HSC Physics · Module 7 flashcards

The Nature of Light

84 cards. Work out each answer first, then open the card to check it and see the working.

  1. Maxwell's equations predicted that electromagnetic waves travel through a vacuum at a speed determined by …Show answer

    Answer: the electric and magnetic constants of free space

    Maxwell showed c = 1/√(μ0ε0), which gave a value matching the measured speed of light.

    From Practice Exam 1, question 11 →

  2. A metal has a work function of 2.3 eV. What is its threshold frequency?Show answer

    Answer: 5.6 × 1014 Hz

    f0 = W/h = (2.3 × 1.602 × 10-19)/(6.626 × 10-34) = 5.6 × 1014 Hz.

    From Practice Exam 1, question 13 →

  3. A muon at rest has a lifetime of 2.2 μs. What lifetime would an observer measure for a muon moving at 0.98c?Show answer

    Answer: 11 μs

    γ = 1/√(1 − 0.98²) = 5.03. t = γt0 = 5.03 × 2.2 = 11 μs.

    From Practice Exam 1, question 14 →

  4. Hertz's experiments with radio waves were important because they showed that the waves …Show answer

    Answer: could be reflected, refracted and polarised like light

    Hertz generated and detected radio waves and showed they had the same wave properties as light, confirming Maxwell's prediction of electromagnetic waves.

    From Practice Exam 2, question 13 →

  5. A spaceship is 100 m long in its own rest frame. It passes Earth at 0.60c. What length does an observer on Earth measure?Show answer

    Answer: 80 m

    γ = 1/√(1 − 0.36) = 1.25. L = L0/γ = 100/1.25 = 80 m.

    From Practice Exam 2, question 16 →

  6. The null result of the Michelson–Morley experiment …Show answer

    Answer: provided no evidence for the existence of the aether

    The experiment looked for a change in the speed of light due to Earth's motion through the aether. No shift was found, so no aether motion was detected.

    From Practice Exam 2, question 17 →

  7. Rømer's estimate of the speed of light was based on …Show answer

    Answer: timing variations in the eclipses of Jupiter's moon Io

    Rømer noticed Io's eclipses appeared late when Earth was far from Jupiter and early when near, and attributed the difference to light's travel time across Earth's orbit.

    From Practice Exam 3, question 13 →

  8. A diffraction grating has 500 lines per mm. At what angle does the second-order maximum of 600 nm light appear?Show answer

    Answer: 36.9°

    d = 1/500 000 = 2.0 × 10-6 m. sinθ = mλ/d = 2 × 6.0 × 10-7/2.0 × 10-6 = 0.60, so θ = 36.9°.

    From Practice Exam 3, question 15 →

  9. Light of frequency 8.0 × 1014 Hz falls on a metal of work function 2.0 eV. What is the maximum kinetic energy of the photoelectrons?Show answer

    Answer: 1.3 eV

    Ephoton = hf = 6.626 × 10-34 × 8.0 × 1014 = 5.3 × 10-19 J = 3.3 eV. KEmax = 3.3 − 2.0 = 1.3 eV.

    From Practice Exam 3, question 16 →

  10. A spacecraft travels at 0.80c. The crew measure the journey as taking 10 years. How long does the journey take according to observers on Earth?Show answer

    Answer: 16.7 years

    γ = 1/√(1 − 0.64) = 1.67. The crew measure proper time, so Earth measures t = γt0 = 16.7 years.

    From Practice Exam 3, question 17 →

  11. What is the energy of a photon of wavelength 400 nm?Show answer

    Answer: 5.0 × 10-19 J

    E = hc/λ = 6.626 × 10-34 × 3.0 × 108/4.0 × 10-7 = 5.0 × 10-19 J (about 3.1 eV).

    From Practice Exam 4, question 14 →

  12. In a photoelectric experiment the stopping voltage is 1.5 V. What is the maximum kinetic energy of the photoelectrons?Show answer

    Answer: 2.4 × 10-19 J

    KEmax = eVstop = 1.602 × 10-19 × 1.5 = 2.4 × 10-19 J (that is, 1.5 eV).

