IB Physics C.3: Wave phenomena worked solutions

The source sets the frequency, and a stationary boundary cannot change it. The speed falls to c/n, so \lambda=v/f falls by the same factor 1.50. B changes the frequency at the boundary; D makes the wavelength grow while the wave slows.

The heavy string barely moves, so the junction behaves almost like a fixed end: the reflection is inverted. The transmitted pulse is always on the same side as the incident pulse, smaller and slower. A and C invert the wrong pulse; D describes a free end with no second string.

The angle of incidence is measured from the normal: 90^\circ-50^\circ=40^\circ. Then \sin r=\sin40^\circ/1.50=0.429, r=25.4^\circ. B puts 50^\circ into Snell's law; C forgets to refract; D gives the refracted ray's angle to the surface.

(a) At the critical angle the refracted ray runs along the surface: n\sin48.0^\circ=1\times\sin90^\circ, so n=1/\sin48.0^\circ=1.35 (A1). (b) v=c/n=3.00\times10^{8}\times\sin48.0^\circ=2.23\times10^{8}\,\mathrm{m\,s^{-1}} (A1).

Spreading through a gap is diffraction; it is strongest when gap and wavelength are of similar size. A is the trap: short waves pass almost straight through.

At t=2.5\,\mathrm{s} pulse X covers 3.5 to 5.5\,\mathrm{cm} and pulse Y covers 4.5 to 6.5\,\mathrm{cm}. At x=4.0 only X is present: +2.0\,\mathrm{cm}. At x=5.0 both overlap: +2.0+(-1.0)=+1.0\,\mathrm{cm}. B adds sizes and ignores the sign.

One generator drives both speakers, so they are coherent (B is wrong) and P is a minimum. A minimum is silent only if the two amplitudes there are equal; P is nearer one speaker, so the cancellation is incomplete. A reverses the rule; D is false: sound cancels as well as any wave.

Path difference =35.5-31.0=4.5\,\mathrm{cm}=1.5\lambda, a half-odd number of wavelengths: the waves arrive in antiphase, so P is a minimum. D adds the two paths instead of subtracting them.

s=\lambda D/d, so the new spacing is (\tfrac12)(2)/(2)=\tfrac12 of s. D puts d on the top line; C assumes that doubling lengths always cancels.

Seven bright fringes enclose six spacings: s=12.6/6=2.10\,\mathrm{mm} (M1). \lambda=sd/D=(2.10\times10^{-3})(0.50\times10^{-3})/2.40=4.38\times10^{-7}\,\mathrm{m} (A1). Dividing by 7 gives 375\,\mathrm{nm} and loses the A1.

The first minimum is at \theta=\lambda/b, so a smaller b widens the central maximum. Less light now gets through and it is spread over a wider angle, so the peak falls (the centre amplitude is proportional to b, so halving b quarters the peak). A keeps the peak height fixed: the commonest wrong sketch.

The angles obey n\lambda=d\sin\theta, which does not contain the number of slits. Fewer slits means less light and less complete cancellation just off each maximum, so each maximum is broader and dimmer. A and B move the maxima; C describes adding slits.

(a) The ray is perpendicular to the wavefronts and the normal is perpendicular to the boundary; turning both lines by 90^\circ keeps the angle between them, so the ray meets the normal at 35^\circ (R1). (b) v=f\lambda: 12\times0.024=0.288\,\mathrm{m\,s^{-1}} and 12\times0.016=0.192\,\mathrm{m\,s^{-1}} (A1 each); the frequency is the same on both sides. (c) \sin\theta_2=\sin35^\circ\times(0.192/0.288)=0.382 (M1), \theta_2=22.5^\circ (A1). The wavefront angle to the boundary equals the ray angle to the normal.

