Waves: practice questions with answers

Answer λ = 12.2 cm

λ = c ÷ f = 3.00 × 10⁸ ÷ 2.45 × 10⁹ = 0.1224 m = 12.2 cm.

• Got 122 cm? You used 2.45 × 10⁸ Hz. 1 GHz = 10⁹ Hz, so f = 2.45 × 10⁹ Hz and λ = c ÷ f = 12.2 cm.
• Got 1.22 cm? You used 2.45 × 10¹⁰ Hz. 1 GHz = 10⁹ Hz, so λ = 3.00 × 10⁸ ÷ 2.45 × 10⁹ = 12.2 cm.

Answer w = 4.56 mm

w = λD ÷ a = (633 × 10⁻⁹ × 1.8) ÷ (0.25 × 10⁻³) = 4.56 × 10⁻³ m = 4.56 mm.

• Got 4560 mm? You are out by a factor of 1000, probably mixing mm and m. λ = 633 × 10⁻⁹ m and a = 0.25 × 10⁻³ m, so w = λD ÷ a = 4.56 mm.
• Got 2.28 mm? You halved the result. The equation is w = λD ÷ a with a = 0.25 × 10⁻³ m, giving 4.56 mm.

Answer θ = 45 °

d = 1 ÷ 600 000 = 1.667 × 10⁻⁶ m. nλ = d sin θ gives sin θ = 2 × 589 × 10⁻⁹ ÷ 1.667 × 10⁻⁶ = 0.707, so θ = 45°.

• Got 20.7 °? You used n = 1. For the second order n = 2, so sin θ = 2λ ÷ d = 0.707 and θ = 45°.
• Got 0.785 °? Your calculator was in radians. Switch to degrees: θ = sin⁻¹(0.707) = 45°.

Answer v = 176 m s⁻¹

Third harmonic: L = 3λ/2, so λ = 2 × 0.80 ÷ 3 = 0.533 m. v = fλ = 330 × 0.533 = 176 m s⁻¹.

• Got 264 m s⁻¹? You used λ = L. In the third harmonic there are 1.5 wavelengths on the string, so λ = 2L ÷ 3 = 0.533 m and v = fλ = 176 m s⁻¹.
• Got 528 m s⁻¹? You used λ = 2L, which is the fundamental. For the third harmonic λ = 2L ÷ 3, so v = fλ = 176 m s⁻¹.

Answer B: Path difference of λ/2

An odd number of half wavelengths puts the waves in antiphase, giving destructive interference.

• Why not A? A path difference of a whole number of wavelengths gives constructive interference.
• Why not C? Two wavelengths is a whole number, so the waves arrive in phase and reinforce.

Answer C: Microwaves

Microwaves are transverse electromagnetic waves, so their oscillations can be restricted to one plane.

• Why not A? Sound is a longitudinal wave, so it cannot be polarised.
• Why not B? Ultrasound is longitudinal, so it cannot be polarised.