Reflection, Refraction & Optics — Print · Grade 12 Science
Chapter: Waves, Sound & OpticsReflection, Refraction & Optics
Learning goals
- I can apply the law of reflection and use the index of refraction (n = c/v) and Snell's law to describe how light bends.
- I can predict image formation by lenses and mirrors using the thin-lens equation and magnification.
Readings for this unit
Before this unit
Angles from the normal, refractive index, and geometry.
Use the detailed parent plan: two core 50–60-minute sessions per focused unit, with a third assigned for selected calculation/modeling topics. First study the core explanation and Parts 1–2 with their figures/checks; next study Parts 3–4, the whole-unit application, worked cases, varied practice and exit check. Attempt reasoning before opening guidance. Extend time when actual understanding requires it; investigations and chapter assessments have separate sessions.
Start with something to explain
Why does a lens form an inverted image in one position and an upright one in another?
Illustrative evidence case. The real object moves from beyond a converging lens’s focus to inside it.
Notice and wonder
Record two observations using the stated quantities or relationships. Keep observations separate from possible causes. Write one question that a new measurement could help answer.
Why does a lens form an inverted image in one position and an upright one in another?
Your first model
Trace parallel and central rays and distinguish actual rays from backward extensions.
Build the language from the idea
- reflection
- The bouncing of light off a surface; the angle of incidence equals the angle of reflection, both measured from the normal.
- normal
- The imaginary line drawn perpendicular to a surface at the point where a ray strikes it; all angles are measured from it.
- refraction
- The bending of light as it changes speed passing from one medium into another.
- index of refraction (n)
- How much a medium slows light: n = c ÷ v, where v is light's speed in the medium; larger n means slower light and more bending.
- total internal reflection
- When light in a higher-index medium strikes a boundary beyond the critical angle and is entirely reflected back, as in a fiber-optic cable.
- focal length (f)
- The distance from a lens or mirror to the point where parallel rays converge (or appear to diverge from).
- real vs. virtual image
- A real image forms where light rays actually converge and can be projected on a screen; a virtual image only appears to come from where rays seem to originate.
When light strikes a surface, the simplest thing it can do is bounce off — reflection. The rule is exact: the angle of incidence equals the angle of reflection, with both angles measured from the normal, the line drawn perpendicular to the surface. A smooth mirror reflects parallel rays in a single direction (specular reflection) and forms a clear image; a rough wall scatters them every which way (diffuse reflection), which is why you can read this page from any angle but cannot see your face in it. The same law governs a flat bathroom mirror and the curved dish of a telescope.
When light passes from one transparent medium into another, it changes speed, and that speed change bends its path — refraction. The key quantity is the index of refraction, n = c ÷ v, comparing light's vacuum speed c to its slower speed v inside the material; glass has n ≈ 1.5, water n ≈ 1.33, diamond n ≈ 2.42. The bending follows Snell's law, n₁ sin θ₁ = n₂ sin θ₂. A useful rule of thumb: entering a higher-index medium, light slows and bends *toward* the normal; leaving into a lower-index medium, it speeds up and bends *away*. This is why a straw looks broken at the water's surface and why a swimming pool always looks shallower than it is.
Refraction changes light’s speed and wavelength at a stationary boundary, while frequency remains continuous. Index n=c/v is optical behavior, not mass density. Snell’s law n₁sinθ₁=n₂sinθ₂ uses angles from the normal. Entering a higher-index medium bends an oblique ray toward the normal; entering a lower-index medium bends it away.
Total internal reflection requires propagation from a higher-index to a lower-index medium and incidence angle greater than the critical angle. The critical-angle relationship is sinθc=n₂/n₁ for n₁>n₂. A normally incident ray does not bend; reflection/refraction depend on geometry and wavelength-dependent material properties.
Curved mirrors form images by reflection in a closely parallel way. A concave mirror curves inward and can converge light to a real, inverted image — the principle behind a reflecting telescope and a makeup mirror's magnified view up close. A convex mirror curves outward, always producing a small, upright virtual image with a wide field of view, which is why it is used for car side mirrors and store security mirrors (and why 'objects in mirror are closer than they appear'). The same equation, 1 ÷ f = 1 ÷ dₒ + 1 ÷ dᵢ, works for mirrors, and the magnification m = − dᵢ ÷ dₒ tells you the image's size and orientation.
