Mathematics for light and current electricity

Overview

Light and current electricity both use mathematical relationships to describe physical situations. In light, mathematics helps learners compare angles, distances, object size, image size, and the behaviour of rays. In current electricity, mathematics helps learners connect charge, current, time, potential difference, resistance, power, and energy.

This chapter is not mainly about memorising many formulas. It teaches how to read a relationship, choose suitable units, rearrange an equation, interpret a table or graph, and explain what a result means physically. The same mathematical habits used in Form I measurement, formula work, and graphing are now applied to Form II optics and electricity.

The main idea is simple: a Physics formula is a compact statement about quantities. A learner should be able to say what each symbol means, what unit belongs to it, how it changes when another quantity changes, and whether the numerical answer is reasonable.

Prerequisites

• Physical quantities and SI units - Learners should know that a physical quantity has a numerical value and a unit.
• Measurement - Accurate measurements of length, time, angle, current, and potential difference support the relationships in this chapter.
• Mathematical relationships among physical quantities - Formula rearrangement, substitution, direct proportion, inverse relationships, and unit discipline are the main mathematical tools used here.
• Graphical presentation of experimental results - Tables, axes, scales, plotted points, gradients, and graph interpretation are needed for light and electricity data.
• Graphs and mathematical relationships in Physics - Form II graph work extends the same relationship-finding skill.
• Light - The light topic supplies the physical ideas behind angles, ray diagrams, image position, and optical instruments.
• Current electricity - The current electricity topic supplies the physical ideas behind electric current, potential difference, resistance, power, and domestic circuits.

Learning Scope

This chapter covers mathematical skills used in light and current electricity. It includes symbols and units, angle relationships for reflection, proportional reasoning in optical examples, simple magnification ideas, electric current as charge per time, potential difference, resistance, Ohm's law, electric power, energy transfer, table interpretation, and graph interpretation.

This page does not replace the full conceptual chapters on Light or Current electricity. It also does not teach advanced optics, wave optics, semiconductor electronics, detailed domestic installation rules, or full experimental report writing. CSEE_FORMATS_2022 is kept as assessment-only context, and no reviewed Physics exam mappings are claimed on this page.

Subtopics

Mathematical Habits For Light And Electricity

A good calculation in Form II Physics normally follows a careful order:

1. Identify the physical situation.
2. List the quantities given and the quantity required.
3. Write the formula or relationship.
4. Check and convert units where needed.
5. Rearrange the formula if necessary.
6. Substitute values with units.
7. Calculate and write the answer with a unit.
8. Interpret the answer in words.

Key insight: The number alone is not the answer in Physics. The unit and meaning complete the answer.

For example, if a calculation gives $2\ \text{A}$, the learner should know that the result is a current, not a resistance, voltage, power, or charge.

Symbols And Units Used In This Chapter

The table below gathers common quantities used in mathematical work on light and current electricity.

| Quantity | Common symbol | Meaning | Common unit | | --- | ---: | --- | --- | | angle of incidence | $i$ | angle between incident ray and normal | degree, $^\circ$ | | angle of reflection | $r$ | angle between reflected ray and normal | degree, $^\circ$ | | object distance | $u$ | distance from object to mirror or lens | $\text{m}$ or $\text{cm}$ | | image distance | $v$ | distance from image to mirror or lens | $\text{m}$ or $\text{cm}$ | | object height | $h_o$ | height of object | $\text{m}$ or $\text{cm}$ | | image height | $h_i$ | height of image | $\text{m}$ or $\text{cm}$ | | magnification | $M$ | ratio of image size to object size | no unit | | charge | $Q$ | quantity of electric charge | $\text{C}$ | | current | $I$ | rate of flow of charge | $\text{A}$ | | time | $t$ | duration | $\text{s}$ | | potential difference | $V$ | energy transferred per unit charge | $\text{V}$ | | resistance | $R$ | opposition to current | $\Omega$ | | power | $P$ | rate of energy transfer | $\text{W}$ | | energy | $E$ | energy transferred or used | $\text{J}$ |

Some symbols depend on the context. For example, $V$ can mean potential difference, while $v$ may mean image distance in an optics formula. Read the question carefully and define each symbol before substituting numbers.

