AQA A Level Physics Waves Questions And Answers

Longitudinal wave - ANSWER-Particle vibration is parallel to direction of wave propagation Examples of a longitudinal wave - ANSWER-Sound waves, seismic p-waves Transverse wave - ANSWER-Particle vibration is perpendicular to direction of wave propagation Only transverse waves can be polarised Examples of a transverse wave - ANSWER-Electromagnetic radiation, seismic s-waves Particle displacement - ANSWER-The distance of a particle from its equilibrium position in given direction Amplitude - ANSWER-the maximum displacement of a particle (wave) from its equilibrium (or rest) position Frequency - ANSWER-Number of oscillations (of a particle) per second Time period - ANSWER-The time for one complete oscillation Wavelength (for non stationary wave) - ANSWER-Shortest distance between two points in phase wavelength for stationary wave - ANSWER-Distance between alternate nodes or distance from peak to peak/ trough to trough Diffraction - ANSWER-Spreading out of a wave (when it passes through a gap or past the edge of an object) Refraction - ANSWER-Wave bends/changes direction when its speed changes Polarisation - ANSWER-(transverse) wave oscillation is in one plane Application of polarisation in sunglasses - ANSWER-• Light reflected from surfaces is (weakly) polarised in one plane (horizontal) • Polaroid in sunglasses can be orientated to remove this reflected light • Reducing glare Application of polarisation in tv transmitters and aerials - ANSWER-• Signals from tv transmitter (radio waves) are polarised • Aerials need to be orientated (rotated) so they are in same plane as the transmitted signal • For maximum signal strength Superposition - ANSWER-Where two or more waves meet, the resultant displacement equals the vector sum of the individual displacements Conditions for formation of stationary waves - ANSWER-• Two waves travelling past each other in opposite directions • With the same wavelength (or frequency) • Similar amplitudes Monochromatic - ANSWER-Single wavelength Safety with a laser - ANSWER-• Avoid looking along the beam of a laser • Wear laser safety goggles • Avoid reflections • Put up a warning sign that a laser is in use Properties of laser light - ANSWER-• Monochromatic - only a single wavelength • Coherent - waves have a constant phase difference • Collimated - produces an approximately parallel beam Sketches of stationary waves for first 4 harmonics - ANSWER-Length of string = L 1st Harmonic λ = 2 L 2nd Harmonic λ = L 3rd Harmonic λ = 2/3 L 4th Harmonic λ = ½ L Coefficient of friction µ - ANSWER-µ = p A = mass/length = density x Area How does frequency change when very high tension is used on fixed string first harmonic experiment? - ANSWER-diameter of string reduces as Tension increases this means there is a lower mass per unit length so frequency is higher than expected Nodes and antinodes - ANSWER-Nodes - points of no oscillation / zero amplitude Antinodes - points of maximum amplitude Coherent sources - ANSWER-waves (from two sources) that have: • a constant phase difference • same wavelength (or frequency) Appearance of interference fringes from two vertical slit illuminated with yellow light - ANSWER-• Vertical or parallel • Equally spaced • Black and yellow bands Fringe width, w, changes - ANSWER-Slits closer together w - increases Screen further away w - increases Shorter wavelength (eg blue light) w - decreases Width of slits is reduced (Young's slits) - ANSWER-• More interference fringes (more of the pattern) are observed • Fringe spacing remains the same because • Greater amount of diffraction of light through each slit • So bigger region of overlap of the diffracted light from each slit • So more fringes can be observed Explanation of formation of fringes with Young's slits - ANSWER-• Interference fringes formed • Where light from two slits overlaps • The light from the two slits is coherent • Bright fringes formed where constructive interference • because light from the two slits is in phase (path difference equals a whole number of wavelengths) • Dark fringes formed where destructive interference • Because light from the two slits is in anti-phase (path difference equals a whole number + 0.5 wavelengths) Appearance of white light through Young's slits - ANSWER-• Central fringe would be white • Side fringes are (continuous) spectra • Bright fringe would be blue on the side nearest the central fringe. • Bright fringes merge further away from centre. appearance of diffraction pattern from a single slit - ANSWER-• Central bright fringe has twice width of other bright fringes • The other bright fringes have a much lower intensity • and are equally spaced Single slit pattern changes - ANSWER-Narrower slit width • Wider pattern / increased separation • Reduced intensity Shorter wavelength • Narrower pattern / reduced separation Lines per mm of a grating - ANSWER-Spacing, d, of slits on a diffraction grating given by: d = 1/(number of lines per mm) in mm Derivation of nλ = d sinθ - ANSWER-Between slits a and b: Path difference to nth maximum = nλ From trigonometry sin θ = nλ / d Hence nλ = d sinθ NOTE Path difference between light from slits a and c to nth maximum is nλ + nλ = 2nλ Applications of gratings to spectral analysis of light from stars - ANSWER-• Dark lines in spectrum from a star (absorption spectrum) • Reveal the composition of (elements present in) the star's atmosphere How does light change moving from air to glass - ANSWER-• speed - decreases (slows down) • wavelength - decreases (gets shorter) • frequency - remains constant (stays the same) Conditions for total internal reflection - ANSWER-• Angle of incidence is greater than the critical angle • The refractive index of the material light is going from is greater than the refractive index of the material the light is going to. Total internal reflection - ANSWER-Where all the light is reflected back into the material Critical angle - ANSWER-Angle of incidence which produces an angle of refraction of 90 degrees. Structure of an optical fibre - ANSWER-Central core, surrounded by cladding. Refractive index of core must be greater than refractive index of cladding (to ensure total internal reflection) Purpose of cladding - ANSWER-• prevents crossover of signal/data to other fibres • prevents scratching of the core • reduces pulse broadening/dispersion Use of optical fibres - ANSWER-• Communication - improve transmission of data/high speed internet • Endoscopes - improved medical diagnosis How do pulses of light change travelling down optical fibres - ANSWER-• reduced amplitude due to absorption/energy loss and scattering within fibre • pulse broadening due to multipath dispersion from rays taking different paths and different times to travel down same fibre How is multipath dispersion reduced - ANSWER-Core of fibre is made very narrow/thin.