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Reflection, refraction, total internal reflection, lenses and images, colour, and light as a wave, from a glass block to an optical fibre and a pair of glasses. Each section follows one lesson: what you need to know, key terms, the equations with units, a worked example, common mistakes and an exam tip.
1 The Bending of Light
What you need to know
Into the Block
- Refraction is the change in direction of light where its speed changes, at the boundary between two materials.
- Angles are measured from the normal, a line at 90° to the surface where the ray hits. A ray travelling along the normal does not bend.
- Going from air into glass or water, light bends towards the normal; coming out into air, it bends away from the normal.
- A ray through a parallel-sided block leaves parallel to the ray that went in, shifted sideways. A little light is also reflected at each face.
Slower in Glass
- Light is slower in glass, water and oils than in air. The refractive index n says how many times slower: n = c ÷ v. It has no unit and is never less than 1.
- A bigger change of speed gives a bigger bend. If light has the same speed in two materials it does not bend at all, which is why a glass rod almost vanishes in an oil of the same refractive index.
- Optically denser means a bigger refractive index, not a bigger density: oil floats on water but slows light more.
The Quickest Path
- Light takes the quickest path between two points, so it bends where its speed changes (Fermat's principle; beyond GCSE).
- On a ray diagram, light is travelling more slowly in the material where the ray is nearer the normal.
Key terms
- Refraction
- the change of direction of a wave where its speed changes at a boundary
- Normal
- a line drawn at 90° to a surface at the point where a ray meets it
- Refractive index
- the speed of light in a vacuum divided by its speed in the material; no unit
- Optically denser
- having a larger refractive index, so light travels more slowly in it
Equations
| Equation | What it says, symbols and units |
|---|---|
n = speed of light in a vacuum (or air)speed of light in the material | Refractive index says how many times slower light travels in the material.
|
c = 3.0 × 108 m/s | All electromagnetic waves travel at this speed in a vacuum, and at about the same speed in air.
Light in air travels at almost exactly this speed, so air counts as n = 1. |
Diagram to know
A ray through a rectangular glass block: the normal drawn dashed at each face, the ray bending towards the normal on entry and away from it on exit, and leaving parallel to the incoming ray. Label i and r at the first face.
Worked example
A pulse of laser light takes 12.0 ns (12.0 × 10−9 s) to cross a tank of liquid 2.5 m long. Find the speed of light in the liquid and its refractive index. Take c = 3.0 × 108 m/s.
- Speed = distance ÷ time = 2.5 ÷ (12.0 × 10−9) = 2.08 × 108 m/s.
- n = c ÷ v = (3.0 × 108) ÷ (2.08 × 108) = 1.44.
- Check: n is more than 1, and the light is slower than in air, as it must be.
Answer: 2.08 × 108 m/s; n = 1.44 (no unit)
Common mistakes
- Measuring the angles from the surface of the block instead of from the normal.
- Saying light bends towards the normal as it leaves glass. Speeding up, it bends away.
- Giving the refractive index a unit, or getting a value less than 1 by dividing the wrong way round.
Exam tip
In the glass-block practical, mark the ray with crosses or pins well apart, draw the normal with a protractor, and measure both angles from it. Use a range of angles of incidence, not just one.
AP Physics and NGSS: Optics is not in AP Physics 1: AP Physics 2 (Unit 13) covers refraction, writing n = c/v in the same way. NGSS links it to HS-PS4-1, the relationship between wave speed, frequency and wavelength in different media.
2 The Law of Reflection
What you need to know
Mirror law
- Law of reflection: the angle of incidence equals the angle of reflection, both measured from the normal.
- The incident ray, the reflected ray and the normal all lie in the same plane.
- Turning a mirror through an angle turns the reflected ray through twice that angle.
Rough or smooth
- A smooth surface gives specular reflection: parallel rays stay parallel after reflecting, so you can see an image.
- A rough surface gives diffuse reflection: the rays scatter in many directions and there is no image. Each ray still obeys i = r at its own tiny bit of surface.
- You see most objects, such as this page, by diffuse reflection.
