Leaving Certificate Physics

All Mandatory Experiments 🔬

Step-by-step guides, diagrams, and calculations for every mandatory experiment you need to know for Section A of the exam.

🧪 Section A Questions 📊 Graphs & Calculations

The Leaving Cert physics practical exam tests your knowledge of 28 mandatory experiments. You need to know apparatus, procedure, sources of error, and how to do the calculations. This guide covers every single one!

      📚 The 5 categories:
      ⚙️ Mechanics (8 experiments) — motion, forces, gas laws
🌡️ Heat (6 experiments) — specific heat capacity, latent heat
🔊 Sound (3 experiments) — speed of sound, strings
💡 Light (6 experiments) — mirrors, lenses, refraction, wavelength
⚡ Electricity (5 experiments) — current, resistance, I-V curves

    

🎯 What the Exam Tests

Theory Questions

Name the apparatus used
Describe the procedure step by step
State sources of error / precautions
Describe the expected graph
What does the graph prove / show?

Calculation Questions

Use given data to calculate the quantity being measured
Find the slope of a graph
Calculate average values
Convert units
Apply the relevant formula

💡 General Tips for Every Experiment

⚠️ Always Say These!

1. Repeat and average — always improves accuracy
2. Avoid parallax error — view scales at right angles
3. Use lagging/insulation for heat experiments to reduce heat loss
4. Zero the instrument first (thermometer, joulemeter, ohmmeter)
5. A straight line through the origin verifies proportionality

Mechanics — Exp 1

⚙️ Measurement of Velocity

Using a ticker timer to measure constant velocity

What We're Doing

We adjust a tilted runway until a trolley moves at constant velocity. The ticker timer records dots on tape; equally-spaced dots = constant velocity. We measure distance and time to calculate v = s/t.

Apparatus

Ticker timer + tape, low-voltage a.c. power supply, trolley, runway, laboratory jack or stand, metre stick

Diagram

Procedure (Step by Step)

1

Connect the ticker timer to a low-voltage a.c. power supply and attach tape to the trolley.

2

Tilt the runway slightly using the jack/stand.

3

Give the trolley a gentle push and let it run down the tape.

4

Adjust the tilt until the dots on the tape are equally spaced — this means constant velocity (friction is compensated by the slope).

5

Measure the length s of 10 adjacent spaces on the tape with a metre stick.

6

Time t = 10 × 0.02 = 0.2 s (timer makes 50 dots/second, so 1 dot = 0.02 s).

7

Calculate velocity: v = s ÷ t. Repeat with different push strengths.

📐 Formula

v = s / t

s = distance of 10 spaces (metres) | t = 10 × 0.02 = 0.2 s

Example Calculation

📐 Worked Example

Given

10 spaces measure s = 0.12 m

Time

t = 10 × 0.02 = 0.2 s

Velocity

v = s/t = 0.12 / 0.2 = 0.6 m/s

⚠️ Key Points

• Ignore the first 5–6 dots (trolley may not be at constant velocity yet)
• Use precarbonated tape (less friction)
• Some timers make 100 dots/s (time interval = 0.01 s) — check your timer!

Mechanics — Exp 2

⚙️ Measurement of Acceleration

Using a ticker timer to find initial velocity, final velocity, and acceleration

What We're Doing

A trolley rolls down a tilted runway and accelerates. The dots on the ticker tape get further apart as it speeds up. We measure speed at the start and end of the tape, and the distance between those points, to calculate acceleration using a = (v² − u²) / 2s.

Apparatus

Ticker timer + tape, low-voltage a.c. power supply, dynamics trolley, runway, laboratory jack

Diagram

Procedure

1

Set up the tilted runway with ticker timer. Allow trolley to roll freely (don't compensate for friction this time — you want acceleration).

2

Mark out 5 adjacent spaces near the beginning of tape. Measure length s₁. Time = 5 × 0.02 = 0.1 s. Initial velocity: u = s₁ / 0.1

3

Mark out 5 adjacent spaces near the end of tape. Measure length s₂. Final velocity: v = s₂ / 0.1

4

Measure distance s from the centre point of the first group to the centre point of the second group.

5

Calculate: a = (v² − u²) / 2s

6

Repeat by changing the tilt angle. Tabulate results.

📐 Formulas

u = s₁ / t₁ v = s₂ / t₂

a = (v² − u²) / 2s

t₁ = t₂ = 5 × 0.02 = 0.1 s each

Example Calculation

📐 Worked Example

Given

s₁ = 0.03 m, s₂ = 0.09 m, s = 0.50 m

Initial velocity

u = 0.03 / 0.1 = 0.3 m/s

Final velocity

v = 0.09 / 0.1 = 0.9 m/s

Acceleration

a = (0.9² − 0.3²) / (2 × 0.5) = (0.81 − 0.09) / 1.0 = 0.72 m/s²

⚠️ Common Mistakes

• Ignore first 5–6 dots (trolley still being pushed)
• Measure s from centre of s₁ group to centre of s₂ group — NOT from the start of the tape

Mechanics — Exp 3

⚙️ Newton's 2nd Law: a ∝ F

Showing that acceleration is proportional to force for a constant mass

What We're Doing

We use an air track (frictionless) with a vehicle pulled by different weights. Two photogates measure speed before and after, giving us acceleration. We plot a vs F — a straight line through the origin proves a ∝ F.

