You know the content. Now let’s learn how Cambridge examiners test it.
Challenge questions are not about harder facts. They test the same facts you already know — but wrapped in unfamiliar contexts, combined in unexpected ways, or phrased to exploit common misconceptions.
This guide will teach you three things:
1. Where students go wrong — the traps examiners set and how to spot them.
2. How to think through tricky questions — step-by-step reasoning, not guessing.
3. How to tell similar questions apart — because one word can change the answer completely.
Waves is one of the most diagram-heavy and calculation-rich topics in IGCSE Physics. From Snell’s law to oscilloscope traces, from critical angles to diffraction patterns — this topic rewards students who think carefully about direction, medium, and measurement. Work through each section carefully. By the end, you will not just know the content — you will know how to apply it under pressure.
These are the beliefs that feel true but are not. Examiners love to write wrong answers that match these misconceptions — if you hold the misconception, the wrong answer looks perfect.
Challenge questions require chaining multiple ideas together. These walkthroughs show you exactly how to break down complex problems step by step — the way top students think through them.
Two parts — calculate the critical angle using Snell’s law, then explain the role of cladding. Need to connect the calculation to the practical application.
At the critical angle, the refracted angle is 90°. Using Snell’s law: n1 sin(c) = n2 sin(90°). For glass to air: 1.5 × sin(c) = 1 × 1. So sin(c) = 1/1.5 = 0.667. Therefore c = sin−1(0.667) = 41.8° ≈ 42°.
For TIR to work in the fibre, light must hit the core-cladding boundary at an angle GREATER than the critical angle. The cladding has a lower refractive index than the core, so TIR can occur at this boundary. Without cladding, light could leak out wherever the fibre touches another surface or gets scratched.
A — cladding is not just for physical protection; its optical properties are what matter. C — wrong critical angle (48°) and wrong explanation. D — fibre thickness is not the point of cladding.
Tricky wording — “11 consecutive compressions.” Need to figure out how many wavelengths this represents. This is where most students go wrong.
11 consecutive compressions span 10 wavelengths, not 11. Think of it like fence posts: 11 posts, 10 gaps. The distance between compression 1 and compression 11 = 10 wavelengths. So λ = 8.25 / 10 = 0.825 m.
v = f × λ = 440 × 0.825 = 363 m/s. The known speed is 330 m/s. 363 is not equal to 330, so the measurement is NOT consistent (it is about 10% too high, suggesting a measurement error).
A — uses 8.25 as the wavelength (forgot to divide by 10). C — gets the right speed but uses a wrong wavelength (0.75, not 0.825). D — recognises the wrong speed but uses the wrong wavelength.
Object is at 15 cm, focal length is 10 cm. So F = 10 cm and 2F = 20 cm. The object is between F and 2F. Need to use the lens equation or ray diagram rules.
Object between F and 2F produces a real, inverted, magnified image beyond 2F on the other side. This is how projectors work — like the projectors used in classrooms at your school.
Using 1/f = 1/u + 1/v: 1/10 = 1/15 + 1/v. So 1/v = 1/10 − 1/15 = 3/30 − 2/30 = 1/30. Therefore v = 30 cm. Magnification = v/u = 30/15 = 2. Image height = 2 × 4 = 8 cm.
B — virtual only when object is inside F. C — wrong image distance and characteristics. D — same size only when object is at 2F (20 cm), but object is at 15 cm.
Match wave type to use to reason. Need to know EM spectrum properties, uses, and WHY each type is suited to its use. All three columns must be correct for the row to be right.
A — X-rays for bone imaging: correct use. X-rays pass through soft tissue but are absorbed by dense bone, creating contrast on the image. This reason is correct.
B — Infrared for sterilising: incorrect. Gamma rays are used for sterilisation, not infrared. Infrared is used for thermal imaging and remote controls.
C — UV for imaging organs: incorrect. UV does not pass through skin safely — it causes skin damage. Ultrasound (not UV, and not even an EM wave) is used for internal imaging.
D — Gamma for communication: incorrect. Radio waves are used for communication. Gamma rays are far too dangerous for routine use.
B, C, D are all wrong because they pair the wrong wave with the wrong use. Only A correctly matches wave, use, and reason.
Two-part problem. First, use the original experiment to find the speed of sound. Then, use that speed to predict the echo time at a new distance. The key is remembering that echo = sound there AND back.
The sound travels TO the cliff and BACK. Total distance = 2 × 170 = 340 m. Time = 1.0 s. Speed = distance/time = 340 / 1.0 = 340 m/s.
At 85 m from the cliff, total distance = 2 × 85 = 170 m. Time = distance/speed = 170 / 340 = 0.5 s.
A (0.25 s) — this would be the time for sound to travel 85 m one way, forgetting the return trip. C (1.0 s) — same time as before, ignoring the distance change. D (2.0 s) — doubling instead of halving.
These paired questions look almost identical but require completely different answers. One changed word, one swapped direction — and the answer flips entirely. Train your eye to catch these differences before they catch you.
