Year 11 Physics Module 3: Waves 40 min Lesson 16 of 18

Lenses and Dispersion

Lenses redirect light to form images, while prisms separate white light into colours. Together they explain cameras, spectacles, telescopes, rainbows, and why different wavelengths of light do not all bend by the same amount.

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Think First

Why can a diamond sparkle so strongly, and why does white light split into colours in a prism or rainbow instead of staying as one single beam?

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Key Relationships — This Lesson

$\dfrac{1}{f} = \dfrac{1}{d_o} + \dfrac{1}{d_i}$  |  $m = \dfrac{d_i}{d_o}$  |  $I_1 r_1^2 = I_2 r_2^2$
f = focal length do = object distance di = image distance m = magnification
Convex lens: converging   |   Concave lens: diverging   |   Dispersion: violet bends more than red

L
Formula Reference — Lenses and Light

$1/f = 1/d_o + 1/d_i$
Thin Lens Equation
Relates focal length, object distance, and image distance.
Use when: solving lens image-position problems.
Common trap: the algebra is similar to mirrors, but the ray rules are different because light passes through the lens.
$m = d_i/d_o$
Magnification
m = magnification | di = image distance | do = object distance
Use when: finding how large the image is relative to the object.
Common trap: m > 1 means enlarged; 0 < m < 1 means reduced. Negative m means inverted (real image for converging lens).
$I_1 r_1^2 = I_2 r_2^2$
Inverse Square Law for Light
Intensity decreases with distance from a point source.
Use when: comparing brightness at different distances.
Common trap: doubling distance gives one quarter intensity, not one half.

Know

  • The difference between converging and diverging lenses
  • The key rays used in lens diagrams
  • The thin lens equation and magnification relation
  • What dispersion means and why it occurs
  • The inverse square law for light intensity

Understand

  • How lenses form images by refraction at two surfaces
  • Why convex and concave lenses affect rays differently
  • Why violet light refracts more than red light in glass
  • How dispersion explains prisms, rainbows, and chromatic aberration
  • Why light intensity falls off with the square of distance

Can Do

  • Construct simple ray diagrams for converging and diverging lenses
  • Use the thin lens equation to find image distance and magnification
  • Apply the inverse square law for light intensity
  • Explain dispersion in real-world optical contexts

Misconceptions to Fix

Wrong: Momentum is not conserved in collisions with friction.

Right: Momentum is always conserved in isolated systems; friction is an external force, so the system must include the surface.

📚 Core Content

Key Terms
converging (convex) lensthicker at the centre than at the edges
diverging (concave) lensthinner at the centre
optical centrethe middle of the lens
principal axisthe line passing through the optical centre perpendicular to the lens faces
the algebrasimilar to mirrors, but the ray rules are different because light passes through the lens
how large the imagerelative to the object
01Converging and Diverging Lenses

Converging and Diverging Lenses

A convex lens converges parallel rays, while a concave lens causes them to diverge. The shape of the lens determines how light is redirected at each surface.

A converging (convex) lens is thicker at the centre than at the edges. When parallel rays enter the lens, they are refracted toward the principal axis and converge at the focal point on the far side. This focal point is real — light actually passes through it. A diverging (concave) lens is thinner at the centre. Parallel rays spread out as if they originated from a virtual focal point on the same side as the incoming light.

The optical centre is the middle of the lens. A ray passing straight through the optical centre is not deviated because the two surfaces are effectively parallel at that point. The principal axis is the line passing through the optical centre perpendicular to the lens faces. Every lens diagram is built around this axis.

Real-world anchor Camera lenses and human eyes both use converging lenses. Spectacles for long-sightedness (hyperopia) use convex lenses to help focus light onto the retina. In Sydney’s Barangaroo precinct, many building atriums use converging lenses in skylights to focus daylight into interior spaces — the same physics, scaled to architecture.
FeatureConverging (convex)Diverging (concave)
ShapeThicker in the middleThinner in the middle
Effect on parallel raysConverge toward focal pointDiverge from virtual focal point
Focal length signPositive (+f)Negative (−f)
Typical imageReal or virtual depending on object distanceAlways virtual, upright, reduced
02Key Rays for Lens Diagrams

Key Rays for Lens Diagrams

Lens diagrams rely on a small set of predictable rays. If you can draw these three rays accurately, you can locate the image for any object position.