Mirror images
- The image in a plane mirror is virtual, upright, the same size as the object and laterally inverted (left and right appear swapped).
- It is as far behind the mirror as the object is in front, on the normal through the object. No light actually reaches it, so a screen cannot catch it.
- To see yourself from head to toe, a plane mirror only needs to be half your height.
White light
- White light is a mixture of colours. A prism refracts each colour by a different amount, so white light spreads into a spectrum: this is dispersion.
- Red is refracted least and violet most: red, orange, yellow, green, blue, indigo, violet. Glass slows violet light slightly more than red.
- Red has the longest wavelength and lowest frequency of visible light; violet the shortest wavelength and highest frequency. Light of a single frequency is monochromatic.
Key terms
- Angle of incidence
- the angle between the incoming ray and the normal
- Specular reflection
- reflection from a smooth surface, where parallel rays stay parallel
- Virtual image
- an image that light only seems to come from; it cannot be caught on a screen
- Dispersion
- the splitting of white light into colours because each colour refracts by a different amount
Equations
| Equation | What it says, symbols and units |
|---|---|
angle of incidence = angle of reflection | Both angles are measured from the normal. |
image distance behind the mirror = object distance in front | For a plane mirror, the image is as far behind the mirror as the object is in front of it. |
Diagram to know
Finding a plane-mirror image: an object in front of a mirror, its image the same distance behind on the normal, and two rays from the object reflecting into an eye. Draw the rays solid in front of the mirror and dashed behind it, back to the image.
Worked example
A ray of light hits a plane mirror at 35° to the mirror's surface. (a) What is the angle of reflection? (b) What is the angle between the incident and reflected rays? (c) The mirror is turned through 10°. Through what angle does the reflected ray turn?
- (a) Angle of incidence from the normal = 90° − 35° = 55°, so the angle of reflection is also 55°.
- (b) The rays are 55° either side of the normal: 2 × 55° = 110° between them.
- (c) Turning the mirror changes the angles of incidence and reflection by 10° each, so the reflected ray turns through 2 × 10° = 20°.
Answer: (a) 55° (b) 110° (c) 20°
Common mistakes
- Measuring the angles from the mirror's surface instead of from the normal.
- Saying a plane-mirror image is real, or that it is on the surface of the mirror.
- Saying a prism adds colour to white light. It only separates the colours already there.
Exam tip
When you draw a reflection, use a ruler and protractor, draw the normal as a dashed line, and put arrows on every ray. Examiners often withhold a mark for a ray with no arrow.
AP Physics and NGSS: AP Physics 1 does not cover optics; AP Physics 2 (Unit 13) treats reflection and plane mirrors with the same law. NGSS MS-PS4-2 asks students to model how light is reflected, absorbed or transmitted.
3 Snell's Law
What you need to know
Snell's Law
- For light going from air into a material, sin i ÷ sin r is the same at every angle. That constant is the material's refractive index: n = sin i ÷ sin r.
- It is the ratio of the sines that stays fixed, not the ratio of the angles: doubling i does not double r.
- A bigger refractive index gives a smaller angle of refraction, so the ray bends more.
Trapped Light
- Light going from a material with a higher n to one with a lower n (glass to air) bends away from the normal. At the critical angle c, the refracted ray runs along the surface, at 90°.
- If the angle of incidence is greater than the critical angle, no light gets out: all of it is reflected back inside. This is total internal reflection.
- Total internal reflection needs both conditions: light going from higher n to lower n, and an angle of incidence greater than c. Below c, some light is still partly reflected.
- For a material with air outside, n = 1 ÷ sin c. A bigger n gives a smaller critical angle.
Optical Fibres
- An optical fibre traps light by total internal reflection: every time the light meets the wall, it is beyond the critical angle.
- A bend that is too tight lets light leak out, because rays meet the wall at less than the critical angle. Real fibres have a cladding of glass with a slightly lower n.
- Fibres carry telephone, television and internet signals as pulses of light, and bundles of fibres in an endoscope let doctors see inside the body.
