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24:[["$","$L33",null,{"feature":"flashcards"}],"$L34",["$","$L35",null,{"href":"/a-level/physics/eduqas/refraction-of-light/flashcards","label":"Refraction of Light","subjectCode":"a_level_physics__eduqas","surface":"flashcards"}],["$","$L36",null,{"runner":["$","div",null,{"className":"min-h-0 flex-1 overflow-hidden bg-background","children":["$","$L37",null,{"decks":[{"id":"ca6aeef8-645e-46d7-a9f4-b8968a868011","title":"Refraction of Light Flashcards","topicId":"71d27fc7-3da9-5330-a2ba-ca2b3ba8f747","topicSlug":"refraction-of-light","cards":[{"id":"b52cc2d5-52ed-42e1-a0c8-4879bb5fa428","cardType":"basic","points":1,"front":"What physical change causes light to refract when crossing a boundary?","back":"A change in the speed of light.","mcqOptions":[{"text":"A change in the speed of light.","correct":true,"explanation":"Refraction occurs because light changes speed when it crosses into a medium with different optical properties. This speed change causes the light ray to change direction unless it enters along the normal."},{"text":"A change in the frequency of light.","correct":false,"explanation":"Frequency remains constant when light crosses a boundary; confusing frequency with speed is a common misconception. Refraction is caused by a change in the speed of light between media."},{"text":"A change in the amplitude of light.","correct":false,"explanation":"Amplitude affects the intensity of light, not its direction at a boundary. The relevant physical change is the light’s speed, which varies between different media."},{"text":"A change in the polarisation of light.","correct":false,"explanation":"Polarisation describes the orientation of the wave’s oscillations and is not the cause of refraction. Light refracts because its speed changes when entering a different medium."}]},{"id":"db7a428a-9005-4d46-a880-78a01e6a07d0","cardType":"basic","points":1,"front":"In refraction, from which line are the angles of incidence and refraction measured?","back":"The normal (an imaginary line at 90 degrees to the boundary).","mcqOptions":[{"text":"The normal (an imaginary line at 90 degrees to the boundary).","correct":true,"explanation":"Angles of incidence and refraction are measured from the normal, an imaginary line perpendicular to the boundary. Measuring from the surface itself would give complementary angles."},{"text":"The boundary (the line separating the two media).","correct":false,"explanation":"Measuring from the boundary instead of the normal gives the complementary angle and does not follow the convention used in Snell’s Law. The correct reference is the perpendicular normal."},{"text":"The incident ray (the ray approaching the boundary).","correct":false,"explanation":"The incident ray determines the direction of the incoming light but is not the reference line for measuring either angle. Both angles are measured from the normal."},{"text":"The refracted ray (the ray travelling into the new medium).","correct":false,"explanation":"The refracted ray is one of the lines forming the angle of refraction, not the line from which that angle is measured. Angles are measured from the normal."}]},{"id":"f78c0dad-1667-41a7-a23e-869e6740b738","cardType":"cloze","points":null,"front":"The refractive index, $n$, of a medium is defined by the formula $$\\text{[...]}$$.","back":"The refractive index, $n$, of a medium is defined by the formula $$\\boxed{n = \\frac{c}{v}}$$.","mcqOptions":[{"text":"$$$n = \\frac{c}{v}$$","correct":true,"explanation":"Refractive index compares the speed of light in vacuum, $c$, with its speed in the medium, $v$. Therefore, $n$ is calculated using the ratio $c/v$."},{"text":"$$$n = \\frac{v}{c}$$","correct":false,"explanation":"This reverses the required ratio and would normally give a value below one for a material. The refractive index is defined as vacuum speed divided by medium speed, $n = c/v$."