Angle Relationships In Light

In reflection from a plane surface, angles are measured from the normal, not from the mirror surface. The normal is a line drawn at right angles to the reflecting surface.

The reflection relationship is:

$$ i = r $$

where:

• $i$ is the angle of incidence
• $r$ is the angle of reflection

If the angle of incidence is $35^\circ$, then:

$$ r = 35^\circ $$

Key insight: The equal angles are measured from the normal. If a question gives an angle from the surface, first convert it to an angle from the normal.

For a plane mirror:

$$ \text{object distance} = \text{image distance} $$

This means an object $0.8\ \text{m}$ in front of a plane mirror appears $0.8\ \text{m}$ behind the mirror. The total apparent distance between object and image is:

$$ 0.8\ \text{m} + 0.8\ \text{m} = 1.6\ \text{m} $$

Distance, Size, And Magnification In Light

Magnification compares image size with object size. A simple size relationship is:

$$ M = \frac{h_i}{h_o} $$

where:

• $M$ is magnification
• $h_i$ is image height
• $h_o$ is object height

If the image height and object height are measured in the same unit, the units cancel. Magnification has no unit.

For similar ray diagrams, magnification may also be represented by a distance ratio:

$$ M = \frac{v}{u} $$

where $v$ is image distance and $u$ is object distance, when the particular optical model being used supports that relationship.

Key insight: Ratios only work cleanly when the compared quantities use the same unit. Do not compare $12\ \text{cm}$ with $0.03\ \text{m}$ until the units are made consistent.

Magnification can be interpreted as follows:

• $M=1$ means image size equals object size.
• $M>1$ means the image is larger than the object.
• $0<M<1$ means the image is smaller than the object.

This chapter uses magnification as a mathematical relationship. The detailed ray behaviour belongs to Light.

Current As Charge Per Time

Electric current is the rate of flow of charge:

$$ I = \frac{Q}{t} $$

where:

• $I$ is current in amperes, $\text{A}$
• $Q$ is charge in coulombs, $\text{C}$
• $t$ is time in seconds, $\text{s}$

Rearranged forms:

$$ \begin{aligned} I &= \frac{Q}{t} \\ Q &= I t \\ t &= \frac{Q}{I} \end{aligned} $$

The unit relationship is:

$$ 1\ \text{A} = 1\ \text{C/s} $$

Key insight: Current is not the same as charge. Current tells how quickly charge passes a point in a circuit.

If current is constant, then charge is directly proportional to time:

$$ Q \propto t $$

Doubling the time doubles the charge that passes, if the current remains the same.

Potential Difference, Energy, And Charge

Potential difference is linked with energy transferred per unit charge:

$$ V = \frac{E}{Q} $$

where:

• $V$ is potential difference in volts, $\text{V}$
• $E$ is energy in joules, $\text{J}$
• $Q$ is charge in coulombs, $\text{C}$

Rearranged forms:

$$ \begin{aligned} V &= \frac{E}{Q} \\ E &= VQ \\ Q &= \frac{E}{V} \end{aligned} $$

The unit relationship is:

$$ 1\ \text{V} = 1\ \text{J/C} $$

Key insight: A larger potential difference means more energy is transferred by each coulomb of charge.

For example, if $12\ \text{C}$ of charge passes through a device at $6\ \text{V}$:

$$ E = VQ = 6\ \text{V} \times 12\ \text{C} = 72\ \text{J} $$

Resistance And Ohm's Law

Resistance measures how strongly a component opposes the flow of current. A common relationship for an ohmic conductor at constant physical conditions is:

$$ V = IR $$

where:

• $V$ is potential difference
• $I$ is current
• $R$ is resistance

Rearranged forms:

$$ \begin{aligned} V &= IR \\ I &= \frac{V}{R} \\ R &= \frac{V}{I} \end{aligned} $$

The unit of resistance is the ohm, $\Omega$:

$$ 1\ \Omega = 1\ \text{V/A} $$

Key insight: If resistance is constant, current is directly proportional to potential difference. If potential difference is constant, current decreases when resistance increases.

