Wave phenomena

Lesson C.3 Wave Phenomena

Lesson Notes, Worked Examples and Exam Practice

Waves can reflect, change direction at a boundary, spread through gaps and combine with other waves. These behaviours explain mirrors, optical fibres, interference fringes and diffraction spectra. This lesson connects each observation to a physical explanation and the appropriate equation.

Learning objectives

Describe wavefronts and rays, and sketch their behaviour at boundaries and apertures.
Explain reflection, transmission, refraction, diffraction and superposition.
Apply Snell’s law, critical-angle and double-slit equations.
For HL, interpret single-slit and grating patterns, including the diffraction envelope.

Lesson map

SL and HL: Sections C.3.1–C.3.6 cover wavefronts, boundaries, qualitative diffraction, superposition, two-source interference and Young’s experiment.

Additional HL: Sections C.3.7–C.3.9 cover quantitative single-slit diffraction, multiple slits, gratings and the combined diffraction/interference pattern.

IB notation and working conventions

Symbol	Meaning
$x,\quad x_0$	Displacement and amplitude; subscripts distinguish individual waves.
$n_1,\quad n_2$	Refractive indices of media 1 and 2.
$\Delta$	Path difference between two waves.
$s$	Spacing between adjacent double-slit bright fringes.
$d$	Centre-to-centre slit separation, or spacing between adjacent grating lines.
$D$	Distance from slits to screen.
$b$	Width of one slit.
$n$	Integer interference order when used in interference or grating equations.

The meaning of $n$ depends on context: refractive index in refraction, integer order in interference. Use $c=3.00\times10^8\,\mathrm{m\,s^{-1}}$ and $n_{\mathrm{air}}=1.00$ unless stated otherwise. Measure ray angles from the normal, convert lengths to metres, retain working digits and give answers to 2–3 significant figures.

Assume stationary boundaries and linear wave behaviour. Slit and grating calculations below use normal incidence. Small-angle approximations require angles in radians.

C.3.1 Wavefronts and Rays — SL and HL

A wavefront joins points of equal phase, such as points along a crest. A ray indicates the direction of wave propagation and energy transfer. In an isotropic medium, whose properties are the same in all directions, rays are perpendicular to wavefronts.

Plane wavefronts are parallel straight lines in a two-dimensional sketch; the rays are parallel.
A point source produces circular wavefronts in a two-dimensional sketch and spherical wavefronts in three dimensions; rays point radially outwards.
The distance between consecutive wavefronts representing successive crests is one wavelength, $\lambda$ .

How to sketch a wave diagram

Draw several wavefronts with consistent spacing, then add ray arrows perpendicular to them. At a boundary, draw the normal before measuring any angle. After a change of medium, show any change in both wavefront spacing and direction.

C.3.2 Reflection, Transmission and Refraction — SL and HL

Reflection and transmission

Reflection sends wave energy back into the original medium. Transmission carries wave energy across a boundary. At many boundaries, part of the energy is reflected and part is transmitted; absorption may also occur.

$\displaystyle \theta_i=\theta_r$

The incident ray, reflected ray and normal lie in the same plane. Both angles are measured from the normal, not from the surface. Reflection in the same unchanged medium preserves frequency, speed and wavelength.

Worked example 1 — Reflection angles

A ray meets a plane mirror at $35^\circ$ to the normal. Find its angle of reflection and its angle to the mirror surface.

Show answer and explanation

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

$\displaystyle \text{Angle to surface}=90^\circ-35^\circ=55^\circ$

The normal is perpendicular to the surface, so these two angles are complementary.

Refraction: Why the direction changes

Refraction occurs when a wave changes speed as it enters another medium. At oblique incidence, one part of a wavefront reaches the new medium before another part. The change in speed rotates the wavefront and changes the ray direction. At normal incidence, speed and wavelength can change without a change in direction.

$\displaystyle v=f\lambda$

The frequency remains unchanged across a stationary boundary because the transmitted oscillations must follow the incoming oscillations. Wavelength therefore changes in proportion to wave speed.

Change on entering medium 2	Ray behaviour at oblique incidence	Wavelength
Speed decreases	Bends towards the normal	Decreases
Speed increases	Bends away from the normal, if a transmitted ray exists	Increases
Speed unchanged	No refraction of direction	Unchanged

Refractive index and Snell’s law

$\displaystyle n=\frac{c}{v}$

For light, refractive index compares its speed in vacuum with its speed in the material. A larger refractive index means a smaller light speed.

