← Study Hub
Topic 3: Waves — Progress 0 / 80 questions answered (0 correct)

Topic 3: Waves

Cambridge IGCSE Physics 0625 — Extended
General wave properties, light (reflection, refraction, lenses, dispersion), electromagnetic spectrum, and sound. 4 subtopics with 80 practice MCQs.

Hey Tara! Welcome to the world of Waves and Light. This is one of the most beautiful topics in physics -- you experience waves every single day without even realising it. Every time you hear the horn of a BMTC bus, see the ripples when rain falls into a puddle on Brigade Road, or watch the shimmer of light off Ulsoor Lake, you are seeing waves in action. This guide will take you through everything step by step, with lots of Indian examples to help it all click. You have got this!

3.1 General Properties of Waves
Your Score 0 / 20
Wave Properties - IGCSE Physics
Cognito
Watch
Transverse and Longitudinal Waves
Science Shorts
Watch
Ripple Tank Experiments - IGCSE Physics
Free Science Lessons
Watch

What Are Waves?

Imagine you are sitting at Cubbon Park and you throw a stone into the lake. Ripples spread out from where the stone lands, moving outward in circles. Now here is the key insight: the water itself does not travel outward. If you placed a small leaf floating on the water, it would bob up and down as the wave passes, but it would not be carried to the edge of the lake. The wave moves outward, but the water stays roughly where it is.

This is the fundamental idea behind all waves:

⚠ Exam Tip

Waves transfer energy without transferring matter. This is one of the most tested definitions in the entire IGCSE course. Memorise it word for word. The examiner loves to test whether you understand that the medium (water, air, rope) vibrates but does not actually travel with the wave.

Think about doing a "Mexican wave" at an IPL cricket match at M. Chinnaswamy Stadium. Each person stands up and sits down (they vibrate in place), and the wave pattern moves around the stadium. But no person actually runs around the stadium! The wave carries energy (it takes effort to stand up), but no matter (no person) is transferred from one end to the other.

Waves can travel through different media:

  • Ropes and strings -- flick one end of a skipping rope and watch a wave pulse travel to the other end. The rope vibrates but the fibres do not move along it.
  • Springs (Slinky) -- push and pull a Slinky and you can see compression waves travel along it. The coils vibrate but stay in roughly the same position.
  • Water -- waves on a lake, the ocean, or even in your chai cup when you tap the side. The water particles move in small circles but the wave energy spreads outward.
  • Air -- sound waves travel through air. When an auto-rickshaw honks, the air molecules vibrate back and forth, passing the sound energy to your ear, but the air itself does not fly from the auto to you.

Wave Features -- The Key Terms

To describe waves precisely, we need specific vocabulary. Let us go through each term carefully.

Wavefront

A wavefront is a line joining all the points on a wave that are at the same phase (same point in their vibration cycle). Think of the circular ripples spreading out when you drop a stone in water -- each ring is a wavefront. For waves approaching a beach, the wavefronts are roughly straight lines parallel to the shore.

Crest and Trough

A crest is the highest point of the wave -- the peak. A trough is the lowest point -- the dip. If you watch waves at Marina Beach or any Indian coastline, the crests are the tops of the waves where surfers ride, and the troughs are the valleys between them.

Amplitude

The amplitude is the maximum displacement of a wave from its rest position (the undisturbed, equilibrium position). It is measured from the rest position to the crest (or from the rest position to the trough) -- NOT from crest to trough.

⚠ Exam Tip

A very common mistake is measuring amplitude from crest to trough. That distance is actually twice the amplitude. If the distance from crest to trough is 8 cm, the amplitude is 4 cm. Always measure from the rest position (the middle line) to the crest OR trough.

Amplitude tells you how much energy a wave carries. A gentle ripple in your chai has a small amplitude; a tsunami has a massive amplitude carrying devastating energy.

Wavelength (λ)

The wavelength is the distance between two consecutive points that are in phase -- for example, the distance from one crest to the next crest, or one trough to the next trough. We use the Greek letter lambda (λ) to represent it, measured in metres (m).

🧠 Memory Trick

λ looks like a wave! The Greek letter lambda (λ) actually looks like a tiny wave with a peak -- perfect for remembering it represents wavelength.

Frequency (f)

The frequency is the number of complete waves passing a point per second. It is measured in hertz (Hz), where 1 Hz = 1 wave per second.

Think about it this way: if you stand on a bridge over a canal and count how many wave crests pass under you in one second, that number is the frequency. If 5 crests pass every second, the frequency is 5 Hz.

Wave Speed (v)

The wave speed is how fast the wave pattern moves through the medium, measured in metres per second (m/s). This is NOT the speed of the particles vibrating -- it is the speed at which the energy travels.

All of these quantities are connected by one of the most important equations in wave physics:

v = f × λ
v = wave speed (m/s) f = frequency (Hz) λ = wavelength (m)
🧠 Memory Trick

"Very Fine Lamb" -- v = f × λ. Think of a very fine lamb grazing peacefully in the Nilgiri Hills. V for Very, F for Fine, λ (lambda / Lamb) for wavelength. Or use the triangle: put v on top, f and λ on the bottom. Cover what you want to find!

Worked Example A loudspeaker at M. Chinnaswamy Stadium produces sound waves with a frequency of 440 Hz. The speed of sound in air is 330 m/s. What is the wavelength of the sound?
Step 1: Write down what you know
f = 440 Hz, v = 330 m/s, λ = ?
Step 2: Write the formula and rearrange
v = f × λ, so λ = v / f
Step 3: Substitute and calculate
λ = 330 / 440 = 0.75 m
✅ Answer: The wavelength is 0.75 m
Worked Example Waves on Ulsoor Lake have a wavelength of 2.5 m and travel at a speed of 5.0 m/s. How many waves pass a fixed point (a buoy) every second?
Step 1: Write down what you know
λ = 2.5 m, v = 5.0 m/s, f = ?
Step 2: Rearrange the formula
v = f × λ, so f = v / λ
Step 3: Substitute and calculate
f = 5.0 / 2.5 = 2.0 Hz
✅ Answer: 2.0 waves pass the buoy every second (frequency = 2.0 Hz)
Worked Example Radio Mirchi FM broadcasts at a frequency of 98.3 MHz. Radio waves travel at 3.0 × 10⁸ m/s. Calculate the wavelength of Radio Mirchi's signal.
Step 1: Write down what you know (convert units!)
f = 98.3 MHz = 98.3 × 10⁶ Hz = 9.83 × 10⁷ Hz
v = 3.0 × 10⁸ m/s
Step 2: Rearrange
λ = v / f
Step 3: Substitute and calculate
λ = (3.0 × 10⁸) / (9.83 × 10⁷) = 3.05 m
✅ Answer: The wavelength is approximately 3.05 m (about the length of an auto-rickshaw!)
⚠ Exam Tip

Always convert units before substituting into v = f × λ. If frequency is given in kHz, convert to Hz (multiply by 1000). If in MHz, multiply by 10⁶. The examiner gives units like kHz or MHz to test whether you can convert correctly.

Transverse Waves

In a transverse wave, the vibration of the particles is perpendicular (at right angles) to the direction the wave travels.

Imagine shaking a rope side-to-side while your friend holds the other end. Your hand moves up and down, but the wave pulse travels horizontally along the rope. The vibration (up-down) is perpendicular to the wave direction (horizontal). That is a transverse wave.

Important examples of transverse waves:

  • Water surface waves -- the water moves up and down while the wave moves horizontally across the surface
  • All electromagnetic (EM) radiation -- light, radio waves, microwaves, X-rays, UV, infrared, gamma rays. These are ALL transverse waves
  • S-waves (secondary seismic waves) -- these travel through the Earth during an earthquake. India sits on the Indo-Australian tectonic plate, so understanding seismic waves matters! S-waves can only travel through solids, which is how scientists discovered the Earth's outer core is liquid
  • Waves on ropes and strings -- like the strings of a sitar or veena vibrating
🧠 Memory Trick

"S for Side-to-side and Secondary" -- S-waves are transverse because the particles move Side-to-side (perpendicular) to the wave direction. Also, S-waves arrive Second at a seismometer (after P-waves), just like "S" comes after "P" in the alphabet.

Longitudinal Waves

In a longitudinal wave, the vibration of the particles is parallel to (along the same direction as) the direction the wave travels.

Imagine pushing and pulling one end of a Slinky spring. The coils bunch together (compressions) and spread apart (rarefactions) as the wave travels along. The coils vibrate back-and-forth in the same direction the wave moves. That is a longitudinal wave.

Instead of crests and troughs, longitudinal waves have:

  • Compressions -- regions where particles are squashed close together (like a traffic jam on Silk Board junction!)
  • Rarefactions -- regions where particles are spread far apart (like the empty road at 4 AM)

Important examples of longitudinal waves:

  • Sound waves -- when you speak, your vocal cords vibrate and push air molecules back and forth, creating compressions and rarefactions that travel to the listener's ear
  • P-waves (primary seismic waves) -- these are the fastest seismic waves and arrive first at seismometers. They can travel through solids AND liquids, which is why they pass through the Earth's liquid outer core
  • Compression waves in springs -- pushing and pulling a Slinky
⚠ Exam Tip

How to tell them apart in the exam: The key question is "which direction do the particles vibrate relative to the wave direction?" If perpendicular → transverse. If parallel → longitudinal. Sound is ALWAYS longitudinal. Light is ALWAYS transverse. These are the two that come up most often.

Wave Behaviours: Reflection, Refraction, and Diffraction

All waves -- whether water, sound, or light -- show three key behaviours. Understanding these is essential for your exam.

Reflection

Reflection occurs when a wave hits a surface and bounces back. Think of an echo -- when you shout near a large building (like the walls of Bangalore Fort), the sound wave bounces off the wall and comes back to you. Water waves in a ripple tank bounce off a straight barrier in the same way.

The wave keeps the same speed, wavelength, and frequency after reflection. Only the direction changes.

Refraction

Refraction occurs when a wave changes speed as it passes from one medium to another (or from one depth to another), causing it to change direction.

Think about waves approaching a beach at an angle. Where the water becomes shallower, the waves slow down and bend so they arrive almost parallel to the shore. In a ripple tank, when water waves pass from a deep region to a shallow region, they slow down and the wavelength decreases. If they enter at an angle, they change direction.

Key point: the frequency stays the same during refraction (it depends on the source, not the medium), but the speed and wavelength change.

Diffraction

Diffraction is the spreading out of waves when they pass through a gap or around an edge.

You experience diffraction every day! When you are standing outside a classroom and can hear the teacher speaking even though you cannot see them through the doorway, that is because sound waves diffract (spread out) as they pass through the door opening. Sound has a long wavelength, so it diffracts a lot around everyday obstacles.

In a ripple tank, you can observe all three behaviours:

  • Reflection -- place a straight barrier in the tank and waves bounce off it
  • Refraction -- place a glass plate under part of the tank to make the water shallower. Waves slow down and change direction when they cross from deep to shallow water
  • Diffraction through a gap -- place two barriers with a gap between them. Waves spread out after passing through the gap
  • Diffraction at an edge -- place a single barrier. Waves bend around the edge of the barrier into the "shadow" region
⚠ Exam Tip

When drawing diffraction diagrams, the wavelength must stay the SAME on both sides of the gap. Many students draw the waves with a different spacing after the gap -- this loses marks. Also, the waves spread out in a circular pattern from the gap, and they curve around the edges.

Supplement (Extended)

How Wavelength and Gap Size Affect Diffraction

The amount of diffraction depends on the relationship between the wavelength and the size of the gap:

  • When the gap is much larger than the wavelength, there is very little diffraction -- the waves pass straight through with only slight bending at the edges
  • When the gap is about the same size as the wavelength, maximum diffraction occurs -- the waves spread out in a wide semicircular pattern
  • When the gap is smaller than the wavelength, very little wave energy passes through

This is why you can hear someone talking around a corner (sound wavelength is about 0.3-3 m, similar to doorway width) but you cannot see them around the corner (light wavelength is about 500 nm = 0.0000005 m, far smaller than any doorway).

Diffraction at an Edge

When waves meet an edge (rather than a gap), they bend around it. Longer wavelengths diffract more around the edge than shorter wavelengths. This is why you can hear low-pitched sounds (long wavelength) from behind a wall more easily than high-pitched sounds (short wavelength).

⚠ Exam Tip

For maximum marks on diffraction questions, state that "most diffraction occurs when the gap width is approximately equal to the wavelength." At an edge, state that "longer wavelengths diffract more than shorter wavelengths."

⚠ Exam Tip

Ripple tank experiments are a favourite in Paper 6 (practical). Know that: (1) a lamp above the tank projects wave patterns onto a white screen below, (2) you use a stroboscope to "freeze" the wave pattern, and (3) the depth of water controls the wave speed (shallower = slower).

🔎 Apply It: Real-World Physics
Cambridge examiners LOVE testing familiar concepts in unfamiliar situations. Can you spot the physics hiding in these real-world scenarios? Tap each one to reveal the answer.
1
During an IPL cricket match at M. Chinnaswamy Stadium in Bangalore, a "Mexican wave" travels around the stadium. Fans stand up and sit down in sequence. The wave takes 40 seconds to travel around the 170-metre-circumference stadium.
What is the speed of the stadium wave? Is this a transverse or longitudinal wave, and why?
Identify the Physics
This is the wave speed equation: v = d / t. We also need to classify the wave type based on the direction of oscillation vs. direction of travel.
Work It Out
Speed = distance / time = 170 m / 40 s = 4.25 m/s. The wave travels horizontally around the stadium, but each fan moves up and down (vertically). Since the oscillation is perpendicular to the direction of wave travel, this is a transverse wave.
💡 The Aha! Moment
A stadium wave is a perfect real-world example of a transverse wave. The "particles" (fans) move up and down, but the wave energy travels sideways. This is exactly what happens with water waves and light waves too.
2
An earthquake strikes near Latur, Maharashtra. A seismograph 150 km away detects P-waves (longitudinal, speed 6000 m/s) and S-waves (transverse, speed 3500 m/s) arriving at different times.
How much earlier do the P-waves arrive compared to the S-waves?
Identify the Physics
Wave speed equation rearranged for time: t = d / v. We calculate the travel time for each wave type separately, then find the difference.
Work It Out
Distance = 150 km = 150,000 m.
Time for P-waves: t = 150,000 / 6000 = 25 s.
Time for S-waves: t = 150,000 / 3500 = 42.9 s.
Difference = 42.9 - 25 = 17.9 seconds.
The P-waves arrive about 18 seconds before the S-waves.
💡 The Aha! Moment
Seismologists use this time delay to calculate how far away the earthquake is. The bigger the gap between P-wave and S-wave arrival, the further away the earthquake. This is real physics saving lives.
3
A tsunami wave in the Indian Ocean has a wavelength of 200 km and travels at 800 km/h. In deep water, its amplitude is only 0.5 m (barely noticeable on ships), but as it approaches the shore and enters shallow water, the amplitude increases to 10 m.
Calculate the frequency of the tsunami. Why does the amplitude increase so dramatically near shore?
Identify the Physics
Wave speed equation: v = f x λ, rearranged to f = v / λ. Also the relationship between wave speed, amplitude, and energy transfer.
Work It Out
f = v / λ = 800 km/h / 200 km = 4 waves per hour = 4/3600 = 0.0011 Hz.

