Wave phenomena HL Paper 1&2

C.3 Wave Phenomena Papers 1&2

Extended Question Bank: Multiple Choice and Extended Responses

Reflection, Refraction, Superposition, Interference and Diffraction

These original IB-style practice questions are not official IB examination questions. Work through the SL questions first, then attempt the questions marked HL if appropriate. Show your working, include units and explain your reasoning. Attempt each question before opening its answer.

Instructions

Use degrees for angles unless stated otherwise.
Measure incidence, reflection and refraction angles from the normal.
Convert all lengths to consistent units before substituting.
Give numerical answers to 2–3 significant figures.
Use $c=3.00\times10^8\,\mathrm{m\,s^{-1}}$ where required.
Take the refractive index of air as $1.00$ .
For small-angle formulas, use $\sin\theta\approx\tan\theta\approx\theta$ , with $\theta$ in radians.
Assume linear wave behaviour and normal incidence on slits and gratings unless stated otherwise.

Part A — Multiple-Choice Questions

A1 — Wavefronts and rays (SL)

In an isotropic medium, what is the relationship between a ray and a wavefront?

A. The ray is parallel to the wavefront.
B. The ray always makes an angle of 45° to the wavefront.
C. The ray is perpendicular to the wavefront.
D. The ray points along the crest.

Show answer and explanation

Answer: C. A wavefront joins points of equal phase. A ray shows the propagation direction and is perpendicular to wavefronts in an isotropic medium.

A2 — Light entering glass (SL)

Light enters glass from air through a stationary boundary. Which statement is correct?

A. Frequency decreases and wavelength is unchanged.
B. Frequency is unchanged and wavelength decreases.
C. Frequency increases and wavelength decreases.
D. Frequency and wavelength both remain unchanged.

Show answer and explanation

Answer: B. Frequency remains fixed by the source. Speed decreases in glass, so wavelength decreases according to $\lambda=v/f$ .

A3 — Total internal reflection (SL)

Light travels from glass of refractive index $1.50$ towards air. The angle of incidence is $50^\circ$ . What happens?

A. The refracted angle is 50°.
B. The ray refracts towards the normal.
C. The refracted ray travels along the boundary.
D. Total internal reflection occurs.

Show answer and explanation

Answer: D. The critical angle is $\theta_c=\sin^{-1}(1/1.50)=41.8^\circ$ . Since $50^\circ>\theta_c$ and light travels from a higher to a lower refractive index, total internal reflection occurs.

A4 — Diffraction (SL)

For the same wavelength, which gap produces the greatest spreading of waves?

A. A gap whose width is comparable to the wavelength.
B. A gap 100 times the wavelength.
C. A gap 1000 times the wavelength.
D. All gaps produce the same spreading.

Show answer and explanation

Answer: A. Diffraction is most noticeable when the gap width is comparable to the wavelength. A much wider gap produces less angular spreading.

A5 — Coherent sources (SL)

Which condition is required for two sources to be coherent?

A. They must have equal amplitudes.
B. They must have the same frequency and a constant phase difference.
C. They must be exactly in phase at every point in space.
D. They must emit different wavelengths.

Show answer and explanation

Answer: B. Coherent sources have the same frequency and a constant phase difference. Equal amplitudes are not required for coherence, although equal amplitudes allow complete cancellation at a destructive-interference point.

A6 — Double-slit fringe spacing (SL)

The separation of two slits is doubled while wavelength and screen distance remain unchanged. What happens to the fringe spacing?

A. It doubles.
B. It remains unchanged.
C. It halves.
D. It becomes one quarter of its original value.

Show answer and explanation

Answer: C. Using $w=\lambda D/s$ , fringe spacing is inversely proportional to slit separation. Doubling $s$ halves $w$ .

A7 — Single-slit diffraction (HL)

For monochromatic light incident normally on a slit of width $a$ , which equation gives the first diffraction minimum?

A. $a\sin\theta=\lambda/2$
B. $a\sin\theta=2\lambda$
C. $a\cos\theta=\lambda$
D. $a\sin\theta=\lambda$

Show answer and explanation

Answer: D. Single-slit minima satisfy $a\sin\theta=m\lambda$ for $m=1,2,3,\ldots$ . The first minimum has $m=1$ .

