Rae

HomeSubjectsA-Level H2 Physics › The superposition principle, standing waves, interference and single-slit diffraction

A-Level The superposition principle, standing waves, interference and single-slit diffraction

What the A-Level syllabus expects for The superposition principle, standing waves, interference and single-slit diffraction, and how to practise it.

What the syllabus expects

How it's examined

Questions on this topic most often ask you to describe, estimate, find, show. About 7% of the past-paper style questions in Rae's bank for this subject sit in this topic.

Worked examples

Example 1 (3 marks)

(a) Two loudspeakers, M and N, are driven together in phase at equal amplitude and at a frequency of 680 Hz. A point Q lies 18.0 m from M and 20.25 m from N. Sound travels at 340 m s⁻¹. (i) Show that the sound heard at Q has minimum intensity.

Show the worked answer

Wavelength: λ = v/f = 340/680 = 0.50 m. Path difference: NQ − MQ = 20.25 − 18.0 = 2.25 m. In wavelengths: 2.25 / 0.50 = 4.5 = 4½ wavelengths = 9 half-wavelengths. This is an odd number of half-wavelengths, so the two waves arrive exactly in antiphase and interfere destructively, giving minimum (ideally zero) intensity at Q.

Example 2 (4 marks)

(b) Light emitted by a sodium discharge lamp falls normally on a diffraction grating that has 6.00 × 10⁵ lines per metre. Its spectrum includes a closely spaced yellow doublet with wavelengths 589 nm and 590 nm. (i) Find the angular separation between these two lines as seen in the second order spectrum.

Show the worked answer

Grating spacing d = 1/(6.00x10⁵) = 1.667x10⁻⁶ m. Second order: d sin(theta) = 2*lambda. For 589 nm: sin(theta1) = 2*589x10⁻⁹/1.667x10⁻⁶ = 0.7068, theta1 = 44.98 deg. For 590 nm: sin(theta2) = 2*590x10⁻⁹/1.667x10⁻⁶ = 0.7080, theta2 = 45.07 deg. Angular separation = theta2 - theta1 = 0.097 deg (about 1.7x10⁻³ rad).

Example 3 (2 marks)

(iii) With both loudspeakers switched on, the sound intensity measured at point Q equals I₀. Loudspeaker B is then switched off, while loudspeaker A keeps producing sound of the same amplitude and frequency. The intensity at Q now becomes I_A. Work out an estimate for the ratio I_A / I₀.

Show the worked answer

Point Q is a point of constructive interference, where the two waves of equal amplitude a arrive in phase, giving a resultant amplitude of 2a. Intensity is proportional to amplitude squared, so with both speakers on I₀ ∝ (2a)² = 4a². With B switched off, only A contributes amplitude a, so I_A ∝ a². Hence I_A/I₀ = a²/(4a²) = 1/4.

More worked questions on this topic

Ask about The superposition principle, standing waves, interference and single-slit diffractionUse Rae in Telegram

More A-Level H2 Physics topics

Physical quantities, units, measurement uncertainty and vector basics · Types of force, turning effects and conditions for equilibrium · Kinematics, uniformly accelerated motion, momentum and Newton's laws · Energy stores and transfers, work, kinetic and potential energy, fields, power and efficiency · Falling freely, the gravitational potential energy of a uniform field, and how air resistance changes the motion · Impulse and the conservation of momentum and energy · all of A-Level H2 Physics