As the wave enters shallow water, it slows down (the wave speed decreases). But the wave's energy stays roughly the same. Since the energy is related to the amplitude, and the wave is being compressed into a smaller space, the amplitude must increase. The wave "piles up" into a taller wall of water.
💡 The Aha! Moment
A tsunami demonstrates that amplitude and energy are linked. In deep ocean, the energy is spread through a huge depth of water (small amplitude). Near shore, the same energy is squeezed into shallow water, creating a devastating wall. Ships in open ocean barely feel a tsunami pass underneath them.
4
A Radio Mirchi FM station in Bangalore broadcasts at a frequency of 98.3 MHz. Radio waves travel at 3 x 10⁸ m/s (the speed of light).
Calculate the wavelength of the Radio Mirchi signal. Why can you hear the station even when there are buildings between you and the transmitter?
Identify the Physics
Wave speed equation: v = f x λ, rearranged to λ = v / f. Also the concept of diffraction -- waves spreading around obstacles.
Work It Out
f = 98.3 MHz = 98.3 x 10⁶ Hz.
λ = v / f = (3 x 10⁸) / (98.3 x 10⁶) = 3.05 m.

The wavelength is about 3 metres, which is comparable to the size of gaps between buildings. When a wave encounters a gap that is roughly equal to its wavelength, maximum diffraction occurs -- the wave spreads out around the obstacle. That is why FM radio waves can bend around buildings to reach you.
💡 The Aha! Moment
Diffraction is why you can hear FM radio in your room even without a direct line of sight to the transmitter. The wavelength of FM radio (about 3 m) is just right for diffracting around buildings and through windows. This is a perfect exam example of "diffraction occurs most when gap width ≈ wavelength."
5
A physics student uses a CRO (cathode ray oscilloscope) to display the sound wave from a tabla being played. The time-base is set to 5 ms/division and the trace shows 4 complete waves across 8 divisions. When the tabla player hits the drum harder, the trace gets taller but the number of waves stays the same.
Calculate the frequency of the tabla note. What wave property changed when the drummer hit harder?
Identify the Physics
Reading a CRO trace: period from the time-base setting, then f = 1 / T. Amplitude vs. frequency distinction.
Work It Out
Total time shown = 8 divisions x 5 ms/division = 40 ms = 0.040 s.
4 complete waves in 0.040 s, so period T = 0.040 / 4 = 0.010 s.
Frequency f = 1 / T = 1 / 0.010 = 100 Hz.

When the drummer hit harder, the trace got taller -- the amplitude increased. The frequency stayed the same (same number of waves in the same time). Greater amplitude means a louder sound with more energy, but the pitch (frequency) is unchanged.
💡 The Aha! Moment
Amplitude = loudness, frequency = pitch. They are independent. You can hit a tabla harder (louder, more amplitude) without changing the note (same frequency). CRO questions are extremely common in IGCSE -- always check: taller wave = more amplitude = louder; more waves in same space = higher frequency = higher pitch.
Practice Questions: 3.1 General Properties of Waves
20 multiple choice questions -- tap an option to check your answer
Question 1
Waves transfer:
A Matter without energy
B Both energy and matter
C Neither energy nor matter
D Energy without transferring matter
Answer: D -- Waves transfer energy from one place to another without transferring matter. The particles of the medium vibrate but do not travel with the wave.
Question 2
The amplitude of a wave is:
A The distance from crest to trough
B The maximum displacement from the rest position
C The distance from one crest to the next
D The number of waves per second
Answer: B -- Amplitude is the maximum displacement from the rest (equilibrium) position. Crest to trough is double the amplitude. Crest to crest is the wavelength. Waves per second is the frequency.
Question 3
Which of the following is a longitudinal wave?
A Sound
B Light
C Water surface waves
D S-waves
Answer: A -- Sound is a longitudinal wave (particles vibrate parallel to the direction of wave travel). Light, water surface waves, and S-waves are all transverse waves.
Question 4
A wave has a frequency of 50 Hz and a wavelength of 0.4 m. What is its speed?
A 125 m/s
B 50.4 m/s
C 20 m/s
D 200 m/s
Answer: C -- v = f × λ = 50 × 0.4 = 20 m/s.
Question 5
In a transverse wave, the vibrations are:
A Parallel to the direction of wave travel
B In the same direction as the energy transfer
C Perpendicular to the direction of wave travel
D Circular
Answer: C -- In transverse waves, vibrations are perpendicular (at right angles) to the direction the wave travels. Longitudinal waves have vibrations parallel to the wave direction.
Question 6
The wavelength of a wave is the distance between:
A A crest and a trough
B The rest position and a crest
C A compression and a rarefaction
D Two consecutive crests (or two consecutive troughs)
Answer: D -- Wavelength is the distance between two consecutive points in phase, such as crest to crest or trough to trough. Crest to trough is half a wavelength. Rest position to crest is the amplitude.
Question 7
Which of these is NOT a transverse wave?
A Radio waves
B Sound waves
C X-rays
D Visible light
Answer: B -- Sound is a longitudinal wave. All electromagnetic waves (radio, X-rays, visible light, microwaves, UV, infrared, gamma) are transverse.
Question 8
When water waves pass from deep to shallow water, what happens?
A They speed up and the wavelength increases
B They slow down and the wavelength decreases
C They speed up and the frequency changes
D They slow down and the frequency decreases
Answer: B -- In shallower water, waves slow down and the wavelength decreases. The frequency never changes during refraction -- it is set by the source. Since v = fλ, if f is constant and v decreases, λ must also decrease.
Question 9
Diffraction is the:
A Bouncing of waves off a surface
B Changing of speed when entering a new medium
C Spreading of waves when passing through a gap or around an edge
D Splitting of light into colours
Answer: C -- Diffraction is the spreading of waves when they pass through a gap or around an edge. Bouncing is reflection, speed change is refraction, and splitting into colours is dispersion.
Question 10
A wave has a speed of 340 m/s and a frequency of 680 Hz. What is its wavelength?
A 0.5 m
B 2.0 m
C 231 200 m
D 340 m
Answer: A -- λ = v / f = 340 / 680 = 0.5 m. Option C is the result of multiplying instead of dividing.
Question 11
P-waves from an earthquake are:
A Transverse waves that can only travel through solids
B Transverse waves that can travel through solids and liquids
C Longitudinal waves that can travel through solids and liquids
D Longitudinal waves that can only travel through solids
Answer: C -- P-waves (primary waves) are longitudinal and can travel through both solids and liquids. S-waves are transverse and can only travel through solids. This is how scientists know the Earth's outer core is liquid -- S-waves cannot pass through it.
Question 12
In a longitudinal wave, particles vibrate:
A At right angles to the wave direction
B In circles
C Upward only
D Parallel to the direction of wave travel
Answer: D -- In longitudinal waves, particle vibrations are parallel to the wave direction. This creates compressions (particles close together) and rarefactions (particles far apart).
Question 13
A wavefront is:
A The front edge of a wave crest only
B A line joining all points on a wave that are at the same phase
C The distance between two crests
D The maximum height of a wave
Answer: B -- A wavefront is a line or surface joining all points at the same phase of vibration. For circular waves (like ripples from a stone), the wavefronts are circles. For plane waves, the wavefronts are straight lines.
Question 14
When a wave is reflected at a plane surface, which property changes?
A Speed
B Direction
C Frequency
D Wavelength
Answer: B -- During reflection, the wave bounces back so its direction changes. The speed, frequency, and wavelength all stay the same because the wave remains in the same medium.
Question 15
Maximum diffraction through a gap occurs when:
A The gap is much larger than the wavelength
B The gap is much smaller than the wavelength
C The gap is about the same size as the wavelength
D The wave has maximum amplitude
Answer: C -- Maximum diffraction occurs when the gap width is approximately equal to the wavelength. When the gap is much larger, waves pass through almost straight. Amplitude does not affect diffraction.
Question 16
During refraction of a wave, which quantity stays the same?
A Speed
B Wavelength
C Direction
D Frequency
Answer: D -- Frequency never changes during refraction. It is determined by the source and remains constant. Speed and wavelength both change. Direction changes if the wave enters the new medium at an angle.
Question 17
10 water waves pass a post in 5 seconds. The wavelength is 1.2 m. What is the wave speed?
A 12 m/s
B 2.4 m/s
C 6.0 m/s
D 0.6 m/s
Answer: B -- First find f: 10 waves in 5 seconds = 10/5 = 2 Hz. Then v = f × λ = 2 × 1.2 = 2.4 m/s.
Question 18
In a ripple tank, refraction is demonstrated by:
A Placing a barrier in the tank
B Creating a narrow gap between two barriers
C Placing a glass plate under part of the tank to make the water shallower
D Increasing the motor speed
Answer: C -- A glass plate makes the water shallower, causing waves to slow down. This change in speed causes refraction. A barrier demonstrates reflection, and a gap demonstrates diffraction.
Question 19
Which of these is an example of a transverse wave?
A Sound in air
B Compression waves in a spring
C P-waves
D S-waves
Answer: D -- S-waves (secondary seismic waves) are transverse. Sound, compression waves in springs, and P-waves are all longitudinal.
Question 20
A longitudinal wave has regions of compression and rarefaction. What are these equivalent to in a transverse wave?
A Amplitude and wavelength
B Crests and troughs
C Speed and frequency
D Wavefronts and wavelengths
Answer: B -- Compressions (where particles are closest together) correspond to crests, and rarefactions (where particles are furthest apart) correspond to troughs. Both represent the maximum and minimum displacement of particles.
3.2 Light
Your Score 0 / 20
Reflection & Refraction of Light - IGCSE Physics
Cognito
Watch
Total Internal Reflection - IGCSE Physics
Cognito
Watch
Lenses - Converging and Diverging - IGCSE Physics
Free Science Lessons
Watch

3.2.1 Reflection of Light

When light hits a smooth surface like a mirror, it bounces back. This is reflection. You use this every morning when you look in a mirror to get ready for school!

Key Terms for Reflection

  • Normal -- an imaginary line drawn at right angles (90°) to the mirror surface at the point where the light ray hits. This is your reference line for measuring angles
  • Angle of incidence (i) -- the angle between the incoming (incident) ray and the normal
  • Angle of reflection (r) -- the angle between the reflected ray and the normal
⚠ Exam Tip

Angles are ALWAYS measured from the normal, not from the mirror surface. This is a very common mistake. If the question says "a ray hits a mirror at 30° to the surface," then the angle of incidence is actually 90° - 30° = 60° (measured from the normal).

The Law of Reflection

angle of incidence = angle of reflection
i = r Both angles measured from the normal

This law works every single time, without exception. Whether it is a bathroom mirror, the rear-view mirror of an auto-rickshaw, or a lake reflecting the sunset -- the angle of incidence always equals the angle of reflection.

Images in a Plane Mirror

When you look in a flat (plane) mirror, you see an image of yourself. This image has specific properties that the examiner loves to test:

  • Same size as the object
  • Same distance behind the mirror as the object is in front
  • Virtual -- the image cannot be projected onto a screen. The light rays do not actually come from behind the mirror; your brain just interprets them as if they do
  • Laterally inverted -- left and right are swapped. This is why the word "AMBULANCE" is written backwards on the front of Indian ambulances, so it reads correctly in your rear-view mirror!
🧠 Memory Trick

"SALT" for plane mirror image properties: Same size, As far behind the mirror, Laterally inverted, The image is virtual. Remember: "The mirror gives you SALT -- Same size, As far behind, Laterally inverted, and it is a virTual image."

Supplement (Extended)

Constructing Mirror Images

For the extended syllabus, you need to be able to draw accurate ray diagrams showing how a plane mirror forms an image:

  1. Draw the mirror as a straight line with hatching on the back
  2. Place the object in front of the mirror
  3. Draw the normal at the point where each ray hits the mirror
  4. Draw the reflected ray so that the angle of reflection equals the angle of incidence
  5. Extend the reflected rays backwards (as dotted lines) behind the mirror -- they meet at the image position
  6. The image is the same distance behind the mirror as the object is in front

You can also use calculations: if an object is 25 cm in front of a plane mirror, the image is exactly 25 cm behind the mirror. The total distance from object to image is 50 cm.

3.2.2 Refraction of Light

Have you ever noticed that a spoon in a glass of chai looks bent at the surface of the liquid? Or that the bottom of a swimming pool always looks shallower than it actually is? That is refraction -- the bending of light when it passes from one transparent material to another.

Why Does Light Bend?

Light travels at different speeds in different materials. In a vacuum, light travels at its maximum speed (3 × 10⁸ m/s). In glass, it slows down to about 2 × 10⁸ m/s. In water, it is about 2.25 × 10⁸ m/s. When light changes speed at a boundary, it changes direction (unless it hits the boundary at exactly 90°).

