IB Physics HL C.3 Wave phenomena Questions

SLSL HLHL

Use ray and wavefront diagrams to trace boundary changes, calculate interference fringes, and explain how aperture width shapes a diffraction pattern.

357 questions in this syllabus scope, with question parts counted together.

Exam points & syllabus

Exam points

Use rays perpendicular to wavefronts, and draw transmitted fronts consistent with the supplied convergence or boundary geometry.
At a curved reflecting boundary, draw the local normal through the given centre of curvature and use equal incidence and reflection angles.
Use the normal as the angle reference in n₁ sin θ₁ = n₂ sin θ₂; identify both media and the actual direction of travel.
For explicitly stated wavelength-dependent indices, show stronger refraction for the larger index and label both transmitted colours clearly.
Use n = c/v with the index of the medium actually traversed; track the unchanged frequency when relating speed and wavelength.
For travel-time differences, retain enough precision in close refractive indices before subtracting speeds or times.
When a pulse’s speed range and original duration are supplied, calculate the arrival-time spread and add the original duration for its minimum span.
Total internal reflection requires travel from higher to lower refractive index and an incidence angle greater than the critical angle.
Use sin θc = n₂/n₁ for the actual pair of media; a ray at the critical angle travels along the boundary rather than undergoing total reflection.
For supplied fibre geometry, compare path lengths at the common speed: rays closer to the axis can arrive closer together in time.
Apply superposition to signed displacements at the same position and time, rather than adding unsigned amplitudes in every situation.
Read phase offset as a fraction of a complete cycle; a quarter-cycle displacement corresponds to π/2 or the equivalent 3π/2.
Coherence means a constant phase difference; equal frequency or monochromaticity alone does not establish the same initial phase.
For initially in-phase sources, integer-wavelength path differences give constructive interference and half-odd multiples give destructive interference.
When source phases are initially opposite, equal path lengths preserve cancellation; a perpendicular bisector can remain a line of minima.
For equal contributions arriving in phase, amplitudes double and I ∝ A² makes the intensity four times that from one contribution.
Explain two-source fringes through coherent overlap, varying path difference and constructive or destructive superposition.
At a central double-slit maximum, state the actual zero path difference before constructive interference; a generic integer-wavelength condition may not score.
For a moving reflector, one wavelength of round-trip path change needs only half a wavelength of plate movement.
When only one direct source path changes, read adjacent maxima against source displacement; one wavelength of path change restores the phase.
Use s = λD/d for adjacent like fringes and convert all quantities consistently; spacing differs from displacement to an nth-order fringe.
Double a measured adjacent bright-to-dark gap before using the Young spacing equation.
Count intervals between fringe centres: nine successive fringes span eight spacings.
For the same fringe order, angular position scales with wavelength; distinguish the final angle from the requested change.
Do not compare unlike fringe orders by wavelength alone: a blue first bright fringe can lie beyond a red first dark fringe.
Two opposite first minima span a full angular fringe spacing, φ ≈ λ/d; each lies about λ/(2d) from the central maximum.
Adjacent minima have the same separation as adjacent maxima, even when the central point is a minimum.
Identify both reflecting boundaries before explaining interference between their reflected contributions; a missing wavelength needs destructive interference.
At a fixed aperture, shorter wavelengths give less diffraction spreading; a larger aperture also narrows the pattern.
Changing a colour filter changes diffraction-pattern widths while leaving the angular separation of the source centres unchanged.
When a limiting angle is explicitly supplied, use small-angle geometry s ≈ Dθ without importing a separate resolution formula.
Explain appreciable spreading when wavelength is comparable to the opening; diffraction is small when wavelength is much smaller.
For a rectangular slit, θ ≈ λ/b gives the centre-to-first-minimum angle, while the full central screen width is 2Dλ/b.
Use the first minimum of the diffraction envelope rather than an interference minimum when estimating individual slit width or wavelength.
A narrower single slit gives a broader central maximum and lower peak intensity under unchanged illumination.
Compare λ/b when both wavelength and slit width change, and keep full central width separate from its half-width.
A single-slit intensity sketch needs smooth peaks, minima at zero and much smaller secondary maxima in the requested range.
For a first single-slit minimum, pair rays half a slit width apart: bθ/2 = λ/2 gives θ = λ/b.
Explain wave evidence using spreading beyond a geometrical shadow or maxima and minima caused by superposition.
Finite-width slits produce an envelope over interference fringes; a missing maximum can coincide with an envelope minimum.
Keep interference and envelope angular scales distinct: adjacent principal fringes give λ/d, while the first envelope minimum gives λ/b.
Beyond the first envelope minimum, fringes can reappear with lower average intensity and approximately unchanged spacing.
Count the missing interference order before calculating its angle with d sin θ = nλ.
For a grating, use nλ = d sin θ with θ measured from the central direction, and obtain slit spacing from the reciprocal line density.
For maxima on opposite sides, halve the measured total angle before applying the grating equation, or double the one-sided result.
Find the largest integer order allowed by nλ ≤ d, then count both sides plus the central maximum.
Use the longest incident wavelength for the largest complete spectral order; for overlapping orders, equate the two nλ products.
A white-light grating pattern has a white zero order and spectra on both sides, with violet inward and red outward.
More illuminated slits at unchanged spacing make principal maxima brighter and narrower without moving them; secondary maxima become less important.
A plot of sin θ against order has gradient λ/d; smaller λ reduces the slope, while changing screen distance leaves it unchanged.
Increasing lines per unit length decreases slit spacing, spreads spectral angles farther apart and reduces the possible number of orders.

