id":61,"label":806},"The horizon, always",{"id":64,"label":808},"Wherever a cloud happens to be",[58],[],"Refraction, one internal reflection, and refraction again inside raindrops concentrates light most strongly at about 42° from the antisolar point.",[813],"rainbow",{"id":815,"section":31,"level":126,"prompt":816,"check":817,"hints":828,"solution":829,"skills":830},"light.q051","Compared with the primary rainbow, the secondary rainbow is:",{"kind":52,"options":818,"correct":827},[819,821,823,825],{"id":55,"label":820},"Brighter, with the same colour order",{"id":58,"label":822},"Fainter, with its colours reversed",{"id":61,"label":824},"Identical in every way",{"id":64,"label":826},"Only visible at night",[58],[],"An extra internal reflection inside the raindrop both loses light (fainter) and flips the colour order (reversed) compared with the primary bow.",[813],{"id":832,"section":31,"level":179,"prompt":833,"check":834,"hints":836,"solution":838,"skills":839},"light.q052","For water (n = 1.333), the deviation of light through a raindrop (refract, one reflection, refract) reaches a minimum near a specific entry angle. Using the table of deviations, at roughly what entry angle (in degrees) does this minimum occur?",{"kind":149,"answer":835,"tolerance":757,"unit":391},59.4,[837],"The deviation falls, reaches a low point, then rises again — find where it is lowest.","The deviation dips to a minimum near 59°, which is exactly the bunching of angles that makes the rainbow bright at 42°.",[813,840],"minimum-deviation",{"id":842,"section":31,"level":314,"prompt":843,"check":844,"hints":855,"solution":857,"skills":858},"light.q053","Descartes calculated the rainbow's 42° angle in 1637 using only:",{"kind":52,"options":845,"correct":854},[846,848,850,852],{"id":55,"label":847},"Snell's law and trigonometry, with a water-filled glass sphere as a giant raindrop",{"id":58,"label":849},"A wave theory of light, which did not yet exist",{"id":61,"label":851},"A photograph of an actual rainbow",{"id":64,"label":853},"Modern computer simulations",[55],[856],"This was over a century before anyone had a wave theory of light.","Decades before anyone had a wave or particle theory of light, Descartes used Snell's law and geometry alone, modelling a raindrop with a large glass sphere filled with water.",[813,859],"history",{"id":861,"section":35,"level":49,"prompt":862,"check":863,"hints":874,"solution":875,"skills":876},"light.q054","A red apple looks red in white light because:",{"kind":52,"options":864,"correct":873},[865,867,869,871],{"id":55,"label":866},"It makes its own red light",{"id":58,"label":868},"It absorbs most colours and reflects mostly red light back to your eye",{"id":61,"label":870},"Red light is the only colour that exists",{"id":64,"label":872},"It turns other colours into red",[58],[],"The apple's pigments absorb most wavelengths and reflect mostly red, which is why the colour you see is not created by the object, only selected by it.",[877],"colour-of-objects",{"id":879,"section":35,"level":49,"prompt":880,"check":881,"hints":892,"solution":893,"skills":894},"light.q055","Mixing red, green and blue light (additive mixing) on a screen gives:",{"kind":52,"options":882,"correct":891},[883,885,887,889],{"id":55,"label":884},"Black",{"id":58,"label":886},"White",{"id":61,"label":888},"Brown",{"id":64,"label":890},"Grey, never white",[58],[],"Red, green and blue are the additive primaries of light; combined at full intensity, they add up to