DP Physics (last assessment 2024)

Question 19M.3.SL.TZ2.12bi

Date	May 2019	Marks available	[Maximum mark: 2]	Reference code	19M.3.SL.TZ2.12bi
Level	SL	Paper	3	Time zone	TZ2
Command term	Determine	Question number	bi	Adapted from	N/A

bi.

[Maximum mark: 2]

19M.3.SL.TZ2.12bi

The refractive index $n$ of a material is the ratio of the speed of light in a vacuum $c$ , to the speed of light in the material $v$ or $n = \frac{c}{v}$ .

The speed of light in a vacuum $c$ is 2.99792 × 10⁸ m s^-1. The following data are available for the refractive indices of the fibre core for two wavelengths of light:

(bi)

Determine the difference between the speed of light corresponding to these two wavelengths in the core glass.

[2]

Markscheme

$v = \frac{c}{n} = {v_{1299}} = \frac{{2.99792 \times {{10}^8}}}{{1.45061}} = 2.06666 \times {10^8}$ «ms^–1» AND

${v_{1301}} = \frac{{2.99792 \times 10{}^8}}{{1.45059}} = 2.06669 \times {10^8}$ «ms^–1»

$\Delta v = \left( {\frac{1}{{1.45059}} - \frac{1}{{1.45061}}} \right) \times 2.99792 \times {10^8}$ ✔

$\Delta v = 2.85 \times {10^3}$ OR $3 \times {10^3}$ «ms^–1»✔

Examiners report

The calculation of the difference in the speed of light for two different wavelengths was well answered. Candidates often rounded answers to a small number of significant figures when finding the individual speeds.

Syllabus sections

Core

Core » Topic 4: Waves » 4.4 – Wave behaviour

Core » Topic 4: Waves

Question 19M.3.SL.TZ2.12bi

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