C.3 Wave phenomena – IB Physics exam questions with mark scheme

1Standard7 marks

A ray of light in air is incident on a glass block of refractive index 1.52 at an angle of incidence of $40^\circ$.

(a) Calculate the angle of refraction in the glass. [2]

(b) Calculate the speed of light in the glass. [1]

(d) State the two conditions required for total internal reflection to occur. [2]

$n_1\sin\theta_1=n_2\sin\theta_2$: $\sin\theta_2 = \sin40^\circ/1.52$(1)
$\theta_2=25.0^\circ$ (25°)(1)
$v=c/n = 3.00\times10^8/1.52 = 1.97\times10^{8}\ \mathrm{m\,s^{-1}}$ ($1.97\times10^8$)(1)
$\sin\theta_c = n_2/n_1 = 1/1.52$ (= 0.658)(1)
$\theta_c=41.1^\circ$ (accept 41°)(1)
Light travels from a medium of higher refractive index to one of lower refractive index (optically denser to less dense)(1)
Angle of incidence is greater than the critical angle(1)
Answer: (a) 25.0° (b) $1.97\times10^8$ m s$^{-1}$ (c) 41.1° (d) light travels from higher to lower refractive index; angle of incidence > critical angle
Examiner tip: Angles are measured from the normal, and the sine law has $n$ on the side of the matching angle.

2StandardHL only5 marks

Monochromatic light of wavelength $600\ \mathrm{nm}$ passes through a single slit of width $0.10\ \mathrm{mm}$. The diffraction pattern forms on a screen $2.0\ \mathrm{m}$ from the slit.

(a) Calculate the angular position of the first minimum. [2]

(b) Determine the width of the central maximum on the screen. [2]

$\theta=\lambda/b$ (small angle)(1)
$\theta=6.0\times10^{-3}$ rad(1)
Width $=2D\theta$ (central maximum spans both first minima)(1)
Width $=0.024$ m (2.4 cm; $2.4\times10^{-2}$ m)(1)
Width increases because $\theta\propto1/b$ (more diffraction through narrower slit)(1)
Answer: (a) $6.0\times10^{-3}$ rad (b) 0.024 m (c) wider
Examiner tip: The central maximum extends to the first minimum on each side, so double $D\theta$.

3Hard6 marks

In a double-slit experiment, light of wavelength $550\ \mathrm{nm}$ passes through two narrow slits $0.50\ \mathrm{mm}$ apart. The screen is $1.8\ \mathrm{m}$ from the slits.

(a) Calculate the fringe separation on the screen. [2]

(b) The whole apparatus is immersed in water of refractive index 1.33. Determine the new fringe separation. [2]

(c) In the original arrangement (in air), calculate the distance from the centre of the central maximum to the centre of the fourth bright fringe on one side of it. The central maximum is not counted as one of the four. [1]

(d) State, in the original arrangement, the path difference between the waves from the two slits at this fourth bright fringe. [1]

$s=\lambda D/d = 550\times10^{-9}\times1.8/0.50\times10^{-3}$(1)
$s=1.98$ mm (2.0 mm)(1)
Wavelength in water $\lambda'=\lambda/n=550/1.33=413.5$ nm (or $s'=s/n$)(1)
$s'=1.49$ mm (1.5 mm)(1)
$4s = 4\times1.98 = 7.9$ mm (7.92 mm; allow ECF from (a))(1)
$4\lambda=2.2\ \mu$m ($2.2\times10^{-6}$ m; accept “$4\lambda$”)(1)
Answer: (a) 2.0 mm (b) 1.5 mm (c) 7.9 mm (d) $2.2\ \mu$m
Examiner tip: Fringe separation is proportional to wavelength, and the wavelength shortens (frequency unchanged) in water.

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4Core5 marks

The diagram shows straight wavefronts in a ripple tank travelling from deep water towards a straight boundary with shallow water. The wavefronts make an angle of $50^\circ$ with the boundary. In the deep water the wavelength is $2.4\ \mathrm{cm}$ and the wave speed is $0.30\ \mathrm{m\,s^{-1}}$. In the shallow water the wave speed is $0.20\ \mathrm{m\,s^{-1}}$.

