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material and the surrounding medium."},{"text":"Reflection as light passes between materials.","correct":false,"explanation":"Reflection sends light back from a boundary rather than transmitting it through the lens. The direction changes because refraction occurs as light passes between materials."},{"text":"Diffraction as light passes between materials.","correct":false,"explanation":"Diffraction is the spreading of waves around edges or through gaps. A lens instead redirects transmitted light by refraction at boundaries between materials."},{"text":"Dispersion as light passes between materials.","correct":false,"explanation":"Dispersion separates wavelengths because their refractive indices differ, but it is not the general cause of bending. A lens changes ray direction through refraction."}]},{"id":"0fd45a7a-6d5f-42f5-9430-6121bbd34604","cardType":"basic","points":2,"front":"Why does a curved lens surface bend different rays by different amounts?","back":"Different parts of the wavefront meet the surface differently.","mcqOptions":[{"text":"Different parts of the wavefront meet the surface differently.","correct":true,"explanation":"The surface normal changes across a curved lens, so different parts of the wavefront encounter the boundary at different angles and undergo different amounts of refraction."},{"text":"Different parts of the wavefront travel through the lens equally.","correct":false,"explanation":"Equal travel through every part would not explain different bending. The curved surface makes different parts of the wavefront meet the boundary at different angles."},{"text":"Different parts of the wavefront have different frequencies.","correct":false,"explanation":"Rays from the same monochromatic wavefront have the same frequency. They bend differently because different parts of the wavefront meet the curved surface differently."},{"text":"Different parts of the wavefront reflect from the surface differently.","correct":false,"explanation":"This confuses refraction with reflection. A lens transmits and refracts light, with different parts of the wavefront encountering differently oriented sections of its curved surface."}]},{"id":"5de55825-c50e-41e9-a0e0-b169756b4360","cardType":"cloze","points":null,"front":"In the thin lens approximation, all refraction is treated as happening at [ ].","back":"In the thin lens approximation, all refraction is treated as happening at one vertical line through the lens centre.","mcqOptions":[{"text":"one vertical line through the lens centre","correct":true,"explanation":"The thin lens approximation neglects the separation between the two refracting surfaces, representing all refraction at one vertical line through the lens centre."},{"text":"two vertical lines at the lens surfaces","correct":false,"explanation":"Treating the surfaces separately describes a thick lens rather than the thin lens approximation. The approximation combines all refraction at one vertical line through the lens centre."},{"text":"one horizontal line along the principal axis","correct":false,"explanation":"The principal axis describes the lens’s orientation but is not the assumed refraction plane. Refraction is treated as occurring at one vertical line through the lens centre."},{"text":"one point at the principal focus","correct":false,"explanation":"The principal focus is where parallel rays converge, or appear to diverge from, rather than where refraction occurs. Refraction is modelled at one vertical line through the lens centre."}]},{"id":"f17ee048-6808-442c-947a-fede8927db0b","cardType":"basic","points":1,"front":"What happens to a ray through the optical centre of a thin lens?","back":"It continues undeviated.","mcqOptions":[{"text":"It continues undeviated.","correct":true,"explanation":"In the thin lens approximation, a ray directed through the optical centre emerges along the same straight path, so it is drawn as continuing undeviated."},{"text":"It bends towards the principal axis.","correct":false,"explanation":"Rays away from the optical centre may bend towards the principal axis in a converging lens. A ray through the optical centre continues undeviated."},{"text":"It bends through the principal focus.","correct":false,"explanation":"A parallel incident ray is refracted through the principal focus of a converging lens. A ray passing through the optical centre instead continues undeviated."},{"text":"It reflects back along the same path.","correct":false,"explanation":"This confuses lens refraction with reflection from a mirror. Light passes through a thin lens, and a ray through its optical centre continues undeviated."