Cambridge IGCSE0625

Light

Physics 0625 Chapter Notes

What this chapter covers

Light - Reflection of lightLight - Refraction of lightLight - Thin lensesLight - Dispersion of light
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In this chapter

  1. 1Fundamental Properties of Light
  2. 2Reflection and Plane Mirrors
  3. 3Refraction of Light
  4. 4Refractive Index and Snell's Law
  5. 5Total Internal Reflection
  6. 6Converging Lenses and Ray Diagrams
  7. 7The Electromagnetic Spectrum

1. Fundamental Properties of Light

Light is a form of energy that travels as an electromagnetic wave. It can travel through a vacuum, like the space between the Sun and Earth, at an incredibly high speed. We can see objects because they either produce their own light (luminous objects, like the Sun or a lamp) or they reflect light from a source into our eyes (non-luminous objects, like the Moon or this page). A key principle is that light travels in straight lines, which we represent using 'rays' in diagrams. This is why shadows form behind opaque objects.

speed (v) = distance (d) / time (t)

Key term

Luminous: An object that is a source of its own light, such as the Sun or a lit light bulb.

Common pitfall

Confusing luminous objects (which create light) with bright objects (which may just be very good reflectors). The Moon is bright but non-luminous.

Fun fact

The speed of light, c (299,792,458 m/s), is a universal physical constant and the maximum speed at which any conventional matter or energy can travel through space.

Worked example 12 marks

The Sun is approximately 150,000,000 km from Earth. Given that the speed of light in a vacuum is 300,000 km/s, calculate the time it takes for sunlight to reach Earth.

  1. 1

    State the formula: time = distance / speed

  2. 2

    Substitute the given values: time = 150,000,000 km / 300,000 km/s

  3. 3

    Calculate the result: time = 500 s

  4. 4

    The time taken is 500 seconds (or 8 minutes and 20 seconds).

Worked example 23 marks

A student stands in a dark room. Explain why they can see a lit torch, but cannot see a book on the table.

  1. 1

    The lit torch is a luminous object, meaning it emits its own light. These light rays travel in a straight line directly from the torch to the student's eyes, allowing them to see it.

  2. 2

    The book is a non-luminous object. To be seen, it must reflect light from a source into the student's eyes.

  3. 3

    Since the room is dark (apart from the torch beam which may not be pointing at the book), there is no light for the book to reflect, so it cannot be seen.

Recap

  • Light is a form of energy that travels in straight lines.
  • Luminous objects emit their own light, while non-luminous objects are seen by reflected light.
  • Light can travel through a vacuum at a speed of approximately 3 x 10^8 m/s.
  • White objects reflect most wavelengths of light, while black objects absorb most wavelengths.

Quick check

  1. Give one example of a luminous object and one example of a non-luminous object.2 marks
  2. What evidence suggests that light travels in straight lines?1 mark

2. Reflection and Plane Mirrors

Reflection is the process by which light bounces off a surface. When a ray of light strikes a plane (flat) mirror, it obeys the Law of Reflection. This law states that the angle of incidence(i) is equal to the angle of reflection (r). Both angles are measured relative to the 'normal', an imaginary line drawn at 90 degrees to the mirror's surface at the point of incidence. The image formed by a plane mirror is always virtual (cannot be projected onto a screen), upright, laterally inverted (left and right are swapped), and the same size as the object. The image appears to be as far behind the mirror as the object is in front of it.

Angle of incidence (i) = Angle of reflection (r)

Key term

Virtual Image: An image formed where rays of light appear to diverge from, which cannot be formed or projected onto a screen.

Examiner insight

Examiners award marks for accurately drawn ray diagrams, which must include arrows on rays to show the direction of light and use a ruler for straight lines.

Fun fact

In an ambulance, the word 'AMBULANCE' is written in reverse on the front so that drivers viewing it in their rear-view mirror can read it correctly.

Worked example 13 marks

A ray of light strikes a plane mirror. The angle between the incident ray and the mirror surface is 40°. Calculate the angle of reflection.

  1. 1

    The normal is at 90° to the mirror surface.

  2. 2

    The angle of incidence(i) is the angle between the incident ray and the normal.

  3. 3

    Calculate the angle of incidence: i = 90° - 40° = 50°.

  4. 4

    According to the Law of Reflection, angle of reflection(r) = angle of incidence (i).

  5. 5

    Therefore, the angle of reflection is 50°.

Worked example 22 marks

An object is placed 2.5 m in front of a large plane mirror. How far is the image from the object?

  1. 1

    For a plane mirror, the image distance is equal to the object distance.

  2. 2

    Image distance = 2.5 m behind the mirror.

  3. 3

    The total distance between the object and the image is the sum of the object distance and the image distance.

  4. 4

    Total distance = 2.5 m + 2.5 m = 5.0 m.