Key Terms for Refraction

  • Normal -- a line at right angles to the boundary surface at the point where the light enters
  • Angle of incidence (i) -- the angle between the incident ray and the normal
  • Angle of refraction (r) -- the angle between the refracted ray and the normal

The Rules of Refraction

When light goes from a less dense material to a more dense material (e.g., air → glass):

  • Light slows down
  • Light bends towards the normal
  • Angle of refraction is smaller than the angle of incidence

When light goes from a more dense material to a less dense material (e.g., glass → air):

  • Light speeds up
  • Light bends away from the normal
  • Angle of refraction is larger than the angle of incidence
🧠 Memory Trick

"FAST medium = FAT angle" -- In the faster (less dense) medium, the angle from the normal is bigger (fatter). In the slower (more dense) medium, the angle is smaller (thinner). Think of it like Bangalore traffic: on the narrow, crowded MG Road (dense medium), you move slowly and stay close to the centre line (small angle from normal). On the open Mysore Expressway (less dense), you spread out and move fast (large angle from normal).

Experiment: Refraction by a Transparent Block

In this classic experiment, you shine a ray of light into a rectangular glass or Perspex block:

  1. The ray bends towards the normal when entering the block (air → glass, slowing down)
  2. The ray travels straight through the block
  3. The ray bends away from the normal when leaving the block (glass → air, speeding up)
  4. The emerging ray is parallel to the incident ray but shifted sideways (laterally displaced)
⚠ Exam Tip

When drawing refraction through a rectangular block, the emergent ray MUST be parallel to the incident ray. If your diagram shows them at different angles, something is wrong. Also, if light hits the boundary along the normal (at 0°), it passes straight through without bending.

Critical Angle and Total Internal Reflection

When light travels from a denser medium to a less dense medium (like glass to air), something special can happen:

  • At small angles of incidence: some light is refracted (bends away from the normal) and some is reflected back inside. This is partial internal reflection
  • At the critical angle (c): the refracted ray travels exactly along the boundary surface (angle of refraction = 90°). This is the tipping point
  • At angles greater than the critical angle: no light escapes -- ALL of it is reflected back inside the denser medium. This is total internal reflection (TIR)

The critical angle is the angle of incidence in the denser medium for which the angle of refraction is exactly 90°.

⚠ Exam Tip

For total internal reflection to occur, TWO conditions must be met: (1) light must be travelling from a denser medium to a less dense medium (e.g., glass to air, NOT air to glass), and (2) the angle of incidence must be greater than the critical angle. Many students forget condition 1!

Supplement (Extended)

Refractive Index (n)

The refractive index of a material tells you how much it slows down light compared to a vacuum (or air, which is almost the same). It is defined as the ratio of speeds:

n = speed of light in vacuum / speed of light in the material
n = refractive index (no unit -- it is a ratio) n is always ≥ 1 (light is always slowest in a medium, never faster than in vacuum)

Typical values: glass ≈ 1.5, water ≈ 1.33, diamond ≈ 2.42, air ≈ 1.0.

Snell's Law

n = sin i / sin r
n = refractive index of the material i = angle of incidence (in the less dense medium, usually air) r = angle of refraction (in the denser medium)
Worked Example A ray of light enters a glass block at an angle of incidence of 45°. The refractive index of the glass is 1.5. Calculate the angle of refraction.
Step 1: Write down what you know
n = 1.5, i = 45°, r = ?
Step 2: Write and rearrange Snell's law
n = sin i / sin r
sin r = sin i / n
Step 3: Substitute and calculate
sin r = sin 45° / 1.5 = 0.7071 / 1.5 = 0.4714
r = sin⁻¹(0.4714) = 28.1°
✅ Answer: The angle of refraction is 28.1° (the ray bends towards the normal as expected, since it enters a denser medium)
Worked Example Light enters a rectangular glass prism (n = 1.5) at 30° to the normal. What is the angle of refraction inside the glass?
Step 1: Identify values
n = 1.5, i = 30°, r = ?
Step 2: Apply Snell's law
sin r = sin i / n = sin 30° / 1.5 = 0.5 / 1.5 = 0.3333
Step 3: Find the angle
r = sin⁻¹(0.3333) = 19.5°
✅ Answer: The angle of refraction is 19.5°

Refractive Index and Critical Angle

n = 1 / sin c
n = refractive index of the denser medium c = critical angle

This formula connects the refractive index directly to the critical angle. A higher refractive index means a smaller critical angle (so TIR happens more easily). Diamond has n = 2.42, so its critical angle is only about 24° -- this is why diamonds sparkle so brilliantly! Most light that enters gets trapped inside and bounces around before escaping.

Worked Example The refractive index of glass is 1.5. Calculate the critical angle for glass.
Step 1: Write down what you know
n = 1.5, c = ?
Step 2: Rearrange the formula
n = 1 / sin c, so sin c = 1 / n
Step 3: Substitute and calculate
sin c = 1 / 1.5 = 0.6667
c = sin⁻¹(0.6667) = 41.8°
✅ Answer: The critical angle for glass is 41.8°. Any light hitting the glass-air boundary at more than 41.8° will undergo total internal reflection.

Optical Fibres

Total internal reflection is the principle behind optical fibres -- the thin glass or plastic cables that carry internet data all across India and the world. When Tara streams a video on her phone, that data likely travels thousands of kilometres through optical fibres under the Indian Ocean (like the submarine cables connecting India to Singapore and Europe).

How they work: a light signal enters one end of the fibre. The fibre is so thin that light always hits the inner surface at an angle greater than the critical angle, so it undergoes total internal reflection again and again, bouncing along the length of the fibre without escaping. The light (carrying data) can travel enormous distances with very little energy loss.

Advantages of optical fibres over copper cables:

  • Much higher data capacity (bandwidth)
  • No electrical interference
  • Thinner and lighter
  • More secure (harder to tap)
  • Signals travel further before needing amplification

3.2.3 Thin Lenses

Lenses are pieces of transparent material (usually glass or plastic) with curved surfaces. They are everywhere in your daily life -- in your eyes, in cameras, in your spectacles (if you wear them), in the projector at school, and in magnifying glasses.

Two Types of Lenses

Converging lens (convex lens) -- thicker in the middle than at the edges. When a parallel beam of light passes through it, the rays are brought together (converge) to meet at a single point called the principal focus (F).

Diverging lens (concave lens) -- thinner in the middle than at the edges. When a parallel beam of light passes through it, the rays spread out (diverge) as if they came from a point behind the lens. This point is also called the principal focus, but it is virtual (on the same side as the incoming light).

🧠 Memory Trick

Think of Caves: A conCAVE lens has a "cave" shape (thinner in the middle, like a cave entrance) and light diverges. A conVEX lens bulges outward (like a pregnant belly) and brings light together. Also: "con-VEX = con-VERGE" -- they both start with "conv"!

Key Lens Terms

  • Principal axis -- the straight line passing through the centre of the lens, perpendicular to the lens surface
  • Principal focus (F) -- the point where parallel rays converge (for a converging lens) or appear to diverge from (for a diverging lens)
  • Focal length (f) -- the distance from the centre of the lens to the principal focus. A fat, strongly curved lens has a short focal length; a thin, gently curved lens has a long focal length

Ray Diagrams for Converging Lenses -- Real Images

To find where an image forms, you draw at least two of these three standard rays:

  1. Parallel ray -- arrives parallel to the principal axis, then refracts through the principal focus F on the other side
  2. Central ray -- passes straight through the centre of the lens without bending
  3. Focal ray -- passes through the principal focus F on the near side, then refracts to emerge parallel to the principal axis

Where two rays cross on the other side of the lens, that is where the real image forms. A real image can be projected onto a screen (like a cinema screen or the retina of your eye).

Image Characteristics

An image can be described using three properties:

  • Size -- magnified (bigger), diminished (smaller), or same size
  • Orientation -- upright (same way up as the object) or inverted (upside down)
  • Type -- real (can be projected on a screen) or virtual (cannot be projected)

For a converging lens, when the object is beyond the principal focus (more than one focal length away), the image is real and inverted. The further the object is from the lens, the smaller the image becomes.

Virtual Images

When an object is placed between F and the lens (closer than one focal length), the rays diverge after passing through the lens. Your eye traces these diverging rays back to where they appear to meet -- behind the object, on the same side of the lens. This creates a virtual image that is upright, magnified, and on the same side as the object.

A diverging lens always produces a virtual image that is upright, diminished (smaller), and on the same side as the object, no matter where the object is placed.

⚠ Exam Tip

When drawing ray diagrams: (1) always use a ruler, (2) draw at least TWO rays from the top of the object, (3) use solid lines for real rays and dashed lines for virtual rays (extensions behind the lens), (4) mark the image clearly with an arrow. Sloppy diagrams lose marks!

Supplement (Extended)

Ray Diagram for Virtual Image (Converging Lens)

When the object is between F and the lens:

  1. Draw a ray parallel to the principal axis -- it refracts through F on the far side
  2. Draw a ray through the centre of the lens -- it passes straight through
  3. These two refracted rays diverge on the far side. Extend them backwards (dashed lines) to the near side where they appear to meet
  4. That meeting point is where the virtual, upright, magnified image forms

Magnifying Glass

A single converging lens used as a magnifying glass works by placing the object closer than the focal length. The lens produces a virtual, upright, magnified image that you see when you look through the lens. Every shopkeeper at KR Market checking the quality of spices with a magnifying glass is using this principle!

Correcting Vision Defects

Short-sightedness (myopia) -- the eyeball is too long, or the eye lens is too strong, so distant objects are focused in front of the retina. Corrected with a diverging (concave) lens, which spreads out the light before it enters the eye, moving the focal point back onto the retina.

Long-sightedness (hypermetropia) -- the eyeball is too short, or the eye lens is too weak, so close objects are focused behind the retina. Corrected with a converging (convex) lens, which brings the light together more before it enters the eye, moving the focal point forward onto the retina.

🧠 Memory Trick

"Short sight needs a Short lens name" -- Short-sighted people need a diverging (concave) lens. "concave" has fewer letters than "converging" -- short name for short sight! Alternatively: "My DIVe" -- Myopia is corrected with a DIVerging lens.

3.2.4 Dispersion of Light

White light looks, well, white. But it is actually a mixture of many different colours of light. When white light passes through a glass prism, it splits into a beautiful band of colours called a spectrum. This splitting is called dispersion.

Why Does Dispersion Happen?

Different colours of light have slightly different speeds in glass. Red light is the fastest and bends the least; violet light is the slowest and bends the most. Because each colour refracts by a slightly different amount at both surfaces of the prism, they separate into a spectrum.

The Seven Colours (in order)

From least bent to most bent (or from longest wavelength to shortest):

Red -- Orange -- Yellow -- Green -- Blue -- Indigo -- Violet

🧠 Memory Trick

"VIBGYOR" -- Every Indian student knows this! Read the colours backwards (from violet to red): Violet, Indigo, Blue, Green, Yellow, Orange, Red. Or in order from red: "ROY G. BIV" is the English version. Or try: "Richard Of York Gave Battle In Vain".

You see dispersion in nature when sunlight passes through raindrops and creates a rainbow! The raindrop acts like a tiny prism, splitting the white sunlight into its component colours. Bangalore gets some spectacular rainbows during the monsoon season.

⚠ Exam Tip

In the exam, when asked about dispersion through a prism, always state that red is deviated (bent) the least and violet is deviated the most. If drawing a diagram, red must be on the outer edge (least bent) and violet on the inner edge (most bent) of the emerging spectrum.

Supplement (Extended)

Monochromatic Light

Monochromatic light means light of a single frequency (and therefore a single colour and single wavelength). "Mono" means one, "chromatic" means colour.

A laser produces monochromatic light. A sodium street lamp produces nearly monochromatic yellow light. If you pass monochromatic light through a prism, it does NOT split into a spectrum -- it just bends (refracts) as a single colour because there is only one frequency present.

White light is the opposite of monochromatic -- it is polychromatic, containing a mixture of all visible frequencies.

⚠ Exam Tip

If asked "what is monochromatic light?", the precise answer is "light of a single frequency." Do NOT say "single colour" -- while technically similar, the syllabus definition specifically uses "frequency." A laser is the best example of a monochromatic source.

🔎 Apply It: Real-World Physics
Cambridge examiners LOVE testing familiar concepts in unfamiliar situations. Can you spot the physics hiding in these real-world scenarios? Tap each one to reveal the answer.
1
On a hot summer afternoon on the Bangalore-Mysore Expressway, you see what looks like a pool of water on the road ahead. As you drive closer, it disappears. Your friend says "that was a mirage."
Explain why mirages appear, using the physics of refraction. Is the light bending towards or away from the normal?
Identify the Physics
Refraction -- light changing speed and direction as it passes between layers of air at different temperatures (different densities = different optical densities).
Work It Out
The black road surface heats the air just above it. Hot air is less dense (optically less dense) than cool air higher up. Light from the sky travels downward through layers of air that get progressively hotter and less dense.

Each time the light enters a less dense layer, it bends away from the normal (just like light going from glass to air). Eventually the angle of incidence exceeds the critical angle, and total internal reflection occurs -- the light reflects back upward toward your eyes.

Your brain assumes light travels in straight lines, so it thinks the light came from the ground. You see an image of the sky on the road -- which looks like water.
💡 The Aha! Moment
A mirage is actually total internal reflection happening in air, not water. The same physics that keeps light inside fibre optic cables creates the "water on the road" illusion. If the exam asks about mirages, use these key terms: refraction, less dense, away from normal, critical angle, total internal reflection.
2
A jeweller in Jaipur is examining a diamond. Diamonds have a very high refractive index (n = 2.42) and a critical angle of only 24.4 degrees. Glass imitation stones have a much larger critical angle of about 42 degrees.
Why does a real diamond sparkle much more than a glass imitation? Use the concept of total internal reflection.
Identify the Physics
Total internal reflection (TIR) occurs when light hits a boundary at an angle greater than the critical angle, while travelling from a more dense to a less dense medium. The relationship is: sin c = 1/n.
Work It Out
Diamond has a critical angle of just 24.4 degrees. This means any light ray hitting the inside surface at more than 24.4 degrees to the normal gets totally internally reflected -- no light escapes.

Since 24.4 degrees is very small, most light rays that enter the diamond bounce around inside it multiple times before finally escaping through the top face. Each bounce redirects the light, creating intense flashes of brilliance.

Glass imitation (critical angle ~42 degrees) lets much more light escape through the sides and bottom, so fewer internal reflections occur and the stone looks dull in comparison.

