Waves, Sound & Optics — Chapter review · Physics & Environmental Science, Grade 12
Connect wave patterns, sound, electromagnetic radiation, interference, and refraction.
Review the key ideas
Wave Properties
Waves propagate disturbances and energy; material can oscillate locally. Define spatial wavelength, time period and amplitude separately and state the model for v=fλ.
Sound
Sound needs a medium. Frequency strongly influences pitch, while intensity and loudness differ. Decibel ratios, Doppler shifts and standing waves require stated conditions.
Light & the Electromagnetic Spectrum
Light is an electromagnetic wave obeying c = f λ; the spectrum runs from low-energy radio waves to high-energy gamma rays; and light shows wave-particle duality, with each photon carrying energy E = h f.
Reflection, Refraction & Optics
Light reflects at equal angles from the normal and refracts (bends) when its speed changes between media (n = c/v, Snell's law); lenses and mirrors form real or virtual images predicted by 1/f = 1/dₒ + 1/dᵢ.
Superposition, Standing Waves & Resonance
Interference adds local displacement; boundary conditions select standing-wave modes.
Doppler Effect & Relative Motion
Relative motion changes observed frequency; use a formula appropriate to the wave and medium.
Problem 1
Answer and explanation
30 m/s
v = f × λ = 12 × 2.5 = 30 m/s.
Problem 2
Answer and explanation
100 Hz
f = 1 ÷ T = 1 ÷ 0.01 = 100 Hz.
Problem 3
Answer and explanation
4 Hz
f = v ÷ λ = 6 ÷ 1.5 = 4 Hz.
Problem 4
Answer and explanation
340 m
In 2 s the sound makes a round trip: total distance = 340 × 2 = 680 m. The cliff is half of that, since the sound goes there and back: 680 ÷ 2 = 340 m.
Problem 5
Answer and explanation
200 Hz
f = v ÷ λ = 340 ÷ 1.7 = 200 Hz.
Problem 6
Answer and explanation
Microwaves, infrared, ultraviolet, gamma rays
Energy increases as wavelength decreases along the spectrum. Of these four, the order from lowest to highest energy is microwaves → infrared → ultraviolet → gamma rays.
Problem 7
Answer and explanation
1 × 10¹⁵ Hz
f = c ÷ λ = (3 × 10⁸) ÷ (3 × 10⁻⁷) = 1 × 10¹⁵ Hz.
Problem 8
Answer and explanation
2 × 10⁸ m/s
Rearrange n = c ÷ v to v = c ÷ n = (3 × 10⁸) ÷ 1.5 = 2 × 10⁸ m/s.
Problem 9
Answer and explanation
−6 cm: an upright, magnified virtual image (as in a magnifying glass).
1 ÷ dᵢ = 1 ÷ f − 1 ÷ dₒ = 1 ÷ 12 − 1 ÷ 4 = 1 ÷ 12 − 3 ÷ 12 = −2 ÷ 12 = −1 ÷ 6, so dᵢ = −6 cm. The negative image distance means the image is virtual, upright, and enlarged — exactly how a magnifying glass works when the object is inside the focal length.
Problem 10
Answer and explanation
Louder — constructive interference.
When the two waves arrive in phase (crest-on-crest), superposition adds their amplitudes, producing a larger-amplitude — and therefore louder — sound. This reinforcement is called constructive interference.
Problem 11
Answer and explanation
680 Hz
The source approaches, so use the minus sign: f′ = f × v ÷ (v − vₛ) = 600 × 340 ÷ (340 − 40) = 600 × 340 ÷ 300 = 600 × 1.13333 = 680 Hz.
Problem 12
Answer and explanation
v=fλ=10 m/s andT=1/f=.20 s.
Response criteria
- State a scientifically accurate conclusion.
- Show the calculation or mechanism that supports it.
- Include units, conditions, or limits where relevant.
Problem 13
Answer and explanation
f=50 Hz and v=40 m/s. At100 Hz with the same40 m/s speed, λ=0.40 m.
f=50 Hz and v=40 m/s.
At100 Hz with the same40 m/s speed, λ=0.40 m.
Response criteria
- Use the stated measurements or conditions, with correct units.
- Connect the result to the physical or environmental mechanism or evidence relationship.