    From Practice Exam 4, question 16 →

  13. A particle moves at 0.60c. By what factor is its relativistic momentum greater than the classical value mv?Show answer

    Answer: 1.25

    p = γmv. γ = 1/√(1 − 0.36) = 1.25.

    From Practice Exam 4, question 17 →

  14. Fizeau measured the speed of light using …Show answer

    Answer: a rotating toothed wheel and a distant mirror

    Light passed through a gap in a spinning wheel, reflected from a mirror kilometres away, and was blocked by the next tooth at a particular rotation rate, giving the travel time.

    From Practice Exam 5, question 13 →

  15. A hot dense solid such as a tungsten filament produces …Show answer

    Answer: a continuous spectrum

    Closely packed atoms in a hot solid interact so strongly that all wavelengths are emitted. Line spectra come from low-pressure gases.

    From Practice Exam 5, question 15 →

  16. According to special relativity, two events that are simultaneous in one inertial frame …Show answer

    Answer: may not be simultaneous in a frame moving relative to the first

    Because c is the same for all observers, observers in relative motion can disagree about whether spatially separated events occurred at the same time.

    From Practice Exam 5, question 17 →

  17. Maxwell's main contribution to the understanding of light was to …Show answer

    Answer: unify electricity, magnetism and light as electromagnetic waves

    Maxwell's equations predicted self-propagating electric and magnetic waves at a speed equal to that of light, identifying light as an electromagnetic wave.

    From Practice Exam 6, question 13 →

  18. The rotation of a star can be detected from its spectrum because …Show answer

    Answer: the lines are broadened as one edge approaches and the other recedes

    Light from the approaching limb is blue-shifted and from the receding limb red-shifted, smearing each line over a range of wavelengths.

    From Practice Exam 6, question 14 →

  19. In a double-slit experiment, the slit separation is increased while wavelength and screen distance are unchanged. The fringe spacing …Show answer

    Answer: decreases

    Δy = λL/d. Larger d gives closer fringes.

    From Practice Exam 6, question 15 →

  20. The Sun converts 4.3 × 109 kg of mass into energy each second. What is its power output?Show answer

    Answer: 3.9 × 1026 W

    P = (Δm/Δt)c² = 4.3 × 109 × 9.0 × 1016 = 3.9 × 1026 W.

    From Practice Exam 6, question 17 →

  21. In Hertz's experiment, a spark jumped across the gap of a distant loop of wire. This demonstrated that …Show answer

    Answer: electromagnetic waves had travelled from the transmitter to the receiver

    The receiving loop was not connected to the transmitter. The spark showed that an oscillating electromagnetic wave had induced a current in it.

    From Practice Exam 7, question 13 →

  22. A hot gas at low pressure produces …Show answer

    Answer: an emission spectrum of discrete bright lines

    Isolated atoms emit only the wavelengths matching their energy level differences.

    From Practice Exam 7, question 14 →

  23. Light can be polarised. This shows that light is …Show answer

    Answer: a transverse wave

    Only transverse waves have an oscillation direction perpendicular to travel that can be filtered by a polariser.

    From Practice Exam 7, question 15 →

  24. A metal has a work function of 4.5 eV. What is the longest wavelength that will eject electrons? (hc = 1240 eV nm) …Show answer

    Answer: 276 nm

    λmax = hc/W = 1240/4.5 = 276 nm.

    From Practice Exam 7, question 16 →

  25. The proper time between two events is measured by …Show answer

    Answer: an observer for whom the events happen at the same place

    Proper time is the shortest time interval, measured by the clock that is present at both events. Every other observer measures a longer, dilated interval.

    From Practice Exam 7, question 17 →

  26. In an electromagnetic wave, the electric and magnetic fields are …Show answer

    Answer: perpendicular to each other and both perpendicular to the direction of travel

    E, B and the direction of propagation are mutually perpendicular, and E and B oscillate in phase.

    From Practice Exam 8, question 13 →

  27. Compared with a double slit, a diffraction grating with many slits produces …Show answer

    Answer: sharper, brighter maxima

    More slits means more waves adding constructively at the maxima and cancelling more completely between them, giving narrow bright lines.