A8 — Diffraction grating (HL)

A grating is replaced by one with twice as many lines per millimetre. For the same wavelength and order, what happens to the angle of a maximum, provided that order still exists?

A. It increases.
B. It decreases.
C. It remains unchanged.
D. It becomes zero.

Show answer and explanation

Answer: A. Doubling line density halves the grating spacing $d$ . Since $d\sin\theta=m\lambda$ , the value of $\sin\theta$ increases for the same order and wavelength. The order exists only if $m\lambda/d\leq1$ .

Part B — Extended-Response Questions

Question 1 (SL) — Wavefronts and rays

(a) Define a wavefront.
(b) Explain what a ray represents and describe its relationship to wavefronts in an isotropic medium.

Show solution

(a) A wavefront is a line or surface joining points on a wave that are in the same phase, such as points along a crest.

(b) A ray represents the direction of wave propagation and energy transfer. In an isotropic medium, it is perpendicular to the wavefronts.

Question 2 (SL) — Reflection

A light ray is incident on a plane mirror at an angle of incidence of $35^\circ$ . Determine the angle of reflection and explain how the angle is measured.

Show solution

According to the law of reflection:

$\displaystyle \theta_r=\theta_i=35^\circ$

Both angles are measured between the relevant ray and the normal, which is perpendicular to the mirror surface.

Question 3 (SL) — Refraction reasoning

Light enters a transparent medium from air at an oblique angle. Its speed is lower in the medium. State whether the refracted ray bends towards or away from the normal. Explain what happens to its frequency and wavelength.

Show solution

The ray bends towards the normal. Its speed decreases, while its frequency remains unchanged across the stationary boundary.

$\displaystyle v=f\lambda,\qquad \lambda=\frac{v}{f}$

Since frequency is constant, a decrease in speed causes a decrease in wavelength. The refracted wavefronts are therefore more closely spaced.

At oblique incidence, different parts of a wavefront enter the slower medium at different times. This changes the wavefront orientation and bends the ray towards the normal.

Question 4 (SL) — Snell’s law

A ray of light travels from water of refractive index $1.33$ into air of refractive index $1.00$ . The angle of incidence in water is $40^\circ$ . Calculate the angle of refraction in air.

Show solution

Apply Snell’s law:

$\displaystyle n_1\sin\theta_1=n_2\sin\theta_2$

$\displaystyle \sin\theta_2=\frac{1.33\sin40^\circ}{1.00}=0.8549$

$\displaystyle \theta_2=\sin^{-1}(0.8549)=58.7^\circ$

The angle increases because light enters a medium with a lower refractive index and bends away from the normal.

Question 5 (SL) — Critical angle and total internal reflection

Light travels from glass of refractive index $1.50$ towards air of refractive index $1.00$ .

(a) Calculate the critical angle.
(b) State what happens when the angle of incidence exceeds the critical angle.

Show solution

(a) At the critical angle, the refracted angle is $90^\circ$ .

$\displaystyle n_1\sin\theta_c=n_2\sin90^\circ$

$\displaystyle \theta_c=\sin^{-1}\left(\frac{n_2}{n_1}\right)=\sin^{-1}\left(\frac{1.00}{1.50}\right)=41.8^\circ$

(b) Total internal reflection occurs. Light is reflected back into the glass, and no propagating refracted ray enters the air.

The two conditions are propagation from a higher to a lower refractive index and an incidence angle greater than the critical angle.

Question 6 (SL) — Superposition

Two wave pulses overlap at a point. At that instant, pulse 1 produces a displacement of $+3.0\,\mathrm{cm}$ and pulse 2 produces a displacement of $-1.5\,\mathrm{cm}$ . Determine the resultant displacement and state the principle used.

Show solution

The principle of superposition states that the resultant displacement is the algebraic sum of the individual displacements when waves overlap in a linear medium.

$\displaystyle y=y_1+y_2=3.0+(-1.5)=+1.5\,\mathrm{cm}$

The positive sign indicates that the resultant displacement is in the positive direction.