We can verify: sin c = 1/n, so sin c = 1/2.42 = 0.413, and c = sin⁻¹(0.413) = 24.4 degrees. ✔
💡 The Aha! Moment
The smaller the critical angle, the more light gets trapped inside by TIR, and the more the gem sparkles. This is why diamond cutters carefully angle each face -- they are engineering total internal reflection. This is a beautiful exam connection: high refractive index → small critical angle → more TIR → more sparkle.
3
When you stand at the edge of a swimming pool at your school in Bangalore and look down at the tiles on the bottom, the pool looks shallower than it actually is. A 2-metre deep pool appears to be only about 1.5 metres deep.
Explain why the pool appears shallower than it really is, and calculate the refractive index of water using these values.
Identify the Physics
Refraction of light at the water-air boundary. The relationship: n = real depth / apparent depth.
Work It Out
Light from the tiles at the bottom of the pool travels from water (more optically dense) into air (less optically dense). As it crosses the boundary, it bends away from the normal -- it speeds up and refracts away.

Your brain traces the light rays back in straight lines and thinks the tiles are at a shallower position than they really are.

Refractive index n = real depth / apparent depth = 2.0 m / 1.5 m = 1.33.

This matches the known refractive index of water (1.33). ✔
💡 The Aha! Moment
This is why you should NEVER judge the depth of water by looking at it. The pool always looks shallower than it is. The formula n = real depth / apparent depth is the same idea as n = sin i / sin r, just applied to depth instead of angles. Both tell you how much light bends at a boundary.
4
A barber shop in Bangalore has two large plane mirrors facing each other on opposite walls. When you sit in the chair, you can see an infinite tunnel of reflections of yourself stretching into the distance.
Using the law of reflection, explain how the "infinite tunnel" of images forms. Are the images real or virtual?
Identify the Physics
Law of reflection: angle of incidence = angle of reflection. Properties of images in plane mirrors: virtual, same size, laterally inverted, same distance behind mirror as object is in front.
Work It Out
Mirror A creates a virtual image of you (Image 1) behind it. Mirror B then "sees" Image 1 as an object and creates its own virtual image of that image (Image 2) behind Mirror B. Mirror A then reflects Image 2 to create Image 3, and so on.

Each image is formed further behind its respective mirror. The images alternate between being laterally inverted and not (Image 1 is inverted, Image 2 is not, Image 3 is inverted again...). Each successive image appears dimmer because some light is absorbed at each reflection.

All images are virtual -- they cannot be projected onto a screen because the light rays only appear to come from behind the mirror. They never actually pass through the image position.
💡 The Aha! Moment
The "infinite tunnel" effect demonstrates that the image from one mirror can act as the object for another mirror. Each image obeys all the rules: virtual, same size, laterally inverted, and as far behind the mirror as the "object" is in front. This is a favourite exam trick -- asking about multiple mirrors.
5
Jio Fiber delivers internet to homes across India using fibre optic cables. Inside each cable, a thin glass fibre carries data as pulses of light. The light enters one end and travels the entire length of the cable (sometimes hundreds of kilometres) without escaping through the sides.
Explain how total internal reflection keeps the light trapped inside the fibre. Why must the glass core be surrounded by cladding with a lower refractive index?
Identify the Physics
Total internal reflection (TIR). Conditions: light must be in the denser medium, and angle of incidence must exceed the critical angle.
Work It Out
The glass core has a higher refractive index than the cladding around it. When light inside the core hits the core-cladding boundary, it is going from a more dense medium to a less dense medium -- the first condition for TIR is met.

The fibre is designed so that light hits the boundary at an angle greater than the critical angle. This means 100% of the light is reflected back into the core -- none escapes into the cladding. The light zigzags along the entire length of the fibre.

Without cladding, the light would hit a glass-air boundary. While TIR could still occur, any scratches, dirt, or contact with another fibre would disrupt the reflection. The cladding provides a controlled, clean boundary with a precise critical angle, ensuring reliable TIR even when the cable is bent or bundled with other fibres.
💡 The Aha! Moment
Fibre optics is TIR in action. The cladding is NOT there to reflect light (that is a common wrong answer). It is there to provide a less-dense medium around the core so that the conditions for TIR are always met. In the exam, always mention: denser to less dense, angle greater than critical angle, total internal reflection.
Practice Questions: 3.2 Light
20 multiple choice questions -- tap an option to check your answer
Question 1
The angle of incidence is measured between the incident ray and the:
A Mirror surface
B Reflected ray
C Normal
D Horizontal
Answer: C -- Both the angle of incidence and angle of reflection are always measured from the normal (the line perpendicular to the surface), never from the surface itself.
Question 2
An image in a plane mirror is:
A Real, inverted, same size
B Virtual, inverted, magnified
C Virtual, upright, same size
D Real, upright, diminished
Answer: C -- A plane mirror always produces a virtual, upright, same-size, laterally inverted image, located the same distance behind the mirror as the object is in front.
Question 3
When light passes from air into glass, it:
A Speeds up and bends away from the normal
B Slows down and bends towards the normal
C Speeds up and bends towards the normal
D Slows down and bends away from the normal
Answer: B -- Glass is optically denser than air. Light slows down when entering glass and bends towards the normal (the angle of refraction is smaller than the angle of incidence).
Question 4
The critical angle is the angle of incidence in the denser medium for which the angle of refraction is:
A
B 45°
C 90°
D 180°
Answer: C -- At the critical angle, the refracted ray travels along the boundary surface, making an angle of 90° with the normal. Any angle of incidence larger than the critical angle causes total internal reflection.
Question 5
For total internal reflection to occur, light must be travelling:
A From a less dense to a more dense medium
B Along the normal
C From air into glass
D From a more dense to a less dense medium, at an angle greater than the critical angle
Answer: D -- Both conditions must be met: (1) light must travel from a denser to a less dense medium, AND (2) the angle of incidence must exceed the critical angle. Air to glass is the wrong direction.
Question 6
A converging lens is:
A Thinner in the middle than at the edges
B Thicker in the middle than at the edges
C Flat on both sides
D Thicker at the top than the bottom
Answer: B -- A converging (convex) lens is thicker in the middle. It brings parallel light rays together to a focus. A diverging (concave) lens is thinner in the middle and spreads light rays apart.
Question 7
The focal length of a lens is the distance from the centre of the lens to the:
A Object
B Image
C Principal focus
D Edge of the lens
Answer: C -- The focal length is the distance from the centre of the lens to the principal focus, which is where parallel rays converge (for a converging lens) or appear to diverge from (for a diverging lens).
Question 8
When white light passes through a glass prism, it produces a spectrum. This process is called:
A Dispersion
B Diffraction
C Reflection
D Polarisation
Answer: A -- Dispersion is the splitting of white light into its component colours by a prism. Each colour has a different wavelength and is refracted by a slightly different amount, separating the colours.
Question 9
In the visible spectrum, which colour is deviated (bent) the most by a prism?
A Red
B Yellow
C Green
D Violet
Answer: D -- Violet light has the shortest wavelength and is slowed down the most in glass, so it is refracted (deviated) the most. Red has the longest wavelength and is deviated the least.
Question 10
The refractive index of glass is 1.5. The critical angle for glass is approximately:
A 30°
B 42°
C 49°
D 60°
Answer: B -- Using n = 1/sin c: sin c = 1/1.5 = 0.667, so c = sin⁻¹(0.667) ≈ 41.8° ≈ 42°.
Question 11
A ray of light enters a glass block along the normal. What happens?
A It bends towards the normal
B It bends away from the normal
C It passes straight through without bending
D It is totally internally reflected
Answer: C -- When light enters along the normal (angle of incidence = 0°), the angle of refraction is also 0°, so it passes straight through without changing direction. It still slows down, but does not bend.
Question 12
Short-sightedness (myopia) is corrected using a:
A Converging lens
B Diverging lens
C Plane mirror
D Glass prism
Answer: B -- Short-sighted people can see close objects but not distant ones. Their eye focuses too strongly, so the image forms in front of the retina. A diverging (concave) lens spreads the light out slightly before it enters the eye, moving the focus back onto the retina.
Question 13
Optical fibres use which principle to transmit light?
A Refraction
B Diffraction
C Dispersion
D Total internal reflection
Answer: D -- Optical fibres work by total internal reflection. Light enters the fibre and hits the inner surface at angles greater than the critical angle, so it is completely reflected and bounces along the fibre.
Question 14
An object is placed between the principal focus (F) and a converging lens. The image is:
A Real, inverted, diminished
B Virtual, upright, magnified
C Real, upright, magnified
D Virtual, inverted, diminished
Answer: B -- When the object is between F and the lens, the refracted rays diverge. The image is virtual (formed by extending rays backwards), upright, and magnified. This is the magnifying glass arrangement.
Question 15
The law of reflection states that:
A The angle of incidence equals the angle of reflection
B The angle of incidence is twice the angle of reflection
C Light always reflects at 45 degrees
D The reflected ray is parallel to the incident ray
Answer: A -- The law of reflection states that the angle of incidence equals the angle of reflection (i = r). Both angles are measured from the normal. The incident ray, reflected ray, and normal all lie in the same plane.
Question 16
Monochromatic light is light that has:
A A very high intensity
B Multiple wavelengths
C A single frequency
D Been reflected once
Answer: C -- "Mono" means one and "chromatic" means colour. Monochromatic light has a single frequency (and therefore a single wavelength and colour). A laser is a good example of a monochromatic light source.
Question 17
A diverging lens always produces an image that is:
A Real, magnified, inverted
B Real, diminished, upright
C Virtual, magnified, upright
D Virtual, diminished, upright
Answer: D -- A diverging (concave) lens always produces a virtual, upright, diminished image, regardless of where the object is placed. The image forms on the same side as the object.
Question 18
Light enters a glass block (n = 1.5) at an angle of incidence of 30°. Using Snell's law, the angle of refraction is approximately:
A 15°
B 19.5°
C 30°
D 45°
Answer: B -- n = sin i / sin r, so sin r = sin 30° / 1.5 = 0.5 / 1.5 = 0.333. r = sin⁻¹(0.333) ≈ 19.5°. The angle is smaller than the angle of incidence because the light enters a denser medium.
Question 19
When light passes through a rectangular glass block, the emergent ray is:
A At a greater angle to the normal than the incident ray
B Parallel to the incident ray but laterally displaced
C Perpendicular to the incident ray
D Refracted towards the normal
Answer: B -- The emergent ray is always parallel to the incident ray when passing through a rectangular block with parallel sides. It is shifted sideways (laterally displaced) because it bends towards the normal on entering and away from the normal on leaving.
Question 20
Long-sightedness (hypermetropia) is corrected using a:
A Diverging lens
B Plane mirror
C Converging lens
D Concave mirror
Answer: C -- Long-sighted people cannot see close objects clearly because the image forms behind the retina. A converging (convex) lens adds extra focusing power, bringing the image forward onto the retina. Short-sightedness uses the opposite: a diverging lens.
3.3 Electromagnetic Spectrum
Your Score: 0 / 20
Electromagnetic Spectrum - IGCSE Physics
Cognito
Watch
EM Spectrum Explained in 8 Minutes
Science Shorts
Watch
Uses and Dangers of EM Waves - GCSE Physics
Free Science Lessons
Watch

What is the Electromagnetic Spectrum?

Tara, you already use the electromagnetic spectrum every single day -- you just might not know it by that name! When you watch a YouTube video on your phone, your phone receives microwaves from a cell tower. When you change the TV channel with a remote, you are using infrared. When you step out into the Bangalore sunshine, visible light and ultraviolet radiation hit your skin. All of these are electromagnetic (EM) waves.

Here is the key idea: all EM waves are transverse waves that can travel through a vacuum (empty space). They do not need a medium. That is how sunlight reaches Earth across 150 million km of empty space. Pretty amazing, right?

The entire family of EM waves is arranged in a continuous band called the electromagnetic spectrum. They are all the same type of wave -- the only difference is their wavelength and frequency.

⚠ Exam Tip

All EM waves travel at the same speed in a vacuum. This is the speed of light. Do not say "light travels fastest" -- ALL EM waves travel at exactly the same speed in vacuum. The difference between them is wavelength and frequency, NOT speed.

The EM Spectrum in Order

You must know the seven main regions of the spectrum. Here they are, arranged from longest wavelength / lowest frequency to shortest wavelength / highest frequency:

ELECTROMAGNETIC SPECTRUM
Long λ
Short λ
Radio
km - m
Micro-wave
cm
Infra-red
µm
Visible
~500 nm
UV
nm
X-rays
~0.01 nm
Gamma
<0.01 nm
Low f
High f
🧠 Memory Trick

Running Man In Very Ugly Xtra-large Green shorts = Radio, Microwaves, Infrared, Visible, Ultraviolet, X-rays, Gamma rays. This goes from longest wavelength to shortest. Imagine this runner jogging around Cubbon Park -- unforgettable!

