WASSCE Physics · Light as a Wave

Light: straight-line travel, shadows, eclipses and the pinhole camera

By the WASSCE Tutor team

Updated 10 November 2026 · 7 min read · Prepared against the WASSCE syllabus.

You see this page because light from it enters your eyes. Light travels so fast and so straight that for centuries people thought it was instantaneous. This lesson covers the basics: how light travels, how shadows and eclipses form, and the simplest camera of all.

Learning objectives

By the end of this lesson you should be able to:

  • distinguish luminous and non-luminous objects, and transparent, translucent and opaque materials
  • describe rays and beams and demonstrate the rectilinear propagation of light
  • explain the formation of shadows, umbra and penumbra, and eclipses
  • describe the pinhole camera and calculate magnification

Sources of light

  • Luminous objects produce their own light: the Sun, a lamp, a candle flame, a firefly.
  • Non-luminous objects are seen only because they reflect light into our eyes: the Moon, this page, a person.
MaterialLight passes through?Examples
Transparentyes, clearly; objects seen sharplyclear glass, clean water, air
Translucentpartly; objects seen blurredfrosted glass, tracing paper, thin cloth
Opaquenowood, metal, a brick wall

Rays and beams

Definition

A ray is the direction (path) along which light travels, drawn as a straight line with an arrow. A beam is a stream (bundle) of rays.

BeamRays
Parallelrays stay the same distance apart (light from the distant Sun, a torch reflector)
Convergingrays come together to a point (after a convex lens)
Divergingrays spread out from a point (from a lamp or candle)

Light travels in straight lines

Definition

Rectilinear propagation of light: light travels in straight lines in a uniform medium.

Demonstration: three cards with a small hole in each are lined up in front of a candle. The flame is visible through all three holes only when the holes lie on a straight line (checked with a thread pulled tight through them). Move one card sideways and the flame disappears.

Evidence in daily life: sharp shadows; sunbeams through gaps in clouds or a window; we cannot see round corners.

Shadows

A shadow is a region that light cannot reach because an opaque object blocks it.

  • A point source gives a sharp shadow (umbra only).
  • An extended (large) source gives a dark central umbra surrounded by a lighter penumbra.

An extended light source shines on an opaque ball; behind it is a dark umbra where no light reaches, surrounded by a lighter penumbra where only part of the source is blocked

Figure 1. Umbra and penumbra from an extended source. WASSCE Tutor diagram.

Definition
  • Umbra: the region of total shadow, which receives no light from the source.
  • Penumbra: the region of partial shadow, which receives light from part of the source.

Eclipses

Eclipse of the Sun (solar eclipse): the Moon passes between the Sun and the Earth, casting its shadow on the Earth. People in the umbra see a total eclipse; in the penumbra, a partial eclipse. It can happen only at new moon.

Eclipse of the Moon (lunar eclipse): the Earth passes between the Sun and the Moon, and the Moon moves into the Earth's shadow. It can happen only at full moon.

In Sierra Leone

A total solar eclipse crossed parts of West Africa in March 2006, darkening the sky at midday. In Sierra Leone it was seen as a deep partial eclipse. Watching one safely needs proper eclipse filters or a pinhole projector — never look at the Sun directly.

The pinhole camera

A box with a pinhole in front and a screen at the back; rays from the top and bottom of an object cross at the pinhole to form a real, inverted image on the screen

Figure 2. The pinhole camera forms a real, inverted image. WASSCE Tutor diagram.

A light-tight box with a small pinhole at the front and a screen of tracing paper at the back. Rays from each point of the object travel in straight lines through the hole and form an image on the screen.

Characteristics of the image: real, inverted (upside down and laterally inverted), and it can be magnified or diminished.

Magnification

m=image heightobject height=image distanceobject distance=vum = \dfrac{\text{image height}}{\text{object height}} = \dfrac{\text{image distance}}{\text{object distance}} = \dfrac{v}{u}
m=image heightobject height=image distanceobject distance=vu\begin{aligned}m &= \dfrac{\text{image height}}{\text{object height}} \\ &= \dfrac{\text{image distance}}{\text{object distance}} \\ &= \dfrac{v}{u}\end{aligned}
Worked example

A tree 15 m tall is 60 m from a pinhole camera whose screen is 20 cm behind the pinhole. Find the height of the image.

m=vu=0.2060=1300hi=15300=0.05 m=5 cmm = \dfrac{v}{u} = \dfrac{0.20}{60} = \dfrac{1}{300} \qquad h_i = \dfrac{15}{300} = 0.05\ \text{m} = 5\ \text{cm}
m=vu=0.2060=1300hi=15300=0.05 m=5 cm\begin{gathered}\begin{aligned}m &= \dfrac{v}{u} \\ &= \dfrac{0.20}{60} \\ &= \dfrac{1}{300}\end{aligned} \\[6pt] \begin{aligned}h_i &= \dfrac{15}{300} \\ &= 0.05\ \text{m} \\ &= 5\ \text{cm}\end{aligned}\end{gathered}

Effects of changing the pinhole

ChangeEffect on the image
Moving the screen further from the holelarger, but dimmer
Enlarging the holebrighter but blurred (overlapping images)
Making several holesseveral overlapping images
Moving the object closerlarger image

Key points

  • Luminous objects emit light; non-luminous ones reflect it.
  • Light travels in straight lines (rectilinear propagation).
  • An extended source gives an umbra (total shadow) and a penumbra (partial shadow).
  • Solar eclipse: Moon between Sun and Earth; lunar eclipse: Earth between Sun and Moon.
  • Pinhole camera: real, inverted image; m=v/um = v/u.

WASSCE examination focus

WAEC tests this

Describe an experiment showing that light travels in straight lines, explain the formation of umbra and penumbra (with a ray diagram), explain eclipses with diagrams, and describe the pinhole camera — the nature of its image and the effect of enlarging the hole. Magnification calculations are frequent.

Quick check

  1. Give one example of a translucent material.
  2. What is the nature of the image in a pinhole camera?
  3. A pinhole camera 10 cm long views a 2 m tall man 20 m away. Find the image height.

Answers: 1. For example: frosted glass or tracing paper. 2. Real and inverted. 3. m=0.10/20=0.005m = 0.10 / 20 = 0.005, so the image is 2×0.005=0.012 \times 0.005 = 0.01 m = 1 cm tall.

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