Recap

  • The Law of Reflection states that the angle of incidence equals the angle of reflection (i = r).
  • Angles of incidence and reflection are always measured from the normal line.
  • An image in a plane mirror is virtual, upright, laterally inverted, and the same size as the object.
  • The image distance in a plane mirror is equal to the object distance.

Quick check

  1. State the four main characteristics of an image formed in a plane mirror.4 marks

3. Refraction of Light

Refraction is the change in direction of a wave, such as light, as it passes from one medium to another. This happens because the speed of light changes when it enters a different material. For example, light travels more slowly in glass than in air. When light enters a denser medium (like from air to glass) at an angle, it bends towards the normal. When it enters a less dense medium (like from glass to air), it bends away from the normal. If the light ray enters along the normal (angle of incidence = 0°), it does not change direction, although its speed still changes. The angle between the refracted ray and the normal is called the angle of refraction (r).

Key term

Refraction: The bending of a light ray as it passes from one transparent medium to another due to a change in speed.

Common pitfall

Incorrectly showing the ray bending away from the normal when entering a denser medium, or forgetting that the emergent ray from a parallel-sided block is parallel to the incident ray.

Worked example 13 marks

A ray of light travels from air into a rectangular glass block. Draw a diagram to show the path of the ray as it enters the block at an angle and then emerges from the other side.

  1. 1

    Draw a rectangle to represent the glass block and a line for the incoming ray (incident ray) hitting the surface at an angle.

  2. 2

    At the point of entry, draw the normal (a dashed line at 90° to the surface).

  3. 3

    Show the ray bending towards the normal inside the glass block. This is the refracted ray.

  4. 4

    Extend this ray to the opposite face of the block. At the point of exit, draw another normal.

  5. 5

    Show the ray bending away from the second normal as it emerges back into the air. The emergent ray should be parallel to the original incident ray.

Recap

  • Refraction is the bending of light as it changes medium and speed.
  • Light bends towards the normal when entering a denser medium.
  • Light bends away from the normal when entering a less dense medium.
  • No refraction occurs if the light enters a new medium at an angle of incidence of 0°.
  • The change in direction is caused by the change in the speed of light.

Quick check

  1. In which direction does light bend when it passes from water into air?1 mark

4. Refractive Index and Snell's Law

The refractive index(n) of a material is a measure of how much it slows down light and causes it to bend. A higher refractive index means a slower speed of light in that medium and more bending. The refractive index of a vacuum is exactly 1. For air, it's very close to 1. The relationship between the angle of incidence (i), the angle of refraction (r), and the refractive indices of the two media (n1 and n2) is described by Snell's Law. For light entering a material from air (or a vacuum), the law simplifies. The refractive index can also be defined as the ratio of the speed of light in a vacuum(c) to the speed of light in the medium (v).

n = sin(i) / sin(r)

n = c / v

n1 * sin(θ1) = n2 * sin(θ2)

Key term

Refractive Index (n): A dimensionless number that describes how fast light travels through a material, defined as the ratio of the speed of light in a vacuum to the speed of light in the medium.

Examiner insight

Examiners expect you to show the formula, the substitution, and the final answer with correct units (if any). Make sure your calculator is in degrees mode when using sin(i) and sin(r).

Worked example 13 marks

A ray of light enters a glass block from air at an angle of incidence of 45°. The angle of refraction in the glass is 28°. Calculate the refractive index of the glass.

  1. 1

    State Snell's Law for light entering from air: n = sin(i) / sin(r).

  2. 2

    Substitute the given values: n = sin(45°) / sin(28°).

  3. 3

    Calculate the sine values: n = 0.7071 / 0.4695.

  4. 4

    Calculate the final result: n ≈ 1.51.

  5. 5

    The refractive index of the glass is 1.51 (to 3 significant figures).

Worked example 23 marks

The refractive index of diamond is 2.42. The speed of light in a vacuum is 3.0 x 10^8 m/s. Calculate the speed of light in diamond.

  1. 1

    State the formula relating refractive index and speed: n = c / v.

  2. 2

    Rearrange the formula to find the speed in the medium (v): v = c / n.

  3. 3

    Substitute the values: v = (3.0 x 10^8 m/s) / 2.42.

  4. 4

    Calculate the result: v ≈ 1.24 x 10^8 m/s.

  5. 5

    The speed of light in diamond is 1.24 x 10^8 m/s.

Recap

  • Refractive index (n) is a measure of a material's optical density.
  • Snell's Law relates the angles of incidence and refraction: n = sin(i) / sin(r).
  • Refractive index can also be calculated using speeds: n = (speed of light in vacuum) / (speed of light in medium).
  • A higher refractive index means a slower speed of light and greater bending towards the normal.