}]},{"id":"8d19e139-5b6a-4742-8eb6-d53283a05b63","cardType":"basic","points":1,"front":"What does a converging lens do to parallel rays?","back":"It brings them to a focus.","mcqOptions":[{"text":"It brings them to a focus.","correct":true,"explanation":"A converging lens refracts rays initially parallel to the principal axis so that they meet at the principal focus on the far side of the lens."},{"text":"It spreads them from a focus.","correct":false,"explanation":"Spreading parallel rays as though they came from a focus is the action of a diverging lens. A converging lens brings parallel rays to a focus."},{"text":"It keeps them exactly parallel.","correct":false,"explanation":"Parallel rays would remain parallel only if no net convergence occurred. A converging lens refracts them towards the principal axis until they meet at a focus."},{"text":"It reflects them towards a focus.","correct":false,"explanation":"A lens forms a focus by transmitting and refracting light, not by reflecting it. A converging lens brings parallel transmitted rays to a focus."}]},{"id":"6c7c0f01-8fa7-481d-8338-e9283097ae69","cardType":"basic","points":1,"front":"What does a diverging lens do to parallel rays?","back":"They spread out as if from a focus on the object side.","mcqOptions":[{"text":"They spread out as if from a focus on the object side.","correct":true,"explanation":"A diverging lens refracts parallel rays away from the principal axis. Their backward extensions meet at a virtual principal focus on the same side as the object."},{"text":"They meet at a focus on the image side.","correct":false,"explanation":"Meeting at a focus beyond the lens describes parallel rays through a converging lens. A diverging lens spreads them as if from a focus on the object side."},{"text":"They spread out as if from a focus on the image side.","correct":false,"explanation":"The apparent source of the diverging rays lies behind them on the incident side, not beyond the lens. Their virtual focus is on the object side."},{"text":"They remain parallel to the principal axis.","correct":false,"explanation":"This overlooks the refracting effect of the lens. A diverging lens bends parallel rays away from the axis so they appear to originate from an object-side focus."}]},{"id":"013358f5-129f-44e1-952b-0f6b3235b701","cardType":"cloze","points":null,"front":"A converging lens is [ ]; a diverging lens is [ ].","back":"A converging lens is thicker in the middle; a diverging lens is thinner in the middle.","mcqOptions":[{"text":"A converging lens is thicker in the middle; a diverging lens is thinner in the middle.","correct":true,"explanation":"In the standard case of a lens in air, greater central thickness produces convergence, while reduced central thickness and thicker edges produce divergence."},{"text":"A converging lens is thinner in the middle; a diverging lens is thicker in the middle.","correct":false,"explanation":"This reverses the characteristic shapes. A converging lens is thicker in the middle, whereas a diverging lens is thinner in the middle and thicker near its edges."},{"text":"A converging lens is thicker at the edges; a diverging lens is thinner at the edges.","correct":false,"explanation":"This describes the edge thicknesses in reverse. A converging lens is thicker in the middle, while a diverging lens is thinner in the middle."},{"text":"A converging lens is equally thick throughout; a diverging lens is thicker in the middle.","correct":false,"explanation":"A uniformly thick transparent block does not have the curved lens shape needed for convergence. A converging lens is thicker centrally, and a diverging lens is thinner centrally."}]},{"id":"89e535b8-ecd3-442d-acff-e3f8eec4b33f","cardType":"basic","points":1,"front":"What allows a real image to be projected onto a screen?","back":"Light rays actually meet there.","mcqOptions":[{"text":"Light rays actually meet there.","correct":true,"explanation":"A real image forms where light rays physically converge. Because light reaches that position, a screen placed there intercepts the rays and displays the image."},{"text":"Light rays appear to meet there.","correct":false,"explanation":"Rays that only appear to meet form a virtual image, which cannot be captured directly on a screen. Projection requires the light rays to actually meet."},{"text":"Light rays remain parallel there.","correct":false,"explanation":"Parallel rays have not converged to corresponding image points, so they cannot produce a focused projected image. A real image forms where light rays actually meet."