IB Physics Topic 3.3: Wave phenomena | Notes & Flashcards | Aimnova

IB Physics — Wave phenomena

Topic 3.3 of IB Physics covers Wave phenomena, which is part of Unit 3: Wave behaviour. Students explore key concepts including Refraction, Snell's law and total internal reflection, Interference, path difference and coherence, Double-slit interference, Diffraction. A strong understanding of wave phenomena is essential for IB Physics exams and builds the foundation for connected topics across the syllabus.

Key concepts in Wave phenomena

Key Idea: This topic is four things waves DO at a boundary, a gap or an edge: they refract (bend when they change speed), interfere (add up where two waves meet), spread through a double slit into evenly spaced fringes, and diffract (fan out through a gap). It is examined on both papers. Paper 1A tends to be quick multiple-choice — which way a ray bends, constructive vs destructive from a path difference, what happens to the fringe spacing, which wave diffracts more. Paper 2 is longer structured work — a Snell's-law or critical-angle calculation, a 'show that' on path difference, plugging into s = λD/d, or an explain on coherence, diffraction or energy conservation across the fringes.

📋 Key formulas

Most of these carry the data-booklet badge (look for it). The critical-angle relation and the small-angle fringe angle are not printed separately — they come straight from the given equations, so you remember those.

n_1 \sin\theta_1 = n_2 \sin\theta_2

Snell's law (given). Angles measured FROM THE NORMAL. Into a denser (higher-n) medium the ray bends toward the normal.

$n_1$: refractive index of medium 1 (no units)
$n_2$: refractive index of medium 2 (no units)
$\theta_1$: angle of incidence, measured from the normal (°)
$\theta_2$: angle of refraction, measured from the normal (°)

n = \frac{c}{v}

Refractive index from the speed of light in the medium (given). A higher index means slower light; rearrange to v = c ÷ n.

$n$: refractive index of the medium (no units)
$c$: speed of light in a vacuum, 3.0 × 10⁸ m s⁻¹
$v$: speed of light in the medium (m s⁻¹)

\sin\theta_c = \frac{n_2}{n_1}

Critical angle — NOT printed separately; it is Snell's law with the refraction angle set to 90°. n₁ is the denser medium, n₂ the less-dense one. Above θ_c you get total internal reflection.

$\theta_c$: critical angle (°) — the incidence angle giving a 90° refraction
$n_1$: index of the denser medium the light starts in
$n_2$: index of the less-dense medium it tries to enter

\text{constructive: path difference} = n\lambda \qquad \text{destructive: } \left(n + \tfrac{1}{2}\right)\lambda

The two given path-difference rules. A whole number of wavelengths (nλ) → in step → constructive; an extra half-wavelength → antiphase → destructive.

$\text{path difference}$: extra distance one wave travels to reach the point (m)
$n$: a whole number: 0, 1, 2, 3, …
$\lambda$: wavelength of the waves (m)

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

Double-slit fringe spacing (given). All four lengths in metres. Wider fringes with longer λ or bigger D; narrower with bigger slit separation d.

$s$: fringe spacing — gap between neighbouring bright fringes (m)
$\lambda$: wavelength of the light (m)
$D$: distance from the slits to the screen (m)
$d$: separation of the two slits (m)

\theta \approx \frac{\lambda}{d}

Small-angle angular separation of neighbouring maxima (radians) — from d sinθ = nλ with sinθ ≈ θ. Not given; an angle, so it does not depend on the screen distance D.

$\theta$: angular separation of neighbouring fringes (rad)
$\lambda$: wavelength of the light (m)
$d$: separation of the two slits (m)

v = f\lambda

Wave equation (given, from topic 3.2). Used in diffraction to turn a frequency into a wavelength: a lower frequency means a longer wavelength, which diffracts more.

