How Light Travels - Physics for Pilot Assessments

The energy and waves module established what light is: an electromagnetic wave, one band among several that run from radio at the long-wavelength end to gamma rays at the short end, all of them travelling at the same speed through empty space. This lesson takes all of that as given and asks a narrower, more practical question — how does light get from a source to your eye, and what can you predict once you know? By the end you should be able to explain a shadow from first principles, calculate a travel time or a distance from the speed of light in either direction, and say what happens to light's speed when it enters glass or water. That last fact is the hinge the rest of the module turns on.

Rays, and why straight lines are the right idealisation

A ray is a line drawn to show the direction light is travelling, with an arrowhead on it to say which way along that line the energy is going. It is a piece of bookkeeping rather than a physical object: light spreads out from a source in every direction at once, and a ray is one selected path picked out of that spreading so you can reason about it. A narrow beam — a laser pointer, sunlight through a gap in curtains — is close to a single ray made visible, and you only see it at all because dust or mist in the air scatters a little of it sideways into your eye.

What makes rays useful is that in a single uniform material, light travels in straight lines. Nothing bends it, nothing makes it curve around an obstacle, and it does not seek out corners. Every prediction in this module — where the edge of a shadow falls, where a mirror puts an image, where a lens puts its focus — is made by drawing straight lines and seeing where they cross or where they end up. So straightness is not a detail you can take on trust and move past: it is the entire method, and if it failed, none of the constructions in the later lessons would work.

You can convince yourself of it without equipment. Line up three cards with a small hole punched in each and look through them at a lamp: you see the lamp only when all three holes lie on one straight line, and nudging the middle card sideways by a millimetre blacks it out. Sighting along a rifle barrel, a surveyor's level or the edge of a ruler all rely on the same thing. And the sharpest everyday evidence is the one worth spending a figure on, because you can read the geometry straight off it — a shadow.

Shadows are the evidence

A shadow is not a thing that is cast; it is a region light failed to reach because something solid stood in the way. If light travels in straight lines, then the boundary of that region is fixed entirely by geometry: draw a line from the source past each edge of the obstacle and carry it on until it lands, and those two lines mark where the shadow begins and ends. Nothing else is needed. The size of the shadow therefore depends on where the obstacle sits between source and screen — slide it towards the source and the two lines diverge further before they land, so the shadow grows.

A small source sends one ray path past each edge, so the shadow has a single, sharply defined boundary. A larger source has to be treated as many small sources side by side: the middle of the shadow is hidden from all of them, but the flanks are hidden from some parts of the source and lit by others, which is why a real shadow has a soft border rather than a clean line.

The second panel of that figure is the part worth dwelling on, because it explains something you see every day. Treat a broad lamp as a row of small sources packed side by side; each one casts its own sharp-edged shadow, and the shadows are offset from each other because the sources are in different places. Where every one of those shadows overlaps, no light arrives at all, and you get a fully dark region — the umbra. Around it lies a band that some parts of the source can see and others cannot, so it is partly lit: the penumbra. The bigger the source, or the further the obstacle sits from the screen, the wider that partly-lit band becomes, which is why the shadow of your hand is crisp against the wall when you hold it close and blurred when you hold it away.

Example 1 — the size of a shadow

A small lamp, a card 0.5 metres tall, and a wall. The card stands exactly halfway between the lamp and the wall. How tall is the card's shadow on the wall?

Draw the two rays that fix the shadow: one from the lamp past the top edge of the card, one past the bottom edge, each carried straight on until it lands on the wall.
By the time those rays reach the wall they have travelled twice as far from the lamp as they had when they grazed the card, because the card is halfway. Straight lines spreading from a point separate in proportion to distance, so the gap between the two rays doubles.
The gap at the card is the card's own height, 0.5 m, so the gap at the wall — the shadow — is 2 × 0.5 = 1 metre tall.
The same reasoning runs in reverse: slide the card towards the wall and the rays have less distance left in which to spread, so the shadow shrinks towards the card's own size. Nothing here needed anything beyond straight lines.

