Unit 4 · Unit Review

06Unit 4 Review

Consolidate all five chapters — geometric optics, wave optics, electrostatics, electric circuits, and magnetism — with key equations, worked examples, guided practice, vocabulary, and a full unit quiz.

A thorough unit review consolidates your understanding of optics, electrostatics, circuits, and magnetism — the four pillars of classical electromagnetism — and prepares you for modern physics in Unit 5.

Essential Question: How do the principles of optics, electrostatics, electric circuits, and magnetism connect to form a unified picture of electromagnetic phenomena — and how can mathematical models help us predict and control light, charge, and magnetic fields in real-world technology?

Unit 4 Summary

Optics, Electricity, and Magnetism — Chapters 01 through 05

Unit 4 explores the physics of light and electromagnetic phenomena. In Chapter 01 you studied geometric optics — how mirrors and lenses form images using the thin lens equation and magnification. Chapter 02 extended optics to wave behavior: diffraction and double-slit interference. Chapter 03 introduced electrostatics — Coulomb's law, electric fields, and electric potential. Chapter 04 applied electricity to circuits, covering Ohm's law, power, and series/parallel resistor networks. Chapter 05 completed the unit with magnetism: the magnetic force on moving charges and current-carrying wires, Faraday's law, and electromagnetic induction. Together these chapters reveal how electricity and magnetism are two aspects of the same fundamental force.

Chapter-by-Chapter Summary

Ch 01

Reflection, Refraction & Lenses

  • Law of reflection: θᵢ = θᵣ
  • Snell's law: n₁ sin θ₁ = n₂ sin θ₂
  • Thin lens equation: 1/f = 1/dₒ + 1/dᵢ
  • Magnification: m = −dᵢ/dₒ; |m|>1 → enlarged
Ch 02

Wave Optics & Interference

  • Double-slit: dsinθ = mλ (bright), (m+½)λ (dark)
  • Constructive: path difference = mλ
  • Destructive: path difference = (m+½)λ
  • Diffraction grating: sharper, brighter maxima
Ch 03

Electrostatics

  • Coulomb's law: Fₑ = kq₁q₂/r²
  • Electric field: E = F/q = kq/r²
  • Field lines point away from + toward −
  • Electric potential: V = kq/r (scalar)
Ch 04

Electric Circuits

  • Ohm's law: V = IR
  • Power: P = IV = I²R = V²/R
  • Series: same I, voltages add, Rₛ = ΣRᵢ
  • Parallel: same V, currents add, 1/Rₚ = Σ1/Rᵢ
Ch 05

Magnetism & Electromagnetic Induction

  • Magnetic force: F = qvB sin θ
  • Force on wire: F = BIL sin θ
  • Faraday's law: ε = −ΔΦ/Δt
  • Lenz's law: induced current opposes change in flux

Key Equations Reference Sheet

Thin Lens / Mirror

1/f = 1/dₒ + 1/dᵢ

f = focal length

Magnification

m = −dᵢ / dₒ

m < 0 → inverted

Coulomb's Law

Fₑ = kq₁q₂ / r²

k = 8.99 × 10⁹ N·m²/C²

Electric Field

E = F / q

N/C or V/m

Ohm's Law

V = IR

V in volts, I in amps

Electric Power

P = IV = I²R = V²/R

watts (W)

Series Resistance

Rₛ = R₁ + R₂ + …

same current

Parallel Resistance

1/Rₚ = 1/R₁ + 1/R₂ + …

same voltage

Magnetic Force

F = qvB sin θ

right-hand rule

Faraday's Law

ε = −ΔΦ / Δt

Φ = BA cos θ

Worked Examples

Example 1

A converging lens has a focal length of 10 cm. An object is placed 30 cm from the lens. Find the image distance and the magnification.

Use the thin lens equation: 1/f = 1/dₒ + 1/dᵢ

1/10 = 1/30 + 1/dᵢ

1/dᵢ = 1/10 − 1/30 = 3/30 − 1/30 = 2/30

dᵢ = 30/2 = 15 cm (positive → real image on opposite side)

Magnification: m = −dᵢ/dₒ = −15/30 = −0.5

The image is real, inverted (m < 0), and reduced (|m| < 1).