(a) State the property of the waves that does not change as they cross the boundary. [1]

(b) Calculate the wavelength in the shallow water. [2]

frequency OR period; ignore “phase”; do not accept “speed” or “wavelength”(1)
$f=\dfrac{0.30}{0.024}=12.5\ \mathrm{Hz}$ OR $\lambda_2=\lambda_1\dfrac{v_2}{v_1}$ seen(1)
$\lambda_2=1.6\ \mathrm{cm}$ (0.016 m); award [2] for a bald correct answer(1)
$\sin\theta_2=\dfrac{v_2}{v_1}\sin\theta_1=\dfrac{0.20}{0.30}\sin50^\circ$ (=0.51); accept use of $\lambda_2/\lambda_1$(1)
$\theta_2=31^\circ$ (30.7°); allow ECF from wavelength in (b)(1)
Answer: (a) frequency (b) 1.6 cm (c) 31°
Examiner tip: The angle between wavefronts and the boundary equals the angle between the ray and the normal, so Snell's law can be used directly with the wave speeds.

5Standard6 marks

Two loudspeakers S₁ and S₂ are $1.50\ \mathrm{m}$ apart and are connected to the same signal generator, so they emit sound in phase with a frequency of $1200\ \mathrm{Hz}$. A microphone is placed at point P, $4.20\ \mathrm{m}$ from S₁ and $5.05\ \mathrm{m}$ from S₂. The speed of sound is $340\ \mathrm{m\,s^{-1}}$.

(a) Calculate the wavelength of the sound. [1]

(b) Deduce whether the microphone at P detects a maximum or a minimum of intensity. [2]

(c) The frequency of the signal generator is slowly increased from $1200\ \mathrm{Hz}$. Determine the lowest frequency above $1200\ \mathrm{Hz}$ at which the intensity at P is a minimum. [2]

(d) Suggest why the intensity at a minimum is not zero in this arrangement. [1]

$\lambda=\dfrac{340}{1200}=0.283\ \mathrm{m}$ (0.28 m)(1)
path difference $=5.05-4.20=0.85\ \mathrm{m}$ and $0.85/0.283=3.0$ wavelengths(1)
whole number of wavelengths (sources in phase) so constructive interference: maximum; no mark for “maximum” without the path-difference reasoning(1)
minimum needs $0.85=3.5\lambda$ (next half-integer above 3) so $\lambda=0.243\ \mathrm{m}$(1)
$f=\dfrac{340}{0.243}=1400\ \mathrm{Hz}$; award [0] for 1300 Hz or other non-half-integer conditions(1)
waves arrive with unequal amplitudes (P is further from S₂ so its wave is weaker) OR sound reflected from walls/floor also reaches P; do not accept “sources not coherent”(1)
Answer: (a) 0.283 m (b) maximum: path difference = 3.0λ (c) 1400 Hz (d) unequal amplitudes / reflections
Examiner tip: Always convert the path difference into a number of wavelengths before deciding: whole number means maximum, half-integer means minimum (for in-phase sources).

6Hard6 marks

An optical fibre has a core of refractive index $1.50$ surrounded by cladding of refractive index $1.46$. A ray of light in air enters the flat end of the fibre at an angle of incidence $\theta$ and travels along the core by total internal reflection at the core–cladding boundary.

(a) Calculate the critical angle at the core–cladding boundary. [2]

(b) Show that the largest value of $\theta$ for which the light remains in the core is about $20^\circ$. [3]

$\sin c=\dfrac{1.46}{1.50}$ (=0.973)(1)
$c=76.7^\circ$; accept 77°(1)
angle of refraction at the end face $=90^\circ-76.7^\circ=13.3^\circ$ (geometry linking refraction angle to critical angle)(1)
$1.00\sin\theta=1.50\sin13.3^\circ$ (=0.344)(1)
$\theta=20.1^\circ$, to at least 3 s.f. or with working shown (a “show that” needs more precision than the 20° given); accept $\sin\theta=\sqrt{1.50^2-1.46^2}$(1)
protects the core surface from scratches/dirt that would allow light to escape OR stops light crossing between touching fibres OR the small index difference keeps only rays close to the axis, so there is less (modal/waveguide) dispersion; accept any one(1)
Answer: (a) 76.7° (b) θ = 20.1° (c) e.g. protects core / reduces dispersion
Examiner tip: In (b) the ray must hit the side at exactly the critical angle, so the refraction angle at the end is 90° minus the critical angle, not the critical angle itself.