},{"text":"Light rays begin to diverge there.","correct":false,"explanation":"Divergence alone does not explain why a screen can capture an image. The defining feature is that the rays actually meet at each real image point."}]},{"id":"9b3e1b47-aeb9-4673-9650-aa630bfb4b78","cardType":"basic","points":1,"front":"Why can a virtual image not be projected onto a screen?","back":"Rays only appear to come from it.","mcqOptions":[{"text":"Rays only appear to come from it.","correct":true,"explanation":"The rays do not physically meet at a virtual image; only their backward extensions intersect. A screen cannot capture an image where no light rays actually converge."},{"text":"Rays actually converge at it.","correct":false,"explanation":"Actual convergence is the defining feature of a real image, which can be projected. At a virtual image, rays only appear to come from the image position."},{"text":"Rays are completely absorbed by it.","correct":false,"explanation":"Failure to project is not caused by absorption. A virtual image cannot be caught on a screen because rays only appear to come from its position."},{"text":"Rays emerge parallel from it.","correct":false,"explanation":"Parallel rays are associated with an image effectively at infinity, not with every virtual image. A virtual image is unprojectable because rays only appear to come from it."}]},{"id":"d74c995d-44c6-485d-b46b-bfa977257e7e","cardType":"cloze","points":null,"front":"Real images from a single converging lens are usually [ ], while virtual images are usually [ ].","back":"Real images from a single converging lens are usually inverted, while virtual images are usually upright.","mcqOptions":[{"text":"Real images from a single converging lens are usually inverted, while virtual images are usually upright.","correct":true,"explanation":"Real images formed by a single converging lens are inverted because the rays cross. Virtual images are upright because they arise from backward extensions of diverging rays."},{"text":"Real images from a single converging lens are usually upright, while virtual images are usually inverted.","correct":false,"explanation":"This reverses the usual orientations. Rays cross when forming a real image, making it inverted, whereas a virtual image from a converging lens is upright."},{"text":"Real images from a single converging lens are usually upright, while virtual images are usually upright.","correct":false,"explanation":"The virtual-image orientation is right, but a real image from one converging lens is usually inverted because the refracted rays physically cross before reaching the image."},{"text":"Real images from a single converging lens are usually inverted, while virtual images are usually inverted.","correct":false,"explanation":"The real-image orientation is right, but a virtual image formed by a converging lens is upright because only backward extensions of the refracted rays intersect."}]},{"id":"4a35b60c-6c7a-40a0-9e9f-f1f608decd67","cardType":"basic","points":1,"front":"For a converging lens, where does a ray parallel to the principal axis go?","back":"It refracts through the far focus.","mcqOptions":[{"text":"It refracts through the far focus.","correct":true,"explanation":"A converging lens bends a ray initially parallel to the principal axis so that it passes through the principal focus on the far side of the lens."},{"text":"It refracts through the near focus.","correct":false,"explanation":"Confusing the two focal points reverses the ray rule. A parallel incident ray is refracted through the far focus, on the opposite side of the lens."},{"text":"It continues through the optical centre.","correct":false,"explanation":"A ray already directed through the optical centre continues approximately undeviated. A ray parallel to the principal axis instead refracts through the far focus."},{"text":"It emerges parallel to the principal axis.","correct":false,"explanation":"Emergence parallel to the axis applies to a ray travelling through the near focus before reaching the lens. A parallel incident ray refracts through the far focus."}]},{"id":"491bb58b-d97f-4b87-b517-6ba6fb09f6bf","cardType":"basic","points":1,"front":"For a converging lens, what happens to a ray through the near focus?","back":"It emerges parallel to the principal axis.","mcqOptions":[{"text":"It emerges parallel to the principal axis.","correct":true,"explanation":"The reversibility of light means that the reverse of a parallel ray passing through the far focus is a near-focus ray emerging parallel to the principal axis."},{"text":"It refracts through the far focus.","correct":false,"explanation":"Passing through the far focus is the rule for a ray incident parallel to the axis. A ray through the near focus emerges parallel to the principal axis."