These principles are the working parts of nearly every optical device you rely on. Your eye is a lens focusing a real image on the retina; eyeglasses and contacts nudge that image onto the retina when it lands too soon or too late. Cameras, microscopes, telescopes, projectors, and endoscopes are all careful arrangements of lenses and mirrors, and fiber optics carry data as trapped light. Every one of them is engineered from the same short list of rules — reflect at equal angles, refract by Snell's law, and form images with the thin-lens equation — which is exactly what makes optics such a powerful, predictable branch of physics.
Model check: Optically denser means higher refractive index, not necessarily greater mass density. It slows in the simple model and bends toward the normal. Frequency stays continuous across a stationary boundary; wavelength changes.
Build your understanding
Use the detailed parent plan: two core 50–60-minute sessions per focused unit, with a third assigned for selected calculation/modeling topics. First study the core explanation and Parts 1–2 with their figures/checks; next study Parts 3–4, the whole-unit application, worked cases, varied practice and exit check. Attempt reasoning before opening guidance. Extend time when actual understanding requires it; investigations and chapter assessments have separate sessions.
Part 1 · Measure ray angles from the normal
A normal is a line perpendicular to the boundary at the interaction point. The angle of reflection equals the angle of incidence, both measured from this normal. A transmitted ray can change direction by refraction when phase speed changes across the boundary. The drawing shows transmission toward the normal in a higher-index medium. A ray is a direction model, not a material stream or an extra physical wall at the normal.
Read the visual model
Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.
Is a 30° angle measured from the surface also 30° from the normal?
Visual reasoning — after your attempt
No. It is 60° from the normal for a flat boundary.
For reflection, incident and reflected angles are equal when measured from the surface normal. Refraction uses the normal too. A ray angle measured from the surface is the complementary angle, so using it directly in Snell’s law gives a different and usually incorrect result.

A ray 30° from the surface is 60° from the normal. Draw the local normal before labeling an angle.
Pause and explain
Why must the reference line be explicit?
Check your reasoning after answering
The reflection and refraction equations use angles relative to the normal; a surface-angle convention must first be converted.
Part 2 · Use Snell’s law and its conditions
The refractive index n=c/v describes phase speed relative to vacuum. Snell’s law n₁sinθ₁=n₂sinθ₂ applies to the stated interface model. Entering higher n bends a ray toward the normal except at normal incidence. Total internal reflection can occur when traveling from higher to lower n and the required sine would exceed one; the critical angle satisfies sinθc=n₂/n₁. It does not occur for every ray entering glass from air.
Read the visual model
Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.
For n₁=1.5 and n₂=1.0, estimate the critical angle.
Visual reasoning — after your attempt
θc=arcsin(1/1.5)≈41.8° from the normal.
For suitable media, n₁ sinθ₁=n₂ sinθ₂ relates ray directions at an interface. A ray entering a larger refractive index bends toward the normal in the ordinary model. Total internal reflection requires travel from larger to smaller index and an incidence angle exceeding the critical value.
A ray from air into glass does not meet the larger-to-smaller condition for total internal reflection, even at a large incident angle.
Pause and explain
What two conditions are needed for total internal reflection?
Check your reasoning after answering
The initial medium must have larger refractive index, and the incidence angle measured from the normal must exceed the critical angle.
Part 3 · Use a consistent thin-lens convention
In the stated thin-lens convention, 1/f=1/do+1/di, with f positive for a converging lens, do positive for a real object and di positive for a real image on the far side. Magnification m=−di/do gives image height divided by object height. For f=10 cm and do=20 cm, di=20 cm and m=−1: the real image is inverted and equal in height. State the convention rather than interpreting every negative sign as an error.
Read the visual model
Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.
What happens for this lens with a real object at do=5 cm?
