Reflection, refraction and lenses

Use ray diagrams and explain changing wave direction.

Start here

The key idea

Refraction happens when a wave changes speed as it crosses a boundary.

Reflection Refraction And Lenses
incident rayrefracted rayglass boundary

Use the labels to explain the scientific relationship shown.

Revision notes

The bit that matters

Short notes first. Learn the idea, then use the worked example and questions to check it properly.

1

Reflection of light

When light hits a surface it can be reflected.The law of reflection states that the angle of incidence equals the angle of reflection, both measured from the normal (a line at 90 degrees to the surface).Specular reflection occurs at smooth surfaces and produces a clear image, while diffuse reflection from rough surfaces scatters light in all directions.Angles in reflection and refraction are always measured from the normal, not the surface.

2

Refraction of light

Refraction is the change in direction of light when it passes between media of different optical density and changes speed.g. g. leaving glass into air) it bends away from the normal.If light hits the boundary along the normal it does not bend, but its speed still changes.The wavelength changes during refraction while the frequency stays the same.

3

Convex (converging) lenses

A convex lens is thicker in the middle and brings parallel rays of light together at the principal focus.The focal length is the distance from the centre of the lens to the principal focus.A convex lens can form a real image (rays actually meet, can be projected) when the object is beyond the focal point, or a magnified virtual image (as in a magnifying glass) when the object is closer than the focal length.

4

Concave lenses and magnification

A concave (diverging) lens is thinner in the middle and spreads parallel rays out so they appear to come from a virtual principal focus.It always produces a smaller, upright, virtual image.Magnification = image height / object height (a ratio with no units).A magnification greater than 1 means the image is larger than the object; less than 1 means it is smaller.

Key terms

Definitions to learn

Normal

A construction line drawn at 90 degrees to a surface where a ray meets it.

Angle of incidence

The angle between the incoming ray and the normal.

Refraction

The change in direction of light as it changes speed crossing a boundary.

Principal focus

The point where rays parallel to the axis converge (or appear to diverge from).

Magnification

Image height divided by object height; a ratio with no units.

Worked example

Explain why a light ray bends when it enters glass from air at an angle.

1

Describe the change in speed.

2

Link the speed change to direction.

Final answer

Light slows down in glass, so the ray changes direction towards the normal.

Exam habit

Draw ray diagrams with a ruler and label the normal, incident ray, reflected/refracted ray, and all angles from the normal — not from the surface.For refraction, state both what changes (speed, wavelength) and what stays constant (frequency).

Watch out

Do not say refraction always means bending; a ray along the normal changes speed without changing direction.

Examiner tips

How to score full marks

  • 1Always measure and draw angles from the normal, never from the surface.
  • 2Real images can be projected onto a screen; virtual images cannot — state which when describing an image.
  • 3For ray diagrams use the standard rays: parallel-to-axis through focus, and straight through the lens centre.
Practice questions

Try these yourself

Start with the core skill, then open the answer only after you have attempted the full question.