Key terms
- Snell's law
- for one material, sin i ÷ sin r is constant and equal to its refractive index
- Critical angle
- the angle of incidence inside a material for which the refracted ray travels along the surface
- Total internal reflection
- the complete reflection of light inside a material when the angle of incidence exceeds the critical angle
- Optical fibre
- a thin glass fibre that carries light by total internal reflection
Equations
| Equation | What it says, symbols and units |
|---|---|
n = sin isin r | How strongly a material bends light entering it from air.
|
n = 1sin c | Beyond the critical angle, light inside the material is totally internally reflected.
|
n = speed of light in a vacuum (or air)speed of light in the material | Refractive index says how many times slower light travels in the material.
|
Diagram to know
Three rays inside a semicircular glass block meeting the flat face: one below the critical angle (refracted out, plus a weak reflection), one at the critical angle (refracted along the surface) and one above it (totally internally reflected, with i = r).
Worked example
Light in air enters a glass block of refractive index 1.50 at an angle of incidence of 40°. (a) Find the angle of refraction. (b) Find the critical angle for this glass with air outside.
- (a) sin r = sin i ÷ n = sin 40° ÷ 1.50 = 0.643 ÷ 1.50 = 0.429.
- r = sin−1 0.429 = 25.4°. It is smaller than 40°, so the ray bent towards the normal, as it should going in.
- (b) sin c = 1 ÷ n = 1 ÷ 1.50 = 0.667, so c = sin−1 0.667 = 41.8°.
Answer: (a) 25.4° (b) 41.8°
Common mistakes
- Leaving the calculator in radians. Check that sin 30 gives 0.5 before you start.
- Putting the angles the wrong way up in n = sin i ÷ sin r: i is always the angle in air, the larger one.
- Saying total internal reflection happens when light goes from air into glass. It only happens going from glass (higher n) to air (lower n).
Exam tip
To explain how an optical fibre works, name both conditions: the light is inside the glass going towards air, and it meets the wall at more than the critical angle, so it is totally internally reflected. "The light bounces off the walls" scores nothing.
AP Physics and NGSS: AP Physics 1 does not include optics; AP Physics 2 (Unit 13) writes Snell's law between any two materials as n₁ sin θ₁ = n₂ sin θ₂, so sin c = noutside ÷ ninside. NGSS HS-PS4-5 uses fibres as an example of technology that transmits information with waves.
4 Lenses
What you need to know
Focal Length
- A converging (convex) lens brings rays parallel to the principal axis together at one point, the principal focus F.
- The focal length f is the distance from the centre of the lens to F. A more curved (fatter) lens has a shorter focal length.
- A diverging (concave) lens spreads parallel rays out as if they came from a principal focus in front of it. Light can pass either way, so each lens has a focus on both sides.
Real Images
- Two rays from the top of the object find the image: one parallel to the axis, which bends through F; and one through the centre of the lens, which goes straight on. Where they cross is the top of the image.
- Object beyond 2F: image real, inverted, diminished, between F and 2F (a camera or the eye).
- Object at 2F: image real, inverted, the same size, at 2F on the other side. Object between F and 2F: image real, inverted, enlarged, beyond 2F (a projector).
- A real image forms where rays really meet, so a screen can catch it.
The Magnifying Glass
- An object closer to a converging lens than F gives a virtual, upright, enlarged image on the same side as the object: a magnifying glass.
- To find it on a diagram, extend the two rays leaving the lens backwards as dashed lines; they meet at the image. No screen can catch it.
Fixing Eyesight
- In short sight the eye focuses light from far objects in front of the retina; a diverging lens corrects it.
- In long sight the eye focuses light from near objects behind the retina; a converging lens corrects it.
- Lens power P = 1 ÷ f, in dioptres (D), with f in metres; it is positive for a converging lens and negative for a diverging lens.
Key terms
- Principal focus
- the point where rays parallel to the principal axis meet (or seem to come from) after the lens
- Focal length
- the distance from the centre of a lens to its principal focus
- Real image
- an image formed where rays actually meet; it can be shown on a screen
- Short sight
- far objects look blurred because the eye focuses their light in front of the retina
Equations
| Equation | What it says, symbols and units |
|---|---|
P = 1f | Lens power: a more strongly curved lens has a shorter focal length and a bigger power.