Useful proportional statements:

| Formula | If this is constant | Relationship | | --- | --- | --- | | $V=IR$ | $R$ | $V \propto I$ | | $V=IR$ | $V$ | $I \propto \frac{1}{R}$ | | $I=\frac{Q}{t}$ | $I$ | $Q \propto t$ | | $V=\frac{E}{Q}$ | $Q$ | $V \propto E$ | | $P=VI$ | $V$ | $P \propto I$ |

Always state the quantity that is kept constant. Without that condition, a proportional statement may be incomplete.

Power And Energy In Electric Circuits

Electrical power is the rate at which electrical energy is transferred:

$$ P = \frac{E}{t} $$

where:

• $P$ is power in watts, $\text{W}$
• $E$ is energy in joules, $\text{J}$
• $t$ is time in seconds, $\text{s}$

Power can also be related to potential difference and current:

$$ P = VI $$

When $V=IR$ is combined with $P=VI$, other useful forms may be produced:

$$ \begin{aligned} P &= VI \\ P &= I^2R \\ P &= \frac{V^2}{R} \end{aligned} $$

These forms should be used carefully. First decide which quantities are known, then choose the form that matches them.

Energy transferred by an electrical device can be found from:

$$ E = Pt $$

or, if potential difference and charge are known:

$$ E = VQ $$

Key insight: Power measures how fast energy is transferred. A high-power device uses the same amount of energy in a shorter time than a lower-power device, if both transfer the same total energy.

Tables In Light And Electricity

Tables help learners test patterns and prepare data for graphs.

Example current-voltage table:

| Potential difference, $V$ ($\text{V}$) | Current, $I$ ($\text{A}$) | | ---: | ---: | | $0$ | $0.0$ | | $2$ | $0.5$ | | $4$ | $1.0$ | | $6$ | $1.5$ | | $8$ | $2.0$ |

To test whether resistance is constant, calculate:

$$ R = \frac{V}{I} $$

For the non-zero readings:

$$ \frac{2}{0.5} = \frac{4}{1.0} = \frac{6}{1.5} = \frac{8}{2.0} = 4 $$

So the resistance is constant at:

$$ 4\ \Omega $$

Key insight: A constant ratio of $\frac{V}{I}$ suggests constant resistance for the measured range.

Example light table:

| Object distance ($\text{cm}$) | Image distance ($\text{cm}$) | | ---: | ---: | | $20$ | $20$ | | $30$ | $30$ | | $40$ | $40$ |

For a plane mirror pattern, object distance equals image distance in each row. The important relationship is equality, not a complicated formula.

Graph Interpretation In Light And Electricity

Graphs show whether a relationship is direct, inverse, constant, or changing.

For an ohmic conductor at constant physical conditions, a graph of potential difference $V$ against current $I$ is a straight line through the origin:

$$ V = IR $$

If $V$ is on the vertical axis and $I$ is on the horizontal axis, the gradient gives resistance:

$$ \text{gradient} = \frac{\Delta V}{\Delta I} = R $$

For a graph of charge $Q$ against time $t$ at constant current:

$$ Q = It $$

If $Q$ is on the vertical axis and $t$ is on the horizontal axis, the gradient gives current:

$$ \text{gradient} = \frac{\Delta Q}{\Delta t} = I $$

For a plane mirror distance relationship, a graph of image distance against object distance should show equal values if measurements are ideal:

$$ \text{image distance} = \text{object distance} $$

Key insight: A graph gradient has a unit. For a $V$ against $I$ graph, the gradient unit is $\text{V/A}$, which is $\Omega$.

Choosing The Correct Relationship

Many errors happen because a learner chooses a formula only because it contains a familiar symbol. A better method is to match the words in the problem to the meaning of the formula.