7HardHL only8 marks

Light from a mercury lamp is incident normally on a diffraction grating that has $600$ lines per millimetre. The light includes a green line of wavelength $546.1\ \mathrm{nm}$ and two yellow lines of wavelengths $577.0\ \mathrm{nm}$ and $579.1\ \mathrm{nm}$.

(a) Calculate the angle between the central maximum and the first-order green maximum. [2]

(b) Determine the highest order in which the green line can be observed. [2]

(d) A double slit with the same slit separation is used in place of the grating. Explain why the two yellow lines are much harder to distinguish. [2]

$d=\dfrac{1}{600}\ \mathrm{mm}=1.67\times10^{-6}\ \mathrm{m}$ and $\sin\theta=\dfrac{546.1\times10^{-9}}{1.67\times10^{-6}}$ (=0.328)(1)
$\theta=19.1^\circ$(1)
$n_{\max}=\dfrac{d}{\lambda}=3.05$ OR $\sin\theta\le1$ used(1)
highest order = 3 (must be rounded down; 3.05 → 3); do not award for 4 or “3.05”(1)
$\theta_{577.0}=43.8^\circ$ and $\theta_{579.1}=44.0^\circ$ (both angles from $2\lambda=d\sin\theta$)(1)
separation $=0.20^\circ$; accept 0.2°; allow $3.5\times10^{-3}$ rad(1)
with a double slit the maxima are broad (intensity varies gradually, cos² shape), while a grating with many slits gives very narrow, sharp maxima(1)
so the broad maxima of the two close wavelengths overlap and cannot be resolved; ignore references only to the grating maxima being brighter(1)
Answer: (a) 19.1° (b) 3rd order (c) 0.20° (d) double-slit maxima are broad, so the two lines overlap
Examiner tip: The highest order is the whole number below d/λ; rounding 3.05 to the nearest integer is fine here but rounding 3.7 up to 4 would be wrong.

8StretchHL only8 marks

Monochromatic light is incident normally on two slits. Each slit has width $b$ and the centres of the slits are $0.45\ \mathrm{mm}$ apart. The graph shows how the intensity of the light on a screen $1.50\ \mathrm{m}$ from the slits varies with distance from the centre of the pattern.

(a) Use the graph to determine the wavelength of the light. [2]

(b) Determine $b$. [2]

(d) The width of each slit is halved. The slit separation is unchanged. Describe two changes to the pattern. [2]

fringe separation $s=2.0\ \mathrm{mm}$ read from graph and $\lambda=\dfrac{sd}{D}=\dfrac{2.0\times10^{-3}\times0.45\times10^{-3}}{1.50}$(1)
$\lambda=6.0\times10^{-7}\ \mathrm{m}$ (600 nm)(1)
first minimum of single-slit envelope at $8.0\ \mathrm{mm}$, so $\theta=\dfrac{8.0\times10^{-3}}{1.50}=5.3\times10^{-3}\ \mathrm{rad}$(1)
$b=\dfrac{\lambda}{\theta}=1.1\times10^{-4}\ \mathrm{m}$ (0.11 mm); allow ECF from (a)(1)
the double-slit (interference) condition predicts a maximum at 8 mm (4th order, $d\sin\theta=4\lambda$)(1)
but this coincides with the first minimum of the single-slit diffraction envelope (light from each slit has zero intensity there), so the fringe is missing; accept $d/b=4$ so the 4th order is missing(1)
envelope (central diffraction maximum) becomes twice as wide, so more fringes appear inside it (first missing fringe now at 16 mm / 8th order)(1)
fringe separation unchanged (still 2.0 mm) OR fringes less bright / overall intensity lower; do not credit “fringes further apart”(1)
Answer: (a) 600 nm (b) 0.11 mm (c) interference maximum coincides with diffraction minimum (d) envelope twice as wide; spacing unchanged; dimmer
Examiner tip: Separate the two length scales: the fringe spacing depends on the slit separation d, the envelope width depends on the slit width b.

