Light: sources, rays and shadows
By the end of this lesson you will be able to
- Distinguish luminous from non-luminous objects, and describe light as a form of energy that can be transferred to other stores (solar cells, photosynthesis, warming a surface).
- State that light travels in straight lines at 300 000 km/s in air, and use rays and ray diagrams to represent it.
- Explain how shadows form, why they have a dark umbra and a fuzzy penumbra, and how the same geometry produces solar and lunar eclipses.
- Draw the ray diagram for a pinhole camera and calculate the image height using similar triangles.
- Apply these ideas to an unfamiliar situation and evaluate a claim about it — Criterion A at Year 3 level.
The puzzle
Close the curtains of a room on a bright afternoon until only a single small gap is left — a hole the size of a pea. Wait a minute for your eyes to adjust. On the wall opposite the gap a faint picture appears: the trees outside, a car going past, a person walking — all in colour, all moving, and all upside down. Nothing in the room is drawing this picture. There is no lens, no screen, no electricity. A thousand years ago a scholar in Cairo sat in exactly such a darkened room and worked out, from that inverted picture, the single most important fact about light. What is it?
Predict what would happen to the picture on the wall if you made the hole bigger, then if you moved the wall further from the hole, then if you covered the hole with red cellophane. Then guess: why is the picture upside down?
Compare your answer
Bigger hole: brighter but blurrier — each point outside now lights a small patch on the wall instead of a single spot. Wall further away: the picture grows, but gets dimmer, because the same light is spread over more wall. Red cellophane: the picture turns red — only red light gets through. And it is upside down because light travels in straight lines: light from the top of a tree has to travel downwards to get through the hole, so it lands low on the wall; light from the ground travels upwards and lands high. The crossing at the hole flips everything. Every camera ever made, including the one in your phone, starts from that fact.
The science — structured notes
1.1 Luminous and non-luminous — where light comes from
A luminous object makes its own light: the Sun and every other star, a flame, a lightning flash, a glowing filament, an LED, a firefly, a phone screen. A non-luminous object does not make light; you see it only because it reflects light from a luminous source into your eye. The Moon is the classic example — it shines only with reflected sunlight — and so is everything in this room except the lamp and the screen. This page, your hand, the desk: all non-luminous, all visible only because light from somewhere else bounces off them.
| Source | How the light is produced | Energy store it comes from |
|---|---|---|
| Sun and stars | Nuclear fusion in the core heats the surface to about 5500 °C; hot matter glows | Nuclear |
| Candle, gas flame, paraffin lamp | Burning heats soot particles until they glow yellow-white | Chemical (the fuel) |
| Filament bulb | Current heats a tungsten wire to 2500 °C | Electrical (only 5 % becomes light; the rest is heat) |
| LED, phone screen | Electrons in a semiconductor drop to a lower energy level and release the difference as light | Electrical (up to 40 % becomes light) |
| Firefly, glow-worm, deep-sea fish | A chemical reaction (luciferin + oxygen) releases light without heat — bioluminescence | Chemical |
| Lightning | Air heated to 30 000 °C by the current glows | Electrical (charge separated in the cloud) |
1.2 Light is a form of energy
Light carries energy from the source to whatever absorbs it. You can feel it: stand in the Botswana sun and your skin warms, because the light energy is transferred to the thermal store of your skin. You can measure it: a light meter reads the energy arriving per second per square metre, in W/m², and in Gaborone at noon in October that is about 1000 W/m² — a kilowatt on every square metre of ground. And you can convert it:
1.3 Light travels in straight lines — and fast
In air, water or glass — any single uniform material — light travels in straight lines. You can see it: sunbeams through dusty air, a laser pointer through mist, a torch beam in smoke, all dead straight. This is called rectilinear propagation, and the whole of this unit’s optics rests on it. We represent light with a ray: a straight line with an arrow showing the direction of travel. A beam is a bundle of rays — parallel (a laser, sunlight), diverging (a torch) or converging (after a lens).