Light does not need anything to travel through

Sound, as the waves lesson set out, is a pressure disturbance passed from one particle of a material to the next, so it cannot cross a vacuum: take the air out of a jar containing a ringing bell and the sound dies away while the bell is plainly still shaking. Light is not like that. Put a lamp in the same jar and pump the air out, and the lamp goes on shining just as brightly. An electromagnetic wave does not need particles to pass the disturbance along — it is a self-sustaining oscillation that carries itself. This is why sunlight reaches you across a hundred and fifty million kilometres of near-empty space and why the Sun is silent.

That contrast is worth holding onto in exactly those terms, because it is one of the most reliably examined facts in the whole module and the wrong answer is the intuitive one. Sound needs a medium and travels fastest in solids, slower in liquids, slower still in gases, and not at all in a vacuum. Light needs no medium, travels fastest in a vacuum, and is slowed by every material it passes through. The two behave in opposite directions, and an option that says light travels faster in water than in air is borrowing the rule for sound and applying it to the wrong wave.

The speed of light

In a vacuum, light travels at 299,792,458 metres per second. Nothing in these tests requires that figure; the working value is 3 × 10⁸ metres per second, three hundred million, and the approximation is good to about one part in a thousand. The notation 3 × 10⁸ means 3 followed by eight zeros — the power of ten just counts how many places the decimal point moves to the right — and it is used because writing 300,000,000 out longhand invites miscounting. The symbol for this speed is a lower-case c, and it is the same for every electromagnetic band: radio waves, microwaves, infrared, visible light, ultraviolet, X-rays and gamma rays all cross a vacuum at exactly c. They differ in wavelength and frequency, not in speed.

Because the speed is constant, travel time and distance are tied together by the ordinary speed relationship from the mechanics module, with c in place of the speed.

Distance travelled by light

d = c\,t

$d$: the distance the light travels, in metres (m)
$c$: the speed of light in a vacuum, about 3 × 10⁸ metres per second (m/s)
$t$: the time taken, in seconds (s)

One relationship, two ways round. If a question gives you a distance and wants a time, divide both sides by c: the c's on the right cancel and you are left with $t = \frac{d}{c}$ . If it gives you a time and wants a distance, use it as it stands and multiply. The two examples below do one each, because that is the pair of arrangements items are built from.

Example 2 — how long does sunlight take to reach us?

The Sun is about 1.5 × 10¹¹ metres away. Light leaves its surface at 3 × 10⁸ metres per second. How long is it in transit?

The unknown is the time, so rearrange d = c t by dividing both sides by c, which gives $t = \frac{d}{c}$ .
Substitute: $t = \frac{1.5 \times 10^{11}}{3 \times 10^8}$
Handle the numbers and the powers of ten separately. $\frac{1.5}{3} = 0.5$ , and $\frac{10^{11}}{10^8} = 10^3$ , because dividing powers of ten means subtracting the indices: 11 − 8 = 3.
So t = 0.5 × 10³ = 500 seconds.
Convert to minutes by dividing by 60: $\frac{500}{60} \approx 8.3\ \text{min}$ That is a little over eight minutes. Sunlight you see now left the Sun eight minutes ago.

Example 3 — the same relationship the other way round

Light reflected from the Moon takes about 1.28 seconds to reach the Earth. How far away is the Moon? Here the time is given and the distance is missing, so no rearrangement is needed at all.

Use d = c t directly: $d = 3 \times 10^8 \times 1.28$
3 × 1.28 = 3.84, and the power of ten is untouched, so d = 3.84 × 10⁸ metres.
That is 384 million metres, or about 384,000 kilometres — divide by 1000 to go from metres to kilometres, because a kilometre is a thousand metres.
Check the size against Example 2. The Moon's light takes about a second and the Sun's takes about 500, so the Sun should be roughly 400 times further away: $400 \times 384{,}000\ \text{km} \approx 1.5 \times 10^8\ \text{km}$ That is the 1.5 × 10¹¹ metres we started from. The two answers agree.