},{"text":"$$$n = c v$$","correct":false,"explanation":"Multiplying the two speeds gives incorrect dimensions and does not form a ratio. Refractive index is dimensionless because it is calculated as $c/v$."},{"text":"$$$n = c - v$$","correct":false,"explanation":"Subtracting the speeds produces a quantity with speed units, so it cannot represent refractive index. The definition uses their ratio, $n = c/v$, which has no units."}]},{"id":"0fda0008-c53f-4b1b-a6e3-b46d7fbae90e","cardType":"basic","points":1,"front":"Why does refractive index have no units?","back":"It is a ratio of two speeds.","mcqOptions":[{"text":"It is a ratio of two speeds.","correct":true,"explanation":"Refractive index is calculated by dividing one speed by another, such as $n = c/v$. The units cancel in this ratio, leaving a dimensionless quantity."},{"text":"It is a difference between two speeds.","correct":false,"explanation":"A difference between two speeds would still have units of metres per second. Refractive index has no units because it is a ratio of two speeds, not their difference."},{"text":"It is a product of two speeds.","correct":false,"explanation":"Multiplying two speeds would produce units of metres squared per second squared. The units cancel only when the speeds are divided to form a ratio."},{"text":"It is a ratio of two wavelengths.","correct":false,"explanation":"A wavelength ratio would also be dimensionless, but it is not the definition being used here. Refractive index is specifically the ratio of the speed of light in vacuum to its speed in the medium."}]},{"id":"b04fdcd2-975f-423f-b670-02c170950daa","cardType":"basic","points":1,"front":"When light enters a new medium, which wave property stays constant?","back":"Frequency.","mcqOptions":[{"text":"Frequency.","correct":true,"explanation":"The frequency is fixed by the source and cannot change as light crosses a stationary boundary. The speed and wavelength adjust so that the wave continues to satisfy $v = f\\lambda$."},{"text":"Speed.","correct":false,"explanation":"Speed generally changes when light enters a medium with a different refractive index. Frequency stays constant, while wavelength changes to maintain $v = f\\lambda$."},{"text":"Wavelength.","correct":false,"explanation":"Wavelength changes when the speed changes and frequency remains fixed. It is therefore not the wave property that stays constant across the boundary."},{"text":"Amplitude.","correct":false,"explanation":"Amplitude may change because some light can be reflected or absorbed, but it is not the invariant property asked for. The frequency remains constant at the boundary."}]},{"id":"3c7de278-bba7-4a56-b61e-056f3c9986f4","cardType":"cloze","points":null,"front":"When light enters a medium with a higher refractive index, its speed [ ] and its wavelength decreases.","back":"When light enters a medium with a higher refractive index, its speed decreases and its wavelength decreases.","mcqOptions":[{"text":"decreases","correct":true,"explanation":"A higher refractive index means a lower speed because $n = c/v$. Since frequency remains constant, the reduced speed also produces a shorter wavelength using $v = f\\lambda$."},{"text":"increases","correct":false,"explanation":"This reverses the relationship between refractive index and speed. From $n = c/v$, increasing refractive index means speed decreases, so the correct term is “decreases”."},{"text":"remains constant","correct":false,"explanation":"Speed does not remain constant when light enters a medium with a different refractive index. The higher index corresponds to a lower speed, so the correct term is “decreases”."