Waves for IB Physics HL

Reading cue. The arrow leaves the ambulance pointing at the observer, so 25 m s⁻¹ is a source closing on a still listener: it is subtracted from 340 m s⁻¹, not added.

1. aState State how the wavelength of the sound in the region between the ambulance and the observer compares with the wavelength of the sound behind the ambulance.

   recall2 marks

   Check answer · 2 marks

   1. wavelength is shorter in front of the ambulance, on the side where the observer stands
   2. wavelength is longer behind the ambulance

2. bCalculate Calculate the wavelength of the sound in the air between the ambulance and the observer.

   routine3 marks

   Check answer · 3 marks

   1. in one period the wavefronts advance 340 m s⁻¹ while the source advances 25 m s⁻¹ after them
   2. λ = (340 - 25) / 512
   3. λ = 0.615 m

3. cDetermine Determine the frequency the observer hears once the ambulance has passed and is travelling away along the same road, and hence determine the change in the frequency heard as the ambulance goes by.

   demanding4 marks

   Check answer · 4 marks

   1. receding source: the source speed is added, giving (340 + 25) in the denominator
   2. f = 512 × 340 / 365 = 477 Hz
   3. approaching frequency is 512 × 340 / 315 = 553 Hz
   4. change = 553 - 477 = 76 Hz, heard as a sudden drop in pitch as the ambulance passes

4. dDiscuss The ambulance is instead parked at the roadside with its siren sounding, and the observer runs towards it along the road at 25 m s⁻¹, the same speed as the arrow in the figure. Discuss whether the observer now hears the same frequency as in the situation drawn.

   top of the paper4 marks

   Check answer · 4 marks

   1. moving observer, stationary source: f = 512 x (340 + 25) / 340 = 550 Hz
   2. this is close to, but not equal to, the 553 Hz of the drawn moving-source case
   3. so the shift is not fixed by the relative motion alone
   4. the air is a preferred medium: a moving source alters the wavelength itself, while a moving observer meets unaltered wavefronts at a different rate

Reading cue. Only the middle ray enters the gap, so b is the gap and not the bar; the tall central band reaches the first minimum on both sides, so it spans twice each side band.

1. aState State, in terms of the wavelength and the slit width b marked on the figure, the angle at which the first dark band appears, and state the line on the drawing from which that angle is measured.

   recall2 marks

   Check answer · 2 marks

   1. θ = λ / b
   2. measured from the dashed straight line continuing from the slit to the screen

2. bDetermine The slit width is b = 0.12 mm, the light has wavelength 590 nm, and the screen is 1.8 m from the slit. Determine the width of the tall central band drawn on the screen.

   routine3 marks

   Check answer · 3 marks

   1. first minimum at θ = λ / b = 590 × 10⁻⁹ / 1.2 × 10⁻⁴ = 4.9 × 10⁻³ rad
   2. the central band runs from the first minimum on one side to the first minimum on the other, so its width is 2 λ D / b
   3. width = 2 × 590 × 10⁻⁹ x 1.8 / 1.2 × 10⁻⁴ = 1.8 × 10⁻² m (1.8 cm)

3. cShow (that) Show that the central band should be drawn twice as wide as each of the bands beside it.

   demanding4 marks

   Check answer · 4 marks

   1. minima occur at θ = λ/b, 2 λ/b, 3 λ/b, ... from the centre
   2. the central band lies between the minima at -λ/b and +λ/b, so it spans 2 λ/b
   3. each side band lies between consecutive minima, for example λ/b and 2 λ/b, so it spans λ/b
   4. hence the widths are in the ratio 2 : 1

4. dDiscuss The single slit is narrowed until b is smaller than the wavelength of the light. Discuss what then becomes of the row of bands drawn on the screen.

   top of the paper5 marks

   Check answer · 5 marks

   1. a minimum requires sin θ = λ / b
   2. with b < λ this gives λ/b > 1, which no angle can satisfy
   3. so no dark bands form and the screen is lit right across its width with no minima
   4. far less light passes through, so the illumination is much fainter
   5. the drawn row of separate bands is replaced by a single very broad, faint spread

Reading cue. d = 0.25 mm is centre-to-centre between the slits, and D = 2.40 m runs from the slit plane to the screen, not from the laser drawn further left.