Light is also astonishingly fast. In a vacuum, and very nearly in air, it travels at 300 000 km/s (3.0 × 108 m/s) — seven and a half times round the Earth in one second. Nothing travels faster.
| Journey | Distance | Time for light |
|---|---|---|
| Gaborone to Francistown | 430 km | 0.0014 s — 1.4 milliseconds |
| Once round the Earth | 40 000 km | 0.13 s |
| Moon to Earth | 384 000 km | 1.3 s — you see the Moon as it was 1.3 s ago |
| Sun to Earth | 150 000 000 km | 500 s = 8 min 20 s |
| Nearest star (Proxima Centauri) to Earth | 40 000 000 000 000 km | 4.2 years |
| Across the Milky Way | about 1018 km | about 100 000 years |
The Sun is 150 million km away. Calculate the time light takes to reach the Earth.
time = distance ÷ speed. Convert first: 150 000 000 km and 300 000 km/s are in matching units, so no conversion is needed.
time = 150 000 000 km300 000 km/s = 500 s, which is 8 minutes 20 seconds.
So sunlight arriving now left the Sun 8 min 20 s ago. If the Sun went out this instant you would not know for another eight minutes — and the sky would stay lit until the last light had crossed the gap.
“When you look at a star, you are seeing it as it is now.”
Reveal answer
False. Light travels at a finite speed, 300 000 km/s, and the stars are so far away that their light takes years to reach you — 4.2 years from the nearest, thousands of years from many you can see from a dark Kalahari sky. You see each star as it was when the light left it. A telescope is a time machine; the further it looks, the further back it sees.
1.4 Shadows — umbra, penumbra and eclipses
Because light travels in straight lines, an opaque object (one that light cannot pass through) blocks the rays that hit it, and behind it is a region no light reaches: a shadow. A transparent material (clear glass, water, air) lets light through in straight lines, so you can see through it; a translucent one (frosted glass, tracing paper, thin cloth) lets light through but scatters it, so you see a glow but no image.
A solar eclipse is the Moon’s shadow falling on the Earth. The Moon is an opaque ball lit by an extended source, so its shadow has an umbra — a dark cone only about 100–270 km wide where it touches the Earth, inside which the Sun is completely hidden (a total eclipse) — and a much wider penumbra, thousands of kilometres across, inside which part of the Sun is hidden (a partial eclipse). A lunar eclipse is the reverse: the Moon passes through the Earth’s shadow and turns a dim copper-red, because the only light reaching it has been bent through the Earth’s atmosphere. On 25 November 2030 the umbra of the Moon will cross southern Botswana, passing from Namibia over the Kgalagadi — Tsabong lies close to the centre line — and on into South Africa, and for around two minutes the Sun will vanish completely: the first total eclipse over the country since December 2002.
A 1.5 m tall student stands 2 m from a small, bright lamp at floor level in a dark room; the wall is 6 m from the lamp.
(a) Draw the ray from the lamp past the top of her head and use similar triangles to find the height of her shadow on the wall.
(b) The small lamp is replaced by a large frosted lamp of the same brightness. Describe two changes in the shadow.
(c) She walks towards the wall. Describe what happens to the size and the edge of the shadow.
Reveal answer
(a) The ray from the lamp past the top of her head continues in a straight line to the wall. Triangles: height ÷ distance is the same for both, so shadow height = 1.5 m × 62 = 4.5 m — three times her height, because the wall is three times as far from the lamp as she is.
(b) The edges become blurred: a dark umbra in the middle surrounded by a grey penumbra, because light from different parts of the wide lamp reaches past her head at slightly different angles. The umbra is a little smaller than the sharp shadow was, because the outer parts of the lamp light up its edges.
(c) The shadow shrinks (she blocks a narrower cone of rays as she moves away from the lamp; at the wall it would be exactly her size) and its edge gets sharper, because the penumbra is narrower when the object is close to the screen.