},{"text":"It continues through the optical centre.","correct":false,"explanation":"Approximate undeviated travel applies to a ray directed through the optical centre. A ray through the near focus is refracted to emerge parallel to the principal axis."},{"text":"It emerges perpendicular to the principal axis.","correct":false,"explanation":"A focal ray is not turned through $90^\\circ$. For a converging lens, a ray passing through the near focus emerges parallel to the principal axis."}]},{"id":"822b8de3-6414-4cb2-aac9-8e7d5d4dce08","cardType":"basic","points":1,"front":"When drawing a ray diagram, where should the principal rays start?","back":"From the same object point, usually the top.","mcqOptions":[{"text":"From the same object point, usually the top.","correct":true,"explanation":"Principal rays must originate from one object point so their refracted paths locate the corresponding image point. The top is normally chosen to determine image height and orientation."},{"text":"From different object points, usually the top.","correct":false,"explanation":"Rays from different object points form different image points, so their intersection would not locate one valid image point. Principal rays should start from the same object point."},{"text":"From the optical centre, usually the top.","correct":false,"explanation":"The optical centre is a point on the lens, not the usual starting point for construction rays. Rays should begin at the same object point, usually its top."},{"text":"From the near focal point, usually the top.","correct":false,"explanation":"The near focus is used to define one principal ray’s direction, but it is not where all rays begin. They start from the same object point."}]},{"id":"5424d2b8-367a-4e14-8822-a754fabcaeb5","cardType":"basic","points":1,"front":"For a converging lens, what image forms when the object is beyond $2F$?","back":"Real, inverted and diminished.","mcqOptions":[{"text":"Real, inverted and diminished.","correct":true,"explanation":"With the object beyond $2F$, rays converge between $F$ and $2F$ on the opposite side. Their crossing produces a real, inverted image smaller than the object."},{"text":"Real, inverted and magnified.","correct":false,"explanation":"A real, inverted, magnified image occurs when the object lies between $F$ and $2F$. Beyond $2F$, the image is real, inverted and diminished."},{"text":"Real, inverted and same-sized.","correct":false,"explanation":"An image equal in size to the object forms when the object is at $2F$. Placing it beyond $2F$ produces a real, inverted and diminished image."},{"text":"Virtual, upright and magnified.","correct":false,"explanation":"A virtual, upright, magnified image forms when the object is inside the focal length. An object beyond $2F$ forms a real, inverted and diminished image."}]},{"id":"2ca88437-ca36-4ea9-aa87-8f6ee27b8e6e","cardType":"basic","points":1,"front":"For a converging lens, where is the image if the object is beyond $2F$?","back":"Between $F$ and $2F$ on the other side.","mcqOptions":[{"text":"Between $F$ and $2F$ on the other side.","correct":true,"explanation":"For an object beyond $2F$, the converging rays meet on the opposite side of the lens between $F$ and $2F$, producing a diminished real image."},{"text":"Beyond $2F$ on the other side.","correct":false,"explanation":"An image beyond $2F$ occurs when the object is between $F$ and $2F$. If the object is beyond $2F$, its image lies between $F$ and $2F$."},{"text":"At $2F$ on the other side.","correct":false,"explanation":"The image is exactly at $2F$ only when the object is also at $2F$. An object beyond $2F$ forms an image between $F$ and $2F$."},{"text":"Between $F$ and $2F$ on the same side.","correct":false,"explanation":"The same-side placement suggests a virtual image, which occurs for an object inside $F$. Beyond $2F$, the real image forms between $F$ and $2F$ on the other side."}]},{"id":"30a10a54-c04f-4ddb-9396-f588f3e98275","cardType":"basic","points":1,"front":"What image forms when an object is inside the focal length of a converging lens?","back":"A virtual, upright, magnified image.","mcqOptions":[{"text":"A virtual, upright, magnified image.","correct":true,"explanation":"Inside the focal length, the refracted rays diverge and their backward extensions meet on the object’s side. This produces a virtual, upright and magnified image."},{"text":"A real, inverted, magnified image.","correct":false,"explanation":"A real, inverted, magnified image forms when the object is between $F$ and $2F$, not inside $F$. Inside the focal length, the image is virtual, upright and magnified."