$v$: wave speed — how fast the wave travels (m s⁻¹)
$f$: frequency — waves per second (Hz)
$\lambda$: wavelength — length of one full wave (m)

⚖️ The four phenomena side by side

🌈 Constructive vs destructive — read it off the path difference

Complete cancellation to zero only happens when the two amplitudes are equal. If they differ, in phase gives the sum and antiphase gives the difference. At a dark fringe the energy is not destroyed — it is redistributed into the brighter bright fringes, so the total energy over the whole screen is conserved.

✏️ Worked exam-style questions

IB-style question — critical angle and total internal reflection

Light inside a glass block of refractive index 1.52 meets the boundary with air (index 1.00). (a) Find the critical angle for this glass–air boundary. (b) A ray inside the glass hits the boundary at 50° to the normal — does it escape into the air or is it totally internally reflected?

Solution:

(a) Start with the critical-angle relation (Snell's law with θ₂ = 90°), less-dense air on top:
$\sin\theta_c = \frac{n_2}{n_1} = \frac{1.00}{1.52}$
(a) Evaluate, then take the inverse sine:
$\sin\theta_c = 0.658 \;\Rightarrow\; \theta_c = \sin^{-1}(0.658) = 41^\circ$
(b) Compare the incidence angle with the critical angle:
$\theta_1 = 50^\circ > \theta_c = 41^\circ$
(b) The angle is above the critical angle, so none of the light escapes:
$\Rightarrow \text{total internal reflection}$

Final answer:

(a) θc = 41°. (b) 50° is above the 41° critical angle, so the ray is totally internally reflected — no light escapes into the air. Put the LESS-dense index on top in sinθc = n₂ ÷ n₁, or your sine comes out greater than 1.

IB-style question — refractive index then speed of light

A beam of light travels from air (index 1.00) into a transparent liquid, with an angle of incidence of 58° and an angle of refraction of 36°, both measured from the normal. (a) Find the refractive index of the liquid. (b) Hence find the speed of light in the liquid (c = 3.0 × 10⁸ m s⁻¹).

Solution:

(a) Snell's law from air (n₁ = 1.00), rearranged for n₂:
$n_2 = \frac{n_1 \sin\theta_1}{\sin\theta_2} = \frac{1.00 \times \sin 58^\circ}{\sin 36^\circ}$
(a) Work it out:
$n_2 = \frac{0.848}{0.588} = 1.44$
(b) Rearrange n = c ÷ v to make v the subject:
$v = \frac{c}{n} = \frac{3.0\times10^{8}}{1.44}$
(b) Evaluate — keep the unit:
$v = 2.1\times10^{8}\ \text{m s}^{-1}$

Final answer:

(a) n = 1.44. (b) v = 2.1 × 10⁸ m s⁻¹ — slower than c, as expected for an index above 1. When light starts in air, Snell's law shortcuts to n = sin(incidence) ÷ sin(refraction).

IB-style question — interference from a path difference

Two coherent loudspeakers play the same note of wavelength 0.40 m. At a point P the sound from one speaker has travelled 1.0 m further than from the other. Each speaker alone would give P a wave of amplitude 6.0 units. (a) State whether P is loud or quiet. (b) Find the resultant amplitude at P.

Solution:

(a) Divide the path difference by the wavelength:
$\frac{1.0}{0.40} = 2.5 = \left(2 + \tfrac{1}{2}\right)$
(a) That matches the destructive rule (n + ½)λ with n = 2 → antiphase:
$\text{path difference} = \left(n + \tfrac{1}{2}\right)\lambda \;\Rightarrow\; \text{destructive (quiet)}$
(b) The two amplitudes are EQUAL and the waves are antiphase, so they cancel:
$A = 6.0 - 6.0 = 0$

Final answer:

(a) Quiet — 2.5λ is (n + ½)λ, so the waves arrive antiphase. (b) Resultant amplitude = 0, because the two equal amplitudes cancel completely. Had the amplitudes differed, antiphase would leave their difference, not zero.