Use this guide:

| If the problem asks about | Consider | | --- | --- | | equal angles in reflection | $i=r$ | | object and image distance in a plane mirror | object distance $=$ image distance | | comparing image size with object size | $M=\frac{h_i}{h_o}$ | | charge flowing for a time | $I=\frac{Q}{t}$ | | energy per unit charge | $V=\frac{E}{Q}$ | | current, voltage, and resistance | $V=IR$ | | rate of electrical energy transfer | $P=\frac{E}{t}$ or $P=VI$ |

Key insight: Start from the physical meaning, then choose the formula. Do not start by hunting for any equation that has the same letters.

Key Terms

• Angle of incidence: angle between the incident ray and the normal at the reflecting surface.
• Angle of reflection: angle between the reflected ray and the normal at the reflecting surface.
• Charge: quantity of electricity, measured in coulombs, $\text{C}$.
• Current: rate of flow of charge, measured in amperes, $\text{A}$.
• Direct proportion: relationship where two quantities change in the same ratio when other relevant quantities are constant.
• Electrical energy: energy transferred by an electric circuit or device, measured in joules, $\text{J}$.
• Gradient: change in the vertical quantity divided by change in the horizontal quantity on a graph.
• Magnification: ratio comparing image size with object size; it has no unit.
• Normal: line drawn perpendicular to a surface at the point where a ray meets it.
• Ohm: unit of resistance, symbol $\Omega$.
• Ohm's law: relationship $V=IR$ for an ohmic conductor under constant physical conditions.
• Potential difference: energy transferred per unit charge, measured in volts, $\text{V}$.
• Power: rate of energy transfer, measured in watts, $\text{W}$.
• Resistance: opposition to electric current, measured in ohms, $\Omega$.
• Unit discipline: habit of keeping units visible, consistent, and meaningful during calculations.

Worked Examples

Example 1: Reflection Angle

A ray of light strikes a plane mirror with an angle of incidence of $42^\circ$. Find the angle of reflection.

Use the law of reflection:

$$ i = r $$

Substitute:

$$ \begin{aligned} r &= i \\ &= 42^\circ \end{aligned} $$

Final answer: The angle of reflection is $42^\circ$.

Check: The angle is measured from the normal, not from the mirror surface.

Example 2: Plane Mirror Distance

An object is placed $0.6\ \text{m}$ in front of a plane mirror. Find the image distance behind the mirror and the distance between the object and its image.

For a plane mirror:

$$ \text{image distance} = \text{object distance} $$

So:

$$ \text{image distance} = 0.6\ \text{m} $$

The object and image are on opposite sides of the mirror:

$$ \begin{aligned} \text{object-image distance} &= 0.6\ \text{m} + 0.6\ \text{m} \\ &= 1.2\ \text{m} \end{aligned} $$

Final answer: The image is $0.6\ \text{m}$ behind the mirror, and the object is $1.2\ \text{m}$ from its image.

Example 3: Magnification From Heights

An object is $4\ \text{cm}$ tall. Its image is $10\ \text{cm}$ tall. Find the magnification.

Use:

$$ M = \frac{h_i}{h_o} $$

Substitute:

$$ \begin{aligned} M &= \frac{10\ \text{cm}}{4\ \text{cm}} \\ &= 2.5 \end{aligned} $$

Final answer: The magnification is $2.5$.

Interpretation: The image is $2.5$ times as tall as the object.

Example 4: Current From Charge And Time

A charge of $30\ \text{C}$ passes a point in a circuit in $6\ \text{s}$. Find the current.

Use:

$$ I = \frac{Q}{t} $$

Substitute:

$$ \begin{aligned} I &= \frac{30\ \text{C}}{6\ \text{s}} \\ &= 5\ \text{A} \end{aligned} $$

Final answer: The current is $5\ \text{A}$.

Check: $\text{C/s}$ is equivalent to $\text{A}$.

Example 5: Resistance From Potential Difference And Current

A lamp has a potential difference of $12\ \text{V}$ across it and a current of $3\ \text{A}$ through it. Find its resistance.

Start with:

$$ V = IR $$

Make $R$ the subject:

$$ \begin{aligned} V &= IR \\ R &= \frac{V}{I} \end{aligned} $$

Substitute:

$$ \begin{aligned} R &= \frac{12\ \text{V}}{3\ \text{A}} \\ &= 4\ \Omega \end{aligned} $$

Final answer: The resistance is $4\ \Omega$.

