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Question 18N.3.SL.TZ0.10

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Date November 2018 Marks available [Maximum mark: 8] Reference code 18N.3.SL.TZ0.10
Level SL Paper 3 Time zone TZ0
Command term Calculate, Determine, Show that, Suggest Question number 10 Adapted from N/A
10.
[Maximum mark: 8]
18N.3.SL.TZ0.10

An optic fibre consists of a glass core of refractive index 1.52 surrounded by cladding of refractive index n. The critical angle at the glass–cladding boundary is 84°.

(a.i)

Calculate n.

[2]

Markscheme

« sin θ c = n 1 n 2 » n1 = 1.52 × sin 84.0° ✔

 

n1 = 1.51 ✔

(a.ii)

The refractive indices of the glass and cladding are only slightly different. Suggest why this is desirable.

[1]

Markscheme

to have a critical angle close to 90° ✔

so only rays parallel to the axis are transmitted ✔

to reduce waveguide/modal dispersion ✔

The diagram shows the longest and shortest paths that a ray can follow inside the fibre.

For the longest path the rays are incident at the core–cladding boundary at an angle just slightly greater than the critical angle. The optic fibre has a length of 12 km.

 

(b.i)

Show that the longest path is 66 m longer than the shortest path.

[2]

Markscheme

long path is  12 × 10 3 sin 84 ∘  ✔

= 12066 «m» ✔

«so 66 m longer»

(b.ii)

Determine the time delay between the arrival of signals created by the extra distance in (b)(i).

[2]

Markscheme

speed of light in core is  3.0 × 10 8 1.52 = 1.97 × 10 8 «m s−1» ✔

time delay is  66 1.97 × 10 8 = 3.35 × 10 − 7 «s» ✔

(b.iii)

Suggest whether this fibre could be used to transmit information at a frequency of 100 MHz.

[1]

Markscheme

no, period of signal is 1 × 10−8 «s» which is smaller than the time delay/OWTTE ✔

 

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