9Standard6 marks

A student uses the apparatus shown to determine the wavelength of the light from a laser. The slit separation $d$ is printed on the double-slit slide.

Describe how the student should carry out the experiment and process the data to obtain an accurate value for the wavelength. Include the measurements taken, how they are made and how uncertainties are reduced. [6]

measure $D$ (slits to screen) with a metre rule / tape measure(1)
use a large $D$ (several metres) so that the fringe separation is large and its percentage uncertainty small(1)
measure across many fringes (e.g. 10 fringe spacings) with a ruler/vernier calipers and divide by the number of spacings to find $s$(1)
measure from the centre of one bright fringe to the centre of another (or dark to dark), in a darkened room for clear fringes(1)
calculate $\lambda=\dfrac{sd}{D}$(1)
repeat for several values of $D$ and plot $s$ against $D$; gradient $=\lambda/d$, so $\lambda=d\times$gradient (removes systematic error in the zero of $D$)(1)
repeat readings and take a mean; screen perpendicular to beam(1)
safety: do not look into the laser beam or its reflections (credit within the levels, not essential for Level 3)(1)

Level	Marks	What it looks like
Level 3	5–6	A clear, logical method: both D and s are measured with suitable instruments, s is found over many fringes, λ = sd/D is used, and at least two specific ways of reducing uncertainty are explained (e.g. large D, graph of s against D). A competent student could follow it.
Level 2	3–4	A workable method naming the measurements and the equation λ = sd/D, with at least one way of reducing uncertainty, but some steps lack detail or the order is unclear.
Level 1	1–2	Some relevant points, such as a measurement or the equation, but no coherent method; uncertainty ignored.
No marks	0	Nothing worthy of credit.

Answer: $\lambda = sd/D$; measure s over many fringes, use large D, vary D and use gradient of s against D
Examiner tip: Measuring a single fringe spacing is a common way to lose marks: always measure across many fringes and divide.

10Standard6 marks

A student shines a ray of light from air into a semicircular acrylic block. She measures the angle of incidence $i$ and the angle of refraction $r$ for six rays, then plots $\sin i$ against $\sin r$. The graph shows her points and line of best fit.

(a) Outline how the graph supports Snell's law. [1]

(b) Determine the refractive index of acrylic. [2]

(d) Calculate the critical angle for acrylic in air. [1]

(e) The protractor has an uncertainty of $\pm1^\circ$. Suggest why readings with larger angles of incidence give a more reliable value of the refractive index. [1]

straight line through the origin, so $\sin i\propto\sin r$ ($\sin i/\sin r$ constant)(1)
gradient from a large triangle using the line, e.g. $\dfrac{0.89-0}{0.60-0}$; do not accept a single data pair unless it lies on the line(1)
$n=1.48$; accept 1.45–1.52(1)
$v=\dfrac{3.00\times10^8}{1.48}=2.0\times10^8\ \mathrm{m\,s^{-1}}$; allow ECF from (b)(1)
$\sin c=\dfrac{1}{1.48}$, $c=42.5^\circ$ (43°); accept 41°–44°; allow ECF(1)
the fixed ±1° is a smaller fraction (percentage) of a larger angle, so the percentage uncertainty in the angles (and in $n$) is smaller(1)
Answer: (a) straight line through origin (b) 1.48 (c) $2.0\times10^8\ \mathrm{m\,s^{-1}}$ (d) 42.5° (e) smaller percentage uncertainty
Examiner tip: Take the gradient from the line, not from one of the plotted points, and use a triangle covering most of the line.

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