Waves, sound & optics: practice problems with step-by-step solutions
Also called: geometric optics · Snell’s law · Doppler effect · thin lens equation
Free Waves, sound & optics practice problems with diagrams, hints and full worked solutions. New values every time, at three levels: wave speed, sound level and refraction, standing waves, lenses and Doppler, interference, gratings, mirrors and fibres.
Learn Waves, sound & optics from scratch
Before this: Springs & pendulums
1What a wave is
A wave carries energy from place to place without carrying matter along: the medium only oscillates. On a string or on water the motion is transverse (across the direction of travel); sound in air is longitudinal (along it).
The wavelength λ is the length of one full cycle, the period T the time of one cycle, and the frequency f = 1/T. In one period the wave moves one wavelength, so:
v = fλ string: v = √(T/μ), μ = m/L
Sound in air travels at about 343 m/s, light at c = 3.00 × 10⁸ m/s. The speed depends on the medium, not on the frequency.
Self-check A sound of 686 Hz travels at 343 m/s. What is its wavelength?
- 0.002 m
- 0.5 m
- 2 m
- 235 000 m
Answer: 0.5 m. λ = v/f = 343/686 = 0.5 m. Dividing the other way gives 2 m, and multiplying gives a huge number.
2Sound level and the Doppler effect
Our ears respond to intensity I (W/m²) on a logarithmic scale, so loudness is given in decibels:
β = 10·log₁₀(I/I₀) I₀ = 10⁻¹² W/m²
Ten times the intensity adds 10 dB; two identical sources add about 3 dB. A small source spreads its power over a sphere, I = P/(4πr²), so doubling the distance lowers the level by about 6 dB.
When source and listener approach each other the pitch rises; when they move apart it falls (Doppler effect):
f′ = f·(v ± vo)/(v ∓ vs)
Use the upper signs for motion towards each other. The listener’s speed goes in the numerator, the source’s speed in the denominator.
3Standing waves
Waves reflected at both ends of a string interfere into a standing wave. A string fixed at both ends has nodes at the ends and fits whole half-wavelengths:
L = nλ/2 fₙ = n·v/(2L), n = 1, 2, 3, …
A pipe open at both ends follows the same rule. A pipe closed at one end has a node at the closed end and an antinode at the open end, so only odd harmonics fit: fₙ = n·v/(4L) with n = 1, 3, 5, …
Self-check A pipe closed at one end has its fundamental at 100 Hz. Which of these frequencies can it also produce?
- 150 Hz
- 200 Hz
- 300 Hz
- 50 Hz
Answer: 300 Hz. A closed pipe has only odd harmonics: 100, 300, 500 Hz, … 200 Hz would need n = 2, which a closed pipe cannot hold.
4Refraction and lenses
Light slows down in glass or water by the factor n = c/v (the refractive index). At a boundary the ray changes direction (Snell’s law), with angles measured from the normal:
n₁·sin θ₁ = n₂·sin θ₂ critical angle: sin θc = n₂/n₁
Going from a dense to a less dense medium beyond θc, all light is reflected (total internal reflection). Optical fibres guide light this way.
Thin lenses and curved mirrors obey one equation, with the magnification m = −dᵢ/d₀:
1/f = 1/d₀ + 1/dᵢ
Signs: f > 0 for a converging lens and a concave mirror (f = R/2), f < 0 for a diverging lens and a convex mirror. A positive dᵢ is a real image, a negative dᵢ a virtual one; m < 0 means the image is upside down.
5Interference and diffraction
Light from two narrow slits a distance d apart overlaps on a screen at distance L. Where the path difference d·sin θ is a whole number of wavelengths the waves add: bright fringes.
d·sin θ = mλ yₘ = mλL/d Δy = λL/d
A diffraction grating has many slits: N lines per mm means d = 1/N mm. Its maxima are sharp and far apart, and the highest order is the largest whole m with m ≤ d/λ. A single slit of width a has dark fringes at a·sin θ = mλ, so its central maximum is 2λL/a wide.
6A worked example
Follow every step. Try to predict the next one before you open it.
Worked exampleAn object stands 30 cm in front of a converging lens with f = 10 cm. Find the image distance and the magnification.
11/dᵢ = 1/f − 1/d₀ = 1/10 − 1/30 = 2/30
So dᵢ = 15 cm.
2dᵢ is positive, so the image is real: it forms on the far side of the lens and can be caught on a screen.
3m = −dᵢ/d₀ = −15/30 = −0.5
The image is inverted and half the size of the object.
4A ray parallel to the axis bends through F on the far side; a ray through the centre goes straight. They cross 15 cm behind the lens, below the axis.
Answer dᵢ = 15 cm and m = −0.5: a real, inverted, half-size image.
Self-check An object is 5 cm in front of a converging lens with f = 10 cm. What kind of image forms?
- No image: the rays come out parallel
- Real and inverted, on the far side
- Real and upright, on the same side
- Virtual and upright, on the same side as the object
Answer: Virtual and upright, on the same side as the object. 1/dᵢ = 1/10 − 1/5 = −1/10, so dᵢ = −10 cm: a negative image distance means a virtual image, and m = +2 means upright and enlarged. This is how a magnifying glass works.
7Common mistakes and exam questions
- Measuring angles from the surface instead of from the normal.
- Using 10·log for amplitudes or 20·log for intensities: intensities use 10·log.
- Giving a closed pipe even harmonics.
- Dropping the minus sign of a virtual image or a diverging lens.
- Forgetting to convert nm, mm and cm to metres in interference questions.
- Easy: v = fλ, waves on strings, sound levels, Snell’s law and the speed of light in a material.
- Medium: standing waves on strings and in pipes, the critical angle, thin lenses, the Doppler effect and the inverse-square law.
- Hard: double slits, gratings, single slits, mirrors, two-lens systems, echoes from moving objects and optical fibres.
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