Light

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

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