Not needed for IGCSE 0625; check whether your GCSE board lists it. |
Diagram to know
A ray diagram for a converging lens with the object between F and 2F: the lens as a vertical line with outward arrowheads, F and 2F marked on both sides, one ray parallel to the axis then through F, one ray straight through the centre, and the real, inverted, enlarged image where they cross beyond 2F.
Worked example
An object 3.0 cm tall stands 30 cm from a converging lens of focal length 10 cm. A scale ray diagram (1 square = 1 cm) shows the rays crossing 15 cm behind the lens. How tall is the image, and what kind of image is it?
- The object is beyond 2F (20 cm), so expect a real, inverted, diminished image between F and 2F; 15 cm fits.
- The ray through the centre is a straight line, so heights scale with distances from the lens: image height = 3.0 × 15 ÷ 30 = 1.5 cm.
- Check with the lens equation (beyond GCSE): 1/v = 1/10 − 1/30 = 2/30, so v = 15 cm, as the drawing showed.
Answer: 1.5 cm tall; real, inverted and diminished
Common mistakes
- Drawing the ray through the centre bending at the lens. It goes straight on.
- Describing an image with only one word. Give all three: real or virtual, upright or inverted, enlarged, diminished or the same size.
- Saying short sight is corrected with a converging lens. Short sight needs a diverging lens.
Exam tip
Ray diagrams are marked on accuracy: use a sharp pencil and ruler, mark F on both sides of the lens to scale, and put arrows on the rays. Draw a virtual image and the extended rays as dashed lines.
AP Physics and NGSS: AP Physics 1 does not cover lenses; AP Physics 2 (Unit 13) does, with the thin-lens equation. NGSS HS-PS4-5 asks how devices such as cameras and glasses use waves to capture information.
5 The Lens Equation
What you need to know
Catch the Image
- With the object beyond F, a converging lens gives a real, inverted image that a screen can catch. As the object moves towards F, the image moves further away and gets bigger.
- When the object is at 2F, the image is at 2F on the other side, so the object and image distances are equal: halving that distance gives the focal length.
- A very distant object gives an image at F, which is a quick way to measure the focal length (focus a far window onto a screen).
The Lens Equation
- The lens equation 1/f = 1/u + 1/v links the focal length f, the object distance u and the image distance v, all measured from the centre of the lens (beyond IGCSE and GCSE).
- Sign convention ("real is positive"): u is positive, v is positive for a real image and negative for a virtual image, and f is negative for a diverging lens. Keep all three in the same unit.
- A graph of 1/v against 1/u is a straight line that meets both axes at 1/f.
Magnification
- Magnification = image height ÷ object height. It has no unit; above 1 the image is enlarged, below 1 diminished.
- Beyond GCSE, magnification also equals v ÷ u: the image is as many times taller as it is further from the lens. A projector far from its screen gives a big picture with the slide just beyond F.
Virtual Images
- With the object inside F, the lens equation gives a negative v: a virtual, upright, enlarged image on the same side as the object.
- The nearer the object is to F, the further away and bigger the virtual image. A magnifying glass and a VR headset both hold the object just inside F. With the object exactly at F, the image is at infinity.
Key terms
- Object distance (u)
- the distance from the object to the centre of the lens
- Image distance (v)
- the distance from the centre of the lens to the image; negative for a virtual image
- Magnification
- image height ÷ object height; a ratio with no unit
- Virtual image
- an image the rays only appear to come from; a screen cannot catch it
Equations
| Equation | What it says, symbols and units |
|---|---|
magnification = image heightobject height | How many times bigger the image is than the object; no unit. Beyond GCSE it also equals v ÷ u. AP Physics 2 writes M = −si/so, negative for an inverted image. |
1f = 1u + 1v | The thin-lens equation: any two of f, u and v give the third. Real is positive.
Beyond IGCSE 0625 and GCSE; AP Physics 2 writes it 1/so + 1/si = 1/f. |
Worked example
An object 2.0 cm tall is placed 15 cm from a converging lens of focal length 10 cm. (a) Use the lens equation to find the image distance, the magnification and the image height. (b) Where is the image if the object is moved to 6.0 cm from the lens? Use the convention that real is positive.