(a) \sin\theta_c=1/1.52, \theta_c=41.1^\circ (A1). (b) The light travels from higher to lower refractive index and meets the long face at 45^\circ, which exceeds 41.1^\circ (R1). The reflected ray leaves at 45^\circ on the other side of the normal, so it is turned through 90^\circ (A1). (c) For glass to water \sin\theta_c=1.33/1.52, \theta_c=61.0^\circ (M1). Now 45^\circ<61.0^\circ, so some light refracts into the water (A1): 1.52\sin45^\circ=1.33\sin\theta, \theta=53.9^\circ (A1). Using 1/1.52 again is the glass–air trap.

(a) \lambda=v/f=340/850=0.40\,\mathrm{m}. (b) Path difference 3.60-3.00=0.60\,\mathrm{m} (M1) =1.5\lambda; in-phase sources, so the waves arrive in antiphase and interfere destructively (R1). (c) Resultant amplitude A-0.80A=0.20A (M1); intensity \propto amplitude^2: (0.20)^2=0.040 (A1). Subtracting intensities (1-0.64=0.36) is the trap. (d) Reversing one speaker adds half a cycle, so the waves now arrive in phase: a maximum, amplitude 1.80A, intensity 3.24 times that of S_1 alone (A1).

(a) Each reflected leg is \sqrt{0.60^2+0.25^2}=0.65\,\mathrm{m} (M1); reflected path 1.30\,\mathrm{m}, so the path difference is 0.10\,\mathrm{m} (A1). Using one leg only is the trap. (b) 0.10/0.040=2.5: antiphase, a minimum (A1). (c) The next minimum needs a path difference of 3.5\lambda=0.14\,\mathrm{m}, a reflected path of 1.34\,\mathrm{m} (M1): each leg 0.67\,\mathrm{m}, height \sqrt{0.67^2-0.60^2}=0.298\,\mathrm{m}; the sheet drops 0.048\,\mathrm{m} (A1). (d) Keep the 0.10\,\mathrm{m} path difference and make it 3.5\lambda: \lambda=0.0286\,\mathrm{m} (M1), f=c/\lambda=1.05\times10^{10}\,\mathrm{Hz} (A1).

(a) The two slits must be coherent sources: same frequency and a constant phase difference (R1). Across a laser beam the phase relation is fixed, so the slits sit on the same wavefront. A filament emits from many atoms at random; one narrow slit makes a single small source whose wavefronts reach both slits together (R1). (b) The two dark fringes are at \pm s/2 from the centre, so they are one spacing apart: s=4.8\,\mathrm{mm} (M1). Calling it 2s is the trap. \lambda=sd/D=(4.8\times10^{-3})(0.40\times10^{-3})/3.00=6.4\times10^{-7}\,\mathrm{m} (A1). (c) s=\lambda D/d=(520\times10^{-9})(2.00)/(0.40\times10^{-3}) (M1) =2.60\,\mathrm{mm} (A1).

(a) s=\lambda D/d=(532\times10^{-9})(1.50)/(0.25\times10^{-3})=3.19\times10^{-3}\,\mathrm{m} (A1). (b) Between one maximum and the next the sensor moves one spacing (M1): v=3.19\times10^{-3}/0.040=0.080\,\mathrm{m\,s^{-1}} (A1). Using half a spacing is the trap. (c) s doubles, so 0.080\,\mathrm{s} (A1). (d) Less light comes through the covered slit, so the two waves have unequal amplitudes and cannot cancel completely (R1).

(a) The first minimum is half the central width from the centre: 11.0\,\mathrm{mm} (M1). \theta=0.0110/2.50=4.40\times10^{-3}\,\mathrm{rad} (M1), \lambda=b\theta=(0.15\times10^{-3})(4.40\times10^{-3})=6.6\times10^{-7}\,\mathrm{m} (A1). Using the full 22.0\,\mathrm{mm} doubles the answer. (b) Symmetric curve with the tallest peak at \theta=0, zeros at \pm4.4 and \pm8.8\,\mathrm{mrad} (A1); side maxima about 5% of the central peak and half its width (A1).