Apparatus

Air track + blower, vehicle, two photogates, dual timer, slotted weights (1 N total), black card, metre stick, pulley

Diagram

⚠️ Critical Rule: When you remove a disc from the hanging weights to reduce F, you must add that disc to the vehicle. This keeps the total mass constant — the only variable is the force!

Procedure

1

Level the air track. Measure length l of black card and separation s of photogate beams.

2

Start with F = 1.0 N (all weights hanging). Release vehicle from rest. Record times t₁ and t₂ from the dual timer.

3

Calculate: u = l/t₁, v = l/t₂, then a = (v² − u²) / 2s

4

Remove one 0.1 N disc from the hanging weights and place it on the vehicle. F is now 0.9 N. Repeat.

5

Continue for F = 0.8, 0.7 ... 0.1 N.

6

Plot a graph of a (y-axis) vs F (x-axis).

Expected Graph

Graph of Acceleration (a) vs Force (F) — straight line through origin proves a ∝ F

Conclusion

A straight line through the origin verifies that, for a constant mass, acceleration is proportional to the applied force (Newton's Second Law).

Mechanics — Exp 4

⚙️ Conservation of Momentum

Showing that momentum before collision = momentum after

What We're Doing

Vehicle 1 travels along an air track and collides with stationary Vehicle 2. They stick together (velcro pads). Two photogates measure velocity before and after. We compare m₁u (momentum before) with (m₁+m₂)v (momentum after).

Apparatus

Linear air track, 2 vehicles with velcro pads, blower, 2 photogates, dual timer, black card (length l), metre stick, balance

Key Formulas

📐 Formulas

u = l / t₁ v = l / t₂

p_before = m₁u p_after = (m₁ + m₂)v

If conservation holds: p_before ≈ p_after

Example Calculation

📐 Worked Example

Given

m₁ = 0.5 kg, m₂ = 0.3 kg, l = 0.10 m, t₁ = 0.20 s, t₂ = 0.40 s

Velocity before

u = l/t₁ = 0.10/0.20 = 0.5 m/s

Velocity after

v = l/t₂ = 0.10/0.40 = 0.25 m/s

Momentum before

p = m₁u = 0.5 × 0.5 = 0.25 kg·m/s

Momentum after

p = (m₁+m₂)v = (0.5+0.3) × 0.25 = 0.20 kg·m/s

Verdict

0.25 ≈ 0.20 ✓ (small difference due to air resistance)

⚠️ Precautions

• Level the air track (vehicle should not drift when placed on it)
• Both timers should read equal times when vehicle travels freely (check levelling)
• The velcro pads ensure the vehicles coalesce (stick together) after collision

Mechanics — Exp 5

⚙️ Measurement of g

Using free fall of a metal ball to find acceleration due to gravity

What We're Doing

A metal ball is held by an electromagnet. When released, it falls a measured height h and hits a trapdoor, stopping the millisecond timer. From h = ½gt², we calculate g.

Apparatus

Millisecond timer, metal ball bearing, trapdoor, electromagnet, metre stick

Diagram

Procedure

1

Set up electromagnet at top, trapdoor below. Connect both to the millisecond timer circuit.

2

Measure height h from bottom of ball to trapdoor using metre stick.

3

Switch off the electromagnet — ball falls and hits trapdoor, stopping the timer. Record time t.

4

Repeat 3 times for each h — take the smallest time (avoids reaction delay errors).

5

Repeat for different values of h (e.g. 0.5, 0.7, 0.9, 1.1 m).

6

Calculate g from each h and t. Average all values OR plot h vs t² (slope = g/2).

📐 Formula

h = ½gt² → g = 2h / t²

Or: plot h (y-axis) vs t² (x-axis) → slope = g/2 → g = 2 × slope

Example Graph (h vs t²)

Graph of h vs t² — slope = g/2 ≈ 4.9 → g ≈ 9.8 m/s²

Calculation Example

📐 Worked Example

Given

h = 0.80 m, t = 0.404 s

Calculate g

g = 2h/t² = 2(0.80) / (0.404)² = 1.60 / 0.163 = 9.8 m/s²

⚠️ Precautions

• Take the smallest time for each h (largest times = ball didn't release cleanly)
• Place a piece of paper between ball and electromagnet for quick clean release
• Measure h carefully from the bottom of the ball

Mechanics — Exp 6

⚙️ Verification of Boyle's Law

Showing that pressure × volume = constant for a fixed mass of gas at constant temperature

Boyle's Law

For a fixed mass of gas at constant temperature, pressure is inversely proportional to volume: P ∝ 1/V (so PV = constant)

Apparatus

Boyle's Law apparatus (thick-walled glass tube with oil column), bicycle pump, pressure gauge, volume scale, valve

Procedure

1

Use the pump to increase the pressure on the trapped air column.

2

Close the valve and wait 20 seconds for temperature to reach equilibrium.

3

Read the volume V from the scale and the pressure P from the gauge.

4

Open valve slightly to reduce pressure, wait again, record new P and V.

5

Repeat for at least 6 pairs of P and V readings.

6

Calculate 1/V for each reading. Plot P (y-axis) vs 1/V (x-axis).