Covered syllabus points

C.3.1—Wavefronts and rays C.3.2—Snell’s law C.3.3—Refractive index C.3.4—Total internal reflection C.3.5—Superposition C.3.6—Coherence C.3.7—Path difference C.3.8—Two-source interference C.3.9—Young double-slit equation C.3.10—Reflection, refraction and transmission at boundaries C.3.11—Diffraction around bodies and through apertures C.3.12 (HL)—Single-slit diffraction C.3.13 (HL)—Diffraction envelope C.3.14 (HL)—Diffraction gratings

[Maximum number: 2]

A converging lens is placed between an object and a screen. An image of the object is formed on the screen.

(a)

Draw a ray to locate the focal point of the lens. Label this point with the letter F .

[ 1 ]

(b)

The lens suffers from spherical aberration.

[ 1 ]

(i)

Draw lines to complete the rays in the diagram.

[ 1 ]

[Maximum number: 6]

Monochromatic light enters the base of a plastic beaker that contains water with an oil layer floating on it. A student draws a diagram to show the directions the light takes as it passes through the layers. The student's diagram has one error at position A and one error at position B. The refractive indices of the materials are shown on the diagram.

The light is refracted at an angle of $32^{\circ}$ when it enters the plastic layer as shown.

(a)

Identify, with a reason, the error in the student's diagram for

[ 4 ]

(i)

light crossing the plastic-water interface (position A).

[ 2 ]

the angle of refraction ought to be greater than the angle of incidence OR the ray should refract away from the normal
because ray goes from high refractive index/<<optically>> more dense/slower medium to
low refractive index/optically less dense/faster medium

Marking guidance:

Do not allow use of e.g $n_{1}$
unless medium one is
described e.g. $n_{\text {air }}$

(ii)

light at the water-oil interface (position B).

[ 2 ]

(b)

Calculate the angle of incidence at the air-plastic interface.

[ 2 ]

Use of Snell's Law « $\frac{\sin i}{\sin r}=\frac{1.60}{1}$ »
$i=\sin ^{-1} \ll 1.60 \times \sin 32^{\circ}>=58<^{\circ} \gg$

'Use of' requires a substitution NOT just a statement of a formula

Marking guidance:

Accept 1.0 rad (unit must be included to show a deliberate attempt to use rad rather than a calculator mistake)

(c)

Monochromatic light of wavelength $6.3 \times 10^{-7} \mathrm{~m}$ in air is incident from above at a normal to the oil layer. Rays on the diagram are shown at near-normal incidence for clarity. Three positions X, Y and Z are shown on the diagram.

[ 4 ]

(i)

Identify, with a reason, a position at which there is a phase change of $180^{\circ}$ .

[ 1 ]

(ii)

Determine the minimum thickness of the oil layer for which light is not reflected. State your answer to an appropriate number of significant figures.