IB refraction relation

$\displaystyle \frac{n_1}{n_2}=\frac{\sin\theta_2}{\sin\theta_1}=\frac{v_2}{v_1}$

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

The second equation is the usual rearranged form of the first. Angles belong to the corresponding media and are measured from the normal.

Worked example 2 — Light entering glass

Light of vacuum wavelength $640\,\mathrm{nm}$ enters glass of refractive index $1.60$ from air at $50.0^\circ$ . Calculate its speed, frequency, wavelength in glass and refraction angle.

Show answer and explanation

Speed in glass:

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

Frequency, using the vacuum wavelength:

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

Wavelength in glass:

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

Refraction angle:

$\displaystyle \theta_2=\sin^{-1}\left(\frac{1.00\sin50.0^\circ}{1.60}\right)=28.6^\circ$

Frequency is unchanged. Speed and wavelength decrease, and the ray bends towards the normal.

A parallel-sided glass block

If the external medium is the same on both sides, a ray passing through parallel faces emerges parallel to its original direction, but displaced sideways. The second refraction reverses the angular deviation caused by the first. Inside the block, the wavelength and speed are smaller than in air.

C.3.3 Critical Angle and Total Internal Reflection — SL and HL

When light travels from a higher refractive index to a lower one, the refracted ray bends away from the normal. At the critical angle, its angle of refraction is 90°, so the refracted ray travels along the boundary.

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

$\displaystyle \sin\theta_c=\frac{n_2}{n_1},\qquad n_1>n_2$

Conditions for total internal reflection

The wave travels from a higher to a lower refractive index: $n_1>n_2$ .
The incidence angle exceeds the critical angle: $\theta_i>\theta_c$ .

There is then no propagating refracted ray into the second medium. In the ideal lossless model, all incident energy is reflected. At the critical angle itself, the refracted ray is along the boundary.

Worked example 3 — Glass–air boundary

Light travels from glass of refractive index $1.50$ towards air. Calculate the critical angle and determine what happens at an incidence angle of $50.0^\circ$ .

Show answer and explanation

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

Since $50.0^\circ>41.8^\circ$ and the light travels from higher to lower refractive index, total internal reflection occurs.

Optical fibres: A core with a higher refractive index than its surrounding cladding guides light through repeated total internal reflection. The incidence angles at the core–cladding boundary must exceed the critical angle.

C.3.4 Diffraction — SL and HL

Diffraction is the spreading of waves through an aperture or around an obstacle. It is most noticeable when the aperture width or obstacle size is comparable to the wavelength. For a given wavelength, reducing the gap width increases angular spreading. For a given gap width, increasing the wavelength also increases spreading.

In a wavefront sketch, wide-gap transmission is mostly straight ahead with spreading near the edges. When the gap is comparable to the wavelength, curved wavefronts spread over a much wider range of directions. Keep the wavelength unchanged if the medium and frequency are unchanged.

Examples and comparisons

Sound can spread through a doorway because many audible wavelengths are comparable to ordinary openings.
Visible light has a much shorter wavelength, so diffraction through an ordinary doorway is difficult to notice.
Refraction changes propagation through a change of speed; diffraction occurs through spreading at apertures and edges.

C.3.5 Superposition and Two-Source Interference — SL and HL

Principle of superposition

When waves overlap in a linear medium, the resultant displacement at each point is the algebraic sum of their individual displacements. Add displacements with their signs; do not add intensities directly for coherent waves.

$\displaystyle x=x_1+x_2$

After two pulses pass through one another in an ideal linear medium, they continue with their original shapes. Their temporary overlap does not permanently destroy either pulse.

Coherence and interference

Coherent sources have the same frequency and a constant phase difference. Equal amplitudes are not required for coherence. Interference is the pattern of enhanced and reduced resultant amplitude produced by overlapping waves.

Constructive interference: waves arrive in phase and their displacements reinforce.
Destructive interference: waves arrive in antiphase and their displacements oppose.
Complete cancellation requires equal amplitudes at the observation point as well as antiphase arrival.

Path-difference conditions for sources emitting in phase

$\displaystyle \text{Constructive:}\quad\Delta=n\lambda$

$\displaystyle \text{Destructive:}\quad\Delta=\left(n+\frac12\right)\lambda$

Here $\Delta$ is the magnitude of the path difference and $n=0,1,2,\ldots$ . These conditions assume in-phase emission. If the sources emit in antiphase, the two conditions exchange roles.

