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IGCSE Physics · guided topic map

Light and optics for Cambridge IGCSE Physics

Light and optics for IGCSE Physics, organized into 1 syllabus topic and 5 mapped concept guides.

Syllabus topics
1
Mapped concept guides
5
Educational level
Cambridge IGCSE Core and Extended

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Work in order or jump to the concept named in your specification, course outline, or assignment.

3.2

Light

Waves

5 guides
  1. 01Reflection and ray diagramsMapped lesson
  2. 02Refraction and Snell's lawMapped lesson
  3. 03Total internal reflectionMapped lesson
  4. 04Colour and dispersionMapped lesson
  5. 05Thin converging lensesMapped lesson

Diagrams

Light and optics as IGCSE Physics draws it

The figures from the IGCSE Physics practice papers that sit on these syllabus points — the apparatus, circuits and graphs an exam question actually puts in front of you.

01Fig. 8.1LightIGCSE
Light travelling along an optical fibre by total internal reflectionairairglass core, n = 1.50light innormal

Figure comment

Fig. 8.1A straight length of optical fibre drawn in section as a long horizontal band with parallel walls, shaded to show glass and labelled glass core, n = 1.50, with air labelled outside it. Light entering the flat left-hand end travels up to the upper wall, reflects down to the lower wall, reflects again up to the upper wall and reaches the far end, giving a zig-zag path along the fibre. At one reflection a dashed normal is drawn perpendicular to the wall, and the angle on each side of it is marked with an arc; no angle is given a numerical value.

Read the comment once, then trace every arrow, label, axis or component in the drawing before opening the questions.

3.2 LightOpen IGCSE question · Waves Q8
Guided questions 5 parts

Reading cue. The dashed normal is perpendicular to the fibre wall, so light running nearly along the fibre strikes that wall at a large angle close to 90°, not at a small one.

  1. aState State how the two angles marked at the reflection in Fig. 8.1 compare with each other, and state the line from which each is measured.

    recall2 marks

    Check answer 2 marks
    1. the two angles are equal, since the angle of incidence equals the angle of reflection
    2. both are measured from the normal drawn perpendicular to the wall of the core

  2. bDetermine The core drawn in Fig. 8.1 is replaced by one of refractive index 1.60. Determine the critical angle for the new core, and state whether total internal reflection will now occur for a wider or a narrower range of angles.

    routine3 marks

    Check answer 3 marks
    1. sin C = 1/1.60 = 0.625
    2. C = 39° (38.7°)
    3. the critical angle is smaller than for the 1.50 core, so a wider range of angles exceeds it and gives total internal reflection

  3. cDetermine Light travels at 3.0 × 10⁸ m/s in air. Determine the speed of light inside the core labelled in Fig. 8.1, and the time taken for light to travel 1.0 km along a straight fibre if it runs along the axis without reflecting.

    demanding3 marks

    Check answer 3 marks
    1. v = c/n = 3.0 × 10⁸ / 1.50 = 2.0 × 10⁸ m/s
    2. t = d/v = 1000 / (2.0 × 10⁸)
    3. t = 5.0 × 10⁻⁶ s (5.0 μs)

  4. dExplain Fig. 8.1 shows the light following a zig-zag path rather than running straight along the middle of the core. Explain why a short pulse of light sent into a long fibre arrives at the far end spread out over a longer time than it was sent, and state what this places a limit on.

    top of the paper4 marks

    Check answer 4 marks
    1. light following the zig-zag path travels a greater distance along the fibre than light running along the axis
    2. all of the light travels at the same speed, 2.0 × 10⁸ m/s, in the glass, so the zig-zag light arrives later than the axial light
    3. the pulse therefore arrives stretched out in time
    4. this limits how closely pulses may be sent one after another, and so how much information the fibre can carry each second

Transfer challenge

A prism in a periscope turns light through 90° by reflection at one face, the light meeting that face at 45° to the normal. Explain why the reflection is total for a prism of refractive index 1.50 but fails for one made of a plastic of refractive index 1.35.

Check answer 3 marks
  1. for the glass, sin C = 1/1.50 gives C = 41.8°, and 45° is greater than this, so the light is totally internally reflected
  2. for the plastic, sin C = 1/1.35 gives C = 47.8°
  3. 45° is less than 47.8°, so the light is refracted out through the face instead of being reflected, and the periscope fails
02Fig. 10.1LightIGCSE
A ray traced through a rectangular glass blockglass blocknormalincident rayiremergent ray

Figure comment

Fig. 10.1The rectangular glass block drawn on the paper, seen from above. A ray labelled incident ray comes down from the upper left and meets the top face of the block; a dashed normal is drawn perpendicular to that face at the point where the ray strikes it. The angle between the incident ray and the normal is labelled i, and the angle between the normal and the ray travelling inside the glass is labelled r. The ray crosses the block, leaves through the bottom face and continues as the emergent ray. The optical pins are not shown.

Read the comment once, then trace every arrow, label, axis or component in the drawing before opening the questions.

3.2 LightOpen IGCSE question · Waves Q10
Guided questions 5 parts

Reading cue. Both i and r are measured from the dashed normal, never from the block face, and r is the angle inside the glass, so r is always the smaller of the pair.

  1. aState State how the direction of the emergent ray in Fig. 10.1 compares with that of the incident ray, and give a reason.

    recall2 marks

    Check answer 2 marks
    1. the emergent ray is parallel to the incident ray, though shifted sideways from it
    2. the top and bottom faces are parallel, so the ray bends away from the normal on leaving by the same angle as it bent towards the normal on entering

  2. bCalculate The block in Fig. 10.1 has a refractive index of 1.50 and the angle i is set to 40°. Calculate the angle r.

    routine2 marks

    Check answer 2 marks
    1. sin r = sin i / n = sin 40° / 1.50 = 0.429
    2. r = 25° (25.4°)

  3. cDetermine On the drawing the angle i measures 40° and the angle r measures 25°, each read to the nearest degree. Determine the largest and the smallest values of the refractive index consistent with these two readings.

    demanding3 marks

    Show a hint

    Pair the extremes the right way round: a ratio is largest when its top is largest at the same time as its bottom is smallest.

    Check answer 3 marks
    1. largest value uses the largest i with the smallest r: sin 40.5° / sin 24.5° = 1.57
    2. smallest value uses the smallest i with the largest r: sin 39.5° / sin 25.5° = 1.48
    3. the refractive index therefore lies between 1.48 and 1.57

  4. dSuggest Suggest which of the two angles marked in Fig. 10.1 does more to limit the precision of the refractive index obtained, and justify your answer.

    top of the paper3 marks

    Check answer 3 marks
    1. the angle of refraction r limits it more
    2. r is the smaller angle, so the same reading error of half a degree is a larger fraction of r, and changes sin r by proportionally more than the same error changes sin i
    3. r is also not measured directly: it is drawn between the normal and a line joining the entry and exit points, so errors in the outline drawn round the block and in the marked emergent ray feed into it as well

Transfer challenge

A ray of light inside a glass block of refractive index 1.50 meets the surface at 30° to the normal from within the glass. Calculate the angle at which it leaves the glass, and state what happens instead if that internal angle is increased to 45°.

Check answer 3 marks
  1. sin θ(air) = 1.50 × sin 30° = 0.750
  2. θ(air) = 49° (48.6°)
  3. at 45° the ray exceeds the critical angle of 41.8°, so it is totally internally reflected and no light emerges from that face