Example 6: Energy Used By An Electrical Device

A radio operates at $9\ \text{V}$ with a current of $0.5\ \text{A}$ for $120\ \text{s}$. Find the energy transferred.

First find the power:

$$ \begin{aligned} P &= VI \\ &= 9\ \text{V} \times 0.5\ \text{A} \\ &= 4.5\ \text{W} \end{aligned} $$

Then use:

$$ E = Pt $$

Substitute:

$$ \begin{aligned} E &= 4.5\ \text{W} \times 120\ \text{s} \\ &= 540\ \text{J} \end{aligned} $$

Final answer: The radio transfers $540\ \text{J}$ of energy.

Check: The answer is in joules because $\text{W}=\text{J/s}$.

Example 7: Gradient Of A Current-Voltage Graph

A straight-line graph of potential difference $V$ against current $I$ passes through $(0.5\ \text{A}, 2\ \text{V})$ and $(2.0\ \text{A}, 8\ \text{V})$. Find the resistance represented by the graph.

For a $V$ against $I$ graph:

$$ \text{gradient} = \frac{\Delta V}{\Delta I} = R $$

Calculate:

$$ \begin{aligned} R &= \frac{8\ \text{V} - 2\ \text{V}}{2.0\ \text{A} - 0.5\ \text{A}} \\ &= \frac{6\ \text{V}}{1.5\ \text{A}} \\ &= 4\ \Omega \end{aligned} $$

Final answer: The resistance is $4\ \Omega$.

Interpretation: The straight-line graph suggests constant resistance in the measured range.

Common Mistakes

• Mistake: Measuring reflection angles from the mirror surface. Correction: measure $i$ and $r$ from the normal.
• Mistake: Forgetting that plane mirror image distance equals object distance. Correction: place the image the same distance behind the mirror as the object is in front.
• Mistake: Giving magnification with a unit. Correction: magnification is a ratio, so matching units cancel.
• Mistake: Comparing image height in centimetres with object height in metres. Correction: convert to the same unit before forming a ratio.
• Mistake: Confusing charge and current. Correction: charge is the amount, current is charge per second.
• Mistake: Writing $R=VI$ instead of $R=\frac{V}{I}$. Correction: start from $V=IR$ and rearrange carefully.
• Mistake: Using minutes or hours directly in $I=\frac{Q}{t}$ or $E=Pt$ when the expected SI unit is seconds. Correction: convert time to seconds unless the problem clearly defines another unit system.
• Mistake: Reading a graph gradient without units. Correction: divide the vertical-axis unit by the horizontal-axis unit.
• Mistake: Assuming every $V$-$I$ graph is straight. Correction: straight-line behaviour depends on the component and conditions.
• Mistake: Treating CSEE_FORMATS_2022 as a syllabus source. Correction: use it only as assessment context after review.

Practice Tasks

1. State why units are important in calculations about light and current electricity.
2. A ray has an angle of incidence of $28^\circ$. Find the angle of reflection.
3. An object is $15\ \text{cm}$ in front of a plane mirror. How far behind the mirror is its image?
4. An object is $8\ \text{cm}$ tall and its image is $4\ \text{cm}$ tall. Find the magnification.
5. Explain why magnification has no unit.
6. A charge of $48\ \text{C}$ passes a point in $12\ \text{s}$. Find the current.
7. A current of $2\ \text{A}$ flows for $30\ \text{s}$. Find the charge that passes.
8. A device transfers $180\ \text{J}$ of energy when $15\ \text{C}$ of charge passes through it. Find the potential difference.
9. A resistor has $10\ \text{V}$ across it and $2\ \text{A}$ through it. Find its resistance.
10. A $6\ \Omega$ resistor carries a current of $0.5\ \text{A}$. Find the potential difference across it.
11. A device operates at $12\ \text{V}$ and $1.5\ \text{A}$. Find its power.
12. A $60\ \text{W}$ lamp is used for $300\ \text{s}$. Find the energy transferred.
13. A table of $V$ and $I$ readings gives a constant ratio $\frac{V}{I}=3$. What is the resistance?
14. On a graph of $Q$ against $t$, the gradient is $0.8\ \text{C/s}$. What current does this represent?
15. A learner plots current on the vertical axis and potential difference on the horizontal axis. What physical quantity is represented by the reciprocal of the gradient if the graph is straight?
16. Write one reason why a graph of $V$ against $I$ may fail to remain a straight line for some components.
17. Create a small table for an ohmic conductor of resistance $5\ \Omega$ using currents $0$, $1$, $2$, and $3\ \text{A}$.
18. A question gives an angle of $20^\circ$ between a ray and a mirror surface. Find the angle between the ray and the normal.