- (a) 1/v = 1/f − 1/u = 1/10 − 1/15 = 1/30, so v = 30 cm: positive, so a real image on the far side.
- Magnification = v ÷ u = 30 ÷ 15 = 2.0, so the image is 2.0 × 2.0 = 4.0 cm tall, real and inverted.
- (b) 1/v = 1/10 − 1/6.0 = −1/15, so v = −15 cm: a virtual image 15 cm from the lens on the same side as the object.
- Its magnification is 15 ÷ 6.0 = 2.5, and it is upright: a magnifying glass.
Answer: (a) 30 cm behind the lens, magnification 2.0, 4.0 cm tall (b) v = −15 cm: virtual, upright, 2.5 times taller
Common mistakes
- Subtracting distances (f − u) instead of their reciprocals.
- Forgetting to take the reciprocal at the end: 1/v = 0.0333 means v = 30 cm, not 0.0333 cm.
- Mixing centimetres and metres for u, v and f in the same equation.
Exam tip
When the lens equation gives a negative image distance, do not discard it as an error: say the image is virtual, on the same side of the lens as the object.
AP Physics and NGSS: AP Physics 1 does not include lenses; the thin-lens equation and signed magnification are AP Physics 2 (Unit 13) content. NGSS HS-PS4-5 links lenses to devices that capture and transmit information.
6 Concave Lenses and Light as a Wave
What you need to know
The Spreading Lens
- A diverging lens spreads parallel rays as if they came from the principal focus F in front of it. On a diagram its arrowheads point inwards.
- Wherever the object is, a diverging lens gives a virtual, upright, diminished image between F and the lens, on the same side as the object. A door peephole uses one.
- To draw it: a ray parallel to the axis bends away from the axis, in line with F in front (dash it back to F); a ray through the centre goes straight on. The image is where the dashed line meets the centre ray.
Wavefronts in Glass
- Light is a wave. When a wavefront meets glass at an angle, the end that arrives first slows first, so the wavefront turns and the ray bends towards the normal.
- Along the normal the whole wavefront slows at once, so there is no change of direction.
- The frequency does not change at a boundary, so slower waves have a shorter wavelength: λglass = λair ÷ n. The colour is set by the frequency, so the light stays the same colour.
What Gets Through
- Whether a material transmits, absorbs or reflects light depends on the wavelength. Window glass transmits visible light but absorbs most short-wavelength ultraviolet.
- Clear water transmits visible light but absorbs infrared; black plastic can block visible light yet transmit infrared, as in a TV remote's window.
- Glass fibres carry signals using infrared, which the glass transmits well.
Key terms
- Diverging lens
- a lens, thinner in the middle, that spreads parallel rays apart
- Wavefront
- a line joining points on a wave that are in step, such as the crests
- Wavelength
- the distance from one point on a wave to the same point on the next wave, in m
- Transmit
- to let a wave pass through a material
Equations
| Equation | What it says, symbols and units |
|---|---|
v = fλ | Wave speed is frequency times wavelength, for every kind of wave.
|
λglass = λairn | The frequency stays the same, so the wavelength shrinks by the same factor as the speed.
Higher-tier GCSE explains the turning of wavefronts; this equation is beyond GCSE. |
Diagram to know
Plane wavefronts meeting a glass block at an angle: the fronts closer together inside the glass, turned so the direction of travel (at 90° to the fronts) bends towards the normal.
Worked example
Yellow light of wavelength 590 nm in air enters water, refractive index 1.33. Find (a) the frequency of the light, (b) its speed in the water and (c) its wavelength in the water. Take c = 3.0 × 108 m/s.
- (a) f = c ÷ λ = (3.0 × 108) ÷ (590 × 10−9) = 5.08 × 1014 Hz. The frequency is the same in the water.
- (b) v = c ÷ n = (3.0 × 108) ÷ 1.33 = 2.26 × 108 m/s.
- (c) λ = 590 ÷ 1.33 = 444 nm. Check: v ÷ f, using the unrounded values, gives 4.44 × 10−7 m, the same 444 nm.