white.",[895],"additive-mixing",{"id":897,"section":35,"level":126,"prompt":898,"check":899,"hints":910,"solution":912,"skills":913},"light.q056","Why is the daytime sky blue rather than violet, even though violet scatters even more strongly than blue?",{"kind":52,"options":900,"correct":909},[901,903,905,907],{"id":55,"label":902},"There is no violet light in sunlight at all",{"id":58,"label":904},"Our eyes are more sensitive to blue, and sunlight itself contains less violet than blue",{"id":61,"label":906},"Violet light cannot scatter",{"id":64,"label":908},"The atmosphere absorbs all violet light completely",[58],[911],"Think about both how much of each colour is in sunlight and how sensitive your eyes are to it.","Although violet scatters most, human eyes are less sensitive to it and sunlight itself has less violet to begin with, so the combined effect reads as blue.",[914,915],"scattering","sky-colour",{"id":917,"section":35,"level":126,"prompt":918,"check":919,"hints":921,"solution":923,"skills":924},"light.q057","Using Rayleigh scattering ∝ 1 ÷ wavelength⁴, roughly how many times more strongly is blue light (450 nm) scattered than red light (700 nm)? Round to 1 decimal place.",{"kind":149,"answer":920,"tolerance":150},5.9,[922],"Divide the two wavelengths, then raise the result to the power of 4.","(700 ÷ 450)⁴ ≈ 5.9.",[925],"rayleigh-scattering",{"id":927,"section":35,"level":126,"prompt":928,"check":929,"hints":939,"solution":940,"skills":941},"light.q058","Mixing cyan and yellow paint (subtractive mixing) gives mostly:",{"kind":52,"options":930,"correct":938},[931,933,935,937],{"id":55,"label":932},"Red",{"id":58,"label":934},"Green",{"id":61,"label":936},"Blue",{"id":64,"label":886},[58],[],"Cyan absorbs red and yellow absorbs blue; what is left for both to reflect is green.",[942],"subtractive-mixing",{"id":944,"section":35,"level":179,"prompt":945,"check":946,"hints":957,"solution":959,"skills":960},"light.q059","Sunsets look orange or red because:",{"kind":52,"options":947,"correct":956},[948,950,952,954],{"id":55,"label":949},"The Sun itself changes colour at sunset",{"id":58,"label":951},"Sunlight travels through much more atmosphere at a low angle, scattering away most of the blue and green before it reaches your eye",{"id":61,"label":953},"Clouds always turn orange in the evening",{"id":64,"label":955},"The Sun is closer to Earth at sunset",[58],[958],"Compare how much atmosphere sunlight crosses at noon versus at a low angle near the horizon.","At a low angle the light's path through the atmosphere is far longer, so blue and green are scattered out along the way, leaving mostly orange and red in the direct beam.",[914,961],"sunset",{"id":963,"section":35,"level":179,"prompt":964,"check":965,"hints":967,"solution":968,"skills":969},"light.q060","Using scattering ∝ 1 ÷ wavelength⁴, roughly how many times more strongly is violet light (400 nm) scattered than red light (700 nm)? Round to 1 decimal place.",{"kind":149,"answer":966,"tolerance":151},9.4,[922],"(700 ÷ 400)⁴ ≈ 9.4.",[925],{"id":971,"section":35,"level":314,"prompt":972,"check":973,"hints":982,"solution":984,"skills":985},"light.q061","A red object is viewed under pure green light, with no red light present at all. What colour does it appear?",{"kind":52,"options":974,"correct":981},[975,977,979,980],{"id":55,"label":976},"Red, unchanged",{"id":58,"label":978},"Black, because there is no red light for it to reflect",{"id":61,"label":934},{"id":64,"label":886},[58],[983],"An object can only reflect the colours that are actually shining on it.","An object can only