5.3 Light, Optics, Electricity & Magnetism

Light & Optics: Reflection, Refraction, Lenses & Instruments

Light is an electromagnetic wave that exhibits both wave characteristics (diffraction, interference) and particle properties (photon emission). Geometric optics models light propagation using straight rays.

Reflection & Refraction of Light

Law of Reflection

When light strikes a polished reflective surface (like a plane mirror):

1. The incident ray, reflected ray, and normal all lie in the same plane.
2. The angle of incidence equals the angle of reflection ($\theta_i = \theta_r$).

Snell's Law of Refraction

Refraction is the bending of light as it passes from one medium into another of different optical density, caused by a change in wave speed.

The Index of Refraction ($n$) of a medium is defined as: $n = \frac{c}{v}$ Where $c = 3.0 \times 10^8\text{ m/s}$ (speed of light in vacuum) and $v$ is light speed in the medium.

Snell's Law relates incident and refracted angles: $n_1 \sin\theta_1 = n_2 \sin\theta_2$

Total Internal Reflection: When light travels from a denser medium ($n_1$) to a rarer medium ($n_2$) at an angle greater than the critical angle ($\theta_c$), light is totally reflected back. The critical angle satisfies: $\sin\theta_c = \frac{n_2}{n_1}$ Total internal reflection powers optical fiber communications and prism binoculars.

---

Thin Lenses & Image Formation

Lenses use refraction to converge or diverge light rays to form images.

The Thin Lens Formula & Magnification

$\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$ $m = \frac{h_i}{h_o} = -\frac{d_i}{d_o}$

Where:

• $f$ = focal length (positive for convex, negative for concave)
• $d_o$ = object distance from lens center (always positive for real objects)
• $d_i$ = image distance (positive for real images behind lens, negative for virtual images in front)
• $m$ = linear magnification ($|m| > 1$ magnified, $|m| < 1$ diminished, negative $m$ inverted)

---

Worked Example: Convex Lens Image Calculation

Problem: An object $4\text{ cm}$ tall is placed $15\text{ cm}$ in front of a thin convex lens of focal length $10\text{ cm}$.

1. Find the position of the image ($d_i$).
2. Determine the height ($h_i$) and nature of the image.

Solution:

1. Using the thin lens formula: $\frac{1}{10} = \frac{1}{15} + \frac{1}{d_i} \implies \frac{1}{d_i} = \frac{1}{10} - \frac{1}{15} = \frac{3 - 2}{30} = \frac{1}{30}$ $d_i = 30\text{ cm}$ Since $d_i > 0$, the image is real and formed $30\text{ cm}$ on the opposite side of the lens.

2. Using the magnification formula: $m = -\frac{d_i}{d_o} = -\frac{30}{15} = -2$ $h_i = m \cdot h_o = (-2)(4\text{ cm}) = -8\text{ cm}$ The image is inverted, real, and magnified to a height of $8\text{ cm}$.

Electrostatics, Electric Circuits & Electromagnetism

Electricity and magnetism are unified manifestations of the electromagnetic force, one of the four fundamental interactions of nature.

Electrostatics & Coulomb's Law

Electric charge is a fundamental property of matter, occurring as positive (protons) or negative (electrons). Charge is conserved and quantized in elementary charge units ($e \approx 1.6 \times 10^{-19}\text{ C}$).

Coulomb's Law

The electrostatic force between two stationary point charges is directly proportional to the product of charges and inversely proportional to the square of separation distance: $F = k \frac{|q_1 q_2|}{r^2}$ Where $k = \frac{1}{4\pi\varepsilon_0} \approx 8.99 \times 10^9\text{ N}\cdot\text{m}^2/\text{C}^2$ in vacuum or air. Like charges repel; opposite charges attract.

---

Electric Circuits & Ohm's Law

An electric circuit provides a closed loop through which electric current ($I = \Delta Q / \Delta t$, in Amperes) flows driven by electromotive force or potential difference ($V$, in Volts).

Ohm's Law

At constant temperature, current through a conductor is directly proportional to voltage across it and inversely proportional to resistance ($R$, in Ohms $\Omega$): $V = I R$

Series vs. Parallel Circuits

Electrical Power and Energy

Power dissipated by a resistor: $P = I V = I^2 R = \frac{V^2}{R}$ Electrical energy consumed is $E = P \cdot t$, measured commercially in kilowatt-hours ($1\text{ kWh} = 3.6 \times 10^6\text{ J}$).

---

Electromagnetism & Induction

• Right-Hand Rule: Determines the direction of the magnetic field ($\vec{B}$) around a straight current-carrying wire or solenoid coil.
• Faraday's Law of Induction: An electromotive force ($\mathcal{E}$) is induced in a circuit whenever the magnetic flux ($\Phi_B = B A \cos\theta$) through the circuit changes over time: $\mathcal{E} = -N \frac{\Delta \Phi_B}{\Delta t}$
• Transformers: Devices using electromagnetic induction to step up or step down AC voltage: $\frac{V_p}{V_s} = \frac{N_p}{N_s} = \frac{I_s}{I_p}$

---

Worked Example: Parallel Circuit & Power Calculation

Problem: Two resistors of $R_1 = 6\ \Omega$ and $R_2 = 12\ \Omega$ are connected in parallel across a $24\text{ V}$ battery.

1. Calculate equivalent circuit resistance ($R_{\text{eq}}$).
2. Calculate total current ($I_{\text{total}}$) drawn from the battery and power dissipated in the $6\ \Omega$ resistor.

Solution:

1. Equivalent resistance for parallel combination: $\frac{1}{R_{\text{eq}}} = \frac{1}{6} + \frac{1}{12} = \frac{2 + 1}{12} = \frac{3}{12} = \frac{1}{4} \implies R_{\text{eq}} = 4\ \Omega$

2. Total battery current: $I_{\text{total}} = \frac{V}{R_{\text{eq}}} = \frac{24\text{ V}}{4\ \Omega} = 6\text{ Amperes}$

3. Power dissipated in the $6\ \Omega$ resistor (receiving full $24\text{ V}$): $P_1 = \frac{V^2}{R_1} = \frac{24^2}{6} = \frac{576}{6} = 96\text{ Watts}$