Question 7 (SL) — Interference conditions

Two coherent sources emit waves in phase. At an observation point, their path difference is:
(a) $\Delta=2.0\lambda$ ;
(b) $\Delta=2.5\lambda$ .
For each case, state whether the interference is constructive or destructive.

Show solution

For sources emitting in phase, constructive interference occurs when:

$\displaystyle \Delta=m\lambda$

Destructive interference occurs when:

$\displaystyle \Delta=\left(m+\frac{1}{2}\right)\lambda$

Here, $m=0,1,2,\ldots$ .

(a) $2.0\lambda$ is an integer multiple of the wavelength: constructive interference.

(b) $2.5\lambda$ is an odd multiple of half a wavelength: destructive interference. Complete cancellation requires equal amplitudes at the observation point.

Question 8 (HL) — Diffraction grating

A diffraction grating has 500 lines per millimetre. Light of wavelength $600\,\mathrm{nm}$ is incident normally. Calculate the angle to the second-order maximum, if it exists.

Show solution

Step 1: Convert the line density and calculate the grating spacing.

$\displaystyle N=500\times10^3\,\mathrm{m^{-1}}$

$\displaystyle d=\frac{1}{N}=\frac{1}{500\times10^3}=2.00\times10^{-6}\,\mathrm{m}$

Step 2: Apply the grating equation with $m=2$ .

$\displaystyle d\sin\theta=m\lambda$

$\displaystyle \sin\theta=\frac{2(600\times10^{-9})}{2.00\times10^{-6}}=0.600$

Since $\sin\theta<1$ , the second-order maximum exists.

$\displaystyle \theta=\sin^{-1}(0.600)=36.9^\circ$

Answer: The second-order maxima occur at $36.9^\circ$ on either side of the central maximum.

Question 9 (SL) — Speed, wavelength and refraction

Light of vacuum wavelength $640\,\mathrm{nm}$ enters a transparent material with refractive index $1.60$ . The angle of incidence from air is $50.0^\circ$ .

(a) Calculate the speed of light in the material.
(b) Calculate its frequency.
(c) Calculate its wavelength in the material.
(d) Calculate the angle of refraction.

Show solution

(a)

$\displaystyle v=\frac{c}{n}=\frac{3.00\times10^8}{1.60}=1.88\times10^8\,\mathrm{m\,s^{-1}}$

(b) Use the vacuum wavelength:

$\displaystyle f=\frac{c}{\lambda_0}=\frac{3.00\times10^8}{640\times10^{-9}}=4.69\times10^{14}\,\mathrm{Hz}$

The frequency remains unchanged across the stationary boundary.

(c)

$\displaystyle \lambda=\frac{\lambda_0}{n}=\frac{640}{1.60}=400\,\mathrm{nm}$

(d)

$\displaystyle \sin\theta_2=\frac{1.00\sin50.0^\circ}{1.60}$

$\displaystyle \theta_2=28.6^\circ$

Question 10 (SL; extension in part d) — A parallel-sided glass block

A light ray enters a rectangular glass block from air at $55.0^\circ$ to the normal. The glass has refractive index $1.50$ , and air surrounds the block. Its two refracting faces are parallel.

(a) Calculate the angle of refraction inside the block.
(b) Calculate the angle at which the ray emerges into air.
(c) Describe the relationship between the incident and emergent rays.
(d) Extension: The perpendicular thickness of the block is $20.0\,\mathrm{mm}$ . Calculate the lateral displacement using $h=t\sin(i-r)/\cos r$ , where $t$ is the thickness, $i$ the incidence angle and $r$ the refracted angle.

Show solution

(a)

$\displaystyle r=\sin^{-1}\left(\frac{\sin55.0^\circ}{1.50}\right)=33.1^\circ$

(b) The internal angle of incidence at the second face is also $r$ , because the faces are parallel.

$\displaystyle 1.50\sin r=1.00\sin e$

$\displaystyle e=55.0^\circ$

(c) The emergent ray is parallel to the incident ray but displaced sideways. The two refractions reverse the angular deviation because the external medium is the same on both sides.