1.5 The pinhole camera
The darkened room in the puzzle is a camera obscura — Latin for “dark room”, and the origin of the word camera. A shoebox version with a pinhole in one end and tracing paper over the other is a pinhole camera. Every point on a bright object sends light in all directions, but only one narrow ray from each point can get through the pinhole, and it lands on one spot of the screen. The ray from the top lands at the bottom, the ray from the bottom lands at the top, and the ray from the left lands on the right. The image is therefore inverted, real (it is genuinely on the screen — you can photograph it), dim (the hole lets through very little light) and sharp if the hole is small.
The ray from the tip of the flame passes straight through the pinhole to the screen. It makes two triangles — one outside the box with height H and base u, one inside with height h and base v — and because they share the angle at the pinhole they are similar: same shape, different size. Similar triangles have proportional sides, so:
image height hobject height H = image distance vobject distance u
A candle flame 10 cm tall stands 50 cm in front of a pinhole camera whose box is 20 cm long. Find the height of the image, and what happens to it if the candle is moved to 100 cm.
Similar triangles: h = H × vu = 10 cm × 20 cm50 cm = 10 × 0.4 = 4 cm (40 mm), inverted.
At 100 cm: h = 10 × 20100 = 2 cm. Doubling the object distance halves the image: image height is inversely proportional to object distance, as long as the box length stays the same.
Check the direction: further away, smaller — which matches what your eye sees, because the eye is a pinhole camera with a lens (Lesson 12.5).
Under a leafy tree at midday the ground is covered with small round patches of light. During a partial solar eclipse, every patch becomes a crescent. Explain both observations.
Reveal answer
The tiny gaps between the leaves are pinholes. Each one projects an image of the Sun onto the ground — a disc, because the Sun is a disc — and because the leaves are metres above the ground the images are large enough to see. During a partial eclipse the Sun is a crescent, so every pinhole projects a crescent. The gaps are not round, but a pinhole’s image has the shape of the source, not the hole, as long as the hole is small compared with the distance to the ground. This is the safest way to watch the 2030 eclipse: stand under a tree, or punch a hole in a card, and look at the ground.
A student writes: “The image in a pinhole camera is upside down because the light bounces off the back of the box. If you make the box longer the image gets smaller because the light is weaker by the time it arrives.” Identify three errors.
Reveal answer
The image is inverted because rays travel in straight lines and cross at the pinhole, not because anything bounces — nothing reflects inside the box [1]. A longer box makes the image larger, not smaller: h = H × v ÷ u, and v has increased [2]. “Weaker light” affects brightness, not size — the image in a longer box is indeed dimmer, because the same light is spread over a larger area, but its height is set by geometry alone [3].
Try it — the pinhole camera and the similar triangles
Stand a candle in front of the pinhole box and read the image height on the tracing-paper screen. Move the candle from 20 cm to 100 cm in 10 cm steps, three readings each, and plot image height against distance. Then change the box length (10, 20, 30 cm) and repeat one point; finally swap the small pinhole for a large one and see what changes — and what does not.
Suppose your means with a 20 cm box were: 20 cm → 100 mm, 40 → 51, 60 → 33, 80 → 25, 100 → 20 mm.
(a) Describe the relationship, quoting values, and state whether image height is inversely proportional to object distance.
(b) Explain the relationship using straight-line rays.
(c) A group measured with the large pinhole and got 20 cm → 104 mm, 40 → 47, 60 → 36 mm. Explain why their readings scatter more and whether the mean is affected.
Reveal answer
(a) As object distance increases from 20 to 100 cm the image height falls from 100 mm to 20 mm, steeply at first (100 to 51 mm between 20 and 40 cm) and slowly at the end (25 to 20 mm between 80 and 100 cm). Doubling the distance from 20 to 40 cm halves the height (100 to 51 mm), and from 40 to 80 cm halves it again (51 to 25 mm), and height × distance is roughly constant (2000, 2040, 1980, 2000, 2000 mm cm), so it is inverse proportion within the precision of the ruler.