IB-style question — double-slit wavelength and fringe angle

Two slits 0.30 mm apart are lit by a laser. On a screen 1.8 m away the bright fringes are 3.6 mm apart. (a) Find the wavelength of the light. (b) Find, in radians, the angular separation of two neighbouring bright fringes.

Solution:

(a) Start with the given fringe-spacing formula, rearranged for λ:
$s = \frac{\lambda D}{d} \;\Rightarrow\; \lambda = \frac{s\,d}{D}$
(a) Substitute in metres (s = 3.6×10⁻³, d = 0.30×10⁻³, D = 1.8):
$\lambda = \frac{(3.6\times10^{-3})(0.30\times10^{-3})}{1.8} = 6.0\times10^{-7}\ \text{m}$
(b) Small-angle angular separation of neighbouring maxima:
$\theta \approx \frac{\lambda}{d} = \frac{6.0\times10^{-7}}{0.30\times10^{-3}}$
(b) Work it out — radians have no unit:
$\theta = 2.0\times10^{-3}\ \text{rad}$

Final answer:

(a) λ = 6.0 × 10⁻⁷ m (600 nm). (b) θ ≈ 2.0 × 10⁻³ rad. Get EVERY length into metres first (mm = 10⁻³ m, nm = 10⁻⁹ m); the fringe angle uses λ/d and does not depend on the screen distance D.

🧠 Quick self-check

Tap each card to reveal the answer.

🎯 Exam tips

Exam Tips

Always measure refraction angles from the NORMAL (the dashed line), never from the surface. Into a higher-index (denser, slower) medium the ray bends TOWARD the normal.
Total internal reflection needs BOTH conditions: denser → less-dense, and an incidence angle above the critical angle. Use sinθc = n₂/n₁ with the LESS-dense index on top — a sine above 1 means you flipped the ratio.
For interference, divide the path difference by λ: a whole number is constructive (add amplitudes); a whole number + ½ is destructive. Complete cancellation to zero only happens when the two amplitudes are equal.
Coherent = constant phase difference. State exactly this whenever asked why two sources must be coherent.
For s = λD/d, convert EVERY length to metres first (mm = 10⁻³ m, nm = 10⁻⁹ m). Smaller slit separation d → wider fringes; the fringe ANGLE λ/d is independent of the screen distance D.
Energy is not lost at a dark fringe — it is redistributed into the bright fringes, so the total over the screen is conserved.
Diffraction is greatest when the gap is about the size of the wavelength (gap ≈ λ). Lower frequency → longer wavelength → MORE spreading. (The classic trap is thinking a higher frequency spreads more.)

IB Physics — Wave phenomena

Key concepts in Wave phenomena

Key Idea: This topic is four things waves DO at a boundary, a gap or an edge: they refract (bend when they change speed), interfere (add up where two waves meet), spread through a double slit into evenly spaced fringes, and diffract (fan out through a gap). It is examined on both papers. Paper 1A tends to be quick multiple-choice — which way a ray bends, constructive vs destructive from a path difference, what happens to the fringe spacing, which wave diffracts more. Paper 2 is longer structured work — a Snell's-law or critical-angle calculation, a 'show that' on path difference, plugging into s = λD/d, or an explain on coherence, diffraction or energy conservation across the fringes.

📋 Key formulas

n_1 \sin\theta_1 = n_2 \sin\theta_2

Snell's law (given). Angles measured FROM THE NORMAL. Into a denser (higher-n) medium the ray bends toward the normal.

$n_1$: refractive index of medium 1 (no units)
$n_2$: refractive index of medium 2 (no units)
$\theta_1$: angle of incidence, measured from the normal (°)
$\theta_2$: angle of refraction, measured from the normal (°)

n = \frac{c}{v}

Refractive index from the speed of light in the medium (given). A higher index means slower light; rearrange to v = c ÷ n.