},{"text":"oscillates","correct":false,"explanation":"The speed changes to a new value rather than repeatedly varying in time as it crosses the boundary. A higher refractive index causes speed to decrease."}]},{"id":"bed9c597-284a-4a21-b432-53e9d70dbff8","cardType":"cloze","points":null,"front":"When light enters a medium with a higher refractive index, its speed decreases and its wavelength [ ].","back":"When light enters a medium with a higher refractive index, its speed decreases and its wavelength decreases.","mcqOptions":[{"text":"decreases","correct":true,"explanation":"Frequency remains constant at the boundary, while a higher refractive index reduces the speed. From $v = f\\lambda$, a lower speed with unchanged frequency means the wavelength decreases."},{"text":"increases","correct":false,"explanation":"This would require the speed to increase while frequency stayed constant, contradicting $v = f\\lambda$. A higher refractive index reduces speed, so wavelength also decreases."},{"text":"remains constant","correct":false,"explanation":"Wavelength is not preserved when light enters a different medium. Because frequency stays constant but speed decreases, the wavelength must decrease as well."},{"text":"becomes zero","correct":false,"explanation":"The wavelength becomes shorter but does not disappear; light continues to propagate through the medium. The correct relationship is a decrease in wavelength, not a zero value."}]},{"id":"2653c4cd-6cfb-45f4-a446-26681a89e446","cardType":"basic","points":1,"front":"State Snell's Law.","back":"$$n_1 \\sin \\theta_1 = n_2 \\sin \\theta_2$","mcqOptions":[{"text":"$$n_1 \\sin \\theta_1 = n_2 \\sin \\theta_2$","correct":true,"explanation":"Snell’s Law relates the refractive indices and the angles measured from the normal on each side of a boundary. Their products with the corresponding sines are equal."},{"text":"$$n_1 \\cos \\theta_1 = n_2 \\cos \\theta_2$","correct":false,"explanation":"Using cosine instead of sine is a common error caused by measuring angles from the boundary rather than the normal. Snell’s Law uses the sines of both angles."},{"text":"$$n_1 \\sin \\theta_2 = n_2 \\sin \\theta_1$","correct":false,"explanation":"The angles have been paired with the wrong media, reversing the correspondence between each refractive index and its angle. Each index must multiply the sine of the angle in that medium."},{"text":"$$n_1 \\theta_1 = n_2 \\theta_2$","correct":false,"explanation":"Snell’s Law is not valid with angles alone because the relationship is trigonometric, especially at larger angles. The correct equation uses $\\sin \\theta_1$ and $\\sin \\theta_2$."}]},{"id":"909c63de-d276-4ef5-882b-fd6800977521","cardType":"basic","points":1,"front":"How does light bend when entering a medium with a larger refractive index?","back":"It bends towards the normal.","mcqOptions":[{"text":"It bends towards the normal.","correct":true,"explanation":"Entering a medium with a larger refractive index makes light travel more slowly, so its direction changes towards the normal at the boundary."},{"text":"It bends away from the normal.","correct":false,"explanation":"This reverses the effect of increasing refractive index; light bends away from the normal when it enters a medium with a lower refractive index."},{"text":"It continues parallel to the normal.","correct":false,"explanation":"No change in direction would require the light to enter normally or encounter no change in refractive index. A larger refractive index generally causes bending towards the normal."},{"text":"It bends along the boundary.","correct":false,"explanation":"Light does not usually travel along the boundary after entering the new medium. Increasing refractive index slows it and bends its path towards the normal."}]},{"id":"6ee2069a-d659-4991-b4c1-98f61341230f","cardType":"basic","points":2,"front":"Explain refraction using the wave model.","back":"