    From Practice Exam 8, question 15 →

  28. The maximum kinetic energy of photoelectrons depends on …Show answer

    Answer: the frequency of the light and the work function of the metal

    KEmax = hf − W. Intensity affects how many electrons leave, not their energy.

    From Practice Exam 8, question 16 →

  29. The proper length of an object is measured …Show answer

    Answer: in the frame in which the object is at rest

    Proper length is the longest length. Any observer moving relative to the object measures a contracted length L = L0/γ.

    From Practice Exam 8, question 17 →

  30. A star's spectral lines are blue-shifted. The star is …Show answer

    Answer: moving towards Earth

    A shift to shorter wavelengths indicates a source approaching the observer.

    From Practice Exam 9, question 14 →

  31. In a double-slit experiment, d = 0.50 mm, the screen is 2.0 m away and bright fringes are 2.4 mm apart. The wavelength is …Show answer

    Answer: 600 nm

    λ = dΔy/L = 5.0 × 10-4 × 2.4 × 10-3/2.0 = 6.0 × 10-7 m = 600 nm.

    From Practice Exam 9, question 15 →

  32. A 2.0 W laser emits light of wavelength 500 nm. How many photons does it emit per second?Show answer

    Answer: 5.0 × 1018

    Ephoton = hc/λ = 3.97 × 10-19 J. N = 2.0/3.97 × 10-19 = 5.0 × 1018 per second.

    From Practice Exam 9, question 16 →

  33. Why can no object with mass reach the speed of light?Show answer

    Answer: The energy required becomes infinite as v approaches c

    γ → ∞ as v → c, so the kinetic energy (γ − 1)mc² and the required work become unbounded.

    From Practice Exam 9, question 17 →

  34. Hertz measured the speed of the radio waves he produced by …Show answer

    Answer: setting up standing waves and measuring their wavelength, then using v = fλ

    Reflecting the waves created standing waves. The node spacing gave the wavelength, and the known oscillator frequency gave the speed, which equalled c.

    From Practice Exam 10, question 13 →

  35. A black body at 10 000 K has its peak emission at about (b = 2.898 × 10-3 m K) …Show answer

    Answer: 290 nm

    λmax = b/T = 2.898 × 10-3/10 000 = 2.9 × 10-7 m = 290 nm (ultraviolet).

    From Practice Exam 10, question 14 →

  36. Plane-polarised light of intensity I0 passes through an analyser whose axis is at 45° to the polarisation direction. The transmitted intensity is …Show answer

    Answer: I0/2

    Malus' law: I = I0 cos²45° = I0 × 0.5. (For unpolarised light there would be an extra factor of ½.)

    From Practice Exam 10, question 15 →

  37. In the twin paradox, one twin travels to a distant star at high speed and returns. On return …Show answer

    Answer: the travelling twin is younger

    The travelling twin's clock runs slow relative to Earth. The situation is not symmetric because the traveller changes frames when turning around.

    From Practice Exam 10, question 17 →

  38. The dark Fraunhofer lines in the Sun's spectrum reveal …Show answer

    Answer: the elements present in the Sun's outer atmosphere

    Each element absorbs at its own set of wavelengths, so the pattern of dark lines identifies which elements are present in the cooler outer layers.

    From Practice Exam 11, question 13 →

  39. The SI unit of Planck's constant is …Show answer

    Answer: J s

    E = hf, so h = E/f has units J/Hz = J s.

    From Practice Exam 11, question 14 →

  40. The condition for a bright fringe in a double-slit pattern is …Show answer

    Answer: d sinθ = mλ

    Constructive interference needs a path difference d sinθ equal to a whole number of wavelengths.

    From Practice Exam 11, question 15 →

  41. Light of 300 nm falls on a metal of work function 2.3 eV. The stopping voltage is about (hc = 1240 eV nm) …Show answer

    Answer: 1.8 V

    Ephoton = 1240/300 = 4.13 eV. KEmax = 4.13 − 2.3 = 1.8 eV, so Vstop = 1.8 V.