An important relationship to remember:

  • As wavelength decreases, frequency increases (they are inversely related)
  • As frequency increases, the wave carries more energy
  • So gamma rays (shortest wavelength, highest frequency) are the most energetic and most dangerous
  • Radio waves (longest wavelength, lowest frequency) are the least energetic and least dangerous
Supplement (Extended)

Speed of EM Waves in Vacuum

All electromagnetic waves travel at the same speed in a vacuum:

c = 3.0 × 10⁸ m/s
c = speed of light in vacuum This is approximately 300,000,000 m/s

This speed is often called the speed of light, but remember that ALL EM waves -- not just visible light -- travel at this speed in a vacuum. The wave equation still applies:

v = f × λ
v = wave speed (m/s) -- for EM waves in vacuum, v = c = 3.0 × 10⁸ m/s f = frequency (Hz) λ = wavelength (m)
Worked Example A Bangalore radio station (Radio Mirchi) broadcasts at a frequency of 98.3 MHz. What is the wavelength of the radio waves?
Step 1: Write down what you know
f = 98.3 MHz = 98.3 × 10⁶ Hz = 9.83 × 10⁷ Hz
v = c = 3.0 × 10⁸ m/s
Step 2: Rearrange v = f × λ to find λ
λ = v / f
Step 3: Substitute and calculate
λ = (3.0 × 10⁸) / (9.83 × 10⁷)
λ = 3.05 m
✅ Answer: The wavelength is approximately 3.05 m -- about the length of an auto-rickshaw!
Worked Example A microwave oven in a Bangalore kitchen uses microwaves with a wavelength of 12.2 cm. Calculate the frequency of these microwaves.
Step 1: Write down what you know (convert units!)
λ = 12.2 cm = 0.122 m (always convert to metres)
v = c = 3.0 × 10⁸ m/s
Step 2: Rearrange v = f × λ to find f
f = v / λ
Step 3: Substitute and calculate
f = (3.0 × 10⁸) / 0.122
f = 2.46 × 10⁹ Hz
f ≈ 2.46 GHz
✅ Answer: The frequency is 2.46 × 10⁹ Hz (2.46 GHz) -- this is actually why microwave ovens can sometimes interfere with your Wi-Fi, which also uses 2.4 GHz!
Worked Example An X-ray machine at a hospital in Bangalore produces X-rays with a frequency of 3.0 × 10¹⁸ Hz. What is the wavelength?
Step 1: Write down what you know
f = 3.0 × 10¹⁸ Hz
v = c = 3.0 × 10⁸ m/s
Step 2: Rearrange v = f × λ to find λ
λ = v / f
Step 3: Substitute and calculate
λ = (3.0 × 10⁸) / (3.0 × 10¹⁸)
λ = 1.0 × 10⁻¹⁰ m
λ = 0.1 nm
✅ Answer: The wavelength is 1.0 × 10⁻¹⁰ m (0.1 nm) -- incredibly tiny! That is why X-rays can pass through soft tissue but get stopped by dense bones.
⚠ Exam Tip

When using the wave equation with EM waves, always check your units! Convert wavelengths to metres (not cm or nm) before putting them into the equation. The most common mistake is forgetting to convert MHz to Hz (multiply by 10⁶) or cm to m (divide by 100).

Uses of Each Type of EM Wave

This is a really important section for your exam, Tara. You need to know the typical uses for each region of the spectrum. Let us go through them one by one with examples you will recognise from daily life in Bangalore.

(a) Radio Waves -- longest wavelength, lowest frequency

  • Radio and TV broadcasting -- Radio Mirchi 98.3, Red FM, All India Radio all send radio waves to your receiver. Doordarshan uses radio waves for TV transmission too.
  • Astronomy -- Radio telescopes pick up radio waves from distant stars and galaxies. India's Giant Metrewave Radio Telescope (GMRT) near Pune is one of the world's largest!
  • RFID (Radio-Frequency Identification) -- The FASTag on your family's car uses RFID to automatically pay tolls on the NICE Road or Bangalore-Mysore Expressway. Metro smart cards also use RFID.

(b) Microwaves

  • Satellite TV -- Tata Sky, Airtel Digital TV, and Dish TV all receive signals via microwaves beamed from satellites
  • Mobile phones -- When you make a call or use 4G/5G data on Jio or Airtel, your phone communicates with the cell tower using microwaves
  • Microwave ovens -- The microwave oven in your kitchen heats food by making water molecules vibrate very fast

(c) Infrared (IR)

  • Grills and heaters -- The tandoor at your favourite restaurant uses infrared radiation to cook naan and tikka
  • Remote controls -- Your TV remote sends IR signals (try pointing it at your phone camera -- you can actually see the IR flash!)
  • Intruder alarms / motion sensors -- Security systems in shops on MG Road or Commercial Street detect the IR emitted by a person's warm body
  • Thermal imaging -- Remember during COVID, thermal cameras at Bangalore airport checked passengers' temperatures? Those detect infrared radiation
  • Optical fibres -- Your Jio Fiber or ACT Fibernet broadband uses infrared light travelling through glass fibres to deliver super-fast internet

(d) Visible Light

  • Vision -- The only part of the EM spectrum your eyes can detect! The seven colours of the rainbow: Red, Orange, Yellow, Green, Blue, Indigo, Violet (ROYGBIV)
  • Photography -- Your phone camera captures visible light to take photos
  • Illumination -- LED bulbs, tube lights, and the beautiful Lalbagh flower show lights all produce visible light

(e) Ultraviolet (UV)

  • Security marking -- Invisible ink that only shows up under UV light is used on concert tickets and exam papers to prevent counterfeiting
  • Detecting fake bank notes -- Banks use UV lamps to check if a ₹500 or ₹2000 note is genuine. Real notes have special UV-reactive markings
  • Sterilising water -- UV water purifiers (like Kent or Aquaguard) use UV light to kill bacteria and viruses in drinking water. Very common in Indian homes!

(f) X-rays

  • Medical scanning -- When you break a bone playing basketball or fall off a bicycle, the doctor takes an X-ray because X-rays pass through soft tissue but are absorbed by bones, creating a shadow image
  • Security scanning -- Baggage scanners at Namma Metro stations and Bangalore airport use X-rays to see inside your bags without opening them

(g) Gamma Rays -- shortest wavelength, highest frequency, most energetic

  • Sterilising food and medical equipment -- Gamma rays kill bacteria on surgical instruments and can preserve packaged food without heating it
  • Cancer detection (imaging) -- Doctors inject a gamma-emitting tracer into the patient and use a special camera to detect cancer cells
  • Cancer treatment (radiotherapy) -- Focused beams of gamma rays are aimed at tumours to destroy cancer cells. Major cancer hospitals in Bangalore like Kidwai Memorial use this
🧠 Memory Trick

For remembering uses, think of them in terms of what you encounter in a typical day: Wake up and listen to Radio Mirchi. Call Amma on your Microwave-using mobile. She is cooking with the Infrared tandoor. You See (visible light) the food. Apply UV sunscreen before going out. Get an X-ray at the hospital. Gamma rays treat the illness.

Harmful Effects of EM Radiation

The higher the frequency (and energy) of an EM wave, the more damage it can do to your body. Here are the dangers you need to know:

EM Wave Harmful Effect Why?
Microwaves Internal heating of body tissue Microwaves can penetrate skin and heat water in your cells, like a microwave oven heats food
Infrared Skin burns Too much IR causes burns -- like standing too close to a bonfire during Bhogi / Lohri
Ultraviolet Skin cancer, eye damage UV damages DNA in skin cells (causing cancer) and can damage the cornea and retina of your eyes
X-rays Cell mutation and damage X-rays can ionise atoms in your cells, damaging DNA and potentially causing cancer with excessive exposure
Gamma rays Cell mutation and damage Most penetrating and energetic -- causes the most severe cell damage. That is why radiographers stand behind lead shields
⚠ Exam Tip

Notice the pattern: radio waves and visible light are NOT listed as harmful in the syllabus. The harmful effects start from microwaves and get progressively worse as frequency increases. Also note that X-rays and gamma rays cause the SAME type of harm (cell mutation/damage) -- the exam sometimes tests whether you know both cause the same thing.

Satellite Communication Using Microwaves

When you watch a cricket match on Star Sports via Tata Sky, the signal travels from the stadium to a satellite in space and back down to your dish antenna -- all using microwaves. Let us understand how this works.

(a) Low Earth Orbit Satellites

  • Orbit at heights of about 200-2000 km above Earth
  • They move across the sky quickly (they orbit the Earth in about 90 minutes)
  • Because they are close, signals are stronger and need less power
  • Used for weather monitoring, imaging, and some phone networks
  • You need many satellites to maintain continuous coverage because each one is only overhead for a short time
  • Example: India's ISRO launches many low orbit satellites -- CartoSat for mapping, OceanSat for weather

(b) Geostationary Satellites

  • Orbit at a height of about 36,000 km above the equator
  • They orbit at exactly the same rate as the Earth rotates, so they appear to stay fixed above one spot
  • Your satellite dish can point at one fixed position in the sky and always receive the signal
  • Used for TV broadcasting (Tata Sky, Dish TV), weather (INSAT), and long-distance communication
  • Being much further away means signals are weaker and there is a small time delay
⚠ Exam Tip

A classic exam question is: "Why do satellite TV dishes not need to track the satellite across the sky?" The answer is that TV satellites are geostationary -- they orbit above the equator at the same rate Earth spins, so they stay in the same position relative to the ground. The dish only needs to be pointed once.

Supplement (Extended)

Communication Systems

This extended section covers three important communication technologies you use every day:

(a) Mobile Phones Use Microwaves

Why microwaves and not radio waves? Two reasons:

  • Microwaves can penetrate walls and buildings -- so your Jio or Airtel signal works inside your house, inside a classroom, even inside Mantri Mall
  • Microwave wavelengths are short enough for a short aerial -- the antenna inside your phone is tiny. If phones used radio waves, you would need an antenna several metres long! Imagine walking around with a 3-metre antenna sticking out of your phone.

(b) Bluetooth Uses Radio Waves

When you connect your wireless earbuds or share files between phones using Bluetooth:

  • Bluetooth uses short-range radio waves
  • Radio waves can pass through walls -- so your Bluetooth speaker works from the next room
  • Bluetooth is designed for short distances (typically up to 10 m)

(c) Optical Fibres Use Visible Light or Infrared

Your ACT Fibernet or Jio Fiber uses optical fibres -- thin glass strands that carry light signals:

  • Visible light or infrared travels along the fibre by total internal reflection
  • Optical fibres achieve extremely high data rates -- that is why fibre broadband is so much faster than 4G
  • They can carry far more information than radio waves or microwaves
🧠 Memory Trick

Think of it as a speed ranking: Mobile = Microwaves (both start with M!), Bluetooth = radio waves (B for Both-rooms, because radio passes through walls), Fibre = light/infrared (F for Fastest!).

Digital vs Analogue Signals

Signals can be sent in two ways:

Feature Analogue Signal Digital Signal
What it looks like A smooth, continuous wave that varies in amplitude and frequency A series of ON/OFF pulses (only two values: 0 or 1)
Example Old AM radio, vinyl records, old telephone landlines Digital TV (Tata Sky), 4G/5G mobile data, MP3 music files
Effect of noise Noise distorts the signal and cannot be removed Noise can be identified and removed -- signal regenerated cleanly

Sound as digital or analogue: When you speak, your voice creates an analogue sound wave (a continuous wave). But when you record a song on your phone or send a voice note on WhatsApp, your phone converts it to a digital signal (a series of 0s and 1s). The phone samples the analogue wave many times per second and stores each sample as a number.

Benefits of Digital Signals Over Analogue

  1. Increased data rate -- digital signals can carry much more information per second. That is why 4G streaming works better than old FM radio for music
  2. Increased range -- digital signals can travel further because they can be regenerated (cleaned up) at relay stations without losing quality
  3. Accurate regeneration -- this is the big one. When an analogue signal picks up noise (static, interference), the noise becomes part of the signal and you cannot remove it. But with digital, since the signal is just 0s and 1s, a relay station can read the noisy signal, figure out whether each pulse was meant to be a 0 or a 1, and regenerate a perfect clean copy. It is like photocopying a photocopy versus retyping a text message -- the retyped version stays perfect.
⚠ Exam Tip

The key word for digital signal advantage is "regeneration" not "amplification". Analogue signals can be amplified too, but amplification makes the noise louder as well. Digital signals can be regenerated -- meaning the noise is stripped away and a clean signal is created. Use the word "regenerated" in your exam answer for full marks.

🔎 Apply It: Real-World Physics
Cambridge examiners LOVE testing familiar concepts in unfamiliar situations. Can you spot the physics hiding in these real-world scenarios? Tap each one to reveal the answer.
1
At Bangalore airport security, you place your bag on a conveyor belt and it passes through an X-ray scanner. The screen shows the contents of your bag in different colours -- metals appear blue, organic materials (food, clothes) appear orange, and very dense items appear dark.
Why are X-rays used instead of visible light for security scanning? What property of X-rays makes them suitable for this task but also potentially dangerous?
Identify the Physics
Properties of X-rays from the EM spectrum: short wavelength, high frequency, high energy. X-rays can penetrate soft materials but are absorbed by dense materials like metal and bone.
Work It Out
Visible light cannot pass through the bag material -- it would just show you the outside of the bag. X-rays have very short wavelengths (about 10⁻¹⁰ m) and high energy, which allows them to penetrate soft materials like fabric, plastic, and food.

Different materials absorb different amounts of X-rays. Dense metals absorb more (appear darker/blue), while organic materials absorb less (appear lighter/orange). This creates a contrast image of the bag contents.

The same penetrating ability that makes X-rays useful also makes them dangerous: they can penetrate human tissue and damage or kill living cells. This can cause mutations leading to cancer. That is why the scanner has lead-lined walls (lead absorbs X-rays) and the operator sits behind a protective screen.
💡 The Aha! Moment
The same property that makes an EM wave useful often makes it dangerous. X-rays penetrate -- useful for seeing inside bags and bodies, but dangerous because they also penetrate and damage living cells. In the exam, always link the USE to the DANGER through the same physical property.
2
ISRO's Mars Orbiter Mission (Mangalyaan) communicates with Earth from Mars, a distance of about 225 million km. The communication uses radio waves, which travel at the speed of light (3 x 10⁸ m/s).
Calculate the time delay for a signal sent from Mars to reach ISRO's ground station in Bangalore. Why are radio waves used for deep space communication instead of, say, visible light?
Identify the Physics
All electromagnetic waves travel at the speed of light in a vacuum: v = 3 x 10⁸ m/s. Time = distance / speed. Radio wave properties from the EM spectrum.
Work It Out
Distance = 225 million km = 225 x 10⁶ x 10³ m = 2.25 x 10¹¹ m.
Time = distance / speed = 2.25 x 10¹¹ / 3 x 10⁸ = 750 seconds = 12.5 minutes.

So when ISRO sends a command from Bangalore, it takes 12.5 minutes to reach Mangalyaan, and 12.5 minutes for the response to come back -- a 25-minute round trip.

Radio waves are used because they have the longest wavelength in the EM spectrum, which means they diffract the most around obstacles and can spread out over large distances. They also pass through Earth's atmosphere with minimal absorption, unlike infrared or UV which get absorbed by gases in the atmosphere.
💡 The Aha! Moment
All EM waves travel at the same speed (3 x 10⁸ m/s) in a vacuum, but they are chosen for different jobs based on their wavelength. Radio waves are best for space communication because they diffract well and pass through the atmosphere. This question combines the wave speed equation with EM spectrum knowledge -- a common exam pattern.
3
Your microwave oven at home heats food using microwaves at a frequency of 2.45 GHz (2.45 x 10⁹ Hz). All electromagnetic waves travel at 3 x 10⁸ m/s. The metal mesh on the microwave door has tiny holes about 1 mm across.
Calculate the wavelength of the microwaves. Why can you see your food through the mesh but the microwaves cannot escape?
Identify the Physics
Wave speed equation: v = f x λ. Diffraction depends on the relationship between wavelength and gap size.
Work It Out
λ = v / f = (3 x 10⁸) / (2.45 x 10⁹) = 0.122 m = 12.2 cm.