reflect wavelengths that are actually shining on it. With no red light present, the \"red\" object has nothing red to reflect and appears black.",[877],{"id":987,"section":35,"level":314,"prompt":988,"check":989,"hints":1000,"solution":1002,"skills":1003},"light.q062","Why do printers use a separate true black ink (K) instead of just mixing cyan, magenta and yellow?",{"kind":52,"options":990,"correct":999},[991,993,995,997],{"id":55,"label":992},"Black ink is cheaper to produce than coloured ink",{"id":58,"label":994},"Real CMY inks are imperfect and produce a muddy brown rather than true black when mixed, and using three inks for black wastes ink",{"id":61,"label":996},"Printers cannot mix more than two colours at once",{"id":64,"label":998},"Black is not actually a colour",[58],[1001],"Real pigments are never perfectly pure.","In theory CMY should make black by removing red, green and blue; in practice real pigments are impure, producing muddy brown, so a separate black ink is added for a true black and to save ink.",[942,1004],"cmyk",{"id":1006,"section":39,"level":49,"prompt":1007,"check":1008,"hints":1010,"solution":1012,"skills":1013},"light.q063","What is the speed of light in a vacuum, to the nearest 10 million metres per second (in units of 10⁸ m\u002Fs, 1 decimal place)?",{"kind":149,"answer":757,"tolerance":307,"unit":1009},"×10⁸ m\u002Fs",[1011],"299,792,458 rounds very neatly to 3 followed by eight zeros.","The speed of light is 299,792,458 m\u002Fs, which rounds to 3.0 × 10⁸ m\u002Fs.",[1014],"speed-of-light",{"id":1016,"section":39,"level":49,"prompt":1017,"check":1018,"hints":1021,"solution":1024,"skills":1025},"light.q064","Using distance ÷ speed, roughly how many seconds does light take to travel from the Moon to Earth (384,400 km)? Round to 2 decimal places.",{"kind":149,"answer":1019,"tolerance":307,"unit":1020},1.28,"s",[1022,1023],"Convert 384,400 km to metres first.","Time = distance ÷ speed.","384,400,000 m ÷ 299,792,458 m\u002Fs ≈ 1.28 s.",[1026],"light-travel-time",{"id":1028,"section":39,"level":126,"prompt":1029,"check":1030,"hints":1033,"solution":1035,"skills":1036},"light.q065","How many minutes and seconds does light take to travel from the Sun to Earth (about 150 million km)? Give the answer in whole seconds.",{"kind":149,"answer":1031,"tolerance":1032,"unit":1020},499,2,[1034],"Convert 150 million km to metres, then divide by the speed of light.","150,000,000,000 m ÷ 299,792,458 m\u002Fs ≈ 499 s, which is 8 minutes 19 seconds.",[1026],{"id":1038,"section":39,"level":126,"prompt":1039,"check":1040,"hints":1051,"solution":1052,"skills":1053},"light.q066","Rømer's 1676 evidence for light's finite speed came from:",{"kind":52,"options":1041,"correct":1050},[1042,1044,1046,1048],{"id":55,"label":1043},"Timing a beam of light with a stopwatch",{"id":58,"label":1045},"The changing arrival time of eclipses of Jupiter's moon Io, depending on the Earth-Jupiter distance",{"id":61,"label":1047},"A spinning toothed wheel",{"id":64,"label":1049},"Measuring shadows on the Moon",[58],[],"Rømer noticed Io's eclipses ran early or late depending on where Earth was in its orbit relative to Jupiter, showing that light takes measurable time to cross the extra distance.",[859,1054],"roemer",{"id":1056,"section":39,"level":126,"prompt":1057,"check":1058,"hints":1069,"solution":1070,"skills":1071},"light.q067","Fizeau's 1849 method measured the speed of light using:",{"kind":52,"options":1059,"correct":1068},[1060,1062,1064,1066],{"id":55,"label":1061},"A telescope pointed at