Expected Graph

Graph of P vs 1/V — straight line through origin verifies Boyle's Law

⚠️ Important! Before starting, check the pressure gauge. If it reads 0 with the valve open, you must add atmospheric pressure (1×10⁵ Pa) to every gauge reading to get the true pressure.

Calculation Example

📐 Verifying PV = constant

P (Pa)	V (cm³)	PV
1.0 × 10⁵	40	4.0 × 10⁶
1.5 × 10⁵	27	4.05 × 10⁶
2.0 × 10⁵	20	4.0 × 10⁶

PV ≈ constant ✓ (small differences due to temperature variation)

Mechanics — Exp 7

⚙️ Laws of Equilibrium for Co-planar Forces

Showing that forces up = forces down AND sum of clockwise moments = sum of anticlockwise moments

The Two Conditions for Equilibrium

1. Forces: Sum of upward forces = sum of downward forces
2. Moments: Sum of clockwise moments = sum of anticlockwise moments (about any point)

Apparatus

Two newton balances (0–50 N), metre stick, slotted weights, paperclips, retort stands

Procedure

1

Use one balance to find the weight and centre of gravity of the metre stick.

2

Hang the metre stick horizontally from two newton balances. Hang weights from different positions.

3

Adjust weights until the stick is horizontal and in equilibrium.

4

Record readings on both newton balances (R₁ and R₂).

5

Record the position of every force (weight and balance) on the stick.

6

Check: R₁ + R₂ = sum of all weights (including weight of stick).

7

Calculate moments about the 0 cm mark: Force × distance. Check clockwise = anticlockwise.

📐 Moment Formula

Moment = Force (N) × perpendicular distance (m)

Unit: Newton-metre (N·m)

Example Calculation

📐 Worked Example

Given

Balance R₁ = 3 N at 20 cm, Balance R₂ = 4 N at 80 cm
Weight W₁ = 2 N at 30 cm, Weight W₂ = 3 N at 60 cm, Stick weight = 2 N at 50 cm

Check forces up = forces down

Forces up = R₁ + R₂ = 3 + 4 = 7 N
Forces down = W₁ + W₂ + W_stick = 2 + 3 + 2 = 7 N ✓

Clockwise moments about 0 cm

R₁ × 0.20 + R₂ × 0.80 = 3(0.2) + 4(0.8) = 0.6 + 3.2 = 3.8 N·m

Anticlockwise moments about 0 cm

W₁ × 0.30 + W₂ × 0.60 + W_stick × 0.50 = 0.6 + 1.8 + 1.0 = 3.4 N·m ≈ 3.8 ✓

Mechanics — Exp 8 (HL)

⚙️ Simple Pendulum & Calculation of g

Finding g from the relationship between period and length

The Key Relationship

T² = 4π²l/g so if we plot T² vs l, the slope = 4π²/g, and therefore: g = 4π² / slope

Procedure

1

Clamp thread between two halves of a split cork. Set length l = 1.00 m (from bottom of cork to centre of bob).

2

Set swinging through a small angle (<10°). Measure time for 30 complete oscillations.

3

Divide time by 30 to get the period T.

4

Repeat for l = 0.9, 0.8, 0.7, 0.6, 0.5 m.

5

Plot T² (y-axis) vs l (x-axis). Find slope.

6

Calculate g = 4π² / slope.

📐 Formula

T = 2π√(l/g) → T² = 4π²l/g

g = 4π² / slope (where slope = T²/l)

Expected Graph (T² vs l)

Graph of T² vs l — straight line through origin, slope = 4π²/g ≈ 4.03

Example Calculation

📐 Worked Example

Given l = 1.00 m, time for 30 oscillations = 60.3 s

T = 60.3/30 = 2.01 s T² = 4.04 s²

From graph, slope = T²/l

slope ≈ 4.04/1.00 = 4.04 s²/m

Calculate g

g = 4π²/slope = 4 × 9.87 / 4.04 = 9.77 m/s²

⚠️ Key Precautions

• Use small angles (<10°) — large angles break the simple pendulum formula
• Count 30 swings, not 30 half-swings (one oscillation = there AND back)
• Measure l to the centre of the bob

Heat — Exp 9

🌡️ Calibration of a Thermometer

Creating a calibration curve for a thermistor using a mercury thermometer as reference

What We're Doing

We heat glycerol slowly and record the temperature (from mercury thermometer) AND the resistance of a thermistor at the same time. We plot R vs θ to get a calibration curve — this allows us to use the thermistor to measure temperature later.

Apparatus

Mercury thermometer, thermistor, boiling tube of glycerol, beaker of water (water bath), heat source, ohmmeter/multimeter

Procedure

1

Place both the mercury thermometer and thermistor in the boiling tube of glycerol, which sits in a water bath.

2

Record room temperature θ and resistance R from the ohmmeter.

3

Heat the water bath, increasing temperature by ~5°C each time.

4

At each temperature, record θ and R simultaneously.

5

Take at least 10 pairs of readings up to ~100°C.

6

Plot a graph of R (y-axis) vs θ (x-axis) and draw a smooth curved line.

Expected Graph

Calibration Curve: R vs θ for a Thermistor — CURVED (thermistor resistance decreases with temperature)

💡 Why Glycerol?