The microwave wavelength is 12.2 cm, but the holes in the mesh are only about 1 mm (0.1 cm) across. The holes are much smaller than the wavelength, so the microwaves cannot pass through -- they are reflected by the metal mesh.

Visible light has a wavelength of about 400-700 nm (0.0004-0.0007 mm), which is thousands of times smaller than the 1 mm holes. The holes are much larger than the wavelength of visible light, so light passes straight through without significant diffraction. You can see your food, but the microwaves stay inside.
💡 The Aha! Moment
The mesh acts as a filter based on wavelength. Waves with wavelengths much larger than the gap cannot pass through. Waves with wavelengths much smaller than the gap pass through easily. This is why the same mesh blocks microwaves (λ = 12 cm) but lets visible light through (λ = 0.0005 mm). It is diffraction and EM spectrum knowledge combined in one beautiful real-world example.
4
During the monsoon in Bangalore, the Municipal Corporation uses UV-C lamps to sterilise drinking water in treatment plants. The UV-C light has a wavelength of about 254 nm (254 x 10⁻⁹ m) and kills bacteria and viruses by damaging their DNA.
Calculate the frequency of the UV-C light. Why is UV-C effective at killing microorganisms while visible light is not? Where does UV-C sit in the EM spectrum relative to visible light?
Identify the Physics
Wave speed equation: f = v / λ. Position of UV in the EM spectrum and its properties -- higher frequency (and therefore higher energy) than visible light.
Work It Out
f = v / λ = (3 x 10⁸) / (254 x 10⁻⁹) = 1.18 x 10¹⁵ Hz.

UV-C sits just beyond the violet end of the visible spectrum -- it has a shorter wavelength and higher frequency than visible light. Higher frequency means higher energy per photon.

This higher energy is enough to break chemical bonds in DNA molecules, killing bacteria and viruses. Visible light has lower energy photons that cannot break these bonds -- the light passes through or is absorbed without causing significant damage to DNA.

This is also why UV from the sun causes sunburn -- it damages skin cell DNA. UV-C is the most energetic type of UV and is the most effective at sterilisation.
💡 The Aha! Moment
Moving from radio waves to gamma rays across the EM spectrum: wavelength decreases, frequency increases, and energy increases. UV has more energy than visible light, which is why it can damage cells. This "higher frequency = higher energy = more dangerous" pattern applies across the entire spectrum and is tested constantly in exams.
5
Food irradiation plants in India use gamma rays from Cobalt-60 sources to preserve mangoes and onions for export. The gamma rays kill bacteria, mould, and insects inside the food without making the food radioactive. The TV remote in your living room uses infrared to change channels.
Both gamma rays and infrared are electromagnetic waves. They both travel at 3 x 10⁸ m/s. So why can gamma rays kill bacteria while infrared cannot? What is fundamentally different about them?
Identify the Physics
All EM waves travel at the same speed in a vacuum, but they differ in frequency and wavelength. Energy is proportional to frequency.
Work It Out
Although gamma rays and infrared both travel at 3 x 10⁸ m/s, they have very different frequencies:

Gamma rays: frequency ≈ 10₂⁰ Hz, wavelength ≈ 10⁻¹² m
Infrared: frequency ≈ 10¹² Hz, wavelength ≈ 10⁻⁴ m

Gamma rays have a frequency about 100 million times higher than infrared. Since energy is proportional to frequency, each gamma ray photon carries enormously more energy than an infrared photon.

This energy is enough to ionise atoms (knock electrons off), break molecular bonds, and destroy DNA -- killing bacteria. Infrared photons only have enough energy to make molecules vibrate (causing heating), which is why your remote control is harmless and why infrared is used in heaters, but gamma rays require thick lead or concrete shielding.
💡 The Aha! Moment
Same speed does NOT mean same energy. The key difference between EM waves is their frequency (and wavelength). Higher frequency = more energy per photon = more damage to living cells. This is the single most important idea in the EM spectrum topic. In the exam, never say "gamma rays are stronger" -- say "gamma rays have higher frequency and therefore higher energy."

Practice Questions -- 3.3 Electromagnetic Spectrum

1. Which type of electromagnetic wave has the longest wavelength?
A) Gamma rays
B) X-rays
C) Microwaves
D) Radio waves
Answer: D -- Radio waves have the longest wavelength in the EM spectrum. Remember the order: Radio, Microwaves, Infrared, Visible, UV, X-rays, Gamma -- wavelength decreases along this list.
2. All electromagnetic waves travel at the same speed in a vacuum. This speed is approximately:
A) 3.0 × 10⁶ m/s
B) 3.0 × 10⁷ m/s
C) 3.0 × 10⁸ m/s
D) 3.0 × 10¹⁰ m/s
Answer: C -- All EM waves travel at 3.0 × 10⁸ m/s in a vacuum. This is the speed of light. Be careful with the power of 10 -- it is 10 to the power 8.
3. Your TV remote control uses which type of electromagnetic wave?
A) Infrared
B) Microwaves
C) Ultraviolet
D) Radio waves
Answer: A -- TV remote controls use infrared radiation. Fun fact: you can see the IR LED flash if you point the remote at your phone camera and press a button!
4. Which EM wave is used in a UV water purifier like Kent or Aquaguard to kill bacteria?
A) Infrared
B) Ultraviolet
C) Visible light
D) X-rays
Answer: B -- Ultraviolet (UV) radiation is used to sterilise water. UV light damages the DNA of bacteria and viruses, killing them and making the water safe to drink.
5. A geostationary satellite orbits at a height of approximately:
A) 200 km
B) 2,000 km
C) 36,000 km
D) 360,000 km
Answer: C -- Geostationary satellites orbit at approximately 36,000 km above the equator. At this height, they orbit at exactly the same rate as Earth rotates, so they appear stationary above one point.
6. Excessive exposure to ultraviolet radiation can cause:
A) Internal heating of body tissue
B) Cell mutation only
C) Skin burns only
D) Skin cancer and eye damage
Answer: D -- UV radiation causes skin cancer and eye damage. Internal heating is caused by microwaves, and skin burns by infrared. Cell mutation is caused by X-rays and gamma rays.
7. Which type of EM wave is used for baggage scanning at Namma Metro stations?
A) Gamma rays
B) Ultraviolet
C) X-rays
D) Microwaves
Answer: C -- X-rays are used in security scanners to see inside bags. X-rays pass through soft materials like clothes but are absorbed by denser materials like metals, creating a shadow image.
8. What property is the SAME for all electromagnetic waves travelling in a vacuum?
A) Wavelength
B) Speed
C) Frequency
D) Energy
Answer: B -- All EM waves travel at the same speed in a vacuum (3.0 × 10⁸ m/s). They differ in wavelength, frequency, and energy.
9. Mobile phones use microwaves rather than radio waves because:
A) Microwaves travel faster than radio waves
B) Microwaves can penetrate walls and have short enough wavelengths for a small aerial
C) Microwaves carry less energy
D) Microwaves are less harmful than radio waves
Answer: B -- Microwaves penetrate walls (so signals work indoors) and their short wavelength means the phone's aerial can be very small. All EM waves travel at the same speed in vacuum, so A is wrong.
10. Which EM wave is used for thermal imaging, like the temperature scanners used at Bangalore airport during COVID?
A) Infrared
B) Visible light
C) Ultraviolet
D) Microwaves
Answer: A -- All warm objects emit infrared radiation. Thermal imaging cameras detect this IR to create a heat map. Hotter objects emit more IR, so a person with a fever appears brighter.
11. Gamma rays are used to treat cancer because:
A) They have the longest wavelength
B) They cannot damage healthy cells
C) They travel slower than other EM waves
D) They are highly energetic and can destroy cancer cells
Answer: D -- Gamma rays have the highest energy of all EM waves, which allows them to destroy cancer cells. They CAN damage healthy cells too (B is wrong), which is why treatment is carefully targeted.
12. A key advantage of a geostationary satellite over a low orbit satellite is that:
A) It produces stronger signals
B) It appears stationary relative to the Earth's surface so a dish does not need to track it
C) It orbits closer to Earth
D) It has no time delay for signals
Answer: B -- Geostationary satellites orbit at the same angular speed as Earth's rotation, so they appear fixed in the sky. This means your Tata Sky dish can be pointed at one position permanently. Low orbit satellites are closer (stronger signals) but move across the sky.
13. The harmful effect of microwaves on the human body is:
A) Skin cancer
B) Internal heating of body tissue
C) Skin burns
D) Eye damage
Answer: B -- Microwaves cause internal heating of body tissue because they can penetrate the skin and heat water in cells. Skin cancer is UV, skin burns is infrared, eye damage is UV.
14. Optical fibres carry data using:
A) Radio waves
B) Ultraviolet
C) Microwaves
D) Visible light or infrared
Answer: D -- Optical fibres use visible light or infrared, which travel along the fibre by total internal reflection. This gives very high data rates, which is why fibre broadband (like Jio Fiber) is much faster than 4G.
15. As you move from radio waves to gamma rays across the EM spectrum, what happens to frequency and wavelength?
A) Both increase
B) Frequency increases, wavelength decreases
C) Both decrease
D) Frequency decreases, wavelength increases
Answer: B -- From radio to gamma: frequency increases and wavelength decreases. They are inversely related by v = f × λ. Since speed is constant (in vacuum), if f goes up, λ must go down.
16. The main advantage of digital signals over analogue signals is:
A) Digital signals travel faster
B) Digital signals do not pick up noise
C) Digital signals use less energy
D) Digital signals can be regenerated accurately, removing noise
Answer: D -- Digital signals CAN pick up noise (B is wrong), but the key advantage is that they can be regenerated -- the noise is stripped away and a clean copy is recreated. Analogue signals can only be amplified, which amplifies the noise too.
17. A radio station broadcasts at a frequency of 100 MHz. The wavelength of the radio wave is:
A) 3.0 m
B) 30 m
C) 300 m
D) 0.3 m
Answer: A -- λ = v / f = (3.0 × 10⁸) / (100 × 10⁶) = (3.0 × 10⁸) / (10⁸) = 3.0 m. Remember to convert MHz to Hz first!
18. FASTag on cars uses which technology that relies on radio waves?
A) Bluetooth
B) Wi-Fi
C) RFID (Radio-Frequency Identification)
D) GPS
Answer: C -- RFID uses radio waves to identify and track tags attached to objects. FASTag is an RFID-based electronic toll collection system used on Indian highways.
19. Bluetooth devices (like wireless earbuds) use which type of EM wave?
A) Infrared
B) Microwaves
C) Visible light
D) Radio waves
Answer: D -- Bluetooth uses short-range radio waves. This allows it to work through walls (unlike infrared, which needs line of sight). It operates over short distances, typically up to about 10 metres.
20. Banks check whether a ₹500 note is genuine by placing it under:
A) Infrared light
B) X-ray scanner
C) Ultraviolet light
D) Microwave detector
Answer: C -- Genuine bank notes have special security markings that glow under ultraviolet light. Counterfeit notes lack these markings. This is why UV is listed under "detecting fake notes" in the syllabus.
3.4 Sound
Your Score: 0 / 20
Sound Waves - IGCSE Physics Revision
Cognito
Watch
Ultrasound and Echoes - GCSE Physics
Free Science Lessons
Watch
Sound Waves Explained Simply
Science Shorts
Watch

How is Sound Produced?

Tara, sound is everywhere around you in Bangalore -- the honking of auto-rickshaws on MG Road, the announcements on the Namma Metro, your favourite songs playing on Spotify, the school bell, your teacher's voice. But where does all this sound come from?

Here is the fundamental rule: sound is produced by vibrating sources. No vibration = no sound. Every single sound you have ever heard was made by something vibrating.

  • When you speak, your vocal cords vibrate
  • When a cricket bat hits a ball, the bat vibrates (that satisfying "crack" sound!)
  • A tabla produces sound because the membrane vibrates when the musician strikes it
  • A guitar string vibrates when plucked
  • A loudspeaker has a cone that vibrates back and forth
  • Even a honking auto-rickshaw has a vibrating diaphragm inside its horn

You can feel these vibrations! Touch your throat while humming -- you will feel the vibrations of your vocal cords. Touch a speaker playing loud music and you can feel it shaking.

⚠ Exam Tip

If an exam question asks "How is sound produced?", always use the word "vibrating" or "vibrations" in your answer. Simply saying "by hitting something" is NOT enough. Say: "Sound is produced when an object vibrates, causing the air particles around it to vibrate."

Sound is a Longitudinal Wave

This is a crucial point, Tara. Sound is a longitudinal wave. What does this mean?

In a longitudinal wave, the vibrations of the particles are parallel to (along the same direction as) the direction the wave travels. Think of it this way:

Imagine you are standing in a packed BMTC bus during rush hour. Everyone is squished together. Now the bus brakes suddenly. What happens? The person at the front gets pushed forward, bumps into the person ahead, and a wave of pushing travels through the bus from back to front. The people are moving forward and backward (the same direction the wave moves) -- that is a longitudinal wave!

This is exactly how sound travels through air:

  1. A vibrating object (say, a loudspeaker cone) pushes the air particles in front of it forward
  2. These air particles bump into the particles next to them, pushing them forward
  3. This creates a chain reaction -- a wave of pushes travelling outward from the source
  4. The air particles themselves do not travel far -- they just vibrate back and forth around their normal position
Supplement (Extended)

Compressions and Rarefactions

When a longitudinal sound wave passes through air, it creates two alternating regions:

Think of the BMTC bus analogy again:

Compression = a region where air particles are pushed close together (high pressure). It is like when the bus brakes and everyone at the front is squished together -- bodies pressed tight, no space between people.

Rarefaction = a region where air particles are spread far apart (low pressure). It is like when the bus accelerates again and the people at the back suddenly have too much space -- gaps open up between everyone.