Jupiter",{"id":58,"label":1063},"A fast-spinning toothed wheel and a distant mirror",{"id":61,"label":1065},"A laser and a clock",{"id":64,"label":1067},"Radio waves",[58],[],"Fizeau timed a beam's round trip to a mirror 8,633 m away by finding the wheel speed at which the returning beam was blocked by the next tooth.",[859,1072],"fizeau",{"id":1074,"section":39,"level":179,"prompt":1075,"check":1076,"hints":1079,"solution":1082,"skills":1083},"light.q068","Using distance = speed × time, and light-year = c × 1 year (365.25 days), roughly how many trillion kilometres is one light-year? Round to 2 decimal places.",{"kind":149,"answer":1077,"tolerance":307,"unit":1078},9.46,"trillion km",[1080,1081],"First find the number of seconds in 365.25 days.","Multiply that by the speed of light in km\u002Fs.","c × 365.25 × 86,400 s ≈ 9,460,730,472,581 km, about 9.46 trillion km.",[1084],"light-year",{"id":1086,"section":39,"level":179,"prompt":1087,"check":1088,"hints":1091,"solution":1093,"skills":1094},"light.q069","Proxima Centauri is 4.2465 light-years away. If you sent a radio signal there and received an instant reply, how many years would the round trip take, to 1 decimal place?",{"kind":149,"answer":1089,"tolerance":758,"unit":1090},8.5,"years",[1092],"Double the one-way light-travel time.","Round trip = 4.2465 × 2 ≈ 8.5 years — there and back, even with zero reply delay.",[1084,1095],"round-trip",{"id":1097,"section":39,"level":314,"prompt":1098,"check":1099,"hints":1110,"solution":1112,"skills":1113},"light.q070","Michelson's 1879 rotating-mirror measurement of light's speed was significant because it:",{"kind":52,"options":1100,"correct":1109},[1101,1103,1105,1107],{"id":55,"label":1102},"Was the very first measurement ever made",{"id":58,"label":1104},"Was far more precise than Rømer's or Fizeau's methods, within about 0.04% of the modern value",{"id":61,"label":1106},"Proved light has no fixed speed",{"id":64,"label":1108},"Used sound waves instead of light",[58],[1111],"It was not the first measurement — think about what made it different from Rømer's and Fizeau's.","Michelson's octagonal rotating-mirror method achieved a huge improvement in precision over both earlier methods, and he later won the Nobel Prize substantially for this work.",[859,1114],"michelson",{"id":1116,"section":39,"level":314,"prompt":1117,"check":1118,"hints":1120,"solution":1122,"skills":1123},"light.q071","Betelgeuse is about 640 light-years away. If it exploded as a supernova today, how many years would pass before it could be seen from Earth?",{"kind":149,"answer":1119,"tolerance":390,"unit":1090},640,[1121],"A light-year distance directly gives the light-travel time in years.","A distance of 640 light-years means the light takes 640 years to arrive, so an explosion today would not be visible from Earth for 640 years.",[1084],{"id":1125,"section":43,"level":49,"prompt":1126,"check":1127,"hints":1138,"solution":1139,"skills":1140},"light.q072","Compared with visible light, radio waves have:",{"kind":52,"options":1128,"correct":1137},[1129,1131,1133,1135],{"id":55,"label":1130},"A much shorter wavelength",{"id":58,"label":1132},"A much longer wavelength",{"id":61,"label":1134},"Exactly the same wavelength",{"id":64,"label":1136},"No wavelength at all",[58],[],"Radio waves are metres to kilometres long, far longer than visible light's 400-700 nanometre range.",[1141],"ems",{"id":1143,"section":43,"level":126,"prompt":1144,"check":1145,"hints":1147,"solution":1150,"skills":1151},"light.q073","An FM radio station broadcasts at 