A-LevelPhysicsWaves9702/42 · Feb/Mar 2017 · Q58 Marks

(b) An optic fibre of length 57km is connected between a transmitter and a receiver, as shown in Fig. 5.1. transmitter signal power 15 × 10⁻³ W 57 km optic fibre receiver signal power P noise power 9.0 × 10⁻⁷ W Fig. 5.1 The attenuation per unit length of the optic fibre is 0.50 dB km⁻¹. The transmitter provides an input signal of power 15 × 10⁻³ W to the fibre. The noise power at the receiver is 9.0 × 10⁻⁷ W.

(a)State three advantages of an optic fibre compared to a metal wire for the transmission of a signal.[3]
(b)(i)Show that the signal power P entering the receiver from the optic fibre is 2.1 × 10⁻⁵ W.[2]
(b)(ii)A minimum signal-to-noise ratio of 24 dB is needed at the receiver in order for it to be able to distinguish the signal from the noise. Determine whether the receiver is able to distinguish the signal from the noise.[3]

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Mark scheme

Marking points from the official mark scheme (Feb/Mar 2017). The mark or mark type is shown in brackets.

[B3] any three from:
greater bandwidth
does not suffer from (e.m.) interference/can be used in (e.m.) 'noisy' environments
no/less power/energy radiated/better security/less cross-talk
less attenuation/fewer repeaters/amplifiers needed
less weight/easier to handle/cheaper/occupy less space
[C1] attenuation/gain = 10 log P₁/P₂
[A1] 0.50 × 57 = 10 log (15 × 10⁻³/P) so P = 2.1 × 10⁻⁵ W
[A1] or (0.50 × 57) = 10 log (P/15 × 10⁻³) so P = 2.1 × 10⁻⁵ W
[M1] (calculation of S/N ratio at receiver) S/N ratio = 10 log (2.1 × 10⁻⁵ / 9.0 × 10⁻⁷) or S/N ratio = 14
[A1] 14 < 24 or S/N ratio < minimum S/N ratio
[A1] so not able to distinguish signal from noise
[M1] or (calculation of minimum acceptable power at receiver) 24 = 10 log (P/ 9.0 × 10⁻⁷) or P = 2.3 × 10⁻⁴
[A1] 2.1 × 10⁻⁵ < 2.3 × 10⁻⁴ or power < minimum power
[A1] so not able to distinguish signal from noise

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