Visual reasoning — after your attempt
di=−10 cm and m=+2: an upright enlarged virtual image on the object side.
For a thin lens in the usual real-object convention, 1/f=1/do+1/di. A converging lens has positive f; a positive di denotes a real image on the opposite side. Magnification m=−di/do includes orientation. Name the convention because other texts can define signs differently.
A negative di in this convention denotes a virtual image on the object side. A real image can be projected onto an appropriately placed screen in the model.
Pause and explain
Does a negative image distance mean there is no image?
Check your reasoning after answering
No. It identifies a virtual image under the selected convention; the ray extensions rather than outgoing rays meet at that location.
Part 4 · Check ray geometry against the numerical image
A ray diagram predicts an image by tracing directions from the same object point. In this thin-lens 2f case, a ray parallel to the axis leaves through the far focus, while the ray through the optical center is approximately undeviated. Their actual intersection gives the inverted real image at 2f. For a virtual image, backward extensions intersect instead; the actual outgoing rays do not meet there. A diagram and equation should agree on side, orientation and relative size.
Read the visual model
Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.
Why is this image real rather than virtual?
Visual reasoning — after your attempt
The transmitted rays actually converge at the image point on the far side of the lens.
A ray sketch should match whether the image is real or virtual, enlarged or reduced, and upright or inverted. Thin-lens equations are approximations for suitable paraxial rays and idealized optics. Aberrations, lens thickness, and alignment can change actual image quality.
An object outside a converging lens’s focal distance commonly gives a real inverted image; inside it gives a virtual upright image in this model.
Pause and explain
Why should a correct equation result still be checked with a ray sketch?
Check your reasoning after answering
The sketch exposes sign or orientation errors and makes the model assumptions and image meaning easier to assess.
Apply the whole unit · Predict and test a lens image
Illustrative converging thin lens has f=+10.0 cm; a real object is 30.0 cm in front. Use positive do for a real object and positive di for a real image opposite it.
Your task
Calculate image distance and magnification, describe screen placement and orientation, then predict the qualitative change if the object moves inside the focal distance.
What your response should show
- Define the system, quantities, units, and relevant conditions before selecting a model.
- Show intermediate calculations or a labeled evidence account, including signs or denominators.
- Explain how the supplied evidence supports the conclusion rather than reporting only an answer.
- Identify a meaningful uncertainty, model limit, or measurement needed to test the conclusion.
Response guidance — after your attempt
1/di=1/10.0−1/30.0=1/15.0 cm⁻¹; di=+15.0 cm.
m=−15.0/30.0=−0.500: inverted image with half the object height.
Place a screen about 15.0 cm on the opposite side; inside f, the ideal image becomes virtual and upright.
The sign and scale agree with a converging-lens ray model for an object beyond twice the focal length. A screen tests a real image prediction, but image sharpness also depends on alignment and optical limitations. A virtual image cannot be captured by placing a screen at its apparent location in this simple setup.
Scientific references and source-model foundations
- OpenStax — thin-lens image formation
- OpenSciEd 8.2 — sound-wave foundations
- OpenSciEd public middle-school figure license — CC BY 4.0
Parent / teacher teaching notes
Use a ray sketch and stated signs before equations. Do not use direct Sun imaging as a lens activity; supplied diagrams are sufficient for the prediction. Ask the learner to explain each symbol, axis, unit and arrow before calculating. Calculated examples, supplied illustrative records and actual historical observations are labeled separately. Selected public OpenSciEd middle-school source figures are foundation examples, not complete high-school editions or an endorsement; all advanced explanations are supplied locally.
The key idea
Light reflects at equal angles from the normal and refracts (bends) when its speed changes between media (n = c/v, Snell's law); lenses and mirrors form real or virtual images predicted by 1/f = 1/dₒ + 1/dᵢ.
Law of Reflection
θᵢ = θᵣ
θᵢ is the angle of incidence and θᵣ is the angle of reflection, both measured from the normal (the perpendicular to the surface).
Index of Refraction
n = c ÷ v
n is the index of refraction (no units); c is the speed of light in vacuum; v is the speed of light in the medium. n is always ≥ 1, and larger n means light slows more and bends more.