$n$: refractive index of the medium (no units)
$c$: speed of light in a vacuum, 3.0 × 10⁸ m s⁻¹
$v$: speed of light in the medium (m s⁻¹)

\sin\theta_c = \frac{n_2}{n_1}

Critical angle — NOT printed separately; it is Snell's law with the refraction angle set to 90°. n₁ is the denser medium, n₂ the less-dense one. Above θ_c you get total internal reflection.

$\theta_c$: critical angle (°) — the incidence angle giving a 90° refraction
$n_1$: index of the denser medium the light starts in
$n_2$: index of the less-dense medium it tries to enter

\text{constructive: path difference} = n\lambda \qquad \text{destructive: } \left(n + \tfrac{1}{2}\right)\lambda

The two given path-difference rules. A whole number of wavelengths (nλ) → in step → constructive; an extra half-wavelength → antiphase → destructive.

$\text{path difference}$: extra distance one wave travels to reach the point (m)
$n$: a whole number: 0, 1, 2, 3, …
$\lambda$: wavelength of the waves (m)

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

Double-slit fringe spacing (given). All four lengths in metres. Wider fringes with longer λ or bigger D; narrower with bigger slit separation d.

$s$: fringe spacing — gap between neighbouring bright fringes (m)
$\lambda$: wavelength of the light (m)
$D$: distance from the slits to the screen (m)
$d$: separation of the two slits (m)

\theta \approx \frac{\lambda}{d}

Small-angle angular separation of neighbouring maxima (radians) — from d sinθ = nλ with sinθ ≈ θ. Not given; an angle, so it does not depend on the screen distance D.

$\theta$: angular separation of neighbouring fringes (rad)
$\lambda$: wavelength of the light (m)
$d$: separation of the two slits (m)

v = f\lambda

Wave equation (given, from topic 3.2). Used in diffraction to turn a frequency into a wavelength: a lower frequency means a longer wavelength, which diffracts more.

$v$: wave speed — how fast the wave travels (m s⁻¹)
$f$: frequency — waves per second (Hz)
$\lambda$: wavelength — length of one full wave (m)

⚖️ The four phenomena side by side

🌈 Constructive vs destructive — read it off the path difference

Complete cancellation to zero only happens when the two amplitudes are equal. If they differ, in phase gives the sum and antiphase gives the difference. At a dark fringe the energy is not destroyed — it is redistributed into the brighter bright fringes, so the total energy over the whole screen is conserved.

✏️ Worked exam-style questions

IB-style question — critical angle and total internal reflection

Solution:

(a) Start with the critical-angle relation (Snell's law with θ₂ = 90°), less-dense air on top:
$\sin\theta_c = \frac{n_2}{n_1} = \frac{1.00}{1.52}$
(a) Evaluate, then take the inverse sine:
$\sin\theta_c = 0.658 \;\Rightarrow\; \theta_c = \sin^{-1}(0.658) = 41^\circ$
(b) Compare the incidence angle with the critical angle:
$\theta_1 = 50^\circ > \theta_c = 41^\circ$
(b) The angle is above the critical angle, so none of the light escapes:
$\Rightarrow \text{total internal reflection}$

Final answer:

IB-style question — refractive index then speed of light

Solution:

(a) Snell's law from air (n₁ = 1.00), rearranged for n₂:
$n_2 = \frac{n_1 \sin\theta_1}{\sin\theta_2} = \frac{1.00 \times \sin 58^\circ}{\sin 36^\circ}$
(a) Work it out:
$n_2 = \frac{0.848}{0.588} = 1.44$
(b) Rearrange n = c ÷ v to make v the subject:
$v = \frac{c}{n} = \frac{3.0\times10^{8}}{1.44}$
(b) Evaluate — keep the unit:
$v = 2.1\times10^{8}\ \text{m s}^{-1}$

Final answer:

(a) n = 1.44. (b) v = 2.1 × 10⁸ m s⁻¹ — slower than c, as expected for an index above 1. When light starts in air, Snell's law shortcuts to n = sin(incidence) ÷ sin(refraction).

