PhysicsCore26 min read

Light

Reflection, refraction, total internal reflection and lenses

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01

Reflection and the plane mirror

Definition

Virtual image — An image formed where light rays only appear to come from. It cannot be projected onto a screen, because no light actually passes through it.

Light travels in straight lines, and when it meets a surface it bounces off. The law of reflection is short: the angle of incidence equals the angle of reflection, and both are measured from the normal — the line drawn at right angles to the surface at the point where the ray hits.

Measuring from the normal rather than from the surface itself is a convention, but it is the convention every mark scheme uses. Measuring from the surface gives you the complement of the right answer, which looks plausible and is wrong.

The image in a plane mirror has four properties worth learning as a set: it is the same size as the object, the same distance behind the mirror as the object is in front, laterally inverted (left and right swapped), and virtual.

Real or virtual: one test

A real image forms where light rays actually meet, so it can be caught on a screen — a cinema projection, or the image on a camera sensor. A virtual image forms where rays only appear to originate, so no screen will show it. "Can it be put on a screen?" is the question examiners are testing.

02

Refraction

Light changes speed when it crosses from one transparent material into another, and if it arrives at an angle, that change of speed bends it. This is refraction.

The rule for the direction is worth memorising as a pair. Going into a denser medium — air into glass or water — the light slows and bends towards the normal. Coming out into a less dense medium it speeds up and bends away from the normal.

A ray passing through a rectangular glass block therefore emerges parallel to the way it went in, having been shifted sideways. It bends one way going in and the opposite way coming out, and the two cancel.

Refraction is why a straw in a glass of water looks bent and why a swimming pool looks shallower than it is. Light from the submerged part bends as it leaves the water, but the eye assumes light has travelled in a straight line and traces it straight back — placing the object somewhere it is not.

n = sin i / sin rn = c / vn = 1 / sin cn is the refractive index; for glass about 1.5, for water 1.33
n
refractive index
i
angle of incidence°
r
angle of refraction°
c
critical angle°

Increase the angle of incidence from inside the denser medium and watch the refracted ray bend further from the normal. Past the critical angle it disappears entirely and all the light reflects back inside.

03

Total internal reflection

Send light from glass towards air and increase the angle of incidence. The refracted ray bends further and further from the normal, until at one particular angle it grazes along the boundary itself. That angle is the critical angle.

Push past it and the refracted ray vanishes: all the light is reflected back into the glass. This is total internal reflection, and it happens only when two conditions are met together. The light must be travelling from the denser medium towards the less dense one, and the angle of incidence must exceed the critical angle. Both conditions must be stated to earn full marks.

The critical angle depends on the refractive index. The higher the index, the smaller the critical angle, and the more easily total internal reflection occurs. For glass it is about 42°, for water about 49°, and for diamond only 24° — which is why cut diamonds sparkle so brilliantly.

Worked example 15 marks

Light travels from glass of refractive index 1.50 towards air. Find the critical angle. State and explain what happens to a ray meeting the boundary at 50°.

  1. Use sin c = 1/n.The critical angle is defined for the denser-to-less-dense boundary.
  2. sin c = 1/1.50 = 0.667.
  3. c = 41.8°.About 42° for ordinary glass.
  4. 50° is greater than the critical angle, and the light is going from denser to less dense.Both conditions — state them explicitly.
  5. So total internal reflection occurs and no light emerges.

c = 41.8°, so at 50° the light is totally internally reflected

04

Optical fibres and lenses

An optical fibre is a thin glass core surrounded by cladding of lower refractive index. Light entering at a shallow angle strikes the core–cladding boundary above the critical angle, reflects totally, strikes the other side, reflects again — and so travels the whole length of the fibre even round bends, losing almost nothing.

This is how the internet crosses oceans and how an endoscope lets a surgeon see inside a patient without opening them up. Compared with copper wire, optical fibres carry far more data, lose far less signal over distance, and are immune to electrical interference.

A converging lens is thicker in the middle and brings parallel rays to a focus at the principal focus. The distance from the lens to that point is the focal length. Where the image forms, and whether it is real or virtual, depends entirely on where the object sits relative to the focal length.

Two cases cover most questions. An object further from the lens than the focal length produces a real, inverted image that could be caught on a screen — this is how a camera and the human eye work. An object closer than the focal length produces a virtual, upright, magnified image — this is a magnifying glass.

Key points

  1. Angles are always measured from the normal, never from the surface.
  2. Into a denser medium: slows down, bends towards the normal.
  3. Total internal reflection needs denser-to-less-dense AND an angle above the critical angle.
  4. sin c = 1/n, so a higher refractive index means a smaller critical angle.
  5. Object beyond the focal length gives a real inverted image; inside it, virtual and magnified.