},{"text":"A real, inverted, diminished image.","correct":false,"explanation":"A real, inverted, diminished image is produced by an object beyond $2F$. Inside the focal length, rays do not converge, so the image is virtual, upright and magnified."},{"text":"A virtual, upright, diminished image.","correct":false,"explanation":"Although the image is virtual and upright, a converging lens acts as a magnifier when the object is inside its focal length, so the image is magnified rather than diminished."}]},{"id":"5179f92d-19e6-410d-99f8-f64e8c832b97","cardType":"basic","points":1,"front":"For a converging lens used as a magnifying glass, which side is the virtual image on?","back":"The same side as the object.","mcqOptions":[{"text":"The same side as the object.","correct":true,"explanation":"When the object is inside the focal length of a converging lens, the refracted rays diverge and their backward extensions meet on the object’s side, forming a virtual image."},{"text":"The opposite side from the object.","correct":false,"explanation":"The opposite side is where a converging lens forms a real image when the object lies beyond the focal length. As a magnifying glass, it forms the virtual image on the same side as the object."},{"text":"The focal plane behind the lens.","correct":false,"explanation":"Placing the object at the focal point makes the emerging rays parallel, so no finite image forms. A magnifying glass produces a virtual image on the same side as the object."},{"text":"The lens plane beside the object.","correct":false,"explanation":"The image is not formed in the plane of the lens. Backward extensions of the refracted rays meet on the same side of the lens as the object."}]},{"id":"6992805c-fb38-4439-8d45-c0bf774eea8b","cardType":"basic","points":1,"front":"What image does a diverging lens form for a real object?","back":"A virtual, upright, diminished image.","mcqOptions":[{"text":"A virtual, upright, diminished image.","correct":true,"explanation":"A diverging lens spreads rays from a real object. Their backward extensions meet between the lens and its focal point, producing an image that is always virtual, upright and diminished."},{"text":"A real, inverted, diminished image.","correct":false,"explanation":"A real, inverted, diminished image can be produced by a converging lens with a suitably distant object. A diverging lens instead forms a virtual, upright, diminished image."},{"text":"A virtual, upright, magnified image.","correct":false,"explanation":"Confusing a diverging lens with a converging magnifying glass makes the image seem magnified. For a real object, a diverging lens always produces a virtual, upright, diminished image."},{"text":"A virtual, inverted, diminished image.","correct":false,"explanation":"Virtual images formed by a single diverging lens are upright, not inverted. The backward extensions of the diverging rays produce a virtual, upright, diminished image."}]},{"id":"f6fcd0e4-29be-4521-933e-14d528755101","cardType":"cloze","points":null,"front":"For a virtual image, real refracted rays are drawn solid, then extended backwards with [ ].","back":"For a virtual image, real refracted rays are drawn solid, then extended backwards with dashed lines.","mcqOptions":[{"text":"dashed lines","correct":true,"explanation":"Dashed lines distinguish backward constructions from the paths of real light rays. Their intersection shows the apparent position from which the rays seem to originate."},{"text":"solid lines","correct":false,"explanation":"Solid lines represent the actual paths travelled by light, so using them for backward extensions falsely implies that light reaches the virtual image. The extensions should be dashed lines."},{"text":"ray arrows","correct":false,"explanation":"Arrows indicate the direction in which light travels; they do not distinguish imaginary extensions from real rays. Backward extensions used to locate a virtual image are drawn with dashed lines."},{"text":"normal lines","correct":false,"explanation":"Normal lines are construction lines perpendicular to an optical boundary and are associated with refraction angles. Virtual-ray extensions are instead represented using dashed lines."}]},{"id":"130eea6d-fee6-430e-a1c3-0a0ab8de5851","cardType":"cloze","points":null,"front":"The thin lens equation is [ ].","back":"The thin lens equation is $\\frac{1}{u}+\\frac{1}{v}=\\frac{1}{f}$.","mcqOptions":[{"text":"$$\\frac{1}{u}+\\frac{1}{v}=\\frac{1}{f}$","correct":true,"explanation":"For a thin lens, the reciprocals of the object distance and image distance add to the reciprocal of the focal length, with signs assigned according to the chosen convention."