One side of the wavefront enters the new medium and changes speed first.
This causes the wavefront to pivot.

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

One side of the wavefront enters the new medium and changes speed first.
This causes the wavefront to pivot.

","correct":true,"explanation":"The first part of the wavefront changes speed while the rest continues at its original speed, causing the wavefront to rotate and change the direction of travel."},{"text":"

The entire wavefront enters the new medium and changes speed simultaneously.
This causes the wavefront to remain straight.

","correct":false,"explanation":"If the whole wavefront changed speed simultaneously, there would be no rotation. Refraction occurs because one side changes speed before the other, making the wavefront pivot."},{"text":"

One side of the wavefront enters the new medium and changes frequency first.
This causes the wavefront to pivot.

","correct":false,"explanation":"Frequency remains constant when light crosses a boundary; its speed and wavelength change instead. The speed difference across the wavefront causes the pivot and refraction."},{"text":"

One side of the ray enters the new medium and changes speed first.
This causes the ray to pivot.

","correct":false,"explanation":"Refraction is explained by the wavefront rotating, not by one side of a ray pivoting. One side of the wavefront changes speed first, producing the change in direction."}]},{"id":"8b5b49f2-9630-464f-b62e-19b6602397b1","cardType":"basic","points":2,"front":"In a practical to find refractive index, what should be plotted on the $y$ and $x$ axes to make the gradient the refractive index?","back":"

y-axis: $\\sin \\theta_{\\text{incident}}$
x-axis: $\\sin \\theta_{\\text{refracted}}$

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

y-axis: $\\sin \\theta_{\\text{incident}}$
x-axis: $\\sin \\theta_{\\text{refracted}}$

","correct":true,"explanation":"Snell’s law gives $n_1\\sin\\theta_{\\text{incident}}=n_2\\sin\\theta_{\\text{refracted}}$. For light entering air from the material, plotting incident sine against refracted sine gives gradient equal to the refractive index."},{"text":"

y-axis: $\\sin \\theta_{\\text{refracted}}$
x-axis: $\\sin \\theta_{\\text{incident}}$

","correct":false,"explanation":"These axes reverse the gradient, giving the reciprocal relationship rather than the refractive index. The incident sine must be plotted vertically against the refracted sine horizontally."},{"text":"

y-axis: $\\theta_{\\text{incident}}$
x-axis: $\\theta_{\\text{refracted}}$

","correct":false,"explanation":"Using the angles themselves ignores the sine relationship in Snell’s law, so the graph is not generally linear with the required gradient. The sine of each angle must be used."},{"text":"

y-axis: $\\sin \\theta_{\\text{incident}}$
x-axis: $\\theta_{\\text{refracted}}$

","correct":false,"explanation":"Only one axis using a sine gives the wrong relationship because Snell’s law involves the sine of both angles. Both plotted quantities must be sines."}]},{"id":"9320364c-e9e2-4915-8a8c-0b0983c8cf5c","cardType":"basic","points":1,"front":"Why should very small incident angles be avoided in refraction experiments?","back":"The percentage uncertainty in the angle measurement becomes too large.","mcqOptions":[{"text":"The percentage uncertainty in the angle measurement becomes too large.","correct":true,"explanation":"A fixed absolute uncertainty in measuring a small angle represents a large fraction of that angle, so the percentage uncertainty becomes too large and reduces the reliability of the result."},{"text":"The absolute uncertainty in the angle measurement becomes zero.","correct":false,"explanation":"The measuring instrument still has a finite absolute uncertainty, regardless of the angle. Small angles should be avoided because that fixed uncertainty becomes a large percentage of the measurement."},{"text":"The refractive index of the material becomes too small.","correct":false,"explanation":"The material’s refractive index does not depend on the chosen incident angle. Small angles are avoided because measurement uncertainty becomes large relative to the angle."},{"text":"The angle of refraction becomes impossible to measure.","correct":false,"explanation":"The refracted angle can still be measured at small incidence angles. The problem is that the fixed measurement uncertainty forms too large a percentage of the small angle."}]},{"id":"0796b019-48ca-4f56-9d1c-e03b282d86ec","cardType":"basic","points":2,"front":"What are the two conditions required for total internal reflection (TIR) to occur?","back":"

Light travels from a medium with higher $n$ to one with lower $n$.
The angle of incidence is greater than the critical angle.

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

Light travels from a medium with higher $n$ to one with lower $n$.
The angle of incidence is greater than the critical angle.

","correct":true,"explanation":"Total internal reflection is possible only when light attempts to leave a higher-index medium for a lower-index medium and the incidence angle exceeds the critical angle."},{"text":"

Light travels from a medium with lower $n$ to one with higher $n$.
The angle of incidence is greater than the critical angle.

","correct":false,"explanation":"TIR cannot occur when light travels into a higher-index medium because refraction remains possible at every incidence angle. The light must move from higher $n$ to lower $n$."},{"text":"

Light travels from a medium with higher $n$ to one with lower $n$.
The angle of incidence is less than the critical angle.

","correct":false,"explanation":"Below the critical angle, a refracted ray still leaves the material. TIR begins only when light travels towards lower $n$ and the incidence angle is greater than the critical angle."},{"text":"

Light travels from a medium with lower $n$ to one with higher $n$.
The angle of incidence is less than the critical angle.