“Angle of incidence of 35 degrees” vs “35 degrees to the surface” — one phrase changes the answer by 20 degrees. Always check: is the angle measured from the normal or from the surface?
The direction of travel (air-to-glass vs glass-to-air) completely reverses the direction of bending. Same boundary, opposite refraction. TIR is possible in B (if angle exceeds critical angle) but NEVER possible in A.
“Loudness” = amplitude change (taller/shorter waves). “Pitch” = frequency change (more/fewer waves per second, so closer/further apart on screen). Students who confuse these give the wrong oscilloscope description.
Gap width relative to wavelength determines diffraction amount. Gap >> wavelength = almost no diffraction. Gap ≈ wavelength = maximum diffraction. Same waves, different gap, completely different pattern.
The test for real vs virtual is simple: can it be projected onto a screen? If yes, it is real. If you can only see it by looking into/through the optical device, it is virtual. Converging lenses CAN produce real images; plane mirrors NEVER do.
Challenge questions often combine ideas that students learn separately. These maps show you how the big concepts connect — so when a question bridges two topics, you can see the path between them.
Set by the source. Does NOT change when entering a new medium. Measured in Hz. Think of a tabla player — the drummer sets the frequency, not the air.
Changes when wave enters a new medium. Decreases in denser/slower medium. Measured in metres. Like footprints getting closer together when you walk through sand.
Depends on the medium, not the wave. Sound is faster in solids than gases. Light is slower in glass than air. The medium controls the speed.
If v changes and f is constant, then λ must change proportionally. This single rule explains refraction of light, water waves in ripple tanks, and sound behaviour.
Angle of incidence = angle of reflection. Always happens at any boundary — even when refraction also occurs. Partial reflection is always present.
Bending due to speed change. Toward normal in denser medium, away in less dense. Governed by Snell’s law: n1 sin θ1 = n2 sin θ2.
Angle giving 90° refraction. Only from denser to less dense medium. Calculate: sin c = 1/n. For glass (n=1.5): c = 42°.
100% reflection, zero refraction. Requires denser-to-less-dense AND angle > critical angle. Used in optical fibres, prisms, binoculars, and diamond cutting.
Increasing frequency → Increasing energy → Increasing danger → Decreasing wavelength →
Communication, broadcasting, cooking, satellite links. Low energy, generally safe. Radio masts across India, Jio towers, Wi-Fi routers all use these.
IR: heating, thermal imaging, remote controls, felt as heat. Visible: the only part we can see. Red (lowest frequency) to violet (highest). Sunlight at Marina Beach!
Increasingly penetrating and dangerous. UV: sterilisation, causes sunburn. X-rays: medical imaging. Gamma: cancer treatment, sterilisation of surgical instruments.
Travel at 3 × 108 m/s in a vacuum. Are transverse waves. Do not need a medium. Can be reflected, refracted, and diffracted.
Read the student’s answer. It sounds reasonable, but it is wrong. Can you find the flaw before revealing the correct answer? This is how examiners think when they design wrong options.
“Light enters the optical fibre and bounces off the walls because the glass is shiny. The light keeps bouncing until it reaches the other end, like a ball bouncing inside a tube.”
The student describes simple reflection from a shiny surface, not TIR. TIR is not about shininess — it occurs because the angle of incidence exceeds the critical angle at the boundary between the denser core and less dense cladding. The student does not mention: (1) the denser-to-less-dense boundary, (2) the critical angle, or (3) the role of cladding. This answer would score 0/3.
Light enters the glass core and hits the boundary between the core (higher refractive index) and the cladding (lower refractive index) at an angle greater than the critical angle. Total internal reflection occurs — all the light is reflected back into the core. This repeats along the length of the fibre, keeping the light signal inside the core with minimal loss. The three key marks: core denser than cladding, angle exceeds critical angle, total internal reflection.
“Sound X would show taller waves on the oscilloscope because higher pitch means louder and therefore bigger waves.”
The student confuses pitch with loudness. Higher pitch means higher FREQUENCY, not higher amplitude. Taller waves mean louder (greater amplitude), not higher pitch. Pitch and loudness are independent properties. This is Trap 8 in action — the student equated pitch with loudness.
Sound X (higher pitch) would show waves that are closer together (shorter time period between peaks) because it has a higher frequency. The HEIGHT of the waves (amplitude) tells us about loudness, not pitch. If both sounds are equally loud, the waves would be the same height but Sound X would have more waves crammed into the same time span on the screen.
“Sound travels through walls and doors because it is a powerful type of wave. Light cannot go through walls because it is weaker. That is why you can hear but not see.”
The student talks about waves being “powerful” or “weak,” which is not scientifically meaningful here. The real explanation involves diffraction and wavelength comparison, not wave “power.” Sound can also travel through the gap under the door, not just through the walls. The student misses the entire concept of diffraction.