Snell's Law
n₁ sin θ₁ = n₂ sin θ₂
n₁ and n₂ are the indices of refraction of the two media; θ₁ and θ₂ are the angles from the normal in each medium. At the critical angle for total internal reflection, sin θc = n₂ ÷ n₁.
Thin-Lens / Mirror Equation
1 ÷ f = 1 ÷ dₒ + 1 ÷ dᵢ
f is the focal length; dₒ is the object distance; dᵢ is the image distance (negative for a virtual image). Magnification m = − dᵢ ÷ dₒ, where a negative m means an inverted image.
Worked examples
Example 1
- Use the index of refraction equation n = c ÷ v.
- n = (3 × 10⁸) ÷ (1.25 × 10⁸).
- = 2.4.
Example 2 · Worked application
- Use 1/f=1/d_o+1/d_i with positive real-object convention.
- 1/d_i=1/10−1/30=1/15.
- d_i=15 cm, positive for a real image in this convention.
State the sign convention; a negative image distance has a different interpretation.
Example 3
- Use the thin-lens equation 1 ÷ f = 1 ÷ dₒ + 1 ÷ dᵢ, so 1 ÷ dᵢ = 1 ÷ f − 1 ÷ dₒ.
- 1 ÷ dᵢ = 1 ÷ 10 − 1 ÷ 30 = 3 ÷ 30 − 1 ÷ 30 = 2 ÷ 30 = 1 ÷ 15.
- dᵢ = 15.
Example 4 · Find and fix the mistake
- Test the claim against the stated conditions: State the sign convention; a negative image distance has a different interpretation.
- Objects inside the focal length produce a virtual image in the thin-lens model.
State the sign convention; a negative image distance has a different interpretation.
Example 5 · Reason from the visual model
- In the stated thin-lens convention, 1/f=1/do+1/di, with f positive for a converging lens, do positive for a real object and di positive for a real image on the far side. Magnification m=−di/do gives image height divided by object height. For f=10 cm and do=20 cm, di=20 cm and m=−1: the real image is inverted and equal in height. State the convention rather than interpreting every negative sign as an error.
- di=−10 cm and m=+2: an upright enlarged virtual image on the object side.
Calculated thin-lens cases; paraxial rays and negligible lens thickness are assumed.
Avoid these mistakes
- Optically denser means higher refractive index, not necessarily greater mass density.
Check your understanding
Light enters a higher-index medium obliquely. Explain direction.
Discuss the answer
It slows in the simple model and bends toward the normal. Frequency stays continuous across a stationary boundary; wavelength changes.
Return to the phenomenon
Return to the initial case. Why does a lens form an inverted image in one position and an upright one in another? Revise your drawing or explanation using two named ideas and one stated limitation.
Compare explanations after your attempt
Lens equations and ray intersections change the image side and orientation; inside f the image is virtual and upright.
What should we investigate next?
Which comparison would test a remaining uncertainty? State what you would change, keep comparable and measure.
Investigate, calculate and revise
Keep a first model and an evidence-supported revision. Attempt each task before opening guidance; all required explanations and supplied-data routes are available locally.
Activity 1 · Return to the opening case
Use the opening phenomenon, unit definitions and explained Parts 1–4.
Your task
Return to the initial case. Why does a lens form an inverted image in one position and an upright one in another? Revise your drawing or explanation using two named ideas and one stated limitation.
Success criteria
- Use the stated observations and appropriate units.
- Connect a labeled model with the physical/environmental mechanism.
- State assumptions and an evidence limitation.
- Explain a substantive revision or next test.
Hint
Keep object/material identity, material transfer and energy account distinct.
Response guidance — discuss after your attempt
Lens equations and ray intersections change the image side and orientation; inside f the image is virtual and upright.
Activity 2 · Transfer to a changed case
Cover the worked answer and use the explained models, relationships and notebook.
Your task
A thin converging lens has f=12 cm and a real object at 36 cm. Find image distance and magnification; describe the image.
Success criteria
- Use the stated measurements or conditions, with correct units.