(d)

$\displaystyle h=\frac{20.0\sin(55.0^\circ-33.1^\circ)}{\cos33.1^\circ}=8.90\,\mathrm{mm}$

Question 11 (SL) — Glass–water boundary

Light travels from glass of refractive index $1.50$ towards water of refractive index $1.33$ .

(a) Calculate the critical angle.
(b) Determine what happens at an incidence angle of $60.0^\circ$ and calculate the refracted angle if a refracted ray exists.
(c) State what happens at $70.0^\circ$ .
(d) Explain why total internal reflection cannot occur when light travels from water into glass.

Show solution

(a)

$\displaystyle \theta_c=\sin^{-1}\left(\frac{1.33}{1.50}\right)=62.5^\circ$

(b) Since $60.0^\circ<\theta_c$ , a propagating refracted ray enters the water.

$\displaystyle \theta_2=\sin^{-1}\left(\frac{1.50\sin60.0^\circ}{1.33}\right)=77.6^\circ$

(c) Since $70.0^\circ>\theta_c$ , total internal reflection occurs.

(d) Total internal reflection requires propagation from a higher refractive index to a lower one. Water-to-glass propagation has the opposite direction of refractive-index change.

Question 12 (SL) — Sound interference and unequal amplitudes

Two coherent speakers emit sound in phase at $680\,\mathrm{Hz}$ . The speed of sound is $340\,\mathrm{m\,s^{-1}}$ . At point P, the distances from the speakers are $4.00\,\mathrm{m}$ and $4.75\,\mathrm{m}$ . The individual displacement amplitudes at P are $3.0\,\mu\mathrm{m}$ and $2.0\,\mu\mathrm{m}$ .

(a) Calculate the wavelength.
(b) Determine whether interference at P is constructive or destructive.
(c) Calculate the resultant displacement amplitude at P.
(d) Explain why destructive interference does not always mean zero resultant amplitude.

Show solution

(a)

$\displaystyle \lambda=\frac{v}{f}=\frac{340}{680}=0.500\,\mathrm{m}$

(b)

$\displaystyle \Delta=4.75-4.00=0.75\,\mathrm{m}=1.5\lambda$

The sources emit in phase, so this odd multiple of half a wavelength produces destructive interference. The waves arrive in antiphase.

(c)

$\displaystyle A_{\mathrm{resultant}}=|A_1-A_2|=|3.0-2.0|=1.0\,\mu\mathrm{m}$

(d) Complete cancellation requires equal amplitudes as well as antiphase arrival. Unequal amplitudes leave a non-zero resultant amplitude.

Question 13 (SL challenge) — Sources emitting in antiphase

Two coherent sources emit waves in antiphase, with wavelength $0.040\,\mathrm{m}$ . At point P, the path difference is $0.080\,\mathrm{m}$ . At point Q, it is $0.100\,\mathrm{m}$ .

(a) State the initial phase difference between the sources.
(b) Classify the interference at P.
(c) Classify the interference at Q.
(d) Explain why the usual path-difference rules for in-phase sources must be modified.

Show solution

(a) The initial phase difference is $\pi$ radians.

(b) At P, $\Delta=2\lambda$ . The path difference contributes $4\pi$ radians. Combined with the initial phase difference, the arrival phase difference is equivalent to $\pi$ radians: destructive interference.

(c) At Q, $\Delta=2.5\lambda$ . The path difference contributes $5\pi$ radians. Combined with the initial phase difference, the arrival phase difference is equivalent to zero: constructive interference.

(d) Arrival phase depends on both the initial source phase difference and the phase difference due to unequal paths. For antiphase sources, integer-wavelength path differences produce destructive interference, and half-integer differences produce constructive interference.

Question 14 (SL) — Young’s double-slit experiment

Monochromatic light of wavelength $550\,\mathrm{nm}$ illuminates two narrow slits separated by $0.250\,\mathrm{mm}$ . A screen is $2.00\,\mathrm{m}$ from the slits. Use the small-angle approximation.