(b) The ray from the flame tip travels in a straight line through the pinhole. The angle it makes with the axis is set by the flame height over the object distance; the same angle continues inside the box, so the image height is that angle times the box length. Similar triangles: h ÷ H = v ÷ u, so with H and v fixed, h is proportional to 1 ÷ u.
(c) With a large hole every point of the flame is spread into a blurred disc, so the top and bottom of the image are fuzzy and the reader has to judge where the edge is — a random error of a few millimetres in each reading, different each time. The true height is unchanged (the geometry does not involve the hole), so the mean of many readings is still close to the small-hole value; the range is wider, and the mean is less certain.
Common mistakes to avoid
- “The Moon is a light source” — non-luminous; it reflects sunlight, about 12 % of it.
- “We see because our eyes send out rays” — light enters the eye; nothing leaves it. Ibn al-Haytham settled this a thousand years ago.
- “Light travels instantly” — 300 000 km/s: 1.3 s from the Moon, 8 min 20 s from the Sun, years from the stars.
- Ray arrows missing or pointing the wrong way — every ray needs an arrow from the source towards the eye or screen.
- “A bigger pinhole gives a bigger image” — brighter and blurred; the size depends only on the two distances.
- Umbra and penumbra swapped — umbra is the dark centre (whole source hidden); penumbra is the grey edge (part of the source hidden).
- Forgetting to convert — km and km/s, or m and m/s; a mixed pair is a factor of 1000 out.
How we found out — a dark room in Cairo and a moon of Jupiter
For fifteen hundred years after the Greeks, most scholars believed that we see because the eye sends out invisible rays that touch objects — the “emission theory” of Euclid and Ptolemy. Around 1011, under house arrest in Cairo, the Iraqi-born scholar Ibn al-Haytham (known in Europe as Alhazen) set out to test it. He darkened a room, let light in through a small hole, and studied the inverted picture on the wall. He hung several lamps outside the hole and showed that each produced its own separate spot inside, in a straight line through the hole, and that covering one lamp removed only its spot. Light, he concluded, travels in straight lines from luminous objects, is reflected from non-luminous ones, and enters the eye — the eye emits nothing. His seven-volume Kitab al-Manazir (Book of Optics), finished about 1021, insisted that every claim be tested by a repeatable experiment; it was translated into Latin, read by Kepler and Galileo, and is the reason a modern camera is called a camera.
The speed of light was found by a man who was trying to make a clock. In 1676 the Danish astronomer Ole Rømer, working in Paris, was timing the eclipses of Jupiter’s moon Io as it passed into Jupiter’s shadow every 42.5 hours — a celestial clock that sailors might use to find longitude. The eclipses ran up to 11 minutes early when the Earth was closest to Jupiter and late when it was furthest away. Rømer saw the pattern: the light was taking longer to cross the extra distance. From the size of the Earth’s orbit he estimated that light took about 22 minutes to cross it — the true figure is 16.7 — and so was the first person to show that light has a finite speed at all. Most astronomers, including the great Cassini, refused to believe him; the confirmation came fifty years later from James Bradley, and a laboratory measurement had to wait until Fizeau’s toothed wheel in 1849 gave 313 000 km/s. Rømer had turned a pattern in data that everyone else called an error into the first measurement of the fastest thing there is.
Ibn al-Haytham’s experiment with several lamps is often called the first controlled experiment in optics. Explain what he changed, what he kept the same, and why covering one lamp was the decisive test.