A path difference of one wavelength produces one full cycle of phase difference. Whether waves reinforce depends on both their source phase difference and their different path lengths.

$\displaystyle \text{Phase difference due to paths}=\frac{2\pi\Delta}{\lambda}$

Amplitude and intensity

$\displaystyle x_{0,\mathrm{max}}=x_{0,1}+x_{0,2}$

$\displaystyle x_{0,\mathrm{min}}=|x_{0,1}-x_{0,2}|$

For the same type of wave in the same medium, $I\propto x_0^2$ . Two equal-amplitude coherent waves produce a maximum intensity four times that of either wave alone, and zero intensity at perfect destructive cancellation. Energy is redistributed between bright and dark regions; it is not destroyed.

Worked example 4 — Sound interference and unequal amplitudes

Two coherent speakers emit in phase at $680\,\mathrm{Hz}$ . Sound speed is $340\,\mathrm{m\,s^{-1}}$ . Their distances to P are $4.00\,\mathrm{m}$ and $4.75\,\mathrm{m}$ . Individual displacement amplitudes at P are $3.0\,\mu\mathrm{m}$ and $2.0\,\mu\mathrm{m}$ . Find the interference type and resultant amplitude.

Show answer and explanation

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

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

The path difference is a half-integer multiple of the wavelength, so the waves arrive in antiphase and interfere destructively.

$\displaystyle x_{0,\mathrm{resultant}}=|3.0-2.0|=1.0\,\mu\mathrm{m}$

The resultant amplitude is not zero because the individual amplitudes are unequal.

C.3.6 Young’s Double-Slit Experiment — SL and HL

Monochromatic light illuminates two narrow slits from the same source. The slits act as coherent secondary sources, producing alternating bright and dark fringes on a distant screen. The central bright fringe corresponds to equal path lengths.

IB fringe-spacing equation

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

$s$ is the separation of adjacent bright fringes, $d$ the slit separation and $D$ the screen distance. The same spacing separates adjacent dark fringes. A bright fringe and its neighbouring dark fringe are separated by $s/2$ .

Deriving the fringe spacing

For a screen much farther from the slits than their separation, the path difference is approximately $d\sin\theta$ . At small angles, $\sin\theta\approx\tan\theta\approx y/D$ , where $y$ is distance from the central maximum.

$\displaystyle \Delta\approx\frac{dy}{D}$

$\displaystyle \frac{dy_n}{D}=n\lambda\quad\Rightarrow\quad y_n=\frac{n\lambda D}{d}$

$\displaystyle s=y_{n+1}-y_n=\frac{\lambda D}{d}$

Change, with other quantities fixed	Effect on fringe spacing
Increase wavelength	Spacing increases
Increase screen distance	Spacing increases
Increase slit separation	Spacing decreases

Two independent lamps generally do not maintain a stable relative phase, so a stable fringe pattern is washed out by rapidly changing phase differences. Illuminating both slits coherently avoids this problem.

Worked example 5 — Fringe spacing and order

Light of wavelength $550\,\mathrm{nm}$ illuminates slits separated by $0.250\,\mathrm{mm}$ . The screen is $2.00\,\mathrm{m}$ away. Find the fringe spacing and the distance from the central maximum to the fourth bright fringe.

Show answer and explanation

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

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

The central bright fringe is order zero. Count four complete spacings to the fourth-order bright fringe.

Measuring wavelength accurately

Measure the distance across several fringe intervals and divide by the number of intervals. Seven consecutive bright fringes span six spacings. Measuring a larger span usually reduces the percentage uncertainty for a similar endpoint-reading uncertainty.

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

If the uncertainties in slit separation and screen distance are negligible, the percentage uncertainty in wavelength equals the percentage uncertainty in the measured fringe span. Otherwise, include the uncertainties in all measured factors.

C.3.7 Single-Slit Diffraction — Additional HL

A single rectangular slit illuminated by monochromatic light produces a broad, bright central maximum, dark minima and weaker secondary maxima. Contributions from different parts of the slit interfere with one another.

IB first-minimum approximation

$\displaystyle \theta\approx\frac{\lambda}{b}$

$b$ is slit width. $\theta$ is the angle from the centre to the first minimum on one side, not the full angular width. This small-angle expression gives $\theta$ in radians.