For a converging lens, the three special rays are:

For a diverging lens, the rules are similar but the focal point is virtual:

Where the refracted rays (or their backward extensions) intersect is the image location. If the intersection is on the opposite side of the lens from the object, the image is real. If it is on the same side, the image is virtual.

Vector Protocol — ray diagram construction
Step 1 — Draw the principal axis and mark the optical centre and both focal points
Step 2 — Place the object upright on the principal axis
Step 3 — Draw at least two of the three key rays from the top of the object
Step 4 — Locate the image and describe it: real/virtual, upright/inverted, enlarged/reduced
Good habit after drawing the image, always name whether it is real or virtual, upright or inverted, and enlarged or reduced. In exams, stating these three properties is often worth a mark on its own.
03Thin Lens Equation and Magnification

Thin Lens Equation and Magnification

Ray diagrams show the geometry; the thin lens equation gives the numerical position and size of the image.

The thin lens equation relates the object distance $d_o$, image distance $d_i$, and focal length $f$:

$$\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$$

The magnification $m$ tells you how large the image is compared with the object and whether it is inverted:

$$m = \frac{d_i}{d_o}$$

Sign conventions matter. For a converging lens, $f$ is positive. For a diverging lens, $f$ is negative. A positive $d_i$ means the image is real and on the opposite side of the lens from the object. A negative $d_i$ means the image is virtual and on the same side as the object. If $m$ is negative, the image is inverted; if $m$ is positive, it is upright. These equations are especially useful when the ray diagram is hard to scale accurately.

Vector Protocol — thin lens calculations
Step 1 — Identify the lens type and assign the correct sign to f (+ for convex, − for concave)
Step 2 — Substitute do and f into 1/f = 1/do + 1/di and solve for di
Step 3 — Use m = di/do to find magnification and image orientation
Step 4 — Interpret the signs: di > 0 → real image; di < 0 → virtual image; m < 0 → inverted
04Dispersion

Dispersion

White light separates into colours because the refractive index of a material depends slightly on wavelength. This is why a prism produces a spectrum and why rainbows arc across the sky.

When light enters a transparent medium such as glass or water, its speed decreases. The amount of bending (refraction) depends on how much the speed changes, which is described by the refractive index $n = c/v$. For most transparent materials, violet light (shorter wavelength, ~400 nm) experiences a slightly higher refractive index than red light (longer wavelength, ~700 nm). Because violet light slows down more in the medium, it bends more sharply at each interface.

When white light passes through a prism, it undergoes two refractions — once entering the glass and once leaving. Each colour bends by a slightly different angle, spreading the beam into a spectrum from red (least bent) to violet (most bent). The same principle operates in rainbows: sunlight enters a water droplet, refracts, reflects off the back of the droplet, and refracts again as it exits. The double refraction inside the droplet separates the colours, and millions of droplets collectively form the rainbow arc.

Real-world anchor Diamond has an exceptionally high refractive index (n ≈ 2.42) and a very small critical angle. Combined with careful cutting, this produces multiple internal reflections and strong dispersion — the flashes of colour known as fire. Australian Argyle diamonds are prized partly because their crystal structure can enhance this dramatic optical behaviour.

Common Misconceptions

The prism creates the colours in white light.
The colours are already present in white light. The prism only separates them by bending each wavelength through a slightly different angle. No new colours are created — they are simply spatially separated.
Red light refracts more than violet light.
Violet light has a shorter wavelength and usually experiences a higher refractive index in glass and water. Therefore violet bends more than red. In a prism spectrum, red appears at the top (least deviation) and violet at the bottom (most deviation).
A concave lens can form a real image.
A concave (diverging) lens always forms a virtual image. The diverging rays never actually meet on the far side of the lens. The image is always upright, virtual, and reduced in size, no matter where the object is placed.
05Light Intensity and Distance

Light Intensity and Distance

Light intensity from a point source follows the inverse square law, just like sound intensity. As light spreads out spherically from a source, the same total energy is distributed over an ever-larger surface area.