Reveal answer
He changed one thing at a time — which lamps were lit — and kept the hole, the room and the wall the same. The emission theory predicted nothing about spots on a wall at all; the straight-line theory predicted one spot per lamp, each on the straight line from that lamp through the hole. Covering one lamp and seeing only its spot vanish was decisive because it showed that each patch of light on the wall came along a straight path from one specific source, and no other path — the spots were independent. A theory that makes a prediction which could have failed, and does not, is the pattern of evidence that a scientific claim rests on.
Where it is used now
- Solar energy in Botswana. Light as energy, counted in W/m²: rooftop solar water heaters, the Mmadinare and Jwaneng solar plants, and the village mini-grids that bring electricity to settlements far from the Morupule line. The design begins with the light energy landing on each square metre.
- Photography and cinema. The pinhole became the lens camera, the film camera, the digital sensor and the phone; the projector in a cinema is a pinhole camera run backwards, throwing an image through a lens onto a screen. Both are products of innovation that express ideas — the global context of this unit.
- Medicine. An X-ray radiograph is a shadow: bone is opaque to X-rays, flesh is nearly transparent, and the plate behind the patient records the umbra of the skeleton. A shadow picture that expresses what is inside a body.
- Eclipse science and tourism. The 2030 total eclipse over Botswana is being planned for now: astronomers choose sites along the 150 km wide path of the umbra, and the tourism industry expects tens of thousands of visitors for two minutes of darkness.
- Satellites and remote sensing. BOTSAT-1 photographs the country in reflected sunlight; the length and direction of shadows in its images give the height of buildings and the time of day.
- Sundials and the shadow stick. The oldest clock on Earth, and the Year 2 shadow lab, is the umbra of a stick traced across a day.
Science now — and the scientist you need to be
What is happening in 2026
The pinhole is still the model for every camera, but what happens after the light lands has changed completely. A phone camera in 2026 does not take one photograph; it takes a burst of ten or more and an AI model merges them, sharpening edges, brightening shadows and removing noise — “computational photography”. Astronomers do the same at the largest scale: the James Webb Space Telescope collects infrared light that left galaxies more than 13 billion years ago, and machine-learning software picks the faint, ancient objects out of the noise. Solar power has become the cheapest electricity in history in sunny countries, and the panels now convert 23 % of the light energy they catch; researchers in 2026 are testing tandem cells that reach over 30 % by stacking two materials that absorb different colours. And in the Kalahari, light-pollution maps made from satellite images show one of the darkest skies left on the planet — a resource that astronomers, and the 2030 eclipse, are now counting on.
The mindshift
What you need to be able to do
Three examinable skills from this lesson: draw a ray diagram with a ruler and arrows (pinhole or shadow), calculate a length from similar triangles with a stacked fraction and units, and explain umbra and penumbra with the words “point source” and “extended source”. Criterion A tasks in this unit will hand you an unfamiliar context — a solar eclipse, a camera obscura in a tent, a sundial — and expect you to reach for the same three ideas.
See an example
Tube 25 cm long, pinhole made with a sewing needle. A window 1.2 m tall, 3.0 m away, gave an inverted image 9.5 cm tall on the tracing paper. Calculation: 120 cm × 25 ÷ 300 = 10 cm — within half a centimetre, and the difference came from judging the blurred edge. Widening the hole with a pencil made the image bright enough to photograph but too blurred to measure. The AI said the image inverts because “light rays cross at the pinhole as they travel in straight lines” — the right mechanism.