- Connect the result to the physical or environmental mechanism or evidence relationship.
- Identify the assumption, model limit or further evidence requested.
Hint
Name known quantities and conditions, show intermediate steps and check conservation.
Response guidance — discuss after your attempt
1/di=1/12−1/36=1/18, so di=18 cm and m=−18/36=−0.50. It is real, inverted and half the object height in this paraxial model.
Words to know
- reflection
- The bouncing of light off a surface; the angle of incidence equals the angle of reflection, both measured from the normal.
- normal
- The imaginary line drawn perpendicular to a surface at the point where a ray strikes it; all angles are measured from it.
- refraction
- The bending of light as it changes speed passing from one medium into another.
- index of refraction (n)
- How much a medium slows light: n = c ÷ v, where v is light's speed in the medium; larger n means slower light and more bending.
- total internal reflection
- When light in a higher-index medium strikes a boundary beyond the critical angle and is entirely reflected back, as in a fiber-optic cable.
- focal length (f)
- The distance from a lens or mirror to the point where parallel rays converge (or appear to diverge from).
- real vs. virtual image
- A real image forms where light rays actually converge and can be projected on a screen; a virtual image only appears to come from where rays seem to originate.
Mirages, twinkling stars, and the atmosphere as a lens
Refraction does not need a block of glass — air of different temperatures has slightly different densities, and therefore different indices of refraction, so light bends as it passes through. A hot road heats the air just above it, bending light from the sky upward into your eyes and painting a shimmering 'puddle' — a mirage. The same continuous bending makes stars twinkle as their light threads through turbulent air, lets us see the Sun for a couple of minutes after it has geometrically set, and is why astronomers prize mountaintop and space telescopes above the wobbling atmosphere.
Total internal reflection: how light carries the internet
Beyond the critical angle, light striking the boundary of a higher-index medium is reflected entirely back inside — total internal reflection, an all-or-nothing effect. A fiber-optic cable is a hair-thin glass thread engineered so that light pulses always strike its walls beyond that angle, bouncing thousands of times and staying trapped as they race across continents and under oceans, carrying nearly all the world's internet and phone traffic. The same effect makes a cut diamond sparkle, repeatedly trapping and re-releasing light among its facets, and lets a doctor's endoscope pipe an image out of the body.
Your eye is a lens — and glasses move the image onto the retina
The human eye focuses a real, inverted image onto the retina, using a flexible lens that changes shape to focus near and far. When the eyeball is a touch too long, the image lands in front of the retina and distant objects blur — nearsightedness, corrected by a diverging lens that pushes the focus back. Too short an eyeball does the opposite — farsightedness, corrected by a converging lens. Eyeglasses, contacts, and laser surgery are all just applications of the thin-lens equation, nudging the image onto the retina where it belongs.
Real vs. virtual images: the idea that trips everyone up
The single most confusing point in optics is the difference between the two kinds of image. A real image forms where light rays actually cross, so you can catch it on a screen — the image on a camera sensor, a projector screen, or your retina, and it is inverted. A virtual image forms where rays only appear to originate but never truly meet, so it cannot be projected — your upright reflection in a bathroom mirror, or the enlarged view through a magnifying glass. In the thin-lens equation, a negative image distance is the mathematical signal that the image is virtual.
From Alhazen to the telescope: a short history of light's path
Around 1020, the Arab scholar Ibn al-Haytham (Alhazen) overturned the ancient idea that vision worked by beams shooting out of the eye, arguing instead that light reflects from objects into the eye — and he studied lenses, mirrors, and refraction with careful experiments, earning the title 'father of optics.' Centuries later, spectacle-makers' lenses were combined into the first telescopes; Galileo turned one skyward in 1609 and saw Jupiter's moons, and Newton built a reflecting telescope using a curved mirror instead of a lens. Every modern camera, microscope, and observatory descends from this long refinement of a few simple rules about how light bends and bounces.
In the real world
Every fiber-optic cable carrying internet traffic relies on total internal reflection, trapping pulses of light so they bounce along the glass thread for kilometers without escaping.