","correct":false,"explanation":"Both conditions are reversed here. TIR requires travel from higher to lower refractive index and an incidence angle greater than the critical angle."}]},{"id":"4215ea61-ce88-400e-b01d-de82b7c3ffee","cardType":"basic","points":1,"front":"Define the critical angle.","back":"The angle of incidence for which the angle of refraction is 90 degrees.","mcqOptions":[{"text":"The angle of incidence for which the angle of refraction is 90 degrees.","correct":true,"explanation":"At the critical angle, the refracted ray travels along the boundary, making an angle of 90 degrees to the normal. Any larger incidence angle produces total internal reflection."},{"text":"The angle of incidence for which the angle of refraction is 0 degrees.","correct":false,"explanation":"A refracted angle of 0 degrees means the ray travels along the normal, not along the boundary. The critical angle is defined by refraction at 90 degrees."},{"text":"The angle of refraction for which the angle of incidence is 90 degrees.","correct":false,"explanation":"The critical angle is measured as an angle of incidence, not an angle of refraction. It is the incidence angle that produces a refracted angle of 90 degrees."},{"text":"The angle of incidence for which the angle of refraction is 45 degrees.","correct":false,"explanation":"A 45-degree refracted angle is not the defining condition. The critical angle occurs when the refracted ray is exactly along the boundary, at 90 degrees to the normal."}]},{"id":"2aa1babd-0bd5-4e2d-b9b6-20e5cbfb1cd6","cardType":"cloze","points":null,"front":"The critical angle $\\theta_c$ can be calculated using the equation $$\\text{[...]}$$.","back":"The critical angle $\\theta_c$ can be calculated using the equation $$\\boxed{\\sin \\theta_c = \\frac{n_2}{n_1}}$$.","mcqOptions":[{"text":"$$$\\sin \\theta_c = \\frac{n_2}{n_1}$$","correct":true,"explanation":"At the critical angle, the refracted angle is 90 degrees, so Snell’s law becomes $n_1\\sin\\theta_c=n_2$. Rearranging gives $\\sin\\theta_c=\\frac{n_2}{n_1}$."},{"text":"$$$\\sin \\theta_c = \\frac{n_1}{n_2}$$","correct":false,"explanation":"This reverses the refractive-index ratio in Snell’s law. For light travelling from higher $n_1$ to lower $n_2$, the critical-angle relation is $\\sin\\theta_c=\\frac{n_2}{n_1}$."},{"text":"$$$\\cos \\theta_c = \\frac{n_2}{n_1}$$","correct":false,"explanation":"The critical-angle equation comes from Snell’s law, which uses sines of the angles. Replacing sine with cosine gives the wrong relationship and critical angle."},{"text":"$$$\\sin \\theta_c = n_1 - n_2$$","correct":false,"explanation":"Refractive indices enter the critical-angle equation as a ratio, not a difference. Applying Snell’s law at a 90-degree refracted angle gives $\\sin\\theta_c=\\frac{n_2}{n_1}$."}]},{"id":"8e3ab823-7ffe-4963-a578-492bc52c1308","cardType":"basic","points":1,"front":"How do the refractive indices of the core and cladding in an optical fibre compare?","back":"The core has a slightly higher refractive index than the cladding.","mcqOptions":[{"text":"The core has a slightly higher refractive index than the cladding.","correct":true,"explanation":"The higher refractive index of the core allows light striking the core-cladding boundary at suitable angles to undergo total internal reflection and remain guided through the fibre."},{"text":"The core has a slightly lower refractive index than the cladding.","correct":false,"explanation":"This would prevent the core-cladding boundary from supporting total internal reflection for light travelling in the core. The core must have a slightly higher refractive index."},{"text":"The core and cladding have exactly the same refractive index.","correct":false,"explanation":"Equal refractive indices would remove the refractive-index boundary needed for total internal reflection. Guidance requires the core’s refractive index to be slightly higher."},{"text":"The cladding has a much higher refractive index than the core.","correct":false,"explanation":"A higher-index cladding would not provide the required higher-to-lower index transition at the boundary. The core is slightly higher in refractive index than the cladding."