1. aState State why the two slits are illuminated by the single laser drawn, rather than by two separate lasers of the same wavelength placed one behind each slit.

   recall2 marks

   Check answer · 2 marks

   1. the two slits must act as coherent sources, with a constant phase difference between them
   2. one laser feeding both slits guarantees this; two independent lasers would drift in phase and the fringes would wash out

2. bDetermine The whole arrangement, from slits to screen, is immersed in water of refractive index 1.33. Determine the wavelength of the light in the water and the new separation of adjacent bright fringes.

   routine4 marks

   Check answer · 4 marks

   1. the frequency is unchanged and the speed falls, so λ = 633 / 1.33 = 476 nm
   2. fringe separation s = λ D / d = 476 × 10⁻⁹ x 2.40 / 0.25 × 10⁻³
   3. s = 4.6 × 10⁻³ m (4.6 mm)
   4. the fringes move closer together than the 6.1 mm they are in air

3. cDetermine Keeping the laser and the 2.40 m screen distance as drawn, determine the slit separation that would be needed to space the bright fringes 1.0 cm apart, and state one disadvantage of working with the pattern this produces.

   demanding4 marks

   Check answer · 4 marks

   1. rearrange s = λ D / d to give d = λ D / s
   2. d = 633 × 10⁻⁹ x 2.40 / 0.010
   3. d = 1.5 × 10⁻⁴ m (0.15 mm), so the slits must be closer together than the 0.25 mm drawn
   4. disadvantage: the same light is spread over fewer, wider fringes, so each is fainter and fewer of them fall on the screen

4. dEvaluate A student uses this arrangement to find the laser's wavelength from λ = sd/D. The measurements are d = 0.25 +/- 0.01 mm, D = 2.40 +/- 0.01 m, and the width of ten fringe spacings is 61 +/- 1 mm. Evaluate which measurement limits the precision of the result.

   top of the paper6 marks

   Check answer · 6 marks

   1. for λ = sd/D the fractional uncertainties add: 0.01/0.25 = 4%, 1/61 = 1.6%, 0.01/2.40 = 0.4%
   2. total fractional uncertainty is about 6%
   3. s = 61/10 = 6.1 mm, so λ = 6.1 × 10⁻³ x 0.25 × 10⁻³ / 2.40 = 6.35 × 10⁻⁷ m (635 nm)
   4. 6% of that is 38 nm, so the result is (6.4 +/- 0.4) x 10² nm, which covers the true 633 nm
   5. the slit separation dominates, contributing two thirds of the total; it is the smallest length measured and is quoted to only two significant figures
   6. even a perfect D and a perfect fringe measurement would leave 4%, so only a better measurement of d improves the result appreciably

Reading cue. The horizontal axis is displacement, not time: kinetic energy is an inverted parabola peaking at x = 0 here, not a cosine curve, and the vertical axis carries no scale.

1. aState The axes are drawn to span the whole of the motion, from one extreme to the other. State the values of x, read from the horizontal scale, at which a potential-energy curve drawn on these axes would reach the total energy of the particle, and state the kinetic energy there.

   recall2 marks

   Check answer · 2 marks

   1. at x = -8.0 cm and x = +8.0 cm, the ends of the scale, which are the amplitude
   2. the kinetic energy is zero at those two displacements

2. bDetermine The total energy of the particle is 1.23 × 10⁻² J. Determine its potential energy and its kinetic energy at the gridline x = +4.0 cm.

   routine3 marks

   Check answer · 3 marks

   1. at that gridline x / x0 = 4.0 / 8.0 = 0.50, so x² / x0² = 0.25
   2. potential energy = 0.25 × 1.23 × 10⁻² = 3.1 × 10⁻³ J
   3. kinetic energy = 1.23 × 10⁻² - 3.1 × 10⁻³ = 9.2 × 10⁻³ J

3. cDetermine Determine the displacement, on the scale given, at which the kinetic and potential energies of the particle are equal, and state whether this falls on the +4.0 cm gridline drawn.

   demanding4 marks

   Check answer · 4 marks

   1. equal energies means each is half the total, so x² / x0² = 0.50
   2. x = x0 / sqrt(2) = 8.0 / sqrt(2)
   3. x = +/- 5.7 cm
   4. this lies outside the +/- 4.0 cm gridlines, so the curves do not cross at half the amplitude

4. dExplain The particle is restarted with amplitude 4.0 cm, the gridline on these axes, and the same period. Explain what happens to each of the two energy curves, giving the new total energy.

   top of the paper4 marks

   Check answer · 4 marks

   1. total energy is proportional to x0², so it falls to a quarter: 3.1 × 10⁻³ J
   2. the potential-energy curve keeps exactly the same shape, since potential energy depends on x and not on the amplitude; it is simply followed only out to +/- 4.0 cm
   3. the kinetic-energy curve is a new, lower inverted parabola, still peaking at x = 0 but now at 3.1 × 10⁻³ J
   4. the curves still cross where each is half the total, now at x = 4.0 / sqrt(2) = +/- 2.8 cm