Quick check

  1. If light passes from water (n=1.33) to glass (n=1.5), will it bend towards or away from the normal?1 mark

5. Total Internal Reflection

When light travels from a denser medium to a less dense medium (e.g., from glass to air), it bends away from the normal. As the angle of incidence(i) increases, the angle of refraction(r) also increases, until r reaches 90°. The specific angle of incidence that causes an angle of refraction of 90° is called the critical angle (c). If the angle of incidence is greater than the critical angle, the light does not refract out of the medium at all. Instead, it is completely reflected back into the denser medium. This phenomenon is called Total Internal Reflection (TIR). Two conditions must be met for TIR to occur: 1. Light must be travelling from a denser to a less dense medium. 2. The angle of incidence must be greater than the critical angle.

sin(c) = 1 / n

sin(c) = n2 / n1 (where n1 > n2)

Key term

Critical Angle (c): The angle of incidence in a denser medium for which the angle of refraction in the less dense medium is 90°.

Examiner insight

When asked to explain TIR, always state the two necessary conditions clearly for full marks.

Worked example 13 marks

The refractive index of a type of glass is 1.50. Calculate the critical angle for this glass when in contact with air (n ≈ 1).

  1. 1

    State the formula for the critical angle: sin(c) = 1 / n.

  2. 2

    Substitute the value of n: sin(c) = 1 / 1.50.

  3. 3

    Calculate the value of sin(c): sin(c) = 0.6667.

  4. 4

    Find the angle c by taking the inverse sine: c = sin⁻¹(0.6667).

  5. 5

    Calculate the result: c ≈ 41.8°.

Worked example 23 marks

Explain how total internal reflection is used in optical fibres for telecommunications.

  1. 1

    An optical fibre consists of a very thin, flexible core of high refractive index glass, surrounded by cladding of a lower refractive index glass.

  2. 2

    Light signals (pulses of laser light) are sent into the core at an angle of incidence greater than the critical angle.

  3. 3

    As the light travels down the fibre and strikes the core-cladding boundary, it undergoes total internal reflection repeatedly.

  4. 4

    This traps the light within the core, allowing it to travel over long distances with very little loss of signal strength or information.

Recap

  • Total Internal Reflection (TIR) occurs when light is completely reflected back into a denser medium.
  • Two conditions for TIR: moving from denser to less dense medium, and angle of incidence > critical angle.
  • The critical angle (c) is the angle of incidence that gives an angle of refraction of 90°.
  • The formula for the critical angle is sin(c) = 1/n.
  • Optical fibres and periscopes use TIR to guide light.

Quick check

  1. A ray of light in water (n=1.33) hits the surface with air at an angle of incidence of 55°. The critical angle is 48.8°. What happens to the ray?2 marks

6. Converging Lenses and Ray Diagrams

A lens is a piece of transparent material shaped to refract light in a specific way. A converging lens (or convex lens) is thicker in the middle than at the edges. It causes parallel rays of light to converge (come together) at a single point called the principal focus (or focal point), F. The distance from the center of the lens to the principal focus is the focal length (f). We can predict the position and nature of an image formed by a converging lens by drawing a ray diagram. Three key rays are used: 1. A ray parallel to the principal axis refracts through the lens and passes through the principal focus (F). 2. A ray passing through the center of the lens continues in a straight line without deviation. 3. A ray passing through the principal focus on the object side emerges parallel to the principal axis after refracting through the lens. The image is formed where these rays (or their virtual extensions) intersect.

Key term

Principal Focus (F): The point on the principal axis where rays of light initially parallel to the axis converge after passing through a converging lens.

Examiner insight

For ray diagrams, you must use a ruler and draw at least two of the three standard rays accurately to locate the image. Label the object, image, and principal foci.

Common pitfall

Confusing real and virtual images. A real image is formed where light rays actually meet and can be projected; a virtual image is where they only appear to meet.

Worked example 13 marks

An object is placed in front of a converging lens at a distance greater than twice the focal length (beyond 2F). Describe the characteristics of the image formed.

  1. 1

    A ray diagram would show that rays from the object converge to form an image on the other side of the lens.

  2. 2

    The image is formed between F and 2F on the other side.

  3. 3

    The image is real (it can be projected onto a screen).

  4. 4

    The image is inverted (upside down).

  5. 5

    The image is diminished (smaller than the object).

Worked example 23 marks

Explain how a converging lens can be used as a magnifying glass.

  1. 1

    To act as a magnifying glass, the object must be placed closer to the lens than the principal focus (within the focal length).

  2. 2

    Drawing a ray diagram shows that the refracted rays diverge (spread out) and do not meet on the other side.

  3. 3

    However, when these rays are traced backwards, they appear to come from a point on the same side of the lens as the object.

  4. 4

    This forms a virtual, upright, and magnified image, which is what we see when we look through the lens.