COMPRESSIONS AND RAREFACTIONS
|||||||       |||||||       |||||||       |||||||
C R C R C R C
C = Compression (particles close together)    R = Rarefaction (particles spread apart)
←——— one wavelength (λ) ———→
One wavelength = distance from one compression to the next

A sound wave is a series of compressions and rarefactions travelling outward from the vibrating source, like ripples spreading out -- except instead of going up and down (like water waves), the air particles push back and forth.

🧠 Memory Trick

Compression = Crowded (particles close together, like the crowd in a packed BMTC bus at Majestic). Rarefaction = Room (particles have room to spread out, like an empty bus at midnight). Both start with the same letter!

The Audible Range of Human Hearing

Humans can only hear sounds within a certain range of frequencies:

Audible range: 20 Hz to 20,000 Hz (20 kHz)
Below 20 Hz = infrasound (too low for humans to hear) Above 20,000 Hz = ultrasound (too high for humans to hear)

Some fun facts:

  • A bass guitar produces sounds as low as about 40 Hz
  • The lowest note on a tabla is around 50-80 Hz
  • A normal conversation is around 300-3000 Hz
  • A cricket's chirping (the insect, not the sport!) can be around 3000-8000 Hz
  • Dogs can hear up to about 45,000 Hz -- that is why a dog whistle works even though you cannot hear it
  • Bats can hear up to 100,000 Hz and use ultrasound for navigation
  • As people get older, they lose the ability to hear higher frequencies. Your parents probably cannot hear sounds above 15,000 Hz that you can!

Sound Needs a Medium

This is one of the biggest differences between sound and electromagnetic waves. Sound cannot travel through a vacuum. It MUST have a medium (solid, liquid, or gas) to travel through.

Why? Because sound works by particles bumping into each other. In a vacuum, there are no particles, so there is nothing to bump. It is like trying to pass a message in a game of Chinese Whispers (telephone game) with no people in the chain -- impossible!

That is why in space, which is mostly a vacuum, there is complete silence. In sci-fi movies when spaceships explode with a big "BOOM" -- that is actually wrong! There would be no sound at all in the vacuum of space.

The classic experiment to demonstrate this uses a bell jar:

  1. Place an electric bell inside a glass bell jar
  2. Switch on the bell -- you can hear it ringing
  3. Use a vacuum pump to remove the air from the bell jar
  4. As air is removed, the sound gets quieter and quieter
  5. When most of the air is removed, you can see the bell vibrating but you can barely hear it
  6. You cannot get a perfect vacuum in a school lab, but the demonstration clearly shows that removing the medium reduces the sound
⚠ Exam Tip

Be careful: do NOT say "you hear nothing" in the bell jar experiment. In a school lab, you cannot achieve a perfect vacuum, so you would still hear a very faint sound. Say "the sound becomes very faint" or "almost inaudible" rather than "the sound disappears completely".

Supplement (Extended)

Speed of Sound in Different Media

Sound travels at different speeds in different materials. The key rule is:

Sound is fastest in solids > faster in liquids > slowest in gases

Why? In solids, particles are closest together and have the strongest forces between them, so vibrations pass quickly from one particle to the next. Think of it like a line of people standing shoulder to shoulder passing a nudge along -- it travels really fast because there is no gap. In gases, particles are far apart, so each one has to travel further before bumping into the next one -- like people standing far apart in a line, the nudge takes longer to travel.

Medium Approximate Speed Example
Air (gas) ~330-350 m/s Normal sound around you
Water (liquid) ~1500 m/s Sound underwater in a swimming pool
Steel (solid) ~6000 m/s Sound travelling along railway tracks

This is why you can hear an approaching Indian Railways train by putting your ear to the steel rail long before you can hear it through the air! The sound travels through the solid steel much faster than through the air. (Please do NOT actually try this on a real railway track -- it is extremely dangerous!)

Speed of Sound in Air

Speed of sound in air ≈ 330 - 350 m/s
Use 340 m/s as a good average unless told otherwise This varies slightly with temperature and humidity

For comparison, this is about:

  • About 1,224 km/h (faster than any car, but slower than a passenger aircraft)
  • About 1 km every 3 seconds -- this is useful for estimating how far away a thunderstorm is!
🧠 Memory Trick

Thunder trick: During Bangalore's monsoon thunderstorms, count the seconds between seeing the lightning and hearing the thunder. Divide by 3 to get the distance in km. See lightning, count "1...2...3...4...5...6" then hear thunder? That is 6 ÷ 3 = 2 km away. This works because light is almost instant (3 × 10⁸ m/s) while sound takes about 3 seconds per km.

Experiment: Measuring the Speed of Sound

The IGCSE exam often asks about this experiment. Here is a simple method:

Method 1: Two observers with a large distance

  1. Measure a large distance (at least 100 m) between two points -- you could do this on your school's cricket ground or running track
  2. One person at point A claps two wooden blocks together (or bangs a drum)
  3. A second person at point B starts a stopwatch when they see the clap and stops it when they hear the clap
  4. This works because light travels almost instantly, but sound takes a noticeable time
  5. Repeat several times and take an average to reduce random errors
  6. Calculate: speed = distance / time
v = d / t
v = speed of sound (m/s) d = distance between the two points (m) t = time measured (s)
Worked Example Tara stands at one end of her school cricket ground, 150 m from her friend. She claps two blocks together. Her friend measures a time delay of 0.44 s between seeing and hearing the clap. Calculate the speed of sound.
Step 1: Write down what you know
d = 150 m
t = 0.44 s
Step 2: Use v = d / t
v = 150 / 0.44
v = 341 m/s
✅ Answer: The speed of sound is approximately 341 m/s, which is within the expected range of 330-350 m/s. Looks like a good measurement!
Worked Example A student measures the speed of sound using a 200 m distance. The time measured is 0.59 s. Calculate the speed of sound and suggest one way to improve the accuracy of this experiment.
Step 1: Calculate speed
v = d / t = 200 / 0.59 = 339 m/s
Step 2: Suggest improvement
Repeat the experiment several times and calculate an average time. This reduces random errors caused by human reaction time when starting and stopping the stopwatch.
✅ Answer: Speed = 339 m/s. Improvement: Repeat measurements and take an average to reduce the effect of human reaction time errors.
⚠ Exam Tip

Common exam improvements for this experiment: (1) Use a larger distance to make the time interval bigger and easier to measure accurately. (2) Repeat and average to reduce random errors from reaction time. (3) Use an electronic timer with microphones at each end instead of a stopwatch to remove human reaction time completely.

Amplitude, Loudness, Frequency, and Pitch

Two important relationships you need to know:

Wave Property What You Hear Example
Amplitude (height of wave) Loudness A drummer hitting the tabla harder makes a louder sound (bigger amplitude). Whispering vs shouting.
Frequency (vibrations per second) Pitch A veena's thin string vibrates fast = high frequency = high pitch. The thick string vibrates slowly = low frequency = low pitch.
  • Bigger amplitude = louder sound (more energy). Think of the difference between tapping a tabla gently vs hitting it hard
  • Higher frequency = higher pitch. Think of a baby's cry (high pitch, high frequency) vs a man's deep voice (low pitch, low frequency)
  • These are independent -- you can have a loud, low-pitched sound (like a bass drum) or a quiet, high-pitched sound (like a whistle far away)
🧠 Memory Trick

Amplitude = Awaz (volume/loudness in Hindi). Frequency = how Fast the pitch goes (higher frequency = higher pitch). Two simple letter associations!

Echo: Reflection of Sound

An echo is simply the reflection of sound. When you shout "HELLO!" towards a large building or cliff, you hear "HELLO!" coming back to you a moment later. That is an echo.

How it works:

  1. You produce a sound (shout)
  2. The sound wave travels through the air towards a large, hard surface (a wall, building, cliff)
  3. The sound wave reflects off the surface
  4. The reflected wave travels back to your ears
  5. You hear the original sound again -- that is the echo

For an echo to be heard distinctly, the reflecting surface needs to be at least about 17 metres away. This is because your ears need about 0.1 seconds between the original sound and the echo to hear them as separate sounds. At 340 m/s, sound travels 34 m in 0.1 s, but it has to go there AND back (34 / 2 = 17 m).

Try this: stand in front of a large building (like a wall of your school or a big apartment complex) and clap. If you are far enough away, you will hear the clap echo back!

⚠ Exam Tip

Echo questions often involve calculating distances. Remember that the sound travels to the reflecting surface AND back again, so the total distance = 2 × distance to the wall. A very common mistake is forgetting to divide by 2!

Worked Example Tara stands in front of a large building in Cubbon Park. She claps her hands and hears the echo 0.8 seconds later. If the speed of sound is 340 m/s, how far is she from the building?
Step 1: Calculate total distance travelled by sound
total distance = speed × time = 340 × 0.8 = 272 m
Step 2: The sound travels to the building AND back
distance to building = total distance / 2 = 272 / 2 = 136 m
✅ Answer: Tara is 136 m from the building. Remember: always divide by 2 because the sound makes a round trip!

Ultrasound

Ultrasound is any sound with a frequency greater than 20,000 Hz (20 kHz) -- above the upper limit of human hearing. You cannot hear ultrasound, but it has many useful applications.

Supplement (Extended)

Uses of Ultrasound

1. Non-Destructive Testing (Checking for cracks)

Ultrasound pulses are sent into metal structures like bridge supports, railway tracks, or aircraft parts. If there is a crack inside the metal, the ultrasound reflects off the crack and comes back early. By analysing the reflections, engineers can find hidden defects without cutting the metal open. Indian Railways uses this to check for cracks in rails!

2. Medical Scanning (Ultrasound imaging)

This is the most well-known use. Ultrasound is used to create images of babies in the womb (prenatal scans). It is also used to examine kidneys, the liver, and other internal organs. Ultrasound is preferred over X-rays for this because:

  • It is safe -- no ionising radiation, so it does not damage cells
  • It can show soft tissue clearly (X-rays are better for bones)
  • It can create real-time moving images

3. Sonar (Sound Navigation and Ranging)

Ships and submarines use sonar to measure ocean depth or find objects underwater. A pulse of ultrasound is sent downward, it reflects off the seabed (or a submarine, or a school of fish), and the echo returns. The time taken for the echo to return tells you the distance.

The Indian Navy uses sonar extensively for submarine detection, and fishing boats in coastal Karnataka use it to locate fish!

d = (v × t) / 2
d = depth or distance to object (m) v = speed of sound in the medium (m/s) t = total time for pulse to go and return (s) Divide by 2 because sound travels there AND back
Worked Example A fishing boat off the coast of Mangalore sends an ultrasound pulse towards the seabed. The echo returns after 0.4 seconds. The speed of sound in seawater is 1500 m/s. Calculate the depth of the sea at that point.
Step 1: Write down what you know
v = 1500 m/s
t = 0.4 s
Step 2: Use d = (v × t) / 2
d = (1500 × 0.4) / 2
d = 600 / 2
d = 300 m
✅ Answer: The depth of the sea is 300 m. We divide by 2 because the ultrasound pulse travels down to the seabed and then back up again.
Worked Example An Indian Navy ship uses sonar to detect a submarine. An ultrasound pulse is transmitted and the echo is received 3.2 seconds later. The speed of sound in seawater is 1500 m/s. How far away is the submarine?
Step 1: Write down what you know
v = 1500 m/s
t = 3.2 s
Step 2: Use d = (v × t) / 2
d = (1500 × 3.2) / 2
d = 4800 / 2
d = 2400 m
d = 2.4 km
✅ Answer: The submarine is 2400 m (2.4 km) away. Again, we divide by 2 because the pulse travels to the submarine and back.
Worked Example An engineer testing a steel beam sends an ultrasound pulse into the beam. The beam is 0.5 m thick and the speed of sound in steel is 6000 m/s. The pulse reflects from a crack inside the beam and returns in 0.00012 s. How deep inside the beam is the crack?
Step 1: Write down what you know
v = 6000 m/s
t = 0.00012 s
Step 2: Use d = (v × t) / 2
d = (6000 × 0.00012) / 2
d = 0.72 / 2
d = 0.36 m
✅ Answer: The crack is 0.36 m (36 cm) deep inside the 0.5 m beam. The engineer found the crack without cutting the beam open!

The Wave Equation Applied to Sound

The wave equation works for sound waves too:

v = f × λ
v = speed of sound (m/s) -- about 340 m/s in air f = frequency (Hz) λ = wavelength (m)
Worked Example A tuning fork vibrates at 256 Hz (middle C on a piano). If the speed of sound in air is 340 m/s, what is the wavelength of the sound wave produced?
Step 1: Write down what you know
f = 256 Hz
v = 340 m/s
Step 2: Rearrange v = f × λ to find λ
λ = v / f = 340 / 256 = 1.33 m
✅ Answer: The wavelength is 1.33 m -- about the height of a 10-year-old child. Sound waves can be surprisingly long!
Worked Example The highest note a flute can play has a wavelength of 0.034 m. If the speed of sound is 340 m/s, calculate the frequency. Can a human hear this note?
Step 1: Rearrange to find frequency
f = v / λ = 340 / 0.034 = 10,000 Hz
Step 2: Check audible range
The human audible range is 20 Hz to 20,000 Hz. 10,000 Hz is within this range.
✅ Answer: Frequency = 10,000 Hz. Yes, a human can hear this because it is within the audible range of 20 Hz to 20,000 Hz.
⚠ Exam Tip

When a question asks "Can a human hear this sound?" -- you need to check TWO things: (1) Is the frequency between 20 Hz and 20,000 Hz? (2) Is the sound travelling through a medium (not a vacuum)? Both conditions must be met for a human to hear the sound.

🧠 Memory Trick

For the d = vt/2 sonar equation, think: "Sound goes on a RETURN trip". Like an auto-rickshaw going to the market and coming back -- the total distance on the meter is double the actual distance to the market. So you divide by 2 to get the one-way distance.

🔎 Apply It: Real-World Physics
Cambridge examiners LOVE testing familiar concepts in unfamiliar situations. Can you spot the physics hiding in these real-world scenarios? Tap each one to reveal the answer.
1
During Diwali in Bangalore, you see a firework explode in the sky 680 metres away. You see the bright flash instantly, but the loud BOOM arrives 2 seconds later. The speed of sound in air is 340 m/s.
Verify the distance to the firework using the time delay. Why do you see the flash before hearing the sound?
Identify the Physics
Speed of sound in air (340 m/s) vs. speed of light (3 x 10⁸ m/s). Distance = speed x time.
Work It Out
Distance = speed of sound x time delay = 340 x 2 = 680 m. ✔ This matches.