Glycerol has no dissolved gases, so there's very good thermal contact between it and the thermistor. The water bath limits temperature to 100°C max.

Heat — Exp 10

🌡️ Specific Heat Capacity of Metal (Electrical)

Using a joulemeter to find how much energy heats a metal block by 10°C

Key Formula

Q = mcΔθ
Q = electrical energy (J), m = mass (kg), c = specific heat capacity (J/kg°C), Δθ = temperature rise (°C)

Apparatus

Joulemeter, metal block (with holes for heater and thermometer), heating coil, lagging, thermometer (±0.1°C), glycerol, electronic balance, 12V a.c. supply

Procedure

1

Find the mass of the metal block m using the balance.

2

Insert heating coil and thermometer (in glycerol) into block. Record initial temperature θ₁.

3

Zero the joulemeter. Switch on and allow current to flow until temperature rises by ~10°C.

4

Switch off. Wait for heat to spread through block. Record highest temperature θ₂.

5

Record final joulemeter reading Q.

6

Calculate: c = Q / (m × Δθ)

📐 Formula

Q = mcΔθ → c = Q / (mΔθ)

Δθ = θ₂ − θ₁ (temperature rise)

Example Calculation

📐 Worked Example

Given

m = 1.0 kg, θ₁ = 20°C, θ₂ = 31°C, Q = 3850 J

Temperature rise

Δθ = 31 − 20 = 11°C

Specific heat capacity

c = Q/(mΔθ) = 3850 / (1.0 × 11) = 350 J/kg°C (typical for copper)

⚠️ Sources of Error

• Heat lost to surroundings — use lagging around the block
• Record highest temperature (not the instant you switch off — heat continues to spread)
• Glycerol around thermometer ensures good thermal contact

Heat — Exp 11

🌡️ Specific Heat Capacity of Water (Electrical)

Finding how much energy heats water — must account for the calorimeter too

Key Formula

Q = m_w c_w Δθ + m_cal c_cal Δθ
Energy in = energy gained by water + energy gained by calorimeter
If polystyrene cup used: Q = m_w c_w Δθ (simpler!)

Procedure

1

Weigh the calorimeter m_cal. Fill with water, weigh again m₁. m_water = m₁ − m_cal.

2

Record initial temperature θ₁. Zero joulemeter.

3

Switch on power until temperature rises by ~10°C. Switch off, stir, record highest temperature θ₂.

4

Record joulemeter reading Q. Calculate c_water.

Example Calculation

📐 Worked Example (Polystyrene cup — simple)

Given

m_water = 0.200 kg, Q = 1680 J, Δθ = 2°C... wait let's use realistic numbers.
Q = 8400 J, m_water = 0.200 kg, Δθ = 10°C

c_water = Q/(m_w × Δθ)

c = 8400 / (0.200 × 10) = 4200 J/kg°C ✓

📐 With Copper Calorimeter (c_copper = 390 J/kg°C)

Given

Q = 9180 J, m_w = 0.200 kg, m_cal = 0.100 kg, Δθ = 10°C

Formula

Q = m_w c_w Δθ + m_cal c_cal Δθ
9180 = (0.200)(c_w)(10) + (0.100)(390)(10)
9180 = 2c_w + 390
2c_w = 8790 → c_w = 4395 ≈ 4200 J/kg°C

Heat — Exp 12

🌡️ SHC — Mechanical Method (Cooling)

Hot copper rivets cool down in cold water — heat lost = heat gained

Principle

Hot copper rivets (at 100°C) are added to cool water in a calorimeter. The heat lost by the rivets equals the heat gained by the water and calorimeter.

Key Formula

📐 Heat Lost = Heat Gained

m_co × c_c × Δθ₁ = m_cal × c_c × Δθ₂ + m_w × c_w × Δθ₂

Δθ₁ = fall in temp of rivets = 100 − θ₂
Δθ₂ = rise in temp of water/calorimeter = θ₂ − θ₁

Example Calculation (finding c_copper)

📐 Worked Example

Given

m_co = 0.100 kg, m_cal = 0.050 kg, m_w = 0.150 kg
θ₁ = 15°C, θ₂ = 22°C, rivets started at 100°C
c_water = 4200 J/kg°C (known)

Temperature changes

Δθ₁ (rivets) = 100 − 22 = 78°C Δθ₂ (water) = 22 − 15 = 7°C

Apply formula

0.100 × c_c × 78 = 0.050 × c_c × 7 + 0.150 × 4200 × 7
7.8 c_c = 0.35 c_c + 4410
7.45 c_c = 4410 → c_c = 592 J/kg°C (≈ 390 typical value)

Heat — Exp 13

🌡️ Specific Latent Heat of Fusion of Ice

How much energy does it take to melt ice at 0°C?

Key Idea

Dry crushed ice at 0°C is added to warm water. The ice melts (using the latent heat) and the melted ice then warms up. All the energy comes from the warm water and calorimeter cooling down.

📐 Formula

m_i × l + m_i × c_w × Δθ₁ = m_cal × c_c × Δθ₂ + m_w × c_w × Δθ₂

Left side: energy to melt ice + energy to warm melted ice from 0°C to θ₂
Right side: energy lost by calorimeter + energy lost by warm water
Δθ₁ = rise in temp of ice = θ₂ − 0 = θ₂
Δθ₂ = fall in temp of warm water = θ₁ − θ₂

⚠️ Critical! The ice must be dried first (blot with kitchen paper) so there's no surface water that would add extra mass. Also, warm the water to ~10°C above room temperature first.