Answer:dᵢ = 15 cm (real image); m = −0.5 (inverted, reduced)
Example 2

Two point charges, q₁ = +3.0 μC and q₂ = −2.0 μC, are separated by 0.30 m. Find the magnitude of the electrostatic force between them.

Coulomb's law: Fₑ = k|q₁||q₂| / r²

k = 8.99 × 10⁹ N·m²/C²

|q₁| = 3.0 × 10⁻⁶ C, |q₂| = 2.0 × 10⁻⁶ C, r = 0.30 m

Fₑ = (8.99 × 10⁹)(3.0 × 10⁻⁶)(2.0 × 10⁻⁶) / (0.30)²

Numerator: 8.99 × 10⁹ × 6.0 × 10⁻¹² = 53.94 × 10⁻³ = 0.05394 N·m²

Denominator: (0.30)² = 0.090 m²

Fₑ = 0.05394 / 0.090 = 0.599 N ≈ 0.60 N

The force is attractive (opposite charges).

Answer:Fₑ ≈ 0.60 N (attractive)
Example 3

A 12 V battery is connected to a single resistor of 4 Ω. Find the current through the resistor and the power dissipated.

Ohm's law: V = IR → I = V/R

I = 12 V / 4 Ω = 3 A

Power: P = IV = 3 A × 12 V = 36 W

Verify with P = I²R = (3)² × 4 = 9 × 4 = 36 W ✓

Answer:I = 3 A; P = 36 W
Example 4

Three resistors — R₁ = 6 Ω, R₂ = 3 Ω, R₃ = 6 Ω — are connected in parallel across a 12 V source. Find the equivalent resistance and the total current drawn from the source.

Parallel rule: 1/Rₚ = 1/R₁ + 1/R₂ + 1/R₃

1/Rₚ = 1/6 + 1/3 + 1/6 = 1/6 + 2/6 + 1/6 = 4/6 = 2/3

Rₚ = 3/2 = 1.5 Ω

Total current: I = V/Rₚ = 12/1.5 = 8 A

Check: I₁=12/6=2 A, I₂=12/3=4 A, I₃=12/6=2 A → total=8 A ✓

Answer:Rₚ = 1.5 Ω; I_total = 8 A
Example 5

A proton (charge q = 1.60 × 10⁻¹⁹ C) moves at 2.0 × 10⁶ m/s perpendicular to a uniform magnetic field of 0.50 T. Find the magnitude of the magnetic force on the proton.

Magnetic force: F = qvB sin θ

θ = 90° (perpendicular) → sin 90° = 1

F = (1.60 × 10⁻¹⁹ C)(2.0 × 10⁶ m/s)(0.50 T)

F = 1.60 × 10⁻¹⁹ × 1.0 × 10⁶

F = 1.6 × 10⁻¹³ N

Answer:F = 1.6 × 10⁻¹³ N

Guided Practice

Guided Problem 1

A concave mirror has a focal length of 20 cm. An object is placed 60 cm in front of the mirror. Find the image distance and state whether the image is real or virtual.

Hint: Use 1/f = 1/dₒ + 1/dᵢ with f = 20 cm and dₒ = 60 cm. Solve for 1/dᵢ = 1/20 − 1/60. A positive dᵢ means a real image in front of the mirror.

Guided Problem 2

In a double-slit experiment, the slit separation is d = 0.50 mm and the screen is L = 2.0 m away. The wavelength of light is 600 nm. Find the distance from the central maximum to the first bright fringe.

Hint: For small angles, the fringe spacing is y = mλL/d. Use m = 1, λ = 600 × 10⁻⁹ m, L = 2.0 m, d = 0.50 × 10⁻³ m.

Guided Problem 3

Two charges, q₁ = +5.0 μC and q₂ = +5.0 μC, are 0.20 m apart. Find the electric field at the midpoint between them due to q₁ alone.

Hint: The midpoint is r = 0.10 m from q₁. Use E = kq/r². The direction is away from q₁ (positive charge). Note that by symmetry the fields from both charges cancel at the midpoint.