(a) Calculate the fringe spacing.
(b) Determine the distance from the central bright fringe to the fourth bright fringe.
(c) Determine the new fringe spacing if the screen distance is reduced to $1.00\,\mathrm{m}$ .
(d) Explain why both slits are illuminated by the same light source.

Show solution

(a)

$\displaystyle w=\frac{\lambda D}{s}$

$\displaystyle w=\frac{550\times10^{-9}(2.00)}{0.250\times10^{-3}}=4.40\times10^{-3}\,\mathrm{m}=4.40\,\mathrm{mm}$

(b) The central maximum is order zero.

$\displaystyle y_4=4w=17.6\,\mathrm{mm}$

(c) Fringe spacing is proportional to screen distance. Halving the distance halves the fringe spacing to $2.20\,\mathrm{mm}$ .

(d) Light from the same source produces a stable relative phase at the slits, making them coherent secondary sources and allowing a stable interference pattern.

Question 15 (SL) — Measuring wavelength and uncertainty

Seven consecutive bright fringes are observed in a double-slit experiment. The distance from the first to the seventh is $15.0\pm0.3\,\mathrm{mm}$ . The slit separation is $0.300\,\mathrm{mm}$ and the screen distance is $1.50\,\mathrm{m}$ . Treat uncertainties in slit separation and screen distance as negligible.

(a) Calculate the fringe spacing.
(b) Determine the wavelength.
(c) Estimate the percentage uncertainty and absolute uncertainty in wavelength.
(d) Explain the advantage of measuring across several fringes.

Show solution

(a) Seven bright fringes span six intervals.

$\displaystyle w=\frac{15.0}{6}=2.50\,\mathrm{mm}$

(b)

$\displaystyle \lambda=\frac{ws}{D}$

$\displaystyle \lambda=\frac{2.50\times10^{-3}(0.300\times10^{-3})}{1.50}=5.00\times10^{-7}\,\mathrm{m}=500\,\mathrm{nm}$

(c) Wavelength is directly proportional to the measured span under the stated assumptions.

$\displaystyle \frac{\delta\lambda}{\lambda}=\frac{0.3}{15.0}=0.020=2.0\%$

$\displaystyle \delta\lambda=0.020(500)=10\,\mathrm{nm}$

Result: $\lambda=500\pm10\,\mathrm{nm}$ .

(d) For a similar absolute uncertainty in locating the endpoints, measuring a larger span reduces the percentage uncertainty. Dividing by the number of intervals then gives a more precise fringe spacing.

Question 16 (HL) — Single-slit diffraction

Light of wavelength $600\,\mathrm{nm}$ is incident normally on a slit of width $0.120\,\mathrm{mm}$ . The diffraction pattern is observed on a screen $3.00\,\mathrm{m}$ away.

(a) Calculate the angle to the first minimum.
(b) Estimate the width of the central maximum on the screen, using the small-angle approximation.
(c) Predict the central-maximum width if the slit width is halved.
(d) Describe how the central maximum differs from the adjacent secondary maxima.

Show solution

(a)

$\displaystyle a\sin\theta_1=\lambda$

$\displaystyle \theta_1=\sin^{-1}\left(\frac{600\times10^{-9}}{0.120\times10^{-3}}\right)=0.286^\circ$

(b) The central maximum extends between the first minima on either side.

$\displaystyle W\approx\frac{2D\lambda}{a}$

$\displaystyle W=\frac{2(3.00)(600\times10^{-9})}{0.120\times10^{-3}}=0.0300\,\mathrm{m}=30.0\,\mathrm{mm}$

(c) In the small-angle approximation, $W\propto1/a$ . Halving the slit width doubles the width to $60.0\,\mathrm{mm}$ .

(d) The central maximum is the brightest and is approximately twice the width of each adjacent secondary maximum, measured between successive minima on a distant screen.

Question 17 (HL) — Grating angles and maximum orders

A diffraction grating has 600 lines per millimetre. Blue light of wavelength $450\,\mathrm{nm}$ and red light of wavelength $650\,\mathrm{nm}$ are incident normally.