    From Practice Exam 11, question 16 →

  42. A spacecraft travels at 0.80c to a star 8.0 light-years away (Earth frame). According to the crew the trip takes …Show answer

    Answer: 6.0 years

    Earth frame: t = 8.0/0.80 = 10 years. γ = 1.67, so crew time = 10/1.67 = 6.0 years. (Equivalently the distance contracts to 4.8 ly.)

    From Practice Exam 11, question 17 →

  43. Using the measured values of μ0 and ε0, Maxwell calculated the speed of electromagnetic waves to be about …Show answer

    Answer: 3.0 × 108 m s-1

    c = 1/√(μ0ε0) ≈ 3.0 × 108 m s-1, matching the known speed of light.

    From Practice Exam 12, question 13 →

  44. A hydrogen line with rest wavelength 656.3 nm is observed at 656.5 nm from a star. The star's radial speed is about …Show answer

    Answer: 9 × 104 m s-1

    v = c Δλ/λ = 3.0 × 108 × 0.2/656.3 = 9.1 × 104 m s-1, receding (red shift).

    From Practice Exam 12, question 14 →

  45. A diffraction grating is replaced with one having more lines per millimetre. For the same light, the first-order maximum appears at …Show answer

    Answer: a larger angle

    More lines per mm means smaller slit spacing d. sinθ = λ/d increases.

    From Practice Exam 12, question 15 →

  46. A photon has energy 2.0 eV. Its wavelength is about (hc = 1240 eV nm) …Show answer

    Answer: 620 nm

    λ = hc/E = 1240/2.0 = 620 nm (red).

    From Practice Exam 12, question 16 →

  47. An observer on Earth sees a spaceship's clock running slow. An observer on the spaceship sees Earth's clock …Show answer

    Answer: running slow

    Time dilation is symmetric between inertial frames: each sees the other's clock as slow, because each sees the other as moving.

    From Practice Exam 12, question 17 →

  48. The radial velocity of a star is determined from its spectrum by measuring …Show answer

    Answer: the Doppler shift of the lines

    The amount of shift Δλ/λ = v/c gives the speed towards or away from the observer.

    From Practice Exam 13, question 13 →

  49. Classical physics predicted that a black body should emit unlimited energy at short wavelengths. This failed prediction is called …Show answer

    Answer: the ultraviolet catastrophe

    Planck resolved it by proposing that energy is emitted in packets E = hf, which suppresses high-frequency emission.

    From Practice Exam 13, question 14 →

  50. In Young's experiment, a dark fringe occurs where the path difference from the two slits is …Show answer

    Answer: half a wavelength

    A half-wavelength path difference puts the waves out of phase so they cancel.

    From Practice Exam 13, question 15 →

  51. A 1.0 m rod moves at 0.9c in a direction perpendicular to its length. An observer at rest measures its length as …Show answer

    Answer: 1.0 m

    Length contraction occurs only along the direction of motion. Perpendicular dimensions are unchanged.

    From Practice Exam 13, question 17 →

  52. Hertz showed that his radio waves were polarised by …Show answer

    Answer: rotating the receiving loop and observing the spark strength change

    The spark in the detector depended on the orientation of the receiver relative to the transmitter, which is the signature of a polarised transverse wave.

    From Practice Exam 14, question 13 →

  53. A blue star compared with a red star is …Show answer

    Answer: hotter

    Wien's law: hotter objects peak at shorter wavelengths. Blue stars have surface temperatures above about 10 000 K; red stars below about 3500 K.

    From Practice Exam 14, question 14 →

  54. Unpolarised light passes through a single ideal polarising filter. The transmitted intensity is …Show answer

    Answer: one half

    Averaging cos²θ over all orientations gives ½. Malus' law applies only after the light is already polarised.

    From Practice Exam 14, question 15 →

  55. A muon has a proper lifetime of 2.2 μs and travels at 0.995c (γ = 10). What distance does it travel in the laboratory frame before decaying?Show answer

    Answer: 6600 m

    Lab lifetime = 10 × 2.2 μs = 22 μs. Distance = 0.995 × 3.0 × 108 × 22 × 10-6 = 6.6 × 10³ m.