Example Calculation

📐 Worked Example

Given

m_i = 0.020 kg, m_cal = 0.050 kg, m_w = 0.200 kg
θ₁ = 30°C, θ₂ = 18°C, c_w = 4200, c_c = 390

Temperature changes

Δθ₁ (ice warms) = 18 − 0 = 18°C
Δθ₂ (water cools) = 30 − 18 = 12°C

Apply formula

0.020 l + 0.020(4200)(18) = 0.050(390)(12) + 0.200(4200)(12)
0.020 l + 1512 = 234 + 10080
0.020 l = 10314 − 1512 = 8802
l = 440,100 J/kg ≈ 3.3 × 10⁵ J/kg

Heat — Exp 14

🌡️ Specific Latent Heat of Vaporisation of Water

How much energy does steam release when it condenses?

Key Idea

Dry steam (100°C) is passed into cool water. The steam condenses (releasing latent heat) and cools to θ₂. All the energy heats the water and calorimeter.

📐 Formula

m_s × l + m_s × c_w × Δθ₁ = m_cal × c_c × Δθ₂ + m_w × c_w × Δθ₂

Left: energy from condensing steam + energy from steam cooling from 100°C to θ₂
Right: energy gained by calorimeter + energy gained by water
Δθ₁ (steam) = 100 − θ₂ Δθ₂ (water) = θ₂ − θ₁

⚠️ Use a steam trap! The steam trap removes water droplets so only dry steam enters the calorimeter. Wet steam would carry extra mass that wasn't properly at 100°C.

Example Calculation

📐 Worked Example

Given

m_s = 0.005 kg, m_cal = 0.050 kg, m_w = 0.200 kg
θ₁ = 15°C, θ₂ = 35°C

Temperature changes

Δθ₁ = 100 − 35 = 65°C Δθ₂ = 35 − 15 = 20°C

Apply formula

0.005 l + 0.005(4200)(65) = 0.050(390)(20) + 0.200(4200)(20)
0.005 l + 1365 = 390 + 16800
0.005 l = 15825 → l = 2.2 × 10⁶ J/kg (actual ≈ 2.26 × 10⁶)

Sound — Exp 15

🔊 Measurement of Speed of Sound in Air

Using resonance in an air column and tuning forks

Key Principle

When a tuning fork is held over a tube, the air column resonates at specific lengths. The distance between two resonance points (l₁ and l₂) equals half a wavelength: λ = 2(l₂ − l₁). Then c = fλ.

Apparatus

1000 ml graduated cylinder, resonance tube (able to slide up/down), set of tuning forks (256–512 Hz), metre stick, clamp, wooden block

Diagram

Procedure

1

Strike the highest frequency tuning fork (512 Hz) on a wooden block and hold it horizontally just above the tube opening.

2

Slide the tube slowly until the note is loudest — this is the first resonance position. Measure length l₁.

3

Continue sliding until the second (weaker) resonance. Measure length l₂.

4

Calculate: λ = 2(l₂ − l₁) and c = fλ.

5

Repeat for all tuning forks. Find average speed of sound.

📐 Formulas

λ = 2(l₂ − l₁) c = fλ

Speed of sound in air ≈ 340 m/s

Example Calculation

📐 Worked Example

Given

f = 512 Hz, l₁ = 0.163 m, l₂ = 0.495 m

Wavelength

λ = 2(l₂ − l₁) = 2(0.495 − 0.163) = 2 × 0.332 = 0.664 m

Speed

c = fλ = 512 × 0.664 = 340 m/s

Sound — Exp 16

🔊 String Frequency vs Length

Showing that f ∝ 1/l for a string at constant tension

What We're Showing

The shorter the string, the higher the note (think of a guitar!). We prove this mathematically: frequency is inversely proportional to length (f ∝ 1/l).

Apparatus

Signal generator, U-shaped magnet, sonometer, newton balance (measures tension), tension control key, metre stick, paper rider (or straw)

Procedure

1

Set tension to a constant value (e.g. 20 N) using the tension key. Keep this fixed throughout.

2

Set bridges far apart. Measure length l. Position magnet midway between bridges.

3

Slowly increase signal generator frequency from 0 until wire vibrates maximally (paper rider bounces vigorously).

4

Record the resonant frequency f.

5

Shorten the length (move one bridge inward). Repeat for at least 6 different lengths.

6

Plot f (y-axis) vs 1/l (x-axis). Straight line through origin = f ∝ 1/l.

Expected Graph

Graph of f vs 1/l — straight line through origin verifies f ∝ 1/l

⚠️ Key Points

• Keep tension constant throughout
• Reposition magnet to midway between bridges after each change
• Safety goggles — wire can snap under tension!

Sound — Exp 17 (HL)

🔊 String Frequency vs Tension (HL)

Showing that f ∝ √T for a string of constant length

What We're Showing

Frequency is proportional to the square root of tension: f ∝ √T.
Plot f vs √T → straight line through origin.

Procedure

1

Fix the length at a constant value (e.g. 30 cm). Keep this fixed throughout.