(a) Calculate the grating spacing.
(b) Calculate the first-order angle for each wavelength.
(c) Determine the highest allowed order for each wavelength.
(d) Explain which colour is farther from the central maximum in the same order.

Show solution

(a)

$\displaystyle d=\frac{1}{600\times10^3}=1.67\times10^{-6}\,\mathrm{m}$

(b) Use the unrounded grating spacing and $m=1$ .

$\displaystyle \theta_{\mathrm{blue}}=\sin^{-1}\left(\frac{450\times10^{-9}}{d}\right)=15.7^\circ$

$\displaystyle \theta_{\mathrm{red}}=\sin^{-1}\left(\frac{650\times10^{-9}}{d}\right)=23.0^\circ$

(c) Allowed orders require $m\lambda/d\leq1$ . Choose the greatest integer not exceeding $d/\lambda$ .

For blue light, $d/\lambda=3.70$ , giving a highest order of 3. For red light, $d/\lambda=2.56$ , giving a highest order of 2.

(d) Red light has the larger wavelength. For the same order and grating, $\sin\theta=m\lambda/d$ is larger, so the red maximum lies farther from the central maximum.

Question 18 (HL challenge) — Double-slit fringes inside a diffraction envelope

Two identical slits each have width $a=0.0500\,\mathrm{mm}$ and centre-to-centre separation $s=0.200\,\mathrm{mm}$ . Light of wavelength $500\,\mathrm{nm}$ illuminates them normally. A screen is $2.00\,\mathrm{m}$ away. Use the small-angle approximation for distances on the screen.

(a) Calculate the interference fringe spacing.
(b) Calculate the width of the central single-slit diffraction envelope.
(c) Identify the first positive interference order missing because it coincides with a diffraction minimum.
(d) Determine how many interference bright fringes lie strictly inside the central diffraction envelope.
(e) Explain why their intensities are not all equal.

Show solution

(a)

$\displaystyle w=\frac{\lambda D}{s}=\frac{500\times10^{-9}(2.00)}{0.200\times10^{-3}}=5.00\,\mathrm{mm}$

(b)

$\displaystyle W\approx\frac{2D\lambda}{a}=\frac{2(2.00)(500\times10^{-9})}{0.0500\times10^{-3}}=40.0\,\mathrm{mm}$

The first diffraction minima are at approximately $y=\pm20.0\,\mathrm{mm}$ .

(c) Interference maxima satisfy $s\sin\theta=m\lambda$ . At the first diffraction minimum, $a\sin\theta=\lambda$ . Combining these equations gives:

$\displaystyle m=\frac{s}{a}=\frac{0.200}{0.0500}=4$

The fourth-order maximum is missing on each side.

(d) The orders strictly inside the envelope are $m=-3,-2,-1,0,1,2,3$ : seven bright fringes. Orders $m=\pm4$ coincide with the diffraction minima and are absent.

(e) Each finite-width slit produces a diffraction envelope. This modulates the intensity of the double-slit interference pattern, so the fringes become weaker towards the envelope minima.

Key takeaways

Wavefronts join points of equal phase; rays indicate propagation direction.
Reflection: $\theta_i=\theta_r$ .
Wave equation: $v=f\lambda$ .
Refractive index: $n=c/v$ .
Snell’s law: $n_1\sin\theta_1=n_2\sin\theta_2$ .
Critical angle: $\sin\theta_c=n_2/n_1$ , for propagation from a higher to a lower refractive index.
Superposition: $y=y_1+y_2$ .
Coherent sources have the same frequency and a constant phase difference.
For in-phase sources, constructive interference occurs at $\Delta=m\lambda$ .
For in-phase sources, destructive interference occurs at $\Delta=(m+\frac{1}{2})\lambda$ .
Complete destructive cancellation requires equal amplitudes.
Young’s double-slit fringe spacing: $w=\lambda D/s$ , using the small-angle approximation.
Single-slit minima (HL): $a\sin\theta=m\lambda$ , where $m=1,2,3,\ldots$ .
Central single-slit maximum width (HL): $W\approx2D\lambda/a$ .
Grating principal maxima at normal incidence (HL): $d\sin\theta=m\lambda$ .
Convert grating line density to lines per metre before calculating $d=1/N$ .
Count intervals carefully: seven consecutive bright fringes span six fringe spacings.