Guided Problem 4

A 9 V battery powers two resistors in series: R₁ = 2 Ω and R₂ = 7 Ω. Find the current in the circuit and the voltage drop across each resistor.

Hint: Series: Rₛ = R₁ + R₂ = 9 Ω. Current: I = V/Rₛ. Then V₁ = IR₁ and V₂ = IR₂. Check: V₁ + V₂ should equal 9 V.

Guided Problem 5

A rectangular coil of area 0.040 m² is placed in a uniform magnetic field of 0.30 T perpendicular to the coil. The field drops to zero in 0.020 s. Find the magnitude of the induced EMF.

Hint: Use Faraday's law: |ε| = |ΔΦ/Δt|. Initial flux: Φᵢ = BA = 0.30 × 0.040. Final flux: Φ_f = 0. ΔΦ = Φ_f − Φᵢ.

Key Vocabulary

Focal Length (f)

The distance from a lens or mirror to its focal point — the point where parallel rays converge (converging) or appear to diverge from (diverging).

Example: A converging lens with f = 10 cm focuses parallel light 10 cm behind the lens.

Electric Field (E)

A vector field that describes the force per unit positive charge at every point in space. E = F/q, measured in N/C or V/m.

Example: A +1 μC charge creates a field of 9 × 10⁴ N/C at a distance of 0.30 m.

Electric Potential (V)

The electric potential energy per unit charge at a point in space. It is a scalar quantity measured in volts (V = J/C).

Example: The potential 0.10 m from a +2 μC charge is V = kq/r ≈ 180,000 V.

Resistance (R)

A measure of how strongly a material opposes the flow of electric current. Measured in ohms (Ω). R = V/I.

Example: A 60 W light bulb operating at 120 V has resistance R = V²/P = 240 Ω.

Current (I)

The rate of flow of electric charge through a conductor. I = ΔQ/Δt, measured in amperes (A = C/s).

Example: A 12 V battery connected to a 4 Ω resistor drives a current of 3 A.

Power (P)

The rate at which electrical energy is converted to other forms. P = IV = I²R = V²/R, measured in watts (W).

Example: A 3 A current through a 4 Ω resistor dissipates P = I²R = 36 W.

Magnetic Flux (Φ)

The total magnetic field passing through a surface. Φ = BA cos θ, measured in webers (Wb = T·m²).

Example: A 0.5 T field through a 0.04 m² coil perpendicular to the field gives Φ = 0.02 Wb.

Electromagnetic Induction

The production of an EMF (voltage) in a conductor due to a changing magnetic flux, described by Faraday's law: ε = −ΔΦ/Δt.

Example: A generator converts mechanical rotation into electrical energy via electromagnetic induction.

Mastery Checklist

Unit 4 Mastery Checklist

Ch01

Apply the thin lens equation to find image distance or focal length

Ch01

Calculate magnification and determine if an image is real, virtual, inverted, or upright

Ch02

Use the double-slit formula to find bright and dark fringe positions

Ch02

Distinguish constructive from destructive interference using path difference

Ch03

Apply Coulomb's law to find the electrostatic force between two charges

Ch03

Calculate electric field magnitude and direction from a point charge

Ch04

Apply Ohm's law and power formulas to series and parallel circuits

Ch04

Find equivalent resistance for series and parallel combinations

Ch05

Calculate the magnetic force on a moving charge or current-carrying wire

Ch05

Use Faraday's law to find induced EMF from a changing magnetic flux

Workbook Quiz

Interactive Practice — 5 Questions

1

A converging lens has focal length f = 15 cm. An object is placed 45 cm away. What is the image distance dᵢ?

2

Two charges of +4 μC and −4 μC are separated by 0.20 m. Which statement about the electrostatic force is correct?

3

A 24 V battery is connected to two resistors in series: R₁ = 4 Ω and R₂ = 8 Ω. What is the current in the circuit?

4

Two resistors, 10 Ω and 10 Ω, are connected in parallel. What is the equivalent resistance?

5

An electron (q = 1.60 × 10⁻¹⁹ C) moves at 3.0 × 10⁶ m/s perpendicular to a 0.20 T magnetic field. What is the magnitude of the magnetic force?