},{"text":"$$\\frac{1}{u}-\\frac{1}{v}=\\frac{1}{f}$","correct":false,"explanation":"This introduces a minus sign that does not belong in the stated signed convention. The thin lens equation is $\\frac{1}{u}+\\frac{1}{v}=\\frac{1}{f}$."},{"text":"$$\\frac{1}{u}+\\frac{1}{f}=\\frac{1}{v}$","correct":false,"explanation":"This incorrectly rearranges the lens equation by changing the sign of the focal-length term. The correct relationship is $\\frac{1}{u}+\\frac{1}{v}=\\frac{1}{f}$."},{"text":"$$\\frac{1}{u}+\\frac{1}{v}=f$","correct":false,"explanation":"This equates a reciprocal length to a length, so it is dimensionally inconsistent. The right-hand side must be reciprocal focal length: $\\frac{1}{u}+\\frac{1}{v}=\\frac{1}{f}$."}]},{"id":"d1121e2d-59d7-47c7-85ea-8ff9a3630c81","cardType":"basic","points":1,"front":"Under the common signed convention, what sign does $f$ have for a diverging lens?","back":"Negative: $f<0$.","mcqOptions":[{"text":"Negative: $f<0$.","correct":true,"explanation":"Under the common signed convention, a diverging lens has a negative focal length because its principal focus is virtual and lies on the incident-light side of the lens."},{"text":"Positive: $f>0$.","correct":false,"explanation":"A positive focal length applies to a converging lens under this convention. A diverging lens has a virtual principal focus, so its focal length is negative: $f<0$."},{"text":"Zero: $f=0$.","correct":false,"explanation":"A zero focal length would imply infinite optical power and is not the sign convention for a diverging lens. Its finite focal length is negative: $f<0$."},{"text":"Non-negative: $f\\geq 0$.","correct":false,"explanation":"This excludes the negative value required for a diverging lens and may result from ignoring virtual-focus signs. Under the common convention, its focal length satisfies $f<0$."}]},{"id":"94847b7c-71c7-4f27-8211-58d7fecbf3b8","cardType":"basic","points":1,"front":"Under the common signed convention, what sign does $v$ have for a virtual image?","back":"Negative: $v<0$.","mcqOptions":[{"text":"Negative: $v<0$.","correct":true,"explanation":"Under the common signed convention, a virtual image lies on the same side of the lens as the object, so its image distance is assigned a negative value."},{"text":"Positive: $v>0$.","correct":false,"explanation":"A positive image distance describes a real image on the outgoing-light side under this convention. A virtual image lies on the object side, giving $v<0$."},{"text":"Zero: $v=0$.","correct":false,"explanation":"A zero image distance would place the image in the plane of the lens, not merely indicate that it is virtual. Virtual images have negative image distance: $v<0$."},{"text":"Non-negative: $v\\geq 0$.","correct":false,"explanation":"Treating all distances as unsigned leads to this mistake. The signed convention distinguishes a virtual image by assigning its image distance a negative value, so $v<0$."}]},{"id":"2d89bb29-c1c4-48e2-bc0d-e719f854b9c3","cardType":"basic","points":1,"front":"What is the magnification equation for a thin lens?","back":"$$M=\\frac{\\text{image height}}{\\text{object height}}=\\frac{v}{u}$","mcqOptions":[{"text":"$$M=\\frac{\\text{image height}}{\\text{object height}}=\\frac{v}{u}$","correct":true,"explanation":"Linear magnification is the ratio of image height to object height. For a thin lens, corresponding triangles also make this equal to the ratio of image distance to object distance."},{"text":"$$M=\\frac{\\text{object height}}{\\text{image height}}=\\frac{u}{v}$","correct":false,"explanation":"This takes the reciprocal of both required ratios, a common reversal error. Magnification compares image size with object size, so $M=\\frac{\\text{image height}}{\\text{object height}}=\\frac{v}{u}$."},{"text":"$$M=\\frac{\\text{image height}}{\\text{object height}}=\\frac{u}{v}$","correct":false,"explanation":"The height ratio is written correctly, but the distance ratio has been reversed. Similar triangles give image distance over object distance, so the final ratio must be $\\frac{v}{u}$."},{"text":"$$M=\\frac{\\text{image height}}{\\text{object height}}=\\frac{f}{u}$","correct":false,"explanation":"Using focal length in place of image distance confuses magnification with lens geometry. For a thin lens, the corresponding distance ratio is $\\frac{v}{u}$, not $\\frac{f}{u}$."}]},{"id":"ef8be5fe-24f1-4a8c-9c9e-db9d79a254f2","cardType":"basic","points":1,"front":"What is the unit of magnification?","back":"No unit.","mcqOptions":[{"text":"No unit.","correct":true,"explanation":"Magnification is a ratio of two heights, or two distances, measured in the same unit. Those units cancel, leaving magnification as a dimensionless quantity."