Recap

  • A converging (convex) lens is thicker in the middle and brings parallel rays to a focus.
  • The focal length is the distance from the lens center to the principal focus.
  • An object placed beyond 2F forms a real, inverted, diminished image.
  • An object placed between F and 2F forms a real, inverted, magnified image.
  • An object placed inside F forms a virtual, upright, magnified image (magnifying glass).

Quick check

  1. What type of image is formed by a camera lens on the sensor?1 mark
  2. To get a magnified, virtual image with a convex lens, where must the object be placed?1 mark

7. The Electromagnetic Spectrum

Visible light is just a small part of a much larger family of waves called the electromagnetic (EM) spectrum. All EM waves are transverse waves that transfer energy from a source to an absorber, and they can all travel through a vacuum at the same speed: the speed of light (3.0 x 10^8 m/s). The different types of EM waves are distinguished by their wavelength and frequency. The spectrum, in order of increasing frequency (and decreasing wavelength), is: Radio waves, Microwaves, Infrared, Visible light, Ultraviolet, X-rays, and Gamma rays. The relationship between wave speed (v), frequency (f), and wavelength (λ) is given by the wave equation.

v = f x λ

Key term

Electromagnetic Spectrum: The continuous range of all possible frequencies of electromagnetic radiation, from low-frequency radio waves to high-frequency gamma rays.

Fun fact

Your body emits infrared radiation, which is why you can be seen with thermal imaging cameras even in complete darkness.

Worked example 13 marks

Yellow light has a wavelength of approximately 6 x 10⁻⁷ m. Calculate its frequency. The speed of light is 3 x 10⁸ m/s.

  1. 1

    State the wave equation: v = f x λ.

  2. 2

    Rearrange the formula to find frequency: f = v / λ.

  3. 3

    Substitute the values: f = (3 x 10⁸ m/s) / (6 x 10⁻⁷ m).

  4. 4

    Calculate the result: f = 0.5 x 10¹⁵ Hz, which is 5 x 10¹⁴ Hz.

  5. 5

    The frequency of yellow light is 5 x 10¹⁴ Hz.

Worked example 23 marks

State two properties that are common to all types of electromagnetic waves, and one property that is different between radio waves and X-rays.

  1. 1

    Common property 1: All electromagnetic waves travel at the same speed in a vacuum (the speed of light, 3 x 10⁸ m/s).

  2. 2

    Common property 2: All are transverse waves and can travel through a vacuum.

  3. 3

    Difference: X-rays have a much higher frequency (and therefore higher energy) and a much shorter wavelength than radio waves.

Recap

  • All EM waves travel at the speed of light in a vacuum.
  • The order of the EM spectrum is Radio, Microwave, Infrared, Visible, UV, X-ray, Gamma ray.
  • Frequency increases and wavelength decreases as you move along the spectrum from radio to gamma.
  • The wave equation v = f x λ applies to all EM waves.
  • Higher frequency EM waves, like UV, X-rays and gamma rays, are ionising and can be harmful to cells.

Quick check

  1. Which type of EM wave has the longest wavelength?1 mark
  2. Which type of EM wave is found between microwaves and visible light?1 mark

End-of-chapter exercise

Test yourself on the whole chapter. Work through these before moving on.

  1. A ray of light travels from air into a rectangular glass block with a refractive index of 1.5. The angle of incidence is 60°. Calculate the angle of refraction inside the block. [3]3 marks
  2. An object of height 3 cm is placed 15 cm in front of a converging lens with a focal length of 10 cm. By drawing an accurate ray diagram, determine the position, height, and nature of the image formed. [5]5 marks
  3. Explain the two conditions necessary for total internal reflection to occur. Give one practical application of this phenomenon. [3]3 marks
  4. A person stands 1.2 m away from a plane mirror. They then walk 0.5 m closer to the mirror. What is the new distance between the person and their image? [3]3 marks
  5. Microwaves used in a mobile phone have a frequency of 1.9 x 10⁹ Hz. Calculate their wavelength. (The speed of light is 3.0 x 10⁸ m/s). [3]3 marks
  6. The critical angle of a diamond is 24°. Explain what this means and calculate the refractive index of diamond. [3]3 marks
  7. Distinguish between a real image and a virtual image. Give one example of an optical instrument that produces each type of image. [4]4 marks
  8. A beam of white light is passed through a triangular glass prism. Describe what is observed on a screen placed after the prism and explain why this happens. [4]4 marks
  9. Name three types of electromagnetic radiation that have a higher frequency than visible light. For one of them, state a medical use and a potential danger. [4]4 marks
  10. The speed of light in a transparent liquid is 2.2 x 10⁸ m/s. Calculate the refractive index of the liquid. (The speed of light in a vacuum is 3.0 x 10⁸ m/s). [2]2 marks

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