- Identify the assumption, model limit or further evidence requested.
Problem 14
Answer and explanation
Round-trip 170 m; one-way distance 85 m.
Response criteria
- State a scientifically accurate conclusion.
- Show the calculation or mechanism that supports it.
- Include units, conditions, or limits where relevant.
Problem 15
Answer and explanation
The intensity ratio is10^1.5≈31.6. Echo distance=340×0.30/2=51 m; the two calculations use different models and quantities.
The intensity ratio is10^1.5≈31.6.
Echo distance=340×0.30/2=51 m; the two calculations use different models and quantities.
Response criteria
- Use the stated measurements or conditions, with correct units.
- Connect the result to the physical or environmental mechanism or evidence relationship.
- Identify the assumption, model limit or further evidence requested.
Problem 16
Answer and explanation
f=c/λ=1×10⁸ Hz. Use meters with SI c.
Response criteria
- State a scientifically accurate conclusion.
- Show the calculation or mechanism that supports it.
- Include units, conditions, or limits where relevant.
Problem 17
Answer and explanation
λ=4.00×10⁻⁷ m; f=7.50×10¹⁴ Hz and E=4.97×10⁻¹⁹ J per photon. This is not total beam energy without photon number.
λ=4.00×10⁻⁷ m; f=7.50×10¹⁴ Hz and E=4.97×10⁻¹⁹ J per photon.
This is not total beam energy without photon number.
Response criteria
- Use the stated measurements or conditions, with correct units.
- Connect the result to the physical or environmental mechanism or evidence relationship.
- Identify the assumption, model limit or further evidence requested.
Problem 18
Answer and explanation
It slows in the simple model and bends toward the normal. Frequency stays continuous across a stationary boundary; wavelength changes.
Response criteria
- State a scientifically accurate conclusion.
- Show the calculation or mechanism that supports it.
- Include units, conditions, or limits where relevant.
Problem 19
Answer and explanation
1/di=1/12−1/36=1/18, so di=18 cm and m=−18/36=−0.50. It is real, inverted and half the object height in this paraxial model.
1/di=1/12−1/36=1/18, so di=18 cm and m=−18/36=−0.50.
It is real, inverted and half the object height in this paraxial model.
Response criteria
- Use the stated measurements or conditions, with correct units.
- Connect the result to the physical or environmental mechanism or evidence relationship.
- Identify the assumption, model limit or further evidence requested.
Problem 20
Hints
Write the relationship first; substitute with consistent units.
Answer and explanation
60 Hz
fn=nf 1=3×20.
Problem 21
Answer and explanation
Fixed, free, open, and closed boundaries allow different mode patterns.
Response criteria
- State a scientifically accurate conclusion.
- Show the calculation or mechanism that supports it.
- Include units, conditions, or limits where relevant.
Problem 22
Answer and explanation
f1=v/(2L)=10 Hz; f3=30 Hz. λ3=2L/3=1.0 m, with fixed-end nodes in this ideal string model.
f1=v/(2L)=10 Hz; f3=30 Hz. λ3=2L/3=1.0 m, with fixed-end nodes in this ideal string model.
Response criteria
- Use the stated measurements or conditions, with correct units.
- Connect the result to the physical or environmental mechanism or evidence relationship.
- Identify the assumption, model limit or further evidence requested.
Problem 23
Hints
Write the relationship first; substitute with consistent units.
Answer and explanation
377.78 Hz
400×340/360≈377.78.
Problem 24
Answer and explanation
No. Light requires electromagnetic relativistic Doppler relationships.
Response criteria
- State a scientifically accurate conclusion.
- Show the calculation or mechanism that supports it.
- Include units, conditions, or limits where relevant.
Problem 25
Answer and explanation
Receding:600×330/360=550 Hz. Approaching:600×330/300=660 Hz. The shifts are not symmetric about600 in this finite-speed source model.
Receding:600×330/360=550 Hz.
Approaching:600×330/300=660 Hz.
The shifts are not symmetric about600 in this finite-speed source model.
Response criteria
- Use the stated measurements or conditions, with correct units.
- Connect the result to the physical or environmental mechanism or evidence relationship.
- Identify the assumption, model limit or further evidence requested.