[ 3 ]

Use of $2 d \mathrm{n}=m \lambda$ $2.1 \times 10^{-7}$ «m»

any answer to 2 s.f.

'Use of' requires a substitution
NOT just a statement of a formula

[Maximum number: 1]

A beam of coherent monochromatic light from a distant galaxy is used in an optics experiment on Earth.

The air between the slits and the screen is replaced with water. The refractive index of water is 1.33 .

Calculate the wavelength of the light in water.

$\frac{633.0}{1.33}=476\ \mathrm{nm}$

Exam points

Check both TIR conditions: propagation towards a lower refractive index and incidence greater than the critical angle, measured to the normal.
Calculate sinθc = nreceiving/nincident using the actual boundary media; the receiving index is not automatically one.
Rearrange the critical-angle relation to find the receiving medium’s index, keeping the incident and receiving media clearly identified.
At the critical angle the transmitted ray grazes the boundary; TIR requires a larger incidence angle, rather than equality.
Use boundary geometry to obtain the actual incidence angle before comparing it with θc, and state the reason for any no-emergence conclusion.
As an alternative TIR test, apply Snell’s law and show that the required refracted sine exceeds one, so no real transmitted direction exists.
After TIR, draw equal incidence and reflection angles at that boundary, then apply the correct refraction direction at any subsequent exit face.
For a fibre’s external acceptance angle, find the wall-normal critical angle, take its complement to the axis, then apply Snell’s law at entry.
Keep the internal angle to the fibre axis separate from the external entrance angle: the internal limiting angle is 90° − θc.
Nearly equal core and cladding indices give a large θc, admitting only nearly axial rays; similar path lengths give less arrival-time spread.
For transmission into air, combine sinθc = 1/n with n = c/v to obtain v = c sinθc.
When a boundary medium is changed, use the new pair of indices while retaining only those ray directions explicitly stated to be unchanged.
Use the diagram’s stated angle convention when predicting confinement: an external angle to a surface is complementary to incidence at that entry face.

[Maximum number: 2]

The light is refracted at an angle of $32^{\circ}$ when it enters the plastic layer as shown.

Calculate the critical angle for the plastic-water interface.

$\begin{aligned} \frac{\sin i}{\sin r}=\frac{1.33}{1.60} \text { and } \sin r=1 i=\text { «sin }{ }^{-1} 0.831 »=56 \text { " }^{\circ} \text { » } \end{aligned}$

Accept 0.98 rad (unit required)

[Maximum number: 2]

Two loudspeakers, A and B, are driven in phase and with the same amplitude at a frequency of 850 Hz . Point P is located 22.5 m from A and 24.3 m from B. The speed of sound is $340 \mathrm{~ms}^{-1}$ .

When both loudspeakers are operating, the intensity of sound recorded at Q is $I_{0}$ . Loudspeaker B is now disconnected. Loudspeaker A continues to emit sound with unchanged amplitude and frequency. The intensity of sound recorded at Q changes to $I_{\mathrm{A}}$ .

Estimate $\frac{I_{\mathrm{A}}}{I_{0}}$ .

the amplitude of sound at Q is halved

«intensity is proportional to amplitude squared hence» $\frac{I_{A}}{I_{0}}=\frac{1}{4}$

[Maximum number: 1]

A beam of coherent monochromatic light from a distant galaxy is used in an optics experiment on Earth.

The beam is incident normally on a double slit. The distance between the slits is 0.300 mm . A screen is at a distance D from the slits. The diffraction angle $\theta$ is labelled.

Outline why the beam has to be coherent in order for the fringes to be visible.

[Maximum number: 3]

A beam of coherent monochromatic light from a distant galaxy is used in an optics experiment on Earth.

The beam is incident normally on a double slit. The distance between the slits is 0.300 mm . A screen is at a distance D from the slits. The diffraction angle $\theta$ is labelled.

A series of dark and bright fringes appears on the screen. Explain how a dark fringe is formed.

superposition of light from each slit / interference of light from both slits with path/phase difference of any half-odd multiple of wavelength/any odd multiple of $\pi$ (in words or symbols), producing destructive interference

[Maximum number: 3]

A student investigates how light can be used to measure the speed of a toy train.

Light from a laser is incident on a double slit. The light from the slits is detected by a light sensor attached to the train.