Generated Question Layer

• Direct knowledge questions: Ask learners to define current, charge, potential difference, resistance, power, magnification, normal, and gradient.
• Formula selection questions: Give short situations and ask learners to choose between $i=r$, $M=\frac{h_i}{h_o}$, $I=\frac{Q}{t}$, $V=IR$, $P=VI$, and $E=Pt$.
• Rearrangement questions: Ask learners to make $Q$, $t$, $I$, $V$, $R$, $P$, or $E$ the subject of a formula.
• Unit questions: Ask learners to identify or convert $\text{C}$, $\text{A}$, $\text{s}$, $\text{V}$, $\Omega$, $\text{W}$, $\text{J}$, $\text{cm}$, and $\text{m}$.
• Proportional reasoning questions: Ask how current changes when voltage doubles at constant resistance, or how charge changes when time doubles at constant current.
• Light relationship questions: Use angle, mirror distance, and magnification relationships without requiring advanced optics.
• Table questions: Provide light or electricity readings and ask learners to find a constant ratio, missing value, or pattern.
• Graph questions: Use $V$-$I$, $Q$-$t$, and object-distance/image-distance graphs to ask for gradient, unit, and physical interpretation.
• Error-checking questions: Present incorrect unit conversion, wrong angle reference, reversed formula, or missing graph units and ask learners to correct the work.
• Mixed-skill questions: Combine unit conversion, formula rearrangement, substitution, and interpretation in one light or electricity context.

Learner Aid Opportunities

• diagram: labelled reflection sketch showing incident ray, reflected ray, mirror surface, normal, $i$, and $r$.
• diagram: plane mirror object-image distance sketch showing equal distances from the mirror.
• chart: formula map connecting $I=\frac{Q}{t}$, $V=\frac{E}{Q}$, $V=IR$, $P=VI$, and $E=Pt$.
• graph: sample $V$ against $I$ graph with gradient triangle labelled in $\text{V/A}$.
• graph: sample $Q$ against $t$ graph showing current as gradient.
• interactive: unit-conversion and formula-rearrangement practice for light and circuit quantities.
• interactive: graph-reading task where learners choose axes, calculate gradient, and identify the physical meaning.
• LLM tutor: adaptive hints for choosing the correct formula, checking units, and explaining proportional relationships.

Exam-Derived Signals

• No reviewed past-paper mapping has been attached to this Physics topic in this milestone.
• CSEE_FORMATS_2022 may provide future assessment signals about calculations, graph interpretation, and practical data handling after review.
• CSEE_FORMATS_2022 is assessment-only context. It does not replace or redefine the 2023 Physics syllabus wording used for this page.
• Future exam-derived examples should be clearly marked as reviewed or unreviewed and should not broaden the syllabus scope without maintainer review.

Source And Review Notes

• Official syllabus status: Topic identity, Form II placement, competence, source topic ID, and hub are drawn from the 2023 CSEE Physics syllabus through data/curricula/csee/physics/2023.json.
• Learner expansion status: This chapter is original learner-facing writing based on the official syllabus topic and existing repo context.
• External enrichment status: No external web enrichment was used.
• Textbook status: No textbook wording was used.
• Exam signal status: No past-paper item has been reviewed for this page; CSEE_FORMATS_2022 remains assessment-only context.
• Review risks: Formula selection, symbol conventions, and the exact depth expected for optical magnification, Ohm's law, and electrical power should be checked by a Physics reviewer against local teaching guidance.