Key terms
| Luminous | An object that produces its own light (Sun, flame, LED). Non-luminous objects are seen by reflected light. |
| Ray | A straight line with an arrow that shows the path and direction of light; a beam is a bundle of rays. |
| Rectilinear propagation | Light travels in straight lines in a uniform medium. |
| Speed of light | 300 000 km/s (3.0 × 108 m/s) in a vacuum or air; slower in water and glass. |
| Opaque / transparent / translucent | Blocks light / lets light through in straight lines / lets light through but scatters it. |
| Shadow | The region behind an opaque object that light from the source cannot reach. |
| Umbra | The dark central shadow where the whole source is hidden. |
| Penumbra | The grey edge of a shadow from an extended source, where only part of the source is hidden. |
| Eclipse | Solar: the Moon’s shadow falls on the Earth. Lunar: the Moon passes through the Earth’s shadow. |
| Pinhole camera | A box with a tiny hole; straight rays cross at the hole and form a real, inverted, dim image on a screen. |
| Similar triangles | Triangles of the same shape; their sides are in proportion, so h ÷ H = v ÷ u for a pinhole camera. |
Think like a scientist
Criterion A Read the information, then answer. On 25 November 2030 a total solar eclipse will cross Botswana. The Moon’s umbra — the dark cone of its shadow — will be about 150 km wide where it touches the ground and will sweep in from Namibia across the Kgalagadi, over Tsabong and on into South Africa at about 2000 km/h. Inside the umbra the Sun will be completely hidden for up to 2 minutes 30 seconds; across the rest of southern Africa, inside the penumbra, the Sun will appear as a crescent. A newspaper article claims: “During the eclipse the Sun’s rays become more dangerous than normal, which is why people are warned not to look.”
(a) Explain, with the terms point source and extended source, why the Moon’s shadow has both an umbra and a penumbra, and why the total eclipse is seen only from a narrow strip. [4]
(b) The Moon is 3500 km across and 384 000 km from the Earth; the Sun is 1.4 million km across and 150 million km away. Show, using ratios, why the Moon can just cover the Sun. [3]
(c) Calculate how long the umbra takes to cross a town, if the umbra is 150 km wide and moves at 2000 km/h, and compare it with the stated duration. [2]
(d) A student in Maun plans to watch the eclipse with a pinhole in a card, projecting onto a sheet of paper 1 m away. Explain what she will see and why the method is safe. [3]
(e) Evaluate the newspaper claim. Is the science correct? Give the real reason for the warning and judge whether the warning itself is justified. [3]
Model answer
(a) The Sun is an extended source: light comes from every point across its 1.4 million km disc, not from one point. Behind the Moon there is a cone-shaped region where light from every part of the Sun is blocked — the umbra — and around it a wider region where light from some parts of the Sun gets past the edge of the Moon while light from other parts is blocked — the penumbra. Because the Moon is only just big enough to cover the Sun, the umbra narrows to a tip that barely reaches the Earth, so it is only about 150 km wide on the ground. Only observers inside that strip have the whole Sun hidden; everyone else in the penumbra sees part of the Sun and so a partial eclipse.
(b) The angular size of each is diameter ÷ distance. Moon: 3500 ÷ 384 000 = 0.0091. Sun: 1 400 000 ÷ 150 000 000 = 0.0093. The two ratios are almost equal, so from the Earth the Moon and the Sun appear the same size in the sky — the Moon is 400 times smaller but 400 times nearer. The Moon can therefore just cover the Sun, which is why the umbra only just reaches the Earth and why totality is so brief; when the Moon is at the far point of its orbit its ratio is slightly smaller and a ring of Sun remains visible (an annular eclipse).
(c) Time = distance ÷ speed = 150 km ÷ 2000 km/h = 0.075 h = 4.5 minutes for the whole umbra to pass over a point. The stated maximum of 2 min 30 s is shorter because a town is not at the centre line for the whole crossing, the umbra is an ellipse rather than a band, and the speed given is a rounded figure — but the order of magnitude agrees: a few minutes.
(d) The card with the pinhole is a pinhole camera and the paper is the screen. Light from every point of the Sun travels in a straight line through the hole, so an image of the Sun’s disc forms on the paper — about 1 cm across at 1 m (1 m × 0.0093) — and during the eclipse the image is a crescent that shrinks as the Moon covers more of the Sun. It is safe because she looks at the paper, not the Sun: only the tiny fraction of light that passes through the pinhole reaches the paper, and none of it enters her eye directly from the Sun.
