IGCSE Physics 0625 · Section 3 · Waves — Numerical Solutions
Waves — Numerical Practice: Solutions
A. Wave Speed (v = fλ)
A1
f=50 Hz, λ=4 m. Find speed.
✓200 m/s
A2
v=340 m/s, λ=0.5 m. Find frequency.
✓680 Hz
A3
v=3×10⁸ m/s, f=5×10¹⁴ Hz. Find wavelength.
1
λ = v/f
(3×10⁸) ÷ (5×10¹⁴)
✓6 × 10⁻⁷ m
A4
Period 0.02 s. Find frequency, then wavelength (v=15 m/s).
✓f = 50 Hz; λ = 0.3 m
A5
λ=3 cm, f=4 Hz. Find speed.
✓12 cm/s
A6
f=440 Hz, v=340 m/s. Find wavelength.
✓≈ 0.77 m
B. Refraction and Refractive Index
B1
i=60°, r=35°. Find n.
1
n = sin i / sin r
sin60° ÷ sin35° = 0.866 ÷ 0.574
✓≈ 1.51
B2
n=1.5, i=40°. Find r.
1
sin r = sin i / n
sin40° ÷ 1.5 = 0.643 ÷ 1.5 = 0.429
2
r = sin⁻¹(0.429)
r ≈ 25.4°
✓≈ 25.4°
B3
Critical angle 42°. Find n.
1
n = 1/sin c
1 ÷ sin42° = 1 ÷ 0.669
✓≈ 1.49
B4
n=1.33. Find critical angle.
1
sin c = 1/n
1 ÷ 1.33 = 0.752
2
c = sin⁻¹(0.752)
c ≈ 48.8°
✓≈ 48.8°
B5
n=1.5, i=50° inside glass. Find critical angle, explain what happens.
1
sin c = 1/n
1 ÷ 1.5 = 0.667 → c ≈ 41.8°
✓Critical angle ≈ 41.8°. Since 50° exceeds this, the light undergoes total internal reflection — it does not escape into the air at all, and instead reflects entirely back into the glass.
B6
Why is n always > 1 going from air into a denser medium?
✓Refractive index compares the speed of light in the two media (n = speed in first medium ÷ speed in second medium, for light entering the denser medium). Light always slows down when entering a denser medium, so the ratio of speeds — and hence n — is always greater than 1.
C. Sound Speed, Echoes and Sonar
C1
200 m from wall, echo after 1.2 s. Find speed.
1
Total distance travelled = there and back
2 × 200 = 400 m
2
speed = distance/time
400 ÷ 1.2
✓≈ 333 m/s
C2
Speed 340 m/s, echo 0.5 s. Find distance.
1
distance = (speed × time) / 2
(340 × 0.5) ÷ 2
✓85 m
C3
Sonar returns 0.8 s, speed 1500 m/s. Find depth.
1
depth = (speed × time) / 2
(1500 × 0.8) ÷ 2
✓600 m
C4
Signal 3 s, depth 2250 m. Find speed.
1
speed = (2 × distance) / time
(2 × 2250) ÷ 3
✓1500 m/s
C5
Why is ultrasound used for medical scanning?
✓Ultrasound has a frequency above the range of human hearing, so it doesn’t produce noise, and its short wavelength allows it to be focused into a narrow beam that reflects clearly off boundaries between different body tissues, producing a detailed image without the tissue-damaging effects of radiation like X-rays.
C6
Echo 4 s, target 3000 m away. Find speed.
1
speed = (2 × distance) / time
(2 × 3000) ÷ 4
✓1500 m/s
C7
Audible range and definition of ultrasound.
✓Audible range: approximately 20 Hz to 20000 Hz (20 kHz). Ultrasound is sound with a frequency higher than 20 kHz, above the range of human hearing.
C8
Return time 0.04 s, speed 1450 m/s. Find depth.
1
depth = (speed × time) / 2
(1450 × 0.04) ÷ 2
✓29 m
Mastery Test — Answer Key
- 200 m/s (25 × 8)
- 5000 Hz (1500 ÷ 0.3)
- ≈ 1.53 — n = sin50°/sin30° = 0.766/0.5
- ≈ 45.6° — sin c = 1/1.4 = 0.714 → c = sin⁻¹(0.714)
- ≈ 333 m/s — speed = (2×150)/0.9
- 1800 m — depth = (1500×2.4)/2
- Frequency — a higher frequency produces a higher pitch, and a lower frequency produces a lower pitch.
- 200 Hz — f = 1/T = 1/0.005
- Sound is a mechanical wave, relying on particles of a medium colliding with their neighbours to pass the vibration along. With no particles present (a vacuum), there is nothing to vibrate or collide, so sound cannot travel.
- 6 × 10¹⁴ Hz — f = v/λ = (3×10⁸)/(5×10⁻⁷)