Exit check
Light enters a higher-index medium obliquely. Explain direction.
Check your explanation after answering
It slows in the simple model and bends toward the normal. Frequency stays continuous across a stationary boundary; wavelength changes.
Explain your reasoning without notes. If a key idea is missing, revisit the lesson and try again.
Problem 1
- the horizon
- the edge of the mirror
- the surface itself
- the normal (the perpendicular to the surface)
Answer and explanation
the normal (the perpendicular to the surface)
All angles in optics are measured from the normal, the line perpendicular to the surface at the point of contact.
Problem 2
Answer and explanation
1.5
n = c ÷ v = (3 × 10⁸) ÷ (2 × 10⁸) = 1.5.
Problem 3
- bends toward the normal and slows down
- bends toward the normal and speeds up
- bends away from the normal and speeds up
- passes straight through with no change
Answer and explanation
bends toward the normal and slows down
Entering a denser, higher-index medium, light slows down and bends toward the normal.
Problem 4
Answer and explanation
30 cm
1 ÷ dᵢ = 1 ÷ f − 1 ÷ dₒ = 1 ÷ 15 − 1 ÷ 30 = 2 ÷ 30 − 1 ÷ 30 = 1 ÷ 30, so dᵢ = 30 cm.
Problem 5
Answer and explanation
A real image forms where light rays actually converge, so it can be projected onto a screen (like the image on a camera sensor or a movie screen); it is usually inverted. A virtual image forms where rays only appear to come from — they never actually meet there — so it cannot be projected; it is upright, like the enlarged image seen through a magnifying glass or your reflection in a flat mirror.
Problem 6
Answer and explanation
6 cm
Magnification magnitude = dᵢ ÷ dₒ = 30 ÷ 10 = 3, so the image height = 3 × 2 cm = 6 cm.
Problem 7
Hints
State the sign convention; a negative image distance has a different interpretation.
Answer and explanation
1/d_i=1/10−1/20=1/20; d_i=20 cm.
Problem 8
Hints
State what the supplied observations or conditions establish, and separate that from any assumption.
Answer and explanation
Refraction changes propagation speed and wavelength while frequency is continuous across a stationary boundary.
Problem 9
Answer and explanation
di=−10 cm and m=+2: an upright enlarged virtual image on the object side.
In the stated thin-lens convention, 1/f=1/do+1/di, with f positive for a converging lens, do positive for a real object and di positive for a real image on the far side. Magnification m=−di/do gives image height divided by object height. For f=10 cm and do=20 cm, di=20 cm and m=−1: the real image is inverted and equal in height. State the convention rather than interpreting every negative sign as an error.
di=−10 cm and m=+2: an upright enlarged virtual image on the object side.
Response criteria
- Interpret labels, structures, axes, units and directions correctly.
- Explain the physical or environmental relationship or mechanism.
- Respect stated assumptions and distinguish observation from inference.
Problem 10
Answer and explanation
1/di=1/12−1/36=1/18, so di=18 cm and m=−18/36=−0.50. It is real, inverted and half the object height in this paraxial model.
1/di=1/12−1/36=1/18, so di=18 cm and m=−18/36=−0.50.
It is real, inverted and half the object height in this paraxial model.
Response criteria
- Use the stated measurements or conditions, with correct units.
- Connect the result to the physical or environmental mechanism or evidence relationship.
- Identify the assumption, model limit or further evidence requested.
Problem 11

Answer and explanation
Draw source-to-surface rays, then reflected or transmitted paths that actually reach the observer. Light enters the eye from the path; the eye does not emit rays that create sight. Angles are measured from the boundary normal, reflection uses equal angles and transmitted direction requires the indices and geometry.
Read source labels, scope, axes and intentional omissions.
Use the complete local explanation to draw a justified mechanism or calculate a supported quantity.
State what is observed, what is assumed and what evidence remains needed.
Response criteria
- Interpret the source labels, units and scope accurately.
- Explain the relevant mechanism with a labeled model or supported calculation.
- Identify a meaningful assumption, uncertainty or missing comparison.