2

Set tension to a low value (e.g. 4 N). Find the resonant frequency.

3

Increase tension in steps (e.g. 4 N, 9 N, 16 N, 25 N...). Find new resonant frequency each time.

4

Record T and f. Calculate √T. Plot f vs √T.

📐 Relationship

f ∝ √T

Plot f (y-axis) vs √T (x-axis) → straight line through origin

💡 Tip

Use tension values that are perfect squares (4, 9, 16, 25 N) — makes √T easy to calculate!

Light — Exp 18

💡 Focal Length of a Concave Mirror

Using the mirror formula 1/f = 1/u + 1/v

Formula

1/f = 1/u + 1/v
f = focal length, u = object distance, v = image distance (all in same units)

Apparatus

Concave mirror, screen, lamp-box with crosswire, metre stick

Procedure

1

Approximate f first: Focus a distant object onto a sheet of paper. Distance from mirror to paper ≈ focal length.

2

Place lamp-box well outside the focal length. Move screen until a sharp inverted image of the crosswire appears.

3

Measure u (crosswire to mirror) and v (screen to mirror) with metre stick. Measure from the back of the mirror.

4

Calculate f using 1/f = 1/u + 1/v.

5

Repeat for different values of u. Average all f values.

Example Calculation

📐 Worked Example

Given

u = 30 cm, v = 60 cm

Apply formula

1/f = 1/30 + 1/60 = 2/60 + 1/60 = 3/60 = 1/20

Focal length

f = 20 cm

⚠️ Sources of Error

• Difficulty judging sharpest focus — repeat and average
• Parallax error — view metre stick perpendicular
• Measure from the back of the mirror
• Never place object inside focal length (virtual image can't hit screen!)

Light — Exp 19

💡 Verification of Snell's Law

Showing that sin i / sin r = constant (refractive index)

Snell's Law

n = sin i / sin r
n = refractive index, i = angle of incidence, r = angle of refraction. Both angles measured from the normal.

Apparatus

Glass block (rectangular), lamp-box, 0–360° protractor, protractor/ruler

Procedure

1

Place glass block on the 360° protractor and mark its outline.

2

Shine ray from lamp-box at a set angle of incidence i. Note where ray enters (point A) and exits (point B) the block.

3

Remove block. Join BA and extend to C to find the refracted ray.

4

Measure angle of refraction r using protractor. Calculate sin i / sin r.

5

Repeat for at least 6 different values of i. Plot sin i (y-axis) vs sin r (x-axis).

Expected Graph

sin i vs sin r — straight line through origin, slope = n (refractive index ≈ 1.5 for glass)

Example Calculation

📐 From Exam Data

Given i = 40°, r = 25°

n = sin 40° / sin 25° = 0.643 / 0.423 = 1.52

⚠️ Key Points

• Don't use small angles (below 20°) — large % error
• Look directly down through the block to measure r
• Slope of graph gives refractive index

Light — Exp 20

💡 Refractive Index of a Liquid

Using apparent depth and a pin-and-mirror method

📐 Formula

n = Real Depth / Apparent Depth

The liquid makes the bottom appear closer than it really is.

Procedure

1

Fill a container to the top with water. Place a plane mirror beside it.

2

Place a pin on the bottom of the container (object pin).

3

Adjust the height of a second pin in a cork above the mirror until there is no parallax between its image in the mirror and the image of the pin in the water.

4

Measure: apparent depth = distance from cork pin to back of mirror. Real depth = depth of container.

5

Calculate n = real depth / apparent depth. Repeat for different containers, average.

Example Calculation

📐 Worked Example

Given

Real depth = 15.0 cm, Apparent depth = 11.3 cm

Refractive index

n = 15.0 / 11.3 = 1.33 (water) ✓

Light — Exp 21

💡 Focal Length of a Converging Lens

Same formula as concave mirror — 1/f = 1/u + 1/v

Apparatus

Converging (convex) lens, screen, lamp-box with crosswire, metre stick, retort stand

Procedure

1

Approximate f first: Focus a distant object (e.g. a tree) through the lens onto a screen. Distance = approximate f.

2

Place lamp-box outside focal length. Move screen (on far side of lens) until sharp inverted image appears.

3

Measure u (crosswire to lens centre) and v (lens centre to screen).

4

Calculate f. Repeat for different u values. Average f.

Comparison: Mirror vs Lens Experiment

Feature	Concave Mirror	Convex Lens
Screen position	Same side as object	Opposite side to object
Measure from	Back of mirror	Centre of lens
Image on screen	Inverted ✓	Inverted ✓
Formula	1/f = 1/u + 1/v	1/f = 1/u + 1/v

Example Calculation

📐 Worked Example

Given

u = 20 cm, v = 65 cm

1/f = 1/u + 1/v

1/f = 1/20 + 1/65 = 13/260 + 4/260 = 17/260

f

f = 260/17 = 15.3 cm

Light — Exp 22

💡 Wavelength of Monochromatic Light (Spectrometer)

Using a diffraction grating and sodium lamp

Key Formula

nλ = d sin θ
n = order number (1, 2, 3...), d = slit spacing (d = 1/N where N = slits per metre), θ = diffraction angle

Apparatus

Sodium lamp, spectrometer (with telescope + turntable), diffraction grating (300 lines/mm)

Procedure

1

Focus the telescope for parallel light (no parallax between image and crosswires). Place sodium lamp at the collimator.