IB-style question — interference from a path difference

Solution:

(a) Divide the path difference by the wavelength:
$\frac{1.0}{0.40} = 2.5 = \left(2 + \tfrac{1}{2}\right)$
(a) That matches the destructive rule (n + ½)λ with n = 2 → antiphase:
$\text{path difference} = \left(n + \tfrac{1}{2}\right)\lambda \;\Rightarrow\; \text{destructive (quiet)}$
(b) The two amplitudes are EQUAL and the waves are antiphase, so they cancel:
$A = 6.0 - 6.0 = 0$

Final answer:

IB-style question — double-slit wavelength and fringe angle

Solution:

(a) Start with the given fringe-spacing formula, rearranged for λ:
$s = \frac{\lambda D}{d} \;\Rightarrow\; \lambda = \frac{s\,d}{D}$
(a) Substitute in metres (s = 3.6×10⁻³, d = 0.30×10⁻³, D = 1.8):
$\lambda = \frac{(3.6\times10^{-3})(0.30\times10^{-3})}{1.8} = 6.0\times10^{-7}\ \text{m}$
(b) Small-angle angular separation of neighbouring maxima:
$\theta \approx \frac{\lambda}{d} = \frac{6.0\times10^{-7}}{0.30\times10^{-3}}$
(b) Work it out — radians have no unit:
$\theta = 2.0\times10^{-3}\ \text{rad}$

Final answer:

🧠 Quick self-check

Tap each card to reveal the answer.

🎯 Exam tips

Exam Tips

Always measure refraction angles from the NORMAL (the dashed line), never from the surface. Into a higher-index (denser, slower) medium the ray bends TOWARD the normal.
Total internal reflection needs BOTH conditions: denser → less-dense, and an incidence angle above the critical angle. Use sinθc = n₂/n₁ with the LESS-dense index on top — a sine above 1 means you flipped the ratio.
For interference, divide the path difference by λ: a whole number is constructive (add amplitudes); a whole number + ½ is destructive. Complete cancellation to zero only happens when the two amplitudes are equal.
Coherent = constant phase difference. State exactly this whenever asked why two sources must be coherent.
For s = λD/d, convert EVERY length to metres first (mm = 10⁻³ m, nm = 10⁻⁹ m). Smaller slit separation d → wider fringes; the fringe ANGLE λ/d is independent of the screen distance D.
Energy is not lost at a dark fringe — it is redistributed into the bright fringes, so the total over the screen is conserved.
Diffraction is greatest when the gap is about the size of the wavelength (gap ≈ λ). Lower frequency → longer wavelength → MORE spreading. (The classic trap is thinking a higher frequency spreads more.)

Key concepts in Wave phenomena

📋 Key formulas

⚖️ The four phenomena side by side

🌈 Constructive vs destructive — read it off the path difference

✏️ Worked exam-style questions

IB-style question — critical angle and total internal reflection

IB-style question — refractive index then speed of light

IB-style question — interference from a path difference

IB-style question — double-slit wavelength and fringe angle

🧠 Quick self-check

🎯 Exam tips

Exam Tips

What you'll learn in Topic 3.3

Study resources — 3.3 Wave phenomena

Refraction, Snell's law and total internal reflection

Interference, path difference and coherence

Double-slit interference

Diffraction

Ready to study Wave phenomena?

Ready to practice?

Key concepts in Wave phenomena

📋 Key formulas

⚖️ The four phenomena side by side

🌈 Constructive vs destructive — read it off the path difference

✏️ Worked exam-style questions

IB-style question — critical angle and total internal reflection

IB-style question — refractive index then speed of light

IB-style question — interference from a path difference

IB-style question — double-slit wavelength and fringe angle

🧠 Quick self-check

🎯 Exam tips

Exam Tips

What you'll learn in Topic 3.3

Study resources — 3.3 Wave phenomena

Refraction, Snell's law and total internal reflection

Interference, path difference and coherence

Double-slit interference

Diffraction

Ready to study Wave phenomena?

Ready to practice?