Practice questions

6 questions · 27 marks · full working on every one

Try each one on paper first, then open the working. The marks are shown where they are actually awarded, because that is where they are actually lost.

Short questions

2 · 4 marks

Two marks each, in the style of the short-question section of the paper. Answer in two or three lines.

SQ1[2 marks]
State two differences between a real image and a virtual image.
Model answer

A real image is formed where light rays actually meet and can be caught on a screen. A virtual image is formed where rays only appear to come from and cannot be projected.

Examiner tip. "Can be put on a screen" is the discriminator the mark scheme looks for.

SQ2[2 marks]
Explain why a straw in a glass of water appears bent at the surface.
Model answer

Light from the submerged part refracts as it leaves the water and enters the air, bending away from the normal. The eye assumes light travels in straight lines, so the straw appears displaced.

Examiner tip. The second sentence — that the eye traces the rays straight back — is the mark most often missed.

Long questions

1 · 9 marks

Theory and numerical together, as they appear in the long-question section.

LQ1[9 marks]
A ray of light travels from water of refractive index 1.33 toward the water–air surface.
  1. Calculate the critical angle for the water–air boundary. [3]
  2. The ray meets the surface at an angle of incidence of 60°. State and explain what happens to it. [3]
  3. Explain how an optical fibre uses this effect to carry a signal, and give one advantage of optical fibres over copper wires. [3]
Mark scheme
  1. Uses sin c = 1/n[1]
  2. sin c = 1/1.33 = 0.752[1]
  3. c = 48.8°accept 49°[1]
  4. Total internal reflection occurs[1]
  5. Because 60° is greater than the critical angle of 48.8°[1]
  6. And the light is travelling from the denser medium to the less dense oneboth conditions[1]
  7. Light repeatedly totally internally reflects at the core–cladding boundary, so it follows the fibre even around bends[1]
  8. Advantage: much higher data capacity / less signal loss over distance / no electrical interferenceany one[1]
  9. Second valid point about the mechanism or the advantage[1]

(a) 48.8° (b) total internal reflection (c) repeated TIR at the core–cladding boundary

Examiner tip. When a question gives an angle and asks "state and explain", the explanation must compare that angle with the critical angle you just calculated. Saying only "TIR happens" scores one of three.

Exam questions

3 · 14 marks

Multi-part questions with a full mark scheme.

Q1[6 marks]
A ray of light passes from air into a glass block of refractive index 1.50, striking the surface at an angle of incidence of 40°.
  1. Calculate the angle of refraction.
  2. Calculate the critical angle for this glass.
  3. State the two conditions required for total internal reflection.
Mark scheme
  1. Uses n = sin i / sin r[1]
  2. sin r = sin 40° / 1.50 = 0.643 / 1.50 = 0.4285[1]
  3. r = 25.4°accept 25°[1]
  4. Uses sin c = 1/n = 1/1.50 = 0.667, giving c = 41.8°accept 42°[1]
  5. Light must travel from a denser to a less dense medium[1]
  6. The angle of incidence must exceed the critical angleboth conditions needed[1]

(a) 25.4° (b) 41.8° (c) denser → less dense, and i > c

Examiner tip. Check your calculator is in degrees, and check the direction of the bending: entering glass the ray must bend toward the normal, so r must be smaller than i. If your answer is bigger, you have inverted the formula.

Q2[4 marks]
Explain how an optical fibre carries light around a bend without the light escaping.
Mark scheme
  1. Light strikes the boundary between the core and the cladding[1]
  2. The core is optically denser than the cladding[1]
  3. The angle of incidence is greater than the critical angle[1]
  4. So total internal reflection occurs, repeatedly, and the light follows the fibre[1]

Examiner tip. Name the two media (core and cladding) rather than saying "glass and air". Fibres do not rely on an air gap, and mark schemes look for the cladding.

Q3[4 marks]
An object is placed 15 cm from a converging lens of focal length 10 cm.
  1. State whether the image is real or virtual, and whether it is upright or inverted.
  2. State one device that uses a lens in this way.
  3. State what happens to the image if the object is moved to 5 cm from the lens.
Mark scheme
  1. Real and invertedthe object lies between F and 2F[1]
  2. It is also magnified[1]
  3. A projector (or a film projector / slide projector)[1]
  4. At 5 cm the object is inside F, so the image becomes virtual, upright and magnifieda magnifying glass[1]

(a) real, inverted, magnified (b) projector (c) becomes virtual, upright and magnified

Examiner tip. Compare the object distance with f and 2f before anything else. Here f = 10 and 2f = 20, so 15 cm sits between them — that alone fixes the answer.