2

Place diffraction grating at right angles to beam on turntable.

3

Move telescope to the right until first bright fringe (n=1). Record angle θ_r.

4

Move telescope to first bright fringe on the left. Record angle θ_l.

5

Calculate θ = (θ_r − θ_l) / 2. Then λ = d sin θ / n.

6

Repeat for n = 2, 3... Average all λ values.

📐 Key Formulas

d = 1/N (N = slits per metre)

θ = (θ_r − θ_l) / 2

λ = d sin θ / n

Example Calculation

📐 Worked Example

Given

N = 300 lines/mm = 300,000 lines/m, n = 1, θ_r = 22.5°, θ_l = 337.5° (i.e. 360−22.5)

Calculate d

d = 1/300,000 = 3.33 × 10⁻⁶ m

Calculate θ

θ = (22.5 − (360−22.5−180))/2... simpler: just use θ_r directly if measured from centre. θ = 12°

Calculate λ

λ = d sin θ = (3.33 × 10⁻⁶) × sin 12° = (3.33 × 10⁻⁶) × 0.208 = 5.9 × 10⁻⁷ m = 590 nm (sodium yellow) ✓

Light — Exp 23

💡 Wavelength of Light — Using a Laser

Simpler setup with a laser pointer and diffraction grating

Same Formula

nλ = d sin θ
But this time θ is found from the geometry: tan θ = x/D where x is the distance from centre to bright spot, D is distance from grating to screen.

Apparatus

Laser, diffraction grating (600 lines/mm), 2 metre sticks

⚠️ Safety! Never look directly into a laser beam.

Procedure

1

Clamp a metre stick horizontally. Aim laser at the 50 cm mark at right angles.

2

Place diffraction grating between laser and metre stick.

3

Observe bright spots on the metre stick. Measure distance D from grating to stick.

4

Measure distance x from the central spot (n=0) to the first-order spot (n=1) on each side. Average.

5

Calculate θ using tan θ = x/D. Then λ = d sin θ / n.

📐 Formulas

tan θ = x / D d = 1/N

λ = d sin θ / n

Example Calculation

📐 Worked Example

Given

N = 600 lines/mm = 6 × 10⁵ /m, D = 1.00 m, x = 0.390 m, n = 1

d

d = 1/(6 × 10⁵) = 1.67 × 10⁻⁶ m

θ

tan θ = 0.390/1.00 → θ = 21.3°

λ

λ = d sin θ = (1.67 × 10⁻⁶) × sin 21.3° = (1.67 × 10⁻⁶) × 0.363 = 6.1 × 10⁻⁷ m = 610 nm (red laser)

Electricity — Exp 24

⚡ Verification of Joule's Law (ΔΘ ∝ I²)

Showing that the heat produced in a resistor is proportional to the square of the current

Joule's Law

The heat produced in a resistor is proportional to the square of the current: ΔΘ ∝ I² (at constant resistance and time).

Apparatus

Lagged calorimeter or beaker, heating coil, rheostat, ammeter, thermometer, stopwatch, battery/power supply

Procedure

1

Fill calorimeter with water covering the coil. Record initial temperature θ₁.

2

Start with I = 0.5 A. Allow current to flow for exactly 5 minutes. Stir and record highest temperature θ₂. Calculate ΔΘ = θ₂ − θ₁.

3

Allow to cool back to room temperature. Increase I to 1.0 A. Repeat for same 5 minutes.

4

Repeat for I = 1.5, 2.0, 2.5, 3.0 A.

5

Plot ΔΘ (y-axis) vs I² (x-axis). Straight line through origin = Joule's Law verified.

Expected Graph

ΔΘ vs I² — straight line through origin proves ΔΘ ∝ I²

⚠️ Key Points

• Same time interval (5 min) for each current reading
• Let water cool back to room temperature before next reading
• Keep resistance constant (don't change the coil between trials)
• Lagging reduces heat loss to surroundings

Electricity — Exp 25

⚡ Resistivity of a Wire

Measuring how resistance varies with length, then using wire diameter to find ρ

Key Formula

R = ρl/A → ρ = RA/l
ρ = resistivity (Ω·m), R = resistance (Ω), l = length (m), A = cross-sectional area (m²) = πd²/4

Apparatus

Nichrome/manganin wire (2–3 metres), micrometer, ohmmeter, metre stick, two retort stands

Procedure

1

Note the resistance of the leads (crocodile clips connected together) — this will be subtracted.

2

Stretch wire between stands. Connect crocodile clips a distance l apart. Read total resistance. Subtract leads resistance to get R of wire.

3

Measure l with metre stick. Increase l and repeat for at least 6 lengths.

4

Use micrometer to measure wire diameter d at several points. Note zero error. Find average d.

5

Calculate A = πd²/4. Then ρ = R × A / l for each measurement. Average ρ.