$\displaystyle \text{Full central angular width}\approx\frac{2\lambda}{b}$

At the first minimum, the path difference across the full slit is one wavelength. Dividing the slit into equal halves pairs contributions separated by half a wavelength, giving destructive cancellation.

$\displaystyle b\sin\theta_n=n\lambda,\qquad n=1,2,3,\ldots$

This equation identifies minima. It does not give the positions of the single-slit bright maxima. For a distant screen, the central maximum extends between the first minima on opposite sides.

$\displaystyle W=2D\tan\theta_1\approx\frac{2D\lambda}{b}$

The central maximum is approximately twice as wide as the adjacent secondary maxima on a distant screen.
Reducing slit width broadens the pattern. With incident intensity and illumination fixed, a narrower slit also transmits less power and lowers the central peak intensity.
Increasing wavelength broadens the pattern for unchanged slit width.

Worked example 6 — Central diffraction maximum (HL)

Light of wavelength $600\,\mathrm{nm}$ passes through a slit of width $0.120\,\mathrm{mm}$ . The screen is $3.00\,\mathrm{m}$ away. Find the first-minimum angle and approximate central-maximum width.

Show answer and explanation

$\displaystyle \theta_1\approx\frac{\lambda}{b}=\frac{600\times10^{-9}}{0.120\times10^{-3}}=5.00\times10^{-3}\,\mathrm{rad}$

$\displaystyle \theta_1\approx0.286^\circ$

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

The factor of two is needed because the central maximum extends to the first minimum on each side.

C.3.8 Multiple Slits and Diffraction Gratings — Additional HL

A diffraction grating contains many equally spaced slits or lines. At particular angles, waves from successive slits arrive in phase, producing sharp principal maxima. Adding more illuminated slits at unchanged spacing makes the principal maxima narrower and more intense; their angular positions remain unchanged.

IB grating equation at normal incidence

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

$n=0,1,2,\ldots$ is order and $d$ is grating spacing. The central maximum has $n=0$ ; corresponding non-zero orders occur on both sides.

The path difference between contributions from adjacent slits is $d\sin\theta$ . A principal maximum occurs when that difference is an integer number of wavelengths.

$\displaystyle d=\frac{1}{N}$

$N$ is line density in lines per metre. Multiply lines per millimetre by $10^3$ before taking the reciprocal.

$\displaystyle n\lambda\leq d$

An order is allowed only if $\sin\theta\leq1$ . Choose the greatest integer not exceeding $d/\lambda$ to find the maximum mathematically allowed order. An order at exactly 90° travels along the grating plane and does not reach a finite screen parallel to the grating.

Worked example 7 — Grating angle and highest order (HL)

A grating has 500 lines per millimetre. Light of wavelength $600\,\mathrm{nm}$ is incident normally. Calculate the second-order angle and the highest allowed order.

Show answer and explanation

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

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

$\displaystyle \theta=36.9^\circ$

$\displaystyle \frac{d}{\lambda}=\frac{2.00\times10^{-6}}{600\times10^{-9}}=3.33$

Answer: Second-order maxima are at 36.9° on either side of the centre. The highest allowed order is 3.

White light and spectra

All wavelengths overlap at the central maximum, so it appears white. In each non-zero order, wavelengths separate: violet is nearer the centre and red farther away because longer wavelengths give larger angles. Spectra from different orders can overlap.

Increasing line density reduces grating spacing and increases the angle for a given wavelength and order. Some higher orders may then cease to exist.

C.3.9 The Diffraction Envelope — Additional HL

Real double slits have finite width. Each slit diffracts light, and their waves also interfere with one another. The double-slit fringes sit within a single-slit diffraction envelope, so fringe intensities decrease towards an envelope minimum.

Quantity changed	Main effect, with other quantities fixed
Slit separation $d$	Changes spacing of the interference fringes.
Individual slit width $b$	Changes width of the diffraction envelope.

A missing order occurs when an interference maximum coincides with a diffraction minimum. At the first envelope minimum, combining the two conditions gives:

$\displaystyle d\sin\theta=n\lambda,\qquad b\sin\theta=\lambda$

$\displaystyle n=\frac{d}{b}$

This is a missing interference order only when the ratio is an integer. Distinguish a dark diffraction minimum from a dark fringe between neighbouring interference maxima.

Worked example 8 — Missing orders (HL)

Two slits have width $b=0.0500\,\mathrm{mm}$ and separation $d=0.200\,\mathrm{mm}$ . Light of wavelength $500\,\mathrm{nm}$ forms a pattern on a screen $2.00\,\mathrm{m}$ away. Find the fringe spacing, central-envelope width and number of bright fringes strictly inside the envelope.