(a) d=1/(3.00\times10^{5})=3.33\times10^{-6}\,\mathrm{m} (A1); per millimetre is not per metre. (b) \sin\theta_2=2(546\times10^{-9})/3.33\times10^{-6}=0.328, \theta_2=19.1^\circ (M1); the two second orders are on opposite sides: 38.2^\circ (A1). (c) n\le d/\lambda=6.1, so n_{\max}=6 (M1); orders 0,\pm1,\dots,\pm6 give 13 maxima (A1). Rounding up to 7, or answering 6, are the traps. (d) Second order needs 2\lambda\le d: d=1.09\times10^{-6}\,\mathrm{m} (A1).

(a) Fringe spacing 2.0\,\mathrm{mrad}: d=\lambda/\theta=500\times10^{-9}/2.0\times10^{-3}=2.5\times10^{-4}\,\mathrm{m} (A1). (b) Envelope first minimum at 10\,\mathrm{mrad}: b=\lambda/\theta=5.0\times10^{-5}\,\mathrm{m} (A1). Swapping b and d is the trap. (c) d/b=5: fringes of order 0,\pm1,\dots,\pm4, so 9 (A1); the fifth order falls on the single-slit minimum, where each slit sends no light (R1). (d) Envelope doubles in width; fringes stay 2.0\,\mathrm{mrad} apart, now 19 inside it, each less bright (A1).

d=1.67\times10^{-6}\,\mathrm{m}. (a) Violet: \sin\theta=0.400/1.667, \theta=13.9^\circ; red: \sin\theta=0.700/1.667, \theta=24.8^\circ (M1); width 10.9^\circ (A1). Taking \sin^{-1} of the difference of the sines is the trap. (b) Use the longest wavelength: third order needs \sin\theta=3(0.700)/1.667=1.26>1, impossible (M1); second order red at 57.1^\circ exists, so the second order is the highest complete one (A1). The third order begins (violet at 46.1^\circ) but is cut off, and it overlaps the end of the second. (c) At n=0 every wavelength has \theta=0, so all colours recombine (R1).

(a) Into a faster layer the ray bends away from the normal (the vertical) (R1). (b) \sin\theta_c=v_1/v_2=330/340 (M1), \theta_c=76.1^\circ (A1). 70^\circ is smaller, so the ray does enter the middle layer, at 75.5^\circ. (c) Across parallel layers \sin\theta/v stays constant (M1): \sin\theta_3=\sin70^\circ\times350/330=0.997, \theta_3=85.3^\circ (A1). (d) Turning needs \sin\theta=1: v=330/\sin70^\circ=351\,\mathrm{m\,s^{-1}} (M1). The top layer travels at 350\,\mathrm{m\,s^{-1}}, so it does not turn this ray; a layer faster than 351\,\mathrm{m\,s^{-1}} above would reflect it totally (R1). Claiming the first faster layer must trap the ray is the trap.

(a) The paths are equal, so the waves arrive in phase (R1). (b) At distance x from S_1 the path difference is |(25.0-x)-x|=|25.0-2x|, which cannot exceed 25.0\,\mathrm{cm} (M1). Minima need |25.0-2x|=3.0,\ 9.0,\ 15.0,\ 21.0\,\mathrm{cm}; 27.0 is too large (M1). Each value gives two points, one on each side of the midpoint: 8 minima at the distances listed (A1). (c) Moving \Delta x changes the path difference by 2\Delta x, so successive minima are \lambda/2=3.0\,\mathrm{cm} apart (A1).

(a) \lambda=340/680=0.50\,\mathrm{m} (A1). (b) Path difference at D_1 is 0.75\,\mathrm{m}=1.5\lambda (M1), which alone would put the waves in antiphase. A maximum means the sources add another half cycle: they are in antiphase (R1). Assuming a maximum proves in-phase sources is the trap. (c) At D_2 the path difference is 1.00\,\mathrm{m}=2\lambda, so the paths keep the phase and the antiphase sources give a minimum (A1). It is silent only if the two amplitudes there are equal. (d) \lambda=340/1360=0.25\,\mathrm{m} (M1), not 1.00\,\mathrm{m}; 0.75\,\mathrm{m}=3\lambda, so with antiphase sources D_1 records a minimum (A1).