(e) The science in the claim is wrong: the Sun’s light is not more energetic during an eclipse — the Moon blocks part of it, so less light energy reaches the ground, not more. The real reason for the warning is that people look at the Sun during an eclipse when they would never normally do so; the crescent Sun is still far too bright, its light is focused by the eye’s lens onto the retina and burns it without pain, and the damage is permanent. The warning is fully justified — the risk comes from behaviour, not from the Sun changing — and the article should say so, because a reader who learns that the claim is false might wrongly conclude that the warning is false too.
Probe deeper — exam-style questions
Attempt every question before opening the answer. Marks in brackets show what a full answer is worth.
Classify the Moon, a candle flame, a mirror and a firefly as luminous or non-luminous, and explain how you see a non-luminous object.
Reveal answer
Luminous: candle flame, firefly; non-luminous: Moon, mirror [1]; a non-luminous object is seen because light from a luminous source reflects off it into the eye [1].
Light from the Moon takes 1.28 s to reach the Earth. Calculate the distance to the Moon (speed of light 300 000 km/s), and state how long light would take from a satellite 36 000 km up.
Reveal answer
distance = speed × time = 300 000 × 1.28 = 384 000 km [2: 1 for the method, 1 for the value with unit]; 36 000 ÷ 300 000 = 0.12 s [1].
Explain why the shadow of a football on the ground has a sharp edge when it is on the ground and a blurred edge when it is thrown high into the air on a sunny day.
Reveal answer
The Sun is an extended source, so rays from its different parts pass the ball at slightly different angles [1]; near the ground these rays have not had space to spread, so the umbra has a narrow penumbra and looks sharp [1]; high in the air the rays from opposite edges of the Sun diverge over several metres, so the penumbra widens and the shadow becomes a grey blur — at great enough height the umbra disappears altogether [1].
A pinhole camera 15 cm long is pointed at a tree 12 m tall and 60 m away. Calculate the height of the image, and state two properties of the image apart from its size.
Reveal answer
h = H × v ÷ u = 12 m × 0.15 ÷ 60 = 0.03 m = 3 cm [2: 1 for the method, 1 for the value with unit and consistent conversion]; inverted (and left-right reversed), real, dim [1 for any two].
True or false, with a reason: “In a lunar eclipse the Moon goes dark because the Earth blocks the Sun’s light from reaching it.”
Reveal answer
True [1]: the Moon moves into the Earth’s umbra, so direct sunlight cannot reach it; it does not go completely black because a little sunlight is bent through the Earth’s atmosphere and reddened, which is why an eclipsed Moon looks copper-coloured [1].
A student writes: “Solar panels work because light heats them up. The Moon gives off light at night, so panels also work under moonlight, but less.” Identify three errors.
Reveal answer
Solar cells transfer light energy directly to the electrical store (electrons freed in the semiconductor), not via heating — in fact hot panels produce less [1]; the Moon does not give off light; it reflects sunlight [1]; moonlight is about 400 000 times weaker than sunlight, so the output under it is negligible, not merely “less” [1].
Summary
- Luminous objects make light; non-luminous objects are seen by reflected light. The Moon is non-luminous.
- Light is energy: it can be transferred to electrical (solar cells), chemical (photosynthesis) and thermal stores, and counted in W/m².
- Light travels in straight lines at 300 000 km/s; we draw it as rays with arrows.
- Shadows are where light cannot reach: an umbra (whole source hidden) with a penumbra (part hidden) from an extended source; eclipses are shadows in space.
- A pinhole camera forms a real, inverted, dim image; similar triangles give h ÷ H = v ÷ u.
- Ibn al-Haytham (c. 1021) showed light enters the eye along straight lines; Rømer (1676) found its speed from a pattern of evidence others called error.
Next time → Reflection and mirrors: the two laws, why your reflection is not reversed the way you think, the periscope and the curved mirrors in a car and a torch.