}]},{"id":"6a636472-7c4d-459a-adbe-33f8e9216d5f","cardType":"basic","points":2,"front":"What is multimode dispersion?","back":"Pulse spreading caused by different ray paths (modes) taking different amounts of time to travel through the fibre.","mcqOptions":[{"text":"Pulse spreading caused by different ray paths (modes) taking different amounts of time to travel through the fibre.","correct":true,"explanation":"Different modes follow different ray paths and therefore take different times, spreading an input pulse as it travels through the fibre."},{"text":"Pulse shortening caused by different ray paths (modes) taking different amounts of time to travel through the fibre.","correct":false,"explanation":"Different travel times make pulses spread rather than shorten them. The spreading results specifically from distinct modes taking different times along their paths."},{"text":"Pulse spreading caused by different ray paths (modes) taking the same amount of time to travel through the fibre.","correct":false,"explanation":"Multimode dispersion involves different modes taking different times, not equal travel times. Equal times would prevent this particular pulse-spreading effect."},{"text":"Pulse spreading caused by different wavelengths taking different amounts of time to travel through the fibre.","correct":false,"explanation":"Different wavelengths can cause chromatic dispersion, but multimode dispersion is caused by different ray paths taking different travel times."}]},{"id":"a7b50c41-caae-4576-87fa-98e056169c8e","cardType":"basic","points":2,"front":"What is the consequence of pulses overlapping due to dispersion in optical fibres?","back":"The detector cannot distinguish separate bits of data, limiting the transmission rate.","mcqOptions":[{"text":"The detector cannot distinguish separate bits of data, limiting the transmission rate.","correct":true,"explanation":"Overlapping pulses cannot be reliably separated by the detector, so individual data bits are confused and the maximum transmission rate falls."},{"text":"The detector can distinguish separate bits of data, increasing the transmission rate.","correct":false,"explanation":"Dispersion causes neighbouring bits to overlap, making them harder to distinguish, so it limits rather than increases the transmission rate."},{"text":"The detector cannot distinguish combined bits of data, limiting the signal strength.","correct":false,"explanation":"The detector receives the pulses, but their overlap prevents reliable separation of individual bits. The resulting limit is on transmission rate, not signal strength."},{"text":"The detector cannot distinguish separate bits of data, preventing the signal from travelling.","correct":false,"explanation":"Pulse overlap affects the detector’s ability to resolve separate bits and therefore restricts data rate; it does not directly prevent the signal from travelling."}]},{"id":"83cb9a2b-6bff-4abf-bfa2-436bfa267eb0","cardType":"basic","points":1,"front":"How does a monomode fibre reduce multimode dispersion?","back":"The core is very narrow, so light essentially follows only one path.","mcqOptions":[{"text":"The core is very narrow, so light essentially follows only one path.","correct":true,"explanation":"A very narrow core supports essentially one propagation path, so different ray paths cannot produce significant multimode pulse spreading."},{"text":"The core is very wide, so light essentially follows only one path.","correct":false,"explanation":"A wide core permits many modes and ray paths, increasing multimode dispersion rather than reducing it. The core must be very narrow."},{"text":"The core is very narrow, so light follows many different paths.","correct":false,"explanation":"A narrow core is correct, but allowing many paths would recreate the differing travel times responsible for multimode dispersion."},{"text":"The core is very wide, so light follows many different paths.","correct":false,"explanation":"A wide core generally supports multiple modes, so it does not force light into one path. Monomode behaviour requires a very narrow core."}]},{"id":"0af1b0e8-64e2-4818-a086-4aebf1009152","cardType":"cloze","points":null,"front":"Monomode fibres allow higher data rates because pulses stay [ ] over longer distances.","back":"Monomode fibres allow higher data rates because pulses stay narrower over longer distances.","mcqOptions":[{"text":"narrower","correct":true,"explanation":"With only one main propagation path, pulses undergo less modal spreading and remain narrower, allowing adjacent bits to be resolved over greater distances."