📐 Formulas

A = πd² / 4 ρ = (R/l) × A

Nichrome ρ ≈ 100 × 10⁻⁸ Ω·m | Manganin ρ ≈ 48 × 10⁻⁸ Ω·m

Example Calculation

📐 Worked Example

Given

R = 5.8 Ω, l = 1.00 m, d = 0.27 mm = 2.7 × 10⁻⁴ m

Area

A = π(2.7 × 10⁻⁴)²/4 = π(7.29 × 10⁻⁸)/4 = 5.73 × 10⁻⁸ m²

Resistivity

ρ = RA/l = 5.8 × (5.73 × 10⁻⁸) / 1.00 = 3.3 × 10⁻⁷ Ω·m

⚠️ Key Points

• Always subtract the resistance of the leads (connection wire)
• Measure d at several points along the wire and average
• Note the micrometer zero error and correct for it
• Safety glasses — wire can snap under tension!

Electricity — Exp 26

⚡ Resistance of a Metal Conductor vs Temperature

Metals get more resistive as they heat up

Key Result

The resistance of a metal conductor increases linearly with temperature. Graph: R vs θ = straight line with positive slope.

Apparatus

Coil of wire on frame (in boiling tube of glycerol), beaker of water, heat source, ohmmeter, thermometer

Procedure

1

Place coil in boiling tube of glycerol in a water bath. Connect ohmmeter.

2

Record resistance R and temperature θ (from thermometer in glycerol).

3

Heat the water bath. For every 10°C rise, record R and θ.

4

Plot R (y-axis) vs θ (x-axis). Should be a straight line with positive slope.

Expected Graph

R vs Temperature for Metal — straight line, positive slope (resistance increases with temp)

Electricity — Exp 27

⚡ Resistance of a Thermistor vs Temperature

Thermistors are the opposite — their resistance drops sharply with temperature

Key Result

The resistance of a thermistor decreases rapidly (non-linearly) as temperature increases. Graph: R vs θ = a curve (not a straight line!). This is opposite to a metal conductor.

Apparatus

Thermistor, boiling tube of glycerol or liquid paraffin, beaker of water, heat source, ohmmeter, thermometer

Procedure — same as Metal Resistance experiment

1

Place thermistor in boiling tube of glycerol in water bath. Connect ohmmeter.

2

Record R and θ at room temperature.

3

Heat the water bath. Record R and θ for every 10°C rise.

4

Plot R vs θ. Join points with a smooth curve.

Expected Graph

R vs Temperature for Thermistor — CURVED, resistance falls sharply as temperature rises

Metal Conductor

Resistance INCREASES with temp
Straight line — positive slope

Thermistor (NTC)

Resistance DECREASES with temp
Curved line — gets less steep

Electricity — Exp 28

⚡ I-V Characteristics of Components

Four different components — four different I-V graphs

What We're Doing

We apply different voltages to a component and measure the current. Plotting I vs V shows whether Ohm's Law is obeyed (straight line through origin) or not.

Apparatus (Common to all 4)

Low voltage power supply, rheostat (potential divider), voltmeter, ammeter/milliammeter/microammeter, component

The 4 Components

📌 (a) Metallic Conductor (Nichrome Wire) ▼

A straight line through the origin — Ohm's Law obeyed.

I ∝ V (constant resistance). The slope = 1/R.

I vs V for Nichrome Wire — straight line, I ∝ V, Ohm's Law verified

📌 (b) Filament Bulb ▼

A curved line — Ohm's Law NOT obeyed.

Resistance increases as temperature increases (filament gets hotter at higher currents → higher resistance → current doesn't keep up with voltage).

I vs V for Filament Bulb — CURVED, resistance increases with temperature

📌 (c) Copper Sulfate with Copper Electrodes ▼

A straight line through origin — Ohm's Law obeyed. The solution behaves like a resistor. I ∝ V.

📌 (d) Semiconductor Diode — Forward & Reverse Bias ▼

Forward bias: Very little current until voltage exceeds about 0.6 V (silicon diode). Then current increases rapidly — NOT a straight line.

Reverse bias: Almost no current flows regardless of voltage — the diode blocks current in reverse.

I vs V for Semiconductor Diode — "knee" at 0.6 V forward bias

Summary Table

Component	Graph Shape	Ohm's Law?	Why?
Nichrome wire	Straight line	✅ Yes	Resistance constant
Filament bulb	Curved (flattens)	❌ No	R increases with temp
CuSO₄ solution	Straight line	✅ Yes	Electrolyte acts as resistor
Diode (forward)	Knee at 0.6 V then steep	❌ No	Non-linear semiconductor

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Apparatus

Diagram

Procedure (Step by Step)

Example Calculation

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Diagram

Procedure

Example Calculation

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Diagram

Procedure

Expected Graph

Apparatus

Key Formulas

Example Calculation

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Diagram

Procedure

Example Graph (h vs t²)

Calculation Example

Apparatus

Procedure

Expected Graph

Calculation Example

Apparatus

Procedure

Example Calculation

Procedure

Expected Graph (T² vs l)

Example Calculation

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Procedure

Expected Graph

Apparatus

Procedure

Example Calculation

Procedure

Example Calculation

Key Formula

Example Calculation (finding c_copper)

Example Calculation

Example Calculation

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Diagram

Procedure

Example Calculation

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Procedure

Expected Graph

Procedure

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Procedure

Example Calculation

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Procedure

Expected Graph

Example Calculation

Procedure

Example Calculation

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Procedure

Comparison: Mirror vs Lens Experiment

Example Calculation

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Example Calculation

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Example Calculation

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Procedure

Expected Graph

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