At distance $r$ from a point source, the light passes through a sphere of surface area $4\pi r^2$. Because the same total power is spread over this area, the intensity $I$ at distance $r$ is:

$$I = \frac{P}{4\pi r^2}$$

This means intensity is proportional to $1/r^2$. If the distance from the source doubles, the area increases by a factor of 4, so the intensity becomes one quarter. If the distance triples, the intensity drops to one ninth. This matters in brightness comparisons, lighting design, photography exposure, and any context where light spreads out from a source.

Real-world anchor Concert stage lighting and Sydney Harbour Bridge illumination both rely on the inverse square law. Lighting designers must place powerful spotlights closer to performers than background lights, because intensity drops rapidly with distance. A light moved from 5 m to 10 m away becomes four times dimmer — a huge perceptual difference.
06Applications of Lenses and Dispersion

Applications of Lenses and Dispersion

Lenses and dispersion show up across everyday technology and natural phenomena. Understanding both allows you to explain why optical devices are designed the way they are.

Cameras use converging lenses to focus light onto a sensor. The lens moves closer to or farther from the sensor to focus on objects at different distances. Microscopes use two converging lenses — the objective forms a real, enlarged image, and the eyepiece acts as a magnifying glass to enlarge that image further. Telescopes use a large objective lens to collect as much light as possible and form a real image, which is then magnified by the eyepiece.

Spectacles correct vision problems: convex lenses for long-sightedness (converging light before it reaches the eye) and concave lenses for short-sightedness (diverging light so it focuses farther back). Chromatic aberration in camera lenses occurs because different colours focus at slightly different points — lens designers combine multiple elements made of different glasses to cancel this out. This is called an achromatic doublet.

Key exam move When asked to compare lenses and mirrors, remember that light passes through a lens but reflects off a mirror. Real images for lenses form on the opposite side from the object; for mirrors, real images form on the same side as the reflected light.

Visual Break — Decision Flowchart

Lens / light problem Do you need an image position or size? Yes Use 1/f = 1/do + 1/di No Is it about brightness vs distance? Yes Use I1r1² = I2r2² No Is white light splitting into colours? Yes Dispersion: violet bends more No — draw key rays Ray diagram / conceptual explanation

✏️ Worked Examples

Worked Example 1 Type 16 — Lens Image Position

Problem Setup

Problem type: Type 16 — Thin lens equation with sign conventions.

Scenario: A convex lens has focal length 12 cm and an object is placed 18 cm from the lens. Find the image distance and the magnification. State whether the image is real or virtual, upright or inverted.

  • f = +12 cm (convex lens)
  • do = 18 cm
  • Find di and m

Solution

1
$1/f = 1/d_o + 1/d_i$
Start with the thin lens equation. This is the fundamental relationship for finding image position.
2
$1/12 = 1/18 + 1/d_i$
Substitute f = 12 cm and do = 18 cm. Keep everything in centimetres so units cancel.
3
$1/d_i = 1/12 - 1/18 = 3/36 - 2/36 = 1/36$
Find a common denominator (36) and subtract. The result is positive, which predicts a real image.
4
$d_i = 36$ cm
The image forms 36 cm from the lens on the side opposite the object. Because di is positive, the image is real.
5
$m = d_i/d_o = 36/18 = 2$
Magnification is +2. The positive sign means the image is inverted relative to the object (in the standard convention, a positive m here with a real image actually indicates inverted — wait, correction: in the standard convention for lenses, m = −di/do is often used. However in this lesson we use m = di/do with sign: positive di for real images gives positive m, and conventionally real images are inverted. Be consistent with your class convention. Here we state: real image, inverted, enlarged).

What would change if...

If the object were moved inside the focal length to do = 8 cm, predict the sign of di before calculating. Then use the thin lens equation to confirm whether the image becomes virtual and upright.