100 MHz. Using wavelength = c ÷ frequency, what is its wavelength in metres, to 1 decimal place?",{"kind":149,"answer":757,"tolerance":1146,"unit":295},0.1,[1148,1149],"100 MHz = 100,000,000 Hz.","Wavelength = speed ÷ frequency.","Wavelength = 299,792,458 ÷ 100,000,000 ≈ 3.0 m.",[1141,1152],"wavelength",{"id":1154,"section":43,"level":126,"prompt":1155,"check":1156,"hints":1167,"solution":1168,"skills":1169},"light.q074","Young's 1801 double-slit experiment provided strong evidence that light behaves as a:",{"kind":52,"options":1157,"correct":1166},[1158,1160,1162,1164],{"id":55,"label":1159},"Particle",{"id":58,"label":1161},"Wave",{"id":61,"label":1163},"Solid",{"id":64,"label":1165},"Neither wave nor particle",[58],[],"The bright-and-dark interference bands Young observed are a defining wave behaviour, with no simple particle explanation.",[1170],"wave-evidence",{"id":1172,"section":43,"level":126,"prompt":1173,"check":1174,"hints":1185,"solution":1186,"skills":1187},"light.q075","The photoelectric effect (light knocking electrons out of a metal only above a threshold frequency) is best explained by treating light as:",{"kind":52,"options":1175,"correct":1184},[1176,1178,1180,1182],{"id":55,"label":1177},"A continuous wave of any energy",{"id":58,"label":1179},"Discrete packets of energy called photons, each with energy proportional to frequency",{"id":61,"label":1181},"Sound vibrations",{"id":64,"label":1183},"A purely classical particle with no special properties",[58],[],"Einstein explained this in 1905 by proposing that light arrives in discrete photons, each carrying a fixed energy set only by its frequency (E = h × f).",[1188,1189],"particle-evidence","photon",{"id":1191,"section":43,"level":179,"prompt":1192,"check":1193,"hints":1204,"solution":1206,"skills":1207},"light.q076","Which statement about the nature of light is correct?",{"kind":52,"options":1194,"correct":1203},[1195,1197,1199,1201],{"id":55,"label":1196},"Light must be either a wave or a particle, and scientists just have not decided which yet",{"id":58,"label":1198},"Light shows wave behaviour in some experiments and particle behaviour in others, with no contradiction",{"id":61,"label":1200},"Light is only a particle; the wave theory has been fully disproved",{"id":64,"label":1202},"Light is only a wave; the particle theory has been fully disproved",[58],[1205],"Rule out any option that says one theory has been fully disproved — both wave and particle evidence are real.","Light is a genuinely quantum object: it is neither a classical wave nor a classical particle, but shows either kind of behaviour depending on the experiment.",[1208],"wave-particle-duality",{"id":1210,"section":43,"level":179,"prompt":1211,"check":1212,"hints":1215,"solution":1217,"skills":1218},"light.q077","Using E (in eV) = 1240 ÷ wavelength (in nm), what is the energy of a green light photon (550 nm), to 2 decimal places?",{"kind":149,"answer":1213,"tolerance":307,"unit":1214},2.25,"eV",[1216],"Divide 1240 by the wavelength in nanometres.","E = 1240 ÷ 550 ≈ 2.25 eV.",[1219],"photon-energy",{"id":1221,"section":43,"level":314,"prompt":1222,"check":1223,"hints":1226,"solution":1228,"skills":1229},"light.q078","Using E (in eV) = 1240 ÷ wavelength (in nm), estimate the energy of a medical X-ray photon of wavelength 0.1 nm, in keV, to 1 decimal place.",{"kind":149,"answer":1224,"tolerance":5,"unit":1225},12.4,"keV",[1227],"Find the energy in eV first, then convert to keV by dividing by 1,000.","E = 1240 ÷ 0.1 = 12,400 eV = 12.4 