},{"text":"Metres.","correct":false,"explanation":"Although image and object dimensions may be measured in metres, magnification divides one length by another, so the metres cancel. Magnification therefore has no unit."},{"text":"Dioptres.","correct":false,"explanation":"Dioptres measure lens power, equal to the reciprocal of focal length in metres. Magnification is instead a ratio of like lengths, so it has no unit."},{"text":"Radians.","correct":false,"explanation":"Radians describe angular measure and may be used for angular magnification in other contexts. Linear magnification here is a ratio of like lengths and therefore has no unit."}]},{"id":"c0a2a6a1-31d8-4e12-b4ac-b8edb5545d2c","cardType":"basic","points":1,"front":"A 60 mm object has magnification $0.500$. What is the image height?","back":"30 mm.","mcqOptions":[{"text":"30 mm.","correct":true,"explanation":"Magnification is image height divided by object height, so the image height is $0.500 \\times 60\\ \\text{mm}=30\\ \\text{mm}$."},{"text":"60 mm.","correct":false,"explanation":"This treats the magnification as 1 rather than 0.500. Since the image is half the object’s height, the correct image height is $0.500 \\times 60\\ \\text{mm}=30\\ \\text{mm}$."},{"text":"120 mm.","correct":false,"explanation":"This doubles the object height instead of multiplying by the magnification. The correct calculation is $0.500 \\times 60\\ \\text{mm}=30\\ \\text{mm}$."},{"text":"0.030 mm.","correct":false,"explanation":"This incorrectly converts the result into metres and then labels it as millimetres. Multiplying gives $30\\ \\text{mm}$, which is the correct image height."}]},{"id":"8212b98c-e67d-4699-b341-356a55895d66","cardType":"basic","points":1,"front":"What is lens power in terms of focal length?","back":"$$P=\\frac{1}{f}$, with $f$ in metres.","mcqOptions":[{"text":"$$P=\\frac{1}{f}$, with $f$ in metres.","correct":true,"explanation":"Lens power is the reciprocal of focal length, $P=1/f$, and using focal length in metres gives power in dioptres."},{"text":"$$P=f$, with $f$ in metres.","correct":false,"explanation":"Power increases when focal length decreases, so it is inversely proportional to $f$, not directly proportional. The correct relationship is $P=1/f$ with $f$ measured in metres."},{"text":"$$P=\\frac{1}{f^2}$, with $f$ in metres.","correct":false,"explanation":"Squaring the focal length gives the wrong dependence and incorrect units. Lens power is the reciprocal of focal length: $P=1/f$, with $f$ in metres."},{"text":"$$P=\\frac{1}{f}$, with $f$ in centimetres.","correct":false,"explanation":"The reciprocal relationship is right, but focal length must be converted to metres for power in dioptres. Using centimetres introduces a factor-of-100 error."}]},{"id":"d3e8caa9-4533-4a65-87fe-e524c56cace7","cardType":"cloze","points":null,"front":"The unit of lens power is the [ ], symbol [ ].","back":"The unit of lens power is the dioptre, symbol D.","mcqOptions":[{"text":"The unit of lens power is the dioptre, symbol D.","correct":true,"explanation":"Lens power is measured in dioptres, abbreviated D. One dioptre is equivalent to an inverse focal length of one metre."},{"text":"The unit of lens power is the joule, symbol J.","correct":false,"explanation":"A joule measures energy, not lens power. Lens power is measured in dioptres, with the symbol D, because it is the reciprocal of focal length in metres."},{"text":"The unit of lens power is the watt, symbol W.","correct":false,"explanation":"A watt measures power as a rate of energy transfer, whereas optical lens power uses inverse focal length. The correct unit is the dioptre, symbol D."},{"text":"The unit of lens power is the tesla, symbol T.","correct":false,"explanation":"A tesla measures magnetic flux density and is unrelated to optical power. Lens power is measured in dioptres, abbreviated D."}]},{"id":"2ba5021e-074c-40f1-b72b-b254f5b3615f","cardType":"basic","points":1,"front":"What sign is the power of a converging lens?","back":"Positive.","mcqOptions":[{"text":"Positive.","correct":true,"explanation":"A converging lens has a positive focal length under the usual sign convention. Since $P=1/f$, its lens power is therefore positive."},{"text":"Negative.","correct":false,"explanation":"Negative power corresponds to a diverging lens, which has a negative focal length. A converging lens has positive focal length and therefore positive power."},{"text":"Zero.","correct":false,"explanation":"Zero power would require an infinitely large focal length, meaning the lens produces no convergence. A converging lens has finite positive focal length and positive power."