Light slows down inside a material

The value c is the speed in a vacuum, and it is the fastest light ever goes. Put any material in the way and it travels more slowly: about 2.25 × 10⁸ metres per second in water, about 2 × 10⁸ in ordinary glass, and slower still in diamond at around 1.24 × 10⁸. Air slows it so little — by about three parts in ten thousand — that for every purpose in this module you can treat light in air as travelling at c. The mechanism is that the wave interacts with the electrons in the material, is absorbed and re-radiated a vast number of times on the way through, and the cumulative delay from all that handling shows up as a lower speed for the wave as a whole.

One detail about that slowing matters later, and it follows from the wave equation you already have: wave speed equals frequency times wavelength. When light crosses into glass, its frequency does not change — the wave arriving at the boundary and the wave leaving it oscillate the same number of times per second, because one drives the other. So if the speed falls and the frequency is fixed, the wavelength must fall in the same proportion. Light of wavelength 600 nanometres in air — a nanometre being a billionth of a metre, which is the convenient unit at this scale — has a wavelength of about 400 nanometres inside glass of the sort described above, while remaining, in every sense that matters to your eye, the same orange-red light.

Example 4 — the wavelength inside the glass

Red light of wavelength 660 nanometres in air enters glass, where light travels at 2 × 10⁸ metres per second instead of 3 × 10⁸. What is its wavelength inside the glass?

The frequency cannot change at the boundary — the wave inside the glass is driven by the wave arriving at it, oscillation for oscillation.
Wave speed equals frequency times wavelength, so with the frequency fixed, the wavelength must change in exactly the same proportion as the speed.
The speed falls from 3 × 10⁸ to 2 × 10⁸ metres per second — that is, to two thirds of its value in air.
So the wavelength falls to two thirds as well: 660 × 2/3 = 440 nanometres.
The light is still red, because the eye responds to frequency and the frequency has not changed. Only the spacing of the wave is compressed while it is inside the glass, and it returns to 660 nanometres the moment the light leaves.

This is the fact the next two lessons are built on, so it is worth stating flatly: light changes speed when it crosses from one material into another, and a change of speed at an angled boundary changes the direction of the beam. That is refraction, and it is Lesson 3. Reflection, in Lesson 2, comes first because it needs only the straight-line travel you already have — and because the way angles are measured there is the convention refraction uses too.

Why it matters in aviation

Straight-line travel is the assumption underneath visual flight. A pilot can see a runway, a light or another aircraft only when nothing solid stands on the line between it and the eye, which is why the geometry of an approach is described in terms of what is visible from a given position rather than in terms of what happens to be nearby. It is also why visibility is treated as a limiting quantity at an airfield: runway visual range is a measured figure reported to crews, and it exists because the distance over which light from the runway lights survives the intervening air decides whether an approach may be begun or continued. Cloud, fog and rain do not bend light around themselves; they scatter and absorb it, so what lies behind them simply cannot be seen.

The same straight-line behaviour applies to signals rather than to visible light. Very high frequency radio travels in effectively straight lines too, so it does not follow the curve of the Earth, and the range at which a station can be worked depends on altitude: the higher the aircraft, the further the unobstructed line reaches before the surface gets in the way.

What the test asks

Three misconceptions supply most of the wrong options here. The first is that light, like sound, needs a medium — expect an item about a bell and a lamp in an evacuated jar, or about why space is silent, with an option claiming light cannot cross a vacuum either. The second is the speed ordering: an option will offer glass or water as the medium in which light travels fastest, borrowing the rule that genuinely does hold for sound. Light is fastest in a vacuum, full stop. The third is that different colours travel at different speeds in a vacuum, or that X-rays outrun radio waves; in a vacuum every electromagnetic band travels at exactly c, and only inside a material do the speeds differ at all.