The graph shows the variation with time of the output voltage from the light sensor as the train moves parallel to the slits. The output voltage is proportional to the intensity of light incident on the sensor.

Explain, with reference to the light passing through the slits, why a series of voltage peaks occurs.

[Maximum number: 1]

A student investigates how light can be used to measure the speed of a toy train.

The slits are separated by 1.5 mm and the laser light has a wavelength of $6.3 \times 10^{-7} \mathrm{~m}$ . The slits are 5.0 m from the train track. Calculate the separation between two adjacent positions of the train when the output voltage is at a maximum.

$s=\frac{\lambda D}{d}=\frac{6.3\times10^{-7}\times5.0}{1.5\times10^{-3}}=2.1\times10^{-3}\ \mathrm{m}$

[Maximum number: 6]

The light is refracted at an angle of $32^{\circ}$ when it enters the plastic layer as shown.

(a)

Identify, with a reason, the error in the student's diagram for

[ 4 ]

(i)

light crossing the plastic-water interface (position A).

[ 2 ]

Do not allow use of e.g $n_{1}$
unless medium one is
described e.g. $n_{\text {air }}$

(ii)

light at the water-oil interface (position B).

[ 2 ]

(b)

Calculate the angle of incidence at the air-plastic interface.

[ 2 ]

Accept 1.0 rad (unit must be included to show a deliberate attempt to use rad rather than a calculator mistake)

(c)

[ 4 ]

(i)

Identify, with a reason, a position at which there is a phase change of $180^{\circ}$ .

[ 1 ]

(ii)

Determine the minimum thickness of the oil layer for which light is not reflected. State your answer to an appropriate number of significant figures.

[ 3 ]

Use of $2 d \mathrm{n}=m \lambda$ $2.1 \times 10^{-7}$ «m»

any answer to 2 s.f.

'Use of' requires a substitution
NOT just a statement of a formula

[Maximum number: 3]

This question is about the motion of a ship and observing objects from it.

A security camera on the ship captures an image of two green lamps on the shore. The lamps emit light of wavelength 520 nm .

The camera has a circular aperture of diameter 6.2 mm . The lamps are separated by 1.5 m . Determine the maximum distance between the camera and the lamps at which the images of the lamps can be distinguished.

use of $\theta=\frac{1.22 \lambda}{d}$ ;
$\left(\frac{1.22 \times 520 \times 10^{-9}}{6.2 \times 10^{-3}}\right)=1.02 \times 10^{-4} ;$
distance $=\left(\frac{1.5}{1.02 \times 10^{-4}}=\right) 1.47 \times 10^{4} \mathrm{~m} ;$

Marking guidance:

Award [2 max] if distance is $1.79 \times 10^{4} m$ (ignoring factor of 1.22) in second marking point.

[Maximum number: 2]

A student investigates how light can be used to measure the speed of a toy train.

As the train continues to move, the first diffraction minimum is observed when the light sensor is at a distance of 0.13 m from the centre of the fringe pattern.

(not to scale)

Determine the width of one of the slits.

angular width of diffraction minimum $=\frac{0.13}{5.0}=0.026\ \mathrm{rad}$

slit width $=\frac{\lambda}{d}=\frac{6.3\times10^{-7}}{0.026}=2.4\times10^{-5}\ \mathrm{m}$

[Maximum number: 2]

A student investigates how light can be used to measure the speed of a toy train.

As the train continues to move, the first diffraction minimum is observed when the light sensor is at a distance of 0.13 m from the centre of the fringe pattern.

(not to scale)

Suggest the variation in the output voltage from the light sensor that will be observed as the train moves beyond the first diffraction minimum.

[Maximum number: 3]

Two sources of light produce an interference pattern on a screen.

Determine the largest order of the diffraction pattern in which all the wavelengths in the incident beam are present.

Considers the largest wavelength $\boldsymbol{O} \boldsymbol{R} \quad \theta=90^{\circ}$ $d \sin \theta=n \lambda \Rightarrow n=\frac{10^{-3}}{580} \frac{\sin 90^{\circ}}{650 \times 10^{-9}}=2.65$

So the second order/ n=2

Award [2] if smallest wavelength
used and an answer of n=3.

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