Answer: (a) 5.08 × 1014 Hz (b) 2.26 × 108 m/s (c) 444 nm
Common mistakes
- Saying the frequency changes when light enters glass. Only the speed and the wavelength change.
- Drawing a diverging lens giving a real image. Its image is always virtual, upright and diminished.
- Saying glass is transparent without saying to which wavelengths: it absorbs most ultraviolet.
Exam tip
To explain refraction with wavefronts, say three things in order: one side of the wavefront reaches the boundary first; that side slows down first; so the wavefront changes direction. Each step is usually a mark.
AP Physics and NGSS: AP Physics 1 has no optics; AP Physics 2 (Unit 13) uses λn = λ/n and wavefront diagrams. NGSS HS-PS4-1 is the wave-speed relationship v = fλ across different media, and HS-PS4-4 how absorbed radiation affects matter.
7 The Full Picture of Light
What you need to know
Colour
- An opaque object looks the colour of the light it reflects; it absorbs the other colours. A white object reflects all colours and a black one absorbs them all.
- Under light that contains none of its own colour, an object looks black: a red tomato in pure blue light looks black.
- A filter transmits its own colour and absorbs the rest. Red, green and blue light add: red + green looks yellow, green + blue looks cyan, red + blue looks magenta.
- Transparent materials let light through clearly, translucent ones let it through but scatter it, and opaque ones let none through.
Curved Mirrors
- A concave mirror reflects rays parallel to its axis to a focus F, halfway between the mirror and its centre of curvature: f = R ÷ 2 (beyond GCSE).
- With the object beyond F, the image is real and inverted, in front of the mirror; inside F it is virtual, upright and enlarged, behind the mirror (a make-up or shaving mirror).
- A convex mirror always gives a virtual, upright, diminished image with a wide field of view, as in a car's wing mirror.
Module Finale
- An endoscope combines the topic: a bundle of optical fibres carries light by total internal reflection, a mirror at the tip turns the view through 90°, and a lens forms a real, inverted, diminished image on the end of the bundle.
Key terms
- Opaque
- letting no light through
- Filter
- a transparent material that transmits some colours and absorbs the others
- Concave mirror
- a mirror that curves inwards and brings parallel rays to a focus
- Radius of curvature
- the radius of the sphere the mirror's surface is part of
Equations
| Equation | What it says, symbols and units |
|---|---|
f = R2 | The focus of a curved mirror is halfway to its centre of curvature.
Beyond IGCSE 0625 and GCSE (AP Physics 2). The lens equation also works for curved mirrors. |
Worked example
A stage lamp gives out light that is 50% red, 30% green and 20% blue. It passes through a magenta filter, then a red filter. (a) What percentage of the light reaches a white wall, and what colour does the wall look? (b) What colour does a green leaf look in this light?
- (a) Magenta transmits red and blue: 50 + 20 = 70% gets through the first filter.
- The red filter then transmits only the red: 50% of the lamp's light reaches the wall, and the white wall reflects it, so it looks red.
- (b) The leaf reflects only green. There is no green in the light, so it absorbs everything and looks black.
Answer: (a) 50%, and the wall looks red (b) black
Common mistakes
- Saying a filter adds or changes colour. It can only remove colours from the light that reaches it.
- Saying an object "turns" a different colour under coloured light; it only reflects whichever of its own colours is present, so it may look black.
- Mixing up concave and convex mirrors: the concave one curves in, like a spoon's bowl.
Exam tip
For a colour question, list the colours in the light at each step ("in: red, green, blue; after magenta: red, blue; after red: red"). Writing the list stops you losing marks in your head.
AP Physics and NGSS: AP Physics 1 has no optics; curved mirrors with f = R/2 and the mirror equation are AP Physics 2 (Unit 13). NGSS MS-PS4-2 covers how light is reflected, absorbed or transmitted by different materials.
Written in Physiworld's own words. The course tags come from the same mapping as the syllabus maps; equations are on the IGCSE, GCSE, AP Physics 1 and NGSS formula sheets. Values use g = 9.8 N/kg.
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