},{"text":"wider","correct":false,"explanation":"Dispersion makes pulses broaden as they travel. Monomode fibres reduce this broadening, so the pulses remain narrower rather than wider."},{"text":"closer","correct":false,"explanation":"The key improvement is reduced pulse spreading, not reduced distance between pulses. Therefore the pulses remain narrower over longer distances."},{"text":"fainter","correct":false,"explanation":"Monomode operation reduces modal dispersion and preserves pulse shape. It does not make pulses disappear or become less detectable."}]},{"id":"c521d1d4-916d-40ba-bdd1-3e30b5cc48ac","cardType":"cloze","points":null,"front":"Monomode fibres allow higher data rates because pulses stay narrower over [ ] distances.","back":"Monomode fibres allow higher data rates because pulses stay narrower over longer distances.","mcqOptions":[{"text":"longer","correct":true,"explanation":"Reduced modal dispersion means the pulses broaden less, so they remain narrower while travelling farther before overlapping neighbouring bits."},{"text":"shorter","correct":false,"explanation":"The advantage of monomode fibres is maintaining pulse separation over a greater distance. Shorter distances would not explain their higher data rates."},{"text":"narrower","correct":false,"explanation":"Pulse width is the feature reduced by lower dispersion; the distance over which pulses remain sufficiently separated is longer, not narrower."},{"text":"nearer","correct":false,"explanation":"Monomode fibres preserve pulses over greater distances because only one main path is used. Higher data rates are not explained by nearer distances."}]},{"id":"0957d3e9-10ea-4eb7-a25c-f45f6ca8a33b","cardType":"cloze","points":null,"front":"When using a laser for refraction experiments, the beam should be kept [ ] and directed away from people.","back":"When using a laser for refraction experiments, the beam should be kept low and directed away from people.","mcqOptions":[{"text":"low","correct":true,"explanation":"Keeping the laser beam low reduces the chance of eye exposure while still allowing the refraction experiment to be performed safely."},{"text":"high","correct":false,"explanation":"A high beam is more likely to enter someone’s eyes, increasing the hazard. Laser beams should be kept low and directed away from people."},{"text":"narrow","correct":false,"explanation":"A narrow beam describes its width, not its safe vertical position. The recommended precaution is to keep the beam low."},{"text":"bright","correct":false,"explanation":"The beam’s brightness does not state where it should be aimed. Keeping it low helps prevent accidental eye exposure."}]},{"id":"cece668e-98d0-4634-905b-a534bb9c444f","cardType":"cloze","points":null,"front":"When using a laser for refraction experiments, the beam should be kept low and directed [ ].","back":"When using a laser for refraction experiments, the beam should be kept low and directed away from people.","mcqOptions":[{"text":"away from people","correct":true,"explanation":"Directing the beam away from people reduces the risk of laser light entering someone’s eyes, which can cause serious eye damage."},{"text":"towards people","correct":false,"explanation":"Aiming the beam towards people creates a direct eye-exposure hazard. Safe practice is to direct it away from people."},{"text":"across the room","correct":false,"explanation":"A sideways direction does not guarantee that people are outside the beam. The beam must specifically be directed away from people."},{"text":"at the observer","correct":false,"explanation":"Pointing the beam at the observer increases the chance of eye exposure. It should be directed away from people during the experiment."}]}]}],"examCountdown":{"examDate":"2027-05-21","paperLabel":"Component 1"},"heading":"Refraction of Light","title":"Refraction of Light Flashcards","topbar":{"backHref":"/a-level/physics/eduqas/flashcards"},"topicSidebar":"$L38"}]}],"topicId":"71d27fc7-3da9-5330-a2ba-ca2b3ba8f747","walled":"$L39"}],"$L3a","$L3b"]