Light travels at 3 x 10⁸ m/s, so the flash takes 680 / (3 x 10⁸) = 0.0000023 seconds to reach you -- essentially instant.

Sound travels at only 340 m/s, so the boom takes 680 / 340 = 2 seconds to reach you.

You see the flash before hearing the boom because light travels approximately 1 million times faster than sound. The light arrives almost instantly while the sound takes a noticeable time to cover the same distance.
💡 The Aha! Moment
This is the same principle behind counting the gap between lightning and thunder during monsoon storms. Every 3 seconds of delay means the lightning is about 1 km away (340 m/s x 3 s ≈ 1000 m). The exam loves this type of question -- always use the speed of sound (not light) for the time delay calculation, since light arrives "instantly."
2
A sitar player at a classical music concert in Bangalore plucks a string. She then presses the string against a fret to make the vibrating part of the string shorter, and plucks again. The second note sounds higher-pitched than the first.
Explain why shortening the vibrating length of the string produces a higher pitch. What happens to the frequency and wavelength?
Identify the Physics
Pitch is determined by frequency. The vibrating string produces a standing wave. Shorter string = shorter wavelength = higher frequency (since wave speed on the string stays roughly constant).
Work It Out
When the string vibrates, it forms a standing wave with the wavelength determined by the string length. For the fundamental mode, the wavelength is approximately twice the string length: λ ≈ 2L.

When she presses the fret, the vibrating length L gets shorter, so the wavelength λ also gets shorter.

The wave speed v on the string depends on the string's tension and thickness, which have not changed. Since v = f x λ, and v is constant but λ has decreased, the frequency f must increase.

Higher frequency = higher pitch. That is why the second note sounds higher.

For example, if the full string length is 90 cm and she halves it to 45 cm: the wavelength halves, so the frequency doubles -- producing a note one octave higher.
💡 The Aha! Moment
The sitar, guitar, veena -- they all work the same way. Shorter vibrating length → shorter wavelength → higher frequency → higher pitch. This is v = fλ in action with v held constant. If v is constant, frequency and wavelength are inversely proportional. The exam may describe an unfamiliar instrument -- just apply the same logic.
3
A fishing boat off the coast of Chennai uses sonar to find a school of fish. It sends an ultrasound pulse downward and receives an echo after 0.08 seconds. The speed of sound in seawater is 1500 m/s. The sonar uses a frequency of 50 kHz.
Calculate the depth of the fish. Why is ultrasound used instead of audible sound? Why would this not work in air?
Identify the Physics
Echo/sonar equation: depth = (speed x time) / 2. The division by 2 is because the sound travels down AND back up. Properties of ultrasound (frequency above 20,000 Hz).
Work It Out
Total distance travelled by sound = speed x time = 1500 x 0.08 = 120 m.
But the sound went down to the fish and back up, so: depth = 120 / 2 = 60 m.

Ultrasound (frequency above 20 kHz) is used instead of audible sound because:
1. It can be formed into a narrow, focused beam (less spreading), giving a more precise location
2. It does not disturb the fish or the crew (humans cannot hear it)
3. It has short wavelengths which reflect well off small objects like fish

This would not work well in air because the speed of sound in air is only 340 m/s (much slower) and, more importantly, ultrasound is heavily absorbed by air over short distances. Water is a much better medium for transmitting sound over long distances because the molecules are closer together.
💡 The Aha! Moment
The d = vt/2 formula is one of the most commonly examined calculations in the sound topic. ALWAYS remember to divide by 2 -- the most common mistake is forgetting that the sound makes a return trip. Exam questions may use sonar on ships, bats finding insects, or ultrasound measuring the depth of a crack in a metal pipe -- the physics is identical every time.
4
During a pregnancy scan at a hospital in Bangalore, a doctor places an ultrasound probe on the mother's abdomen. The probe emits ultrasound pulses at 3.5 MHz (3.5 x 10⁶ Hz) and receives reflections from the baby. The speed of sound in body tissue is approximately 1540 m/s.
Calculate the wavelength of the ultrasound inside the body. Why is ultrasound used for baby scans instead of X-rays? Could you use audible sound instead?
Identify the Physics
Wave speed equation: v = f x λ, rearranged to λ = v / f. Safety comparison between ultrasound and X-rays. Properties of ultrasound.
Work It Out
λ = v / f = 1540 / (3.5 x 10⁶) = 4.4 x 10⁻⁴ m = 0.44 mm.

This tiny wavelength means the ultrasound can detect details as small as about 0.44 mm, giving a detailed image of the baby.

Ultrasound is used instead of X-rays because:
1. X-rays are ionising radiation -- they can damage the baby's developing cells and DNA, potentially causing mutations or birth defects
2. Ultrasound is non-ionising -- it is just high-frequency sound waves that reflect off boundaries between different tissues. It causes no known harm to the mother or baby

Audible sound (20 Hz - 20 kHz) would not work because it has much longer wavelengths (e.g., at 20 kHz: λ = 1540/20000 = 0.077 m = 7.7 cm). This wavelength is far too large to detect the fine details of a developing baby -- you need a wavelength comparable to or smaller than the features you want to image.
💡 The Aha! Moment
Ultrasound scans are safe because sound waves (even high-frequency ones) are not ionising -- they cannot damage DNA. X-rays ARE ionising and dangerous to developing babies. This is a classic exam question: "Why ultrasound and not X-rays for prenatal scans?" Always answer: ultrasound is non-ionising so it does not harm the baby's cells.
5
You are standing between two large buildings in Bangalore's central business district, 85 metres from one building and 170 metres from the other. You clap your hands once. The speed of sound in air is 340 m/s. To hear a distinct echo, the reflected sound must arrive at least 0.1 seconds after the original sound.
Will you hear one echo or two separate echoes? Calculate the time for each echo to arrive.
Identify the Physics
Echo calculation: time = (2 x distance to wall) / speed of sound. The factor of 2 accounts for the sound travelling to the wall and back. Minimum time for a distinct echo is 0.1 s.
Work It Out
Echo from the closer building (85 m away):
Time = (2 x 85) / 340 = 170 / 340 = 0.5 seconds. This is greater than 0.1 s, so you hear a distinct echo. ✔

Echo from the farther building (170 m away):
Time = (2 x 170) / 340 = 340 / 340 = 1.0 second. This is also greater than 0.1 s, so you hear this echo too. ✔

The two echoes arrive at different times (0.5 s and 1.0 s), so you will hear two separate, distinct echoes -- the first from the closer building, then the second from the farther building 0.5 seconds later.

If you were only 10 m from a wall: time = (2 x 10) / 340 = 0.059 s. This is less than 0.1 s -- you would NOT hear a distinct echo. The reflected sound would merge with the original sound (this is called reverberation).
💡 The Aha! Moment
The minimum distance for a distinct echo is about 17 metres (giving 0.1 s round trip). Closer than that, the echo blends with the original sound. The exam may ask you to calculate the minimum distance for an echo: d = (speed x minimum time) / 2 = (340 x 0.1) / 2 = 17 m. Always show the "divide by 2" step.

Practice Questions -- 3.4 Sound

1. Sound is produced by:
A) Moving objects
B) Heating objects
C) Vibrating objects
D) Rotating objects
Answer: C -- Sound is always produced by vibrating sources. When a tabla membrane vibrates, a guitar string vibrates, or your vocal cords vibrate, sound is produced.
2. Sound is what type of wave?
A) Transverse
B) Longitudinal
C) Electromagnetic
D) Surface
Answer: B -- Sound is a longitudinal wave. The particles vibrate parallel to (in the same direction as) the direction the wave travels. This is different from transverse waves where vibrations are perpendicular.
3. The audible range of human hearing is:
A) 20 Hz to 20,000 Hz
B) 200 Hz to 20,000 Hz
C) 20 Hz to 2,000 Hz
D) 2 Hz to 200,000 Hz
Answer: A -- Humans can hear frequencies from 20 Hz (very low bass) to 20,000 Hz (very high pitch). Below 20 Hz is infrasound, above 20,000 Hz is ultrasound.
4. Sound cannot travel through:
A) Air
B) Water
C) Steel
D) A vacuum
Answer: D -- Sound needs a medium (solid, liquid, or gas) to travel. In a vacuum, there are no particles to vibrate, so sound cannot travel. This is why there is no sound in outer space.
5. The speed of sound in air is approximately:
A) 3 m/s
B) 340 m/s
C) 33 m/s
D) 3400 m/s
Answer: B -- The speed of sound in air is approximately 330-350 m/s. A good average to use is 340 m/s. This is about 1,224 km/h.
6. In a sound wave, a compression is a region where air particles are:
A) Close together (high pressure)
B) Far apart (low pressure)
C) Not moving at all
D) Moving at the speed of light
Answer: A -- A compression is where particles are pushed close together, creating high pressure. Think of everyone squished together at the front of a crowded BMTC bus when it brakes. The opposite (particles spread apart) is called a rarefaction.
7. Increasing the amplitude of a sound wave makes the sound:
A) Higher in pitch
B) Louder
C) Lower in pitch
D) Faster
Answer: B -- Amplitude determines loudness. Greater amplitude = louder sound. Pitch is determined by frequency, not amplitude. The speed of sound does not change with amplitude.
8. Increasing the frequency of a sound wave makes the sound:
A) Higher in pitch
B) Quieter
C) Louder
D) Lower in pitch
Answer: A -- Frequency determines pitch. Higher frequency = higher pitch. Think of a baby's cry (high frequency, high pitch) versus a man's deep voice (low frequency, low pitch).
9. An echo is caused by the:
A) Refraction of sound
B) Reflection of sound
C) Diffraction of sound
D) Absorption of sound
Answer: B -- An echo is the reflection of sound off a hard surface. The sound wave bounces off the wall or building and returns to your ears.
10. Ultrasound is sound with a frequency:
A) Below 20 Hz
B) Exactly 20,000 Hz
C) Between 20 Hz and 20,000 Hz
D) Above 20,000 Hz
Answer: D -- Ultrasound has a frequency above 20,000 Hz (20 kHz), which is above the upper limit of human hearing. "Ultra" means "beyond" -- beyond the range of human hearing.
11. Sound travels fastest through:
A) Air
B) Water
C) Steel
D) A vacuum
Answer: C -- Sound travels fastest in solids (like steel, ~6000 m/s), then liquids (water, ~1500 m/s), then gases (air, ~340 m/s). Sound cannot travel through a vacuum at all. In solids, particles are closest together so vibrations pass quickly.
12. A ship sends an ultrasound pulse to the seabed. The echo returns after 0.6 s. If the speed of sound in water is 1500 m/s, the depth of the sea is:
A) 900 m
B) 450 m
C) 225 m
D) 1800 m
Answer: B -- d = (v × t) / 2 = (1500 × 0.6) / 2 = 900 / 2 = 450 m. Remember to divide by 2 because the pulse travels down AND back up. If you forgot to divide, you would get 900 m (option A).
13. A person claps 200 m from a wall and hears an echo after 1.18 s. The speed of sound is:
A) 170 m/s
B) 339 m/s
C) 236 m/s
D) 400 m/s
Answer: B -- Total distance = 2 × 200 = 400 m (there and back). Speed = distance / time = 400 / 1.18 = 339 m/s. This matches the expected speed of sound in air.
14. Which of these is NOT a use of ultrasound?
A) Medical scanning of a baby in the womb
B) Measuring ocean depth using sonar
C) Cooking food in a microwave oven
D) Checking for cracks in metal structures
Answer: C -- Microwave ovens use electromagnetic microwaves, not ultrasound. Ultrasound is used for medical scanning, sonar, and non-destructive testing of metals. Do not confuse EM microwaves with sound waves!
15. During a monsoon thunderstorm in Bangalore, Tara sees lightning and hears the thunder 6 seconds later. If the speed of sound is 340 m/s, approximately how far away is the storm?
A) 340 m
B) 1020 m
C) 2040 m
D) 4080 m
Answer: C -- distance = speed × time = 340 × 6 = 2040 m (about 2 km). Note: we do NOT divide by 2 here because thunder is not an echo -- the sound only travels one way, from the storm to Tara.
16. In a longitudinal sound wave, the particles vibrate:
A) Perpendicular to the wave direction
B) In circular motions
C) They do not vibrate at all
D) Parallel to the wave direction
Answer: D -- In a longitudinal wave, particles vibrate parallel to (along the same direction as) the direction of wave travel. In a transverse wave, they vibrate perpendicular. Sound is longitudinal.
17. Why is ultrasound preferred over X-rays for scanning a baby in the womb?
A) Ultrasound provides sharper images
B) Ultrasound does not cause cell damage, unlike X-rays which are ionising
C) Ultrasound is cheaper
D) Ultrasound travels faster than X-rays
Answer: B -- Ultrasound is safe because it is just high-frequency sound -- it does not ionise atoms or damage DNA. X-rays are ionising radiation that can cause cell mutations, making them dangerous for a developing baby.
18. A sound wave has a frequency of 500 Hz and travels at 340 m/s in air. Its wavelength is:
A) 0.68 m
B) 1.47 m
C) 170,000 m
D) 6.8 m
Answer: A -- λ = v / f = 340 / 500 = 0.68 m. Always check your answer makes sense -- sound wavelengths in air typically range from a few cm to several metres.
19. A rarefaction in a sound wave is a region of:
A) High pressure where particles are close together
B) Low pressure where particles are spread apart
C) No particles at all
D) Maximum amplitude
Answer: B -- A rarefaction is a region of low pressure where particles are spread further apart than normal. Think of it as the opposite of a compression. Remember: Rarefaction = Room (particles have room).
20. In an experiment to measure the speed of sound, a student should:
A) Use a very short distance between the two observers
B) Only take one measurement
C) Start the stopwatch when they hear the sound
D) Use a large distance and take multiple readings to find an average
Answer: D -- A large distance gives a bigger time interval, making it easier to measure accurately. Taking multiple readings and averaging reduces the effect of random errors (like human reaction time). Starting the watch when you SEE the clap (not hear it) is correct since light arrives almost instantly.