Following Jesus, the Light of the World

Bible verse — John 8:12 (KJV):
“I am the light of the world: he that followeth me shall not walk in darkness, but shall have the light of life.”

Creation: Remember the Creator of light

Genesis 1:3–4 records that God called light into being and saw that it was good. Hebrews 1:2–3 teaches that God made the universe through His Son, who sustains all things by His powerful word. The light whose behaviour we investigate belongs to the creation of Jesus Christ.

As you apply equations and explain wave phenomena, take time to appreciate the order of creation. Careful observation, sound reasoning and honest calculations are ways to approach your studies responsibly and with gratitude to God.

The Fall: Our need for truth

Sin draws us away from God and into spiritual darkness. First John 1:5–8 teaches that God is light, that there is no darkness in Him, and that we must be honest about our sin. Walking in His light includes honesty in our learning: doing our own work, admitting mistakes and treating others with love.

Physical light is electromagnetic radiation. The statement that God is light reveals His holiness, goodness and truth. The light we measure is created; the Creator Himself is eternal.

Redemption: Come to Jesus

Jesus, the Light of the world, came to rescue us from sin. Through His death and resurrection, He offers forgiveness and new life. First John 1:7–9 teaches that His blood cleanses us from sin and that God forgives those who confess their sins. We can bring our failures to Him and trust His grace.

Restoration: Walk in His light

Revelation 21:23–25 points to the hope of a restored creation illuminated by God’s glory, with the Lamb as its lamp. As we look forward to that future, Jesus calls us to follow Him today in truth, love and obedience.

Your response: Receive Jesus Christ as your personal God, Saviour and Lord. Grow in your relationship with Him by reading and studying the Bible, spending time in prayer, and joining other believers in a local church. Let His light guide both your studies and your daily decisions.

Reflection: How can you demonstrate honesty, humility and love while preparing for examinations and helping others learn?

Prayer: Lord Jesus, thank You for creating light and bringing us the light of life. Forgive my sins and help me follow You. Give me understanding, perseverance and honesty in my studies. Help me grow through Your Word, prayer and fellowship with other believers. Amen.

IB Physics HL Book 2025; Papers 1 & 2 Practice 5 Sections65 Lessons54 Weeks

Curriculum

Wave phenomena HL Paper 1&2

C.3 Wave Phenomena Papers 1&2

Instructions

Part A — Multiple-Choice Questions

A1 — Wavefronts and rays (SL)

A2 — Light entering glass (SL)

A3 — Total internal reflection (SL)

A4 — Diffraction (SL)

A5 — Coherent sources (SL)

A6 — Double-slit fringe spacing (SL)

A7 — Single-slit diffraction (HL)

A8 — Diffraction grating (HL)

Part B — Extended-Response Questions

Question 1 (SL) — Wavefronts and rays

Question 2 (SL) — Reflection

Question 3 (SL) — Refraction reasoning

Question 4 (SL) — Snell’s law

Question 5 (SL) — Critical angle and total internal reflection

Question 6 (SL) — Superposition

Question 7 (SL) — Interference conditions

Question 8 (HL) — Diffraction grating

Question 9 (SL) — Speed, wavelength and refraction

Question 10 (SL; extension in part d) — A parallel-sided glass block

Question 11 (SL) — Glass–water boundary

Question 12 (SL) — Sound interference and unequal amplitudes

Question 13 (SL challenge) — Sources emitting in antiphase

Question 14 (SL) — Young’s double-slit experiment

Question 15 (SL) — Measuring wavelength and uncertainty

Question 16 (HL) — Single-slit diffraction

Question 17 (HL) — Grating angles and maximum orders

Question 18 (HL challenge) — Double-slit fringes inside a diffraction envelope

Key takeaways

Following Jesus, the Light of the World

Creation: Remember the Creator of light

The Fall: Our need for truth

Redemption: Come to Jesus

Restoration: Walk in His light

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