(a) Points equally far above and below the centre are equidistant from O (M1), so each pair arrives in phase; every pair adds constructively and O is the brightest point (R1). (b) Two paired points are b/2 apart, so their rays to a distant point at angle \theta differ in path by (b/2)\sin\theta (M1). Cancellation needs half a wavelength: (b/2)\sin\theta=\lambda/2 (M1). Every point in the upper half has a partner, so all pairs cancel at once: b\sin\theta=\lambda (A1). Pairing only the two edges proves nothing about the rest of the slit. (c) \theta=\lambda/b=3.93\times10^{-3}\,\mathrm{rad}; width 2D\theta=14.2\,\mathrm{mm} (A1). (d) Detector at 7.08\,\mathrm{mm}; new minima at 5.31\,\mathrm{mm} and 10.6\,\mathrm{mm}, so it lies in the first side maximum and receives light (R1).

(a) f=c/\lambda=3.00\times10^{8}/633\times10^{-9}=4.74\times10^{14}\,\mathrm{Hz} (A1). (b) v=c/n=1.88\times10^{8}\,\mathrm{m\,s^{-1}} (A1); \lambda=633/1.60=396\,\mathrm{nm}, with f unchanged (A1). (c) \sin\theta_c=1.33/1.60, \theta_c=56.2^\circ (M1); 60^\circ>56.2^\circ, so the beam is totally internally reflected (A1). (d) s=\lambda D/d=(633\times10^{-9})(2.00)/(0.25\times10^{-3})=5.06\,\mathrm{mm} (A1). In water \lambda=633/1.33=476\,\mathrm{nm} (M1), so s=3.81\,\mathrm{mm} (A1).

(a) The reflected wave travels back through the incident wave and they superpose to form a standing wave (R1); minima are nodes, half a wavelength apart (M1): \lambda=3.00\,\mathrm{cm} (A1). (b) f=c/\lambda=1.00\times10^{10}\,\mathrm{Hz} (A1). (c) Path difference d\sin\theta=\lambda (M1): \sin\theta=3.00/6.0=0.50, \theta=30.0^\circ (A1). (d) s=\lambda D/d assumes \sin\theta\approx\tan\theta\approx\theta, which fails at 30^\circ (R1). n\le d/\lambda=2, but n=2 needs \theta=90^\circ, along the sheet, so only n=0,\pm1: three maxima (A1).

(a) \Delta E=1.89\,\mathrm{eV}=3.02\times10^{-19}\,\mathrm{J} (M1); \lambda=hc/\Delta E=6.58\times10^{-7}\,\mathrm{m} (A1). (b) Different parts of a lamp emit at random, so light from them has no fixed phase relation (R1). One narrow slit acts as a single source; its wavefronts reach both slits, so they act as coherent sources with a constant phase difference (R1). (c) s=\lambda D/d=(6.58\times10^{-7})(1.80)/(0.50\times10^{-3}) (M1) =2.37\,\mathrm{mm} (A1). (d) \Delta E=2.55\,\mathrm{eV}, \lambda=4.88\times10^{-7}\,\mathrm{m} (M1), s=1.76\,\mathrm{mm} (A1). Shorter wavelength, closer fringes.

(a) E_k=qV=(1.60\times10^{-19})(150)=2.40\times10^{-17}\,\mathrm{J} (M1); p=\sqrt{2mE_k}=\sqrt{2(9.11\times10^{-31})(2.40\times10^{-17})}=6.61\times10^{-24}\,\mathrm{kg\,m\,s^{-1}} (A1). (b) \lambda=h/p=6.63\times10^{-34}/6.61\times10^{-24}=1.00\times10^{-10}\,\mathrm{m} (A1). (c) \theta=\lambda/b=2.0\times10^{-3}\,\mathrm{rad} (M1); width 2D\theta=3.2\,\mathrm{mm} (A1). (d) p\propto\sqrt V, so four times the p.d. doubles p and halves \lambda (M1): width 1.6\,\mathrm{mm} (A1). (e) Minima: angles at which no electrons arrive, which particles moving in straight lines could not produce (A1).