Show answer and explanation

$\displaystyle s=\frac{\lambda D}{d}=5.00\,\mathrm{mm}$

$\displaystyle W\approx\frac{2D\lambda}{b}=40.0\,\mathrm{mm}$

$\displaystyle n=\frac{d}{b}=\frac{0.200}{0.0500}=4$

The fourth-order maxima coincide with the first diffraction minima and are missing on both sides. The orders strictly inside the central envelope are −3, −2, −1, 0, 1, 2 and 3: seven bright fringes.

Practice Questions — Expandable Answers

Attempt each question before opening its answer. These are original practice questions, not official IB examination questions.

Question 1 — Reflection and refraction (SL and HL)

A ray makes $25^\circ$ with a mirror surface. (a) Find the incidence and reflection angles. (b) Explain why light entering glass from air at oblique incidence bends towards the normal.

Show answer and explanation

(a) Both angles to the normal are 90° − 25° = 65°.

(b) Light slows in glass. One part of the wavefront enters first and slows first, rotating the wavefront so that the ray bends towards the normal.

Question 2 — Water to air (SL and HL)

Light travels from water, $n_1=1.33$ , into air, $n_2=1.00$ , at an incidence angle of $40.0^\circ$ . Find the refraction angle and describe the changes in frequency and wavelength.

Show answer and explanation

$\displaystyle \theta_2=\sin^{-1}(1.33\sin40.0^\circ)=58.7^\circ$

Frequency remains unchanged. Speed and wavelength increase in air. The ray bends away from the normal.

Question 3 — Glass to water (SL and HL)

Light travels from glass of refractive index $1.50$ towards water of refractive index $1.33$ . Calculate the critical angle and decide whether total internal reflection occurs at $60.0^\circ$ and $70.0^\circ$ .

Show answer and explanation

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

At 60.0°, the angle is below the critical angle, so a refracted ray exists. At 70.0°, total internal reflection occurs.

Question 4 — Superposition and source phase (SL and HL)

(a) Two pulses produce displacements $+3.0\,\mathrm{cm}$ and $-1.5\,\mathrm{cm}$ at the same point. Find the resultant displacement. (b) Two coherent sources emit in antiphase. Classify interference for path differences $2\lambda$ and $2.5\lambda$ .

Show answer and explanation

$\displaystyle x=x_1+x_2=3.0+(-1.5)=+1.5\,\mathrm{cm}$

(b) For antiphase sources, an integer-wavelength path difference preserves antiphase arrival: destructive at 2 wavelengths. A half-integer difference reverses it: constructive at 2.5 wavelengths.

Question 5 — Wavelength and uncertainty (SL and HL)

Seven consecutive bright fringes span $15.0\pm0.3\,\mathrm{mm}$ . Slit separation is $0.300\,\mathrm{mm}$ and screen distance is $1.50\,\mathrm{m}$ . Treat uncertainties in these last two values as negligible. Find the wavelength and its uncertainty.

Show answer and explanation

Seven fringes span six intervals.

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

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

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

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

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

Question 6 — Narrower slit (HL)

A slit produces a central diffraction maximum $30.0\,\mathrm{mm}$ wide on a distant screen. The slit width is halved while wavelength, incident illumination and screen distance are unchanged. Predict the central width and describe the change in its peak intensity.

Show answer and explanation

Since $W\propto1/b$ , the width doubles to 60.0 mm. The narrower slit transmits less power, and the central peak intensity decreases.

Question 7 — Two colours through a grating (HL)

A grating has 600 lines per millimetre. Blue light has wavelength $450\,\mathrm{nm}$ and red light $650\,\mathrm{nm}$ . Calculate their first-order angles and highest allowed orders.

Show answer and explanation

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

Use the unrounded grating spacing in the following calculations.

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

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

For blue, the spacing-to-wavelength ratio is 3.70, so the highest order is 3. For red it is 2.56, so the highest order is 2. Red lies farther from the centre in the same order.

Question 8 — Fringes and envelope (HL)

In a double-slit experiment, (a) slit separation is doubled with slit width unchanged; (b) slit width is halved with slit separation unchanged. Predict the fringe spacing and central-envelope width in each case. Keep wavelength and screen distance fixed.

Show answer and explanation

(a) Fringe spacing halves; envelope width is unchanged.