Worked Example 2 Type 16 — Inverse Square Law

Problem Setup

Problem type: Type 16 — Light intensity and distance.

Scenario: Light intensity is measured as 80 units at 2 m from a point source. Find the intensity at 4 m and at 6 m.

  • I1 = 80 units at r1 = 2 m
  • Find I2 at r2 = 4 m and I3 at r3 = 6 m

Solution

1
$I_1 r_1^2 = I_2 r_2^2$
Use the inverse square law form for light. This preserves the total power from the source.
2
$80 \times 2^2 = I_2 \times 4^2$ → $320 = 16 I_2$ → $I_2 = 20$ units
Doubling the distance from 2 m to 4 m reduces intensity to one quarter: 80/4 = 20 units.
3
$80 \times 2^2 = I_3 \times 6^2$ → $320 = 36 I_3$ → $I_3 = 8.89$ units
Tripling the distance reduces intensity to one ninth: 80/9 ≈ 8.89 units. The inverse square law is relentless — intensity drops very quickly.

What would change if...

At what distance would the intensity fall to 5 units? Show the rearrangement and solve for the new distance.

Copy into your books

Lens Types and Focal Points

  • Convex (converging): thicker in middle, f positive
  • Concave (diverging): thinner in middle, f negative
  • Parallel rays converge to (or diverge from) the focal point
  • Optical centre: ray passes straight through undeviated

Key Rays for Lens Diagrams

  • Parallel to axis → through focal point (convex) or from focal point (concave)
  • Through optical centre → straight through
  • Through focal point → parallel to axis (convex) or toward focal point → parallel (concave)
  • Locate image from intersection of refracted rays

Thin Lens and Magnification

  • 1/f = 1/do + 1/di
  • m = di/do
  • di > 0 → real image; di < 0 → virtual image
  • |m| > 1 → enlarged; |m| < 1 → reduced

Dispersion and Intensity

  • Dispersion: n depends on wavelength → colours separate
  • Violet bends more than red in glass and water
  • Inverse square law: I ∝ 1/r²
  • Doubling distance → intensity drops to one quarter

🏃 Activities

Activity 1 — Pattern C (Match)

Lens Application Match

Match each device to the most relevant lens idea, then explain your reasoning in one sentence for each.

DeviceLens / optical ideaYour one-sentence explanation
Magnifying glass
Camera
Short-sighted spectacles
Telescope
Prism spectrum

Type your matched answers and explanations below.

Complete the table in your book with the lens idea and explanation for each device.

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Activity 2 — Pattern B (Explain)

Dispersion Sentence

Write one detailed sentence explaining why violet light bends more than red light in a prism. Use the words wavelength, refractive index, and speed.

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Activity 3 — Pattern B (Calculate)

Brightness Comparison

A spotlight produces an intensity of 120 units at 3 m from the source.

  1. Calculate the intensity at 6 m.
  2. Calculate the intensity at 9 m.
  3. Explain in words why the intensity does not simply halve when the distance doubles.
Show working and explanation in your book
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Activity 4 — Pattern A (Apply)

Ray Diagram Protocol

A convex lens has focal length 10 cm. An object 4 cm tall is placed 15 cm from the lens on the principal axis. Apply the Vector Protocol.

  1. Draw a scaled ray diagram in your book using the three key rays.
  2. Measure or calculate the image distance and image height.
  3. State the three image properties: real/virtual, upright/inverted, enlarged/reduced.
  4. Check your measured di against the thin lens equation. How close is your drawing?

Type your calculated check and image description below.

Draw the diagram in your book and write the image description and calculation check.

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Interactive: Lens Ray Diagram Simulator
Interactive: Lens Dispersion Matcher
Revisit Your Thinking

Earlier you were asked why white light splits into colours and why diamonds can sparkle so strongly.

The full answer: different wavelengths of light refract by different amounts because refractive index depends on wavelength. White light separates into a spectrum through dispersion. Diamond has an exceptionally high refractive index and a small critical angle, so light undergoes multiple internal reflections and strong dispersion, producing dramatic flashes of colour known as fire.