    From Practice Exam 14, question 17 →

  56. Rømer's observations showed that the timing of Io's eclipses …Show answer

    Answer: varied with the Earth–Jupiter distance

    The eclipses lagged when Earth was further away because the light had further to travel. The lag gave a finite speed for light.

    From Practice Exam 15, question 13 →

  57. The visible emission lines of hydrogen are described by …Show answer

    Answer: the Balmer formula

    Balmer found an empirical formula for the visible lines; Rydberg generalised it and Bohr later derived it.

    From Practice Exam 15, question 14 →

  58. A grating with 300 lines per mm is illuminated with 650 nm light. The first-order maximum appears at …Show answer

    Answer: 11°

    d = 1/300 000 = 3.33 × 10-6 m. sinθ = 6.5 × 10-7/3.33 × 10-6 = 0.195, θ = 11°.

    From Practice Exam 15, question 15 →

  59. The energy equivalent of the electron's rest mass is about …Show answer

    Answer: 8.2 × 10-14 J (0.51 MeV)

    E = mc² = 9.109 × 10-31 × 9.0 × 1016 = 8.2 × 10-14 J ≈ 0.51 MeV.

    From Practice Exam 15, question 17 →

  60. According to Maxwell, a changing electric field …Show answer

    Answer: produces a magnetic field

    This is the key addition Maxwell made to Ampère's law. Together with Faraday's law it allows E and B to regenerate each other as a travelling wave.

    From Practice Exam 16, question 13 →

  61. A star's spectrum peaks at 250 nm. Its surface temperature is about (b = 2.898 × 10-3 m K) …Show answer

    Answer: 11 600 K

    T = b/λ = 2.898 × 10-3/2.5 × 10-7 = 11 600 K.

    From Practice Exam 16, question 14 →

  62. When white light is used in a double-slit experiment, the central fringe is …Show answer

    Answer: white, with coloured fringes either side

    All wavelengths have zero path difference at the centre, so they all add constructively. Away from centre, different wavelengths peak at different positions.

    From Practice Exam 16, question 15 →

  63. Light of 500 nm (2.48 eV) falls on a metal with work function 2.0 eV. The emitted electrons have maximum kinetic energy …Show answer

    Answer: 0.48 eV

    KEmax = 2.48 − 2.0 = 0.48 eV.

    From Practice Exam 16, question 16 →

  64. Einstein's two postulates of special relativity are that the laws of physics are the same in all inertial frames, and that …Show answer

    Answer: the speed of light in a vacuum is the same for all observers

    The constancy of c for all inertial observers, regardless of the motion of the source, leads directly to time dilation and length contraction.

    From Practice Exam 16, question 17 →

  65. Hertz detected waves across his laboratory. This showed that the waves …Show answer

    Answer: travelled through air at a finite speed, later measured as c

    No conductor linked transmitter and receiver. The waves propagated through space, as Maxwell's theory required.

    From Practice Exam 17, question 13 →

  66. A star's spectrum peaks at 700 nm. Its surface temperature is about (b = 2.898 × 10-3 m K) …Show answer

    Answer: 4100 K

    T = 2.898 × 10-3/7.0 × 10-7 = 4100 K — a cool orange-red star.

    From Practice Exam 17, question 14 →

  67. In a double-slit experiment the light source is changed from red to blue, all else unchanged. The fringes become …Show answer

    Answer: closer together

    Δy = λL/d. Blue light has a shorter wavelength, so the spacing decreases.

    From Practice Exam 17, question 15 →

  68. Red light of 700 nm (1.77 eV) shines on a metal with work function 2.3 eV. The result is …Show answer

    Answer: no electrons emitted, however bright the light

    Each photon has less energy than the work function. No single photon can free an electron, and intensity does not change the energy per photon.