1State the law of reflection.
Mark scheme
  1. 1.Compare the two angles measured from the normal.
Angle of incidence = angle of reflection.
2Describe the image produced by a convex lens when the object is beyond the focal length.
Mark scheme
  1. 1.Consider whether the rays meet.
A real, inverted image is formed.
3A wave enters a new material and slows down while frequency stays constant. What happens to wavelength?
Mark scheme
  1. 1.Use v = f lambda.
The wavelength decreases.
4A ray of light hits a mirror at an angle of incidence of 35 degrees. State the angle of reflection and explain your answer.[2 marks]
Mark scheme
  1. 1.Apply law of reflection.
  2. 2.State angle equals incidence.
Angle of reflection = 35 degrees (1) because the angle of incidence equals the angle of reflection, both measured from the normal (1).
5Describe what happens to the direction and speed of a ray of light as it passes from air into glass.[2 marks]
Mark scheme
  1. 1.Light slows down.
  2. 2.Bends towards the normal.
The light slows down as it enters the denser glass (1) and bends towards the normal (1).
6An object 2 cm tall produces an image 6 cm tall through a lens. Calculate the magnification.[2 marks]
Mark scheme
  1. 1.Use magnification = image height / object height.
  2. 2.Substitute.
magnification = image height / object height = 6 / 2 (1) = 3 (no units) (1).
7Explain the difference between a real image and a virtual image, and state which type a magnifying glass produces.[3 marks]
Mark scheme
  1. 1.Define real image.
  2. 2.Define virtual image.
  3. 3.State magnifying glass image.
A real image is formed where light rays actually meet and can be projected onto a screen (1); a virtual image is formed where rays only appear to come from and cannot be projected (1). A magnifying glass produces a magnified virtual image (1).
8An object is placed beyond the focal point of a convex lens. Describe the image formed and explain, using the behaviour of light rays, why it is a real image. Then state how moving the object closer than the focal length changes the image.[4 marks]
Mark scheme
  1. 1.Describe image (real, inverted).
  2. 2.Rays converge and meet.
  3. 3.Real because rays actually meet.
  4. 4.Closer than f gives magnified virtual upright image.
When the object is beyond the focal point the image is real and inverted (1). Rays from the object are refracted by the lens so they actually converge and meet on the far side (1); because the rays genuinely meet, the image can be projected onto a screen, so it is real (1). If the object is moved closer than the focal length, the rays diverge and the lens produces a magnified, upright, virtual image, as in a magnifying glass (1).
9A ray of light in glass strikes the glass-air boundary at an angle of incidence of 30 degrees. The angle of refraction in air is 48 degrees. Explain in which direction the ray bends and why.[3 marks]
Mark scheme
  1. 1.Light travels from glass (denser, slower) to air (less dense, faster).
  2. 2.When light speeds up it bends away from the normal.
  3. 3.Angle in air (48 degrees) is larger than in glass (30 degrees), confirming bending away from normal.
As the ray leaves glass and enters air it speeds up, because air is optically less dense than glass (1); a wave bends away from the normal when it speeds up, so the angle increases from 30 degrees in glass to 48 degrees in air (1); this is consistent with the ray bending away from the normal at the boundary (1)
10Explain what total internal reflection is and state the two conditions needed for it to occur. Give one practical application.[3 marks]
Mark scheme
  1. 1.Define total internal reflection.
  2. 2.Condition 1: light travelling from denser to less dense medium.
  3. 3.Condition 2: angle of incidence exceeds critical angle.
  4. 4.Application: optical fibres or endoscopes.
Total internal reflection occurs when light travelling in a denser medium strikes the boundary with a less dense medium and is completely reflected back into the denser medium, with no transmitted ray (1); the two conditions are: (1) the light must be travelling from a denser medium to a less dense one, and (2) the angle of incidence must be greater than the critical angle for that boundary (1); total internal reflection is used in optical fibres to transmit light (and data) along curved glass cables for internet communications and in medical endoscopes (1)
11A converging lens has a focal length of 10 cm. An object is placed 30 cm from the lens. Describe qualitatively the position and nature of the image formed, and state whether the image is magnified or diminished.[3 marks]
Mark scheme
  1. 1.Object beyond focal length so real inverted image formed on far side.
  2. 2.Object at 30 cm (3f): image forms at 15 cm on far side (using 1/v + 1/u = 1/f gives v = 15 cm).
  3. 3.Image is real, inverted.
  4. 4.Magnification = v/u = 15301\frac{5}{3}0 = 0.5 so diminished.
The object is beyond the focal point, so a real, inverted image is formed on the opposite side of the lens to the object (1); using the lens equation, the image distance is 15 cm from the lens (1); the magnification = image distance / object distance = 15 / 30 = 0.5, so the image is diminished (smaller than the object) (1)
12Compare specular and diffuse reflection. Explain why a smooth mirror gives a clear image but a matte white wall does not, using the behaviour of light rays at each surface.[3 marks]
Mark scheme
  1. 1.Specular: parallel rays remain parallel after reflection.
  2. 2.Diffuse: rough surface reflects rays in all directions.
  3. 3.Mirror: rays from one point all reflect to same point, forming clear image.
  4. 4.Wall: rays scatter randomly, no coherent image.
In specular reflection from a smooth mirror, each incident ray obeys the law of reflection and parallel incident rays remain parallel after reflection; rays from a single point on an object all reflect to a corresponding single point in the image, forming a sharp, clear image (1); in diffuse reflection from a rough surface like a wall, the microscopic bumps mean that neighbouring rays strike the surface at many different angles, so they reflect in many different directions (1); scattered rays from a single object point do not converge to a single image point, so no coherent image is formed — you see the wall uniformly bright because it scatters light towards your eyes from every direction (1)
13Describe how optical fibres use total internal reflection to transmit data as pulses of light. Explain why the fibre must be made of glass with a specific critical angle, and discuss one advantage and one limitation of fibre-optic communication compared with copper cables.[4 marks]
Mark scheme
  1. 1.Light enters one end and hits the glass-cladding boundary at angle > critical angle.
  2. 2.Total internal reflection bounces light along the fibre.
  3. 3.Specific critical angle ensures no light escapes through the sides.
  4. 4.Advantage: high data rate / low signal loss. Limitation: expensive to install / fragile.
A pulse of light is directed into one end of the optical fibre at a shallow angle so that when it strikes the glass-cladding boundary the angle of incidence is greater than the critical angle; total internal reflection occurs and the light bounces along the fibre without escaping through the sides (1); the glass must have a specific refractive index (and thus critical angle) to ensure that rays travelling along the fibre are always above the critical angle even when the fibre bends; the cladding layer (lower refractive index) ensures the critical angle condition is met (1); advantage: optical fibres can carry much higher data rates than copper cables because light can be modulated at very high frequencies, and signal loss per kilometre is very small (1); limitation: glass fibre is more expensive to manufacture and install than copper cable, and the glass is brittle and can break if bent sharply (1)
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