Sound has wavelengths of about 0.02 m to 17 m — similar to or larger than the gap under the door. When the wavelength is comparable to the gap size, significant diffraction occurs, and sound spreads out after passing through the gap, filling the next room. Light has wavelengths of about 400–700 nm (billionths of a metre) — far smaller than the gap. Very little diffraction occurs, so light travels in straight lines through the gap and does not spread around corners. You would only see light directly in line with the gap, not from the side.
“The light refracts away from the normal at an angle of about 50 degrees because it is entering a less dense medium.”
The student correctly identifies that light bends away from the normal in a less dense medium, but fails to check whether the angle exceeds the critical angle. At 50 degrees, the angle of incidence EXCEEDS the critical angle of 42 degrees. Therefore, TIR occurs — no refraction happens at all. The student applied the refraction rule without checking the TIR condition first.
Since the angle of incidence (50°) is greater than the critical angle (42°), and the light is travelling from a denser medium (glass) to a less dense medium (air), total internal reflection occurs. ALL the light is reflected back into the glass at 50° to the normal. No light passes into the air — there is no refracted ray. Always check the TIR conditions BEFORE applying refraction rules.
“The water acts as a magnifying glass and makes the bottom look closer. Water is like a big lens that magnifies everything.”
The student invents a lens analogy that is not correct. Water does not act as a magnifying glass. The pool does not appear magnified — it appears SHALLOWER. The effect is due to refraction of light at the water-air boundary, not magnification. A flat water surface is not a lens.
Light from the bottom of the pool travels from water (denser) into air (less dense) and refracts AWAY from the normal. When your eyes trace the refracted rays back in straight lines (which is how your brain processes light), the rays appear to come from a point that is higher (closer to the surface) than the actual position of the pool bottom. This makes the pool appear shallower than it really is. The formula is: apparent depth = real depth / n. For a 2 m deep pool with water (n = 1.33), the apparent depth is about 1.5 m.
10 questions at Challenge difficulty. Each one is designed to test a specific misconception or require multi-step reasoning. After answering, read the full solution — understanding why each wrong option is wrong matters as much as getting the right answer.
Option D is the biggest trap. It results from applying n = sin(r)/sin(i) instead of n = sin(i)/sin(r) when going from air to glass. Always check: when entering a denser medium, the refracted angle must be SMALLER than the incident angle. If your answer is larger, you have the formula upside down.
Unit conversion is the trap here. “100 MHz” must be converted to “100 × 106 Hz = 108 Hz” before using v = f × λ. Students who forget the “mega” prefix get the wrong power of ten. Option D catches students who multiply instead of divide — always check that your wavelength makes physical sense.
This tests whether you know BOTH conditions for TIR. Many students only remember one condition (angle > critical angle) and forget the other (must be travelling from denser to less dense medium). Option B specifically tests the direction condition.
This requires you to independently assess amplitude (loudness) and period/frequency (pitch). Longer period = lower frequency = lower pitch. Larger amplitude = louder. These are separate characteristics. The trap is forgetting that a longer period means a LOWER frequency (they are inversely related: f = 1/T).
Option A is the most common wrong answer. Students sometimes confuse “radio waves travel further” (in terms of practical range due to diffraction and atmospheric absorption) with “radio waves travel faster.” Distance and speed are different things. All EM waves travel at the same speed in a vacuum; they differ in frequency and wavelength.
The formula is apparent depth = real depth / n. If you get an apparent depth GREATER than the real depth, you have the formula upside down. Common sense check: swimming pools always appear shallower than they are, so apparent depth must be LESS than real depth. Option D fails this common sense check. Always verify your answer against real-world experience.
The key relationship is: maximum diffraction when gap width approximately equals wavelength. Option A might seem like it would give more spreading, but if the gap is too small relative to the wavelength, very little wave energy gets through. Students must know the specific condition: gap ≈ wavelength for maximum diffraction effect.
| Nature | Orientation | Size | |
|---|---|---|---|
| A | Real | Inverted | Magnified |
| B | Virtual | Upright | Magnified |
| C | Virtual | Inverted | Diminished |
| D | Real | Upright | Magnified |
This tests knowledge of the specific case: object inside F. Students must memorise (or derive from ray diagrams) all the cases. The magnifying glass case (object inside F = virtual, upright, magnified) is the one most commonly tested because it is the exception students forget — they associate converging lenses with real, inverted images and forget this special case.
This combines two echo calculations with spatial reasoning. Key steps: (1) divide each echo time by 2 (sound goes and returns), (2) calculate each distance separately, (3) ADD them because the student is between the buildings. Imagine standing on MG Road in Bangalore between two tall buildings — the first echo comes from the closer building, the second from the farther one. Option D is the trap for students who add the times first and then forget to divide correctly.
The fence-post counting trap appears in various forms across IGCSE. Here, “8 crests pass” means only 7 complete wave cycles between the first and eighth crest — just like 8 fence posts have 7 gaps between them. Option C is the classic trap for students who do not subtract 1. Always think: how many GAPS between the posts? This is the same principle as Walkthrough 2 (11 compressions = 10 wavelengths).