},{"text":"Unitless.","correct":false,"explanation":"Lens power has a sign as well as a unit. For a converging lens, the reciprocal of its positive focal length gives positive power in dioptres."}]},{"id":"1d545fd8-c7d9-4f62-8c0c-0154705559ee","cardType":"basic","points":1,"front":"A converging lens has $f=0.250\\ \\text{m}$. What is its power?","back":"+4.00 D.","mcqOptions":[{"text":"+4.00 D.","correct":true,"explanation":"Using $P=1/f$ with $f=0.250\\ \\text{m}$ gives $P=1/0.250=+4.00\\ \\text{D}$. The positive sign indicates convergence."},{"text":"+0.250 D.","correct":false,"explanation":"This uses the focal length itself as the power instead of taking its reciprocal. The correct calculation is $P=1/0.250=+4.00\\ \\text{D}$."},{"text":"-4.00 D.","correct":false,"explanation":"The reciprocal calculation is correct, but the sign is reversed. A converging lens has positive focal length, so its power is $+4.00\\ \\text{D}$."},{"text":"+2.00 D.","correct":false,"explanation":"This is consistent with an incorrect reciprocal or halving slip. Taking the reciprocal accurately gives $1/0.250=+4.00\\ \\text{D}$."}]},{"id":"0c35636e-f1f3-45f0-a14d-76972ece92bf","cardType":"basic","points":2,"front":"Why does focusing a distant object onto a screen estimate $f$ for a converging lens?","back":"For very large $u$, $\\frac{1}{u}\\approx0$, so $v\\approx f$.","mcqOptions":[{"text":"For very large $u$, $\\frac{1}{u}\\approx0$, so $v\\approx f$.","correct":true,"explanation":"The lens equation is $1/f=1/u+1/v$. For a very distant object, $1/u$ becomes negligible, leaving $1/v\\approx1/f$ and therefore $v\\approx f$."},{"text":"For very large $u$, $\\frac{1}{u}\\approx1$, so $v\\approx f$.","correct":false,"explanation":"The reciprocal of a very large distance approaches zero, not one. Therefore the lens equation gives $v\\approx f$, allowing the focal length to be estimated."},{"text":"For very large $u$, $\\frac{1}{u}\\approx0$, so $v\\approx u$.","correct":false,"explanation":"Although the reciprocal term becomes negligible, the remaining equation gives $1/v\\approx1/f$, not $v\\approx u$. Thus the image distance approaches the focal length."},{"text":"For very large $u$, $\\frac{1}{u}\\approx0$, so $v\\approx2f$.","correct":false,"explanation":"Neglecting $1/u$ leaves $1/v\\approx1/f$, so the image distance is approximately one focal length, not two. The distant-object image forms near the focal point."}]},{"id":"7b498108-8b6b-4680-b66e-36088ef21fec","cardType":"basic","points":1,"front":"On a graph of $\\frac{1}{v}$ against $\\frac{1}{u}$, what is the gradient?","back":"Approximately -1.","mcqOptions":[{"text":"Approximately -1.","correct":true,"explanation":"Rearranging the lens equation gives $1/v=-1/u+1/f$. Therefore a graph of $1/v$ against $1/u$ has gradient approximately $-1$."},{"text":"Approximately +1.","correct":false,"explanation":"The positive sign reverses the relationship between the reciprocals. Rearranging gives $1/v=-1/u+1/f$, so the gradient is approximately $-1$."},{"text":"Approximately 0.","correct":false,"explanation":"A zero gradient would mean $1/v$ does not change with $1/u$. The lens equation contains the term $-1/u$, giving a gradient of approximately $-1$."},{"text":"Approximately \\frac{1}{f}.","correct":false,"explanation":"The reciprocal focal length is the y-intercept, not the gradient. From $1/v=-1/u+1/f$, the gradient is approximately $-1$."}]},{"id":"7690748d-d57f-4a20-bc1d-ca43b63adbf5","cardType":"basic","points":1,"front":"On a graph of $\\frac{1}{v}$ against $\\frac{1}{u}$, what is the y-intercept?","back":"$\\frac{1}{f}$.","mcqOptions":[{"text":"$\\frac{1}{f}$.","correct":true,"explanation":"Rearranging the lens equation gives $1/v=-1/u+1/f$. When $1/u=0$, the y-intercept is therefore $1/f$."},{"text":"$f$.","correct":false,"explanation":"The graph uses reciprocal distances, so the intercept must also involve a reciprocal. Setting $1/u=0$ in the rearranged equation gives $1/v=1/f$."},{"text":"0.","correct":false,"explanation":"The intercept is not zero because the lens has finite focal length. The rearranged equation $1/v=-1/u+1/f$ gives an intercept of $1/f$."},{"text":"$-\\frac{1}{f}$.","correct":false,"explanation":"The negative sign belongs to the gradient term, $-1/u$, not the intercept. The y-intercept is the positive reciprocal focal length, $1/f$."}]}]}],"examCountdown":{"examDate":"2027-05-12","paperLabel":"Paper 1"},"heading":"Lenses and optical images","title":"Lenses and optical images Flashcards","topbar":{"backHref":"/as-level/physics/edexcel/flashcards"},"topicSidebar":"$L38"}]}],"topicId":"abef0774-285a-562a-b9d8-bb38f0577a62","walled":"$L39"}],"$L3a","$L3b"]