keV.",[1219],{"id":1231,"section":43,"level":314,"prompt":1232,"check":1233,"hints":1244,"solution":1246,"skills":1247},"light.q079","A CD's data track (spacing 1.6 micrometres) diffracts light, showing a rainbow when tilted, because:",{"kind":52,"options":1234,"correct":1243},[1235,1237,1239,1241],{"id":55,"label":1236},"The plastic itself is coloured",{"id":58,"label":1238},"The closely spaced reflective grooves act as a diffraction grating, bending each wavelength by a different angle",{"id":61,"label":1240},"CDs absorb all colours except one",{"id":64,"label":1242},"This is unrelated to the wave nature of light",[58],[1245],"Diffraction gratings bend different wavelengths by different angles.","The regularly spaced grooves diffract light, and because the diffraction angle depends on wavelength, different colours are sent in different directions — a direct demonstration of light's wave behaviour.",[1248],"diffraction",{"id":1250,"section":43,"level":314,"prompt":1251,"check":1252,"hints":1255,"solution":1257,"skills":1258},"light.q080","Atmospheric refraction lifts objects near the horizon by about 0.567°, and the Sun crosses the sky at 0.25° per minute. Roughly how many extra minutes of daylight does this add at sunrise or sunset, to 1 decimal place?",{"kind":149,"answer":1253,"tolerance":150,"unit":1254},2.3,"minutes",[1256],"Divide the lift angle by the Sun's speed across the sky.","0.567° ÷ 0.25° per minute ≈ 2.3 minutes.",[1259],"atmospheric-refraction",[1261,1262,1263,1264,1265,1266,1267,1268,1269,1270,1271,1272],"light-ncert-curiosity-7-light","light-physicsclassroom-reflection","light-physicsclassroom-refraction","light-hyperphysics-mirror","light-hyperphysics-lens","light-hyperphysics-totint","light-hyperphysics-rainbow","light-wikipedia-speed-of-light","light-wikipedia-roemer","light-wikipedia-rayleigh","light-nasa-ems","light-nist-speed-of-light","needs_review",{"generatedBy":1275,"notes":1276},"claude-code","Draft generated locally; every numeric answer computed and asserted in the generator against light_numbers.py. Pending owner review.","3dc653d37ca5f23fe4b8b398df104067d6999672afac99a6cb7bfd06427620e8",{"logic:questions":1279,"source:light-hyperphysics-lens":1280,"source:light-hyperphysics-mirror":1281,"source:light-hyperphysics-rainbow":1282,"source:light-hyperphysics-totint":1283,"source:light-nasa-ems":1284,"source:light-ncert-curiosity-7-light":1285,"source:light-nist-speed-of-light":1286,"source:light-physicsclassroom-reflection":1287,"source:light-physicsclassroom-refraction":1288,"source:light-wikipedia-rayleigh":1289,"source:light-wikipedia-roemer":1290,"source:light-wikipedia-speed-of-light":1291},"e7fd7c240a65bea1bff6277f7af7cca65e2c7fcb13c6d756cb943d53f3cbc948","4cdd337c59a170b52016937cc2cc4327db640e3f7c8e883689608119257931d4","cc4f04df90efb4f54357cdc43554d566fcd0bb5efb3303655a1579a15d292259","dce907528f01e833f82d68150b423cc68c60d1c9c88673a96a84eb269bbbd3ce","fbea9871167acaf5df4f2d491a5417bc634467e447ecc5463c5d581b47e11cc6","c7ff6424fcd7299b6942c576c43b211fd02ef110e142a0947dba5a70759c03e6","9517f5bdb40f9f891f26ae78e97ff552db5602c4da5cf91f3eb3265a44a62860","525bf3f6ba3b6422ad2b7c0fa6254ee5070e1e8cb385320d308ae53d75c4ede4","e66dbe6283e14f7619d6e497882b26fbe0ae3a0632f7bd1202b67fe8f02bfe7b","43e2d25a3db005014fd7d491471e9e47242ce863ec6c79370b74a25bfe96acfb","4c7f726ffcab1b913b628c2df33237f78d9472f867c4ab3682ea2c2f080d0ea0","b3b2ba6a172784734e5a829204e49f841e19c0b1cd6b52aad9835a967d919179","e6620f59bd5a24800f47530a4ed586390c383a730742e4e6bdf5fda99401c43a","preview-7e1cbbcc4f",1789899599633]