    From Practice Exam 17, question 16 →

  69. A clock on a spaceship moving at 0.50c records one hour. An Earth observer measures the same interval as …Show answer

    Answer: 69 minutes

    γ = 1/√(1 − 0.25) = 1.155. t = 1.155 × 60 = 69 minutes.

    From Practice Exam 17, question 17 →

  70. Arranged in order of increasing frequency, the electromagnetic spectrum runs …Show answer

    Answer: radio, microwave, infrared, visible, ultraviolet, X-ray, gamma

    Radio has the longest wavelength and lowest frequency; gamma the shortest wavelength and highest frequency.

    From Practice Exam 18, question 13 →

  71. The Sun is a class G star with a surface temperature of about …Show answer

    Answer: 5800 K

    Its spectrum peaks near 500 nm, which by Wien's law corresponds to about 5800 K.

    From Practice Exam 18, question 14 →

  72. Plane-polarised light passes through an analyser at 30° to its polarisation direction. The fraction transmitted is …Show answer

    Answer: 0.75

    cos²30° = (0.866)² = 0.75.

    From Practice Exam 18, question 15 →

  73. The energy of a photon is …Show answer

    Answer: inversely proportional to its wavelength

    E = hf = hc/λ.

    From Practice Exam 18, question 16 →

  74. A spaceship is 50 m long at rest. Moving at 0.90c, its length as measured by a stationary observer is …Show answer

    Answer: 22 m

    γ = 1/√(1 − 0.81) = 2.29. L = 50/2.29 = 22 m.

    From Practice Exam 18, question 17 →

  75. The Michelson–Morley experiment was designed to detect …Show answer

    Answer: the motion of Earth through the aether via a shift in interference fringes

    If light moved relative to a stationary aether, rotating the apparatus should have shifted the fringes. No shift was seen.

    From Practice Exam 19, question 13 →

  76. An absorption spectrum appears as …Show answer

    Answer: dark lines on a continuous spectrum

    A cooler gas in front of a continuous source removes specific wavelengths, leaving gaps.

    From Practice Exam 19, question 14 →

  77. A diffraction grating has 600 lines per millimetre. Its slit spacing is …Show answer

    Answer: 1.67 × 10-6 m

    d = 1 mm/600 = 1.0 × 10-3/600 = 1.67 × 10-6 m.

    From Practice Exam 19, question 15 →

  78. In the photoelectric effect, the energy of an absorbed photon is shared between …Show answer

    Answer: the work function and the electron's kinetic energy

    hf = W + KEmax. The work function is the minimum energy to free the electron; the rest is kinetic energy.

    From Practice Exam 19, question 16 →

  79. The energy released when 2.0 g of matter is entirely converted to energy is …Show answer

    Answer: 1.8 × 1014 J

    E = mc² = 2.0 × 10-3 × 9.0 × 1016 = 1.8 × 1014 J.

    From Practice Exam 19, question 17 →

  80. Light passes through two polarisers whose axes are at 90° to each other. The transmitted intensity is …Show answer

    Answer: zero

    After the first polariser the light is polarised. Malus' law: I = I1 cos²90° = 0.

    From Practice Exam 20, question 13 →

  81. Planck's quantum hypothesis was introduced to explain …Show answer

    Answer: the shape of the black body spectrum

    Treating energy as emitted in multiples of hf gave a curve matching experiment and removed the ultraviolet catastrophe.

    From Practice Exam 20, question 14 →

  82. A black body at 3000 K has its peak emission at about (b = 2.898 × 10-3 m K) …Show answer

    Answer: 970 nm

    λmax = 2.898 × 10-3/3000 = 9.7 × 10-7 m = 970 nm (infrared).

    From Practice Exam 20, question 15 →

  83. Einstein's photoelectric equation is …Show answer

    Answer: KEmax = hf − W

    The electron gains hf from one photon and spends W escaping the metal; the rest is kinetic energy.

    From Practice Exam 20, question 16 →

  84. The Lorentz factor γ for an object moving at 0.99c is about …Show answer

    Answer: 7.1

    γ = 1/√(1 − 0.9801) = 1/√0.0199 = 7.1.

    From Practice Exam 20, question 17 →

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