(b) Fringe spacing is unchanged; envelope width doubles.

Slit separation controls interference spacing, while individual slit width controls the diffraction envelope.

Exam readiness checklist

Draw rays perpendicular to wavefronts and measure ray angles from the normal.
At a stationary boundary, keep frequency unchanged and use wave speed to find the wavelength change.
Check both total-internal-reflection conditions before using the critical angle.
State source phase assumptions when applying path-difference rules.
Use $s=\lambda D/d$ with consistent length units and count fringe intervals carefully.
For HL, distinguish first-minimum angle $\lambda/b$ from full central width $2\lambda/b$ .
Convert grating line density before calculating spacing; reject impossible orders.
Keep interference fringe spacing separate from diffraction-envelope width.

In God’s Light

Memory verse — Psalm 36:9 (KJV):
“For with thee is the fountain of life: in thy light shall we see light.”

Creation: Wonder and gratitude

Light makes the world visible and helps us investigate creation. Christians recognise God as the Creator and give thanks for the order that makes scientific learning possible. Studying reflection, refraction and interference invites careful observation and wonder at His works.

The Fall: Honesty about our need

Sin separates us from God and affects our choices. We can be tempted to hide mistakes, copy another person’s work or put achievement above love. God calls us to truth. In our studies, we can practise humility by checking our assumptions, admitting errors and treating others with patience.

Redemption: Come to Jesus Christ

Jesus Christ came to save us from sin. Through His death and resurrection, He offers forgiveness and new life. Turn to Him in repentance and faith, and receive Him as your personal God, Saviour and Lord. Your value is not measured by examination results; you are invited to know His love and follow Him.

Restoration: Grow in hope and service

God promises a restored creation where His people dwell with Him. As you look forward in hope, grow in your faith by reading and studying the Bible, spending time in prayer and connecting with other believers in a local church. Use your learning to serve others with wisdom, honesty and compassion.

Reflection: How can gratitude to God and love for others shape the way you study and help your classmates?

Prayer: Lord Jesus, thank You for Your creation and saving love. Forgive my sins and guide me in Your truth. Give me understanding and perseverance in my studies. Help me grow through Scripture, prayer and fellowship, and use what I learn to serve others. Amen.

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

Curriculum

Wave phenomena

Lesson C.3 Wave Phenomena

Learning objectives

Lesson map

IB notation and working conventions

C.3.1 Wavefronts and Rays — SL and HL

How to sketch a wave diagram

C.3.2 Reflection, Transmission and Refraction — SL and HL

Reflection and transmission

Worked example 1 — Reflection angles

Refraction: Why the direction changes

Refractive index and Snell’s law

IB refraction relation

Worked example 2 — Light entering glass

A parallel-sided glass block

C.3.3 Critical Angle and Total Internal Reflection — SL and HL

Conditions for total internal reflection

Worked example 3 — Glass–air boundary

C.3.4 Diffraction — SL and HL

Examples and comparisons

C.3.5 Superposition and Two-Source Interference — SL and HL

Principle of superposition

Coherence and interference

Path-difference conditions for sources emitting in phase

Amplitude and intensity

Worked example 4 — Sound interference and unequal amplitudes

C.3.6 Young’s Double-Slit Experiment — SL and HL

IB fringe-spacing equation

Deriving the fringe spacing

Worked example 5 — Fringe spacing and order

Measuring wavelength accurately

C.3.7 Single-Slit Diffraction — Additional HL

IB first-minimum approximation

Worked example 6 — Central diffraction maximum (HL)

C.3.8 Multiple Slits and Diffraction Gratings — Additional HL

IB grating equation at normal incidence

Worked example 7 — Grating angle and highest order (HL)

White light and spectra

C.3.9 The Diffraction Envelope — Additional HL

Worked example 8 — Missing orders (HL)

Practice Questions — Expandable Answers

Question 1 — Reflection and refraction (SL and HL)

Question 2 — Water to air (SL and HL)

Question 3 — Glass to water (SL and HL)

Question 4 — Superposition and source phase (SL and HL)

Question 5 — Wavelength and uncertainty (SL and HL)

Question 6 — Narrower slit (HL)

Question 7 — Two colours through a grating (HL)

Question 8 — Fringes and envelope (HL)

Exam readiness checklist

In God’s Light

Creation: Wonder and gratitude

The Fall: Honesty about our need

Redemption: Come to Jesus Christ

Restoration: Grow in hope and service

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