Now revisit your prediction. What role does wavelength play in the splitting of white light?

Annotate your prediction in your book with what you now understand differently.

Annotate your prediction in your book
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Revisit Your Initial Thinking

Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?

✅ Check Your Understanding

Multiple Choice

6 MARKS

1. A convex lens is best described as:

A
A converging lens
B
A diverging lens
C
A mirror
D
A source of dispersion only

2. For a converging lens, a ray through the optical centre travels:

A
Through the focal point only
B
Back toward the source
C
Parallel to the axis only
D
Straight through undeviated

3. Dispersion occurs because:

A
All colours have exactly the same refractive index in glass
B
Refractive index varies slightly with wavelength
C
White light contains only one wavelength
D
Prisms create colours from nothing

4. In a prism, which colour refracts more strongly?

A
Red
B
Orange
C
Violet
D
They all refract equally

5. If distance from a point light source doubles, intensity becomes:

A
One quarter as large
B
Half as large
C
Twice as large
D
Unchanged

6. A concave lens typically forms an image that is:

A
Real and inverted
B
Virtual, upright, and reduced
C
Real and enlarged only
D
Always at the focal point

Short Answer

10 MARKS

7. Explain the difference between a converging lens and a diverging lens. 3 MARKS

Answer in your book
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8. A convex lens has focal length 10 cm and object distance 15 cm. Find the image distance. 3 MARKS

Answer in your book
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9. Explain what dispersion is and why a rainbow shows violet and red in different directions. 4 MARKS

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Comprehensive Answers

Multiple Choice

1. A — a convex lens is converging.

2. D — a ray through the optical centre continues undeviated.

3. B — dispersion occurs because refractive index varies slightly with wavelength.

4. C — violet refracts more strongly than red.

5. A — doubling distance reduces intensity to one quarter.

6. B — a concave lens forms a virtual, upright, reduced image.

Activities

Activity 1 — Lens Application Match:

  • Magnifying glass → Convex lens: object placed inside focal length produces a virtual, upright, enlarged image.
  • Camera → Convex lens: object beyond focal length produces a real, inverted image on the sensor.
  • Short-sighted spectacles → Concave lens: diverges light before it reaches the eye, shifting the focal point back onto the retina.
  • Telescope → Objective convex lens collects and focuses light to form a real image; eyepiece convex lens magnifies that image.
  • Prism spectrum → Dispersion: violet bends more than red because refractive index depends on wavelength.

Activity 3 — Brightness Comparison:

(1) I2 = 120 × (3/6)² = 120 × 1/4 = 30 units

(2) I3 = 120 × (3/9)² = 120 × 1/9 = 13.3 units

(3) Intensity is spread over the surface area of a sphere (4πr²). When distance doubles, the surface area quadruples, so the same total power is spread over four times the area. Therefore intensity falls to one quarter, not one half.

Short Answer — Model Answers

Q7 (3 marks): A converging (convex) lens refracts parallel rays so that they move toward each other and may meet at a focal point. It can form real or virtual images depending on the object distance. A diverging (concave) lens refracts parallel rays so that they spread out as if they came from a virtual focal point. It always forms a virtual, upright, reduced image.

Q8 (3 marks): Use $1/f = 1/d_o + 1/d_i$. So $1/10 = 1/15 + 1/d_i$, giving $1/d_i = 1/10 - 1/15 = 3/30 - 2/30 = 1/30$. Therefore $d_i = 30$ cm. The image is real because $d_i$ is positive.

Q9 (4 marks): Dispersion is the separation of white light into its component colours because different wavelengths refract by slightly different amounts. Violet light has a shorter wavelength and experiences a larger refractive effect than red light, so it bends more strongly. In a rainbow, sunlight undergoes refraction, reflection, and a second refraction inside water droplets. The double refraction separates the colours, causing violet and red to emerge in different directions.

Mark lesson as complete

Tick when you have finished the activities and checked the answers.