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Date May 2014 Marks available 3 Reference code 14M.3.HL.TZ2.14
Level Higher level Paper Paper 3 Time zone Time zone 2
Command term Determine Question number 14 Adapted from N/A

Question

This question is about thin-film interference.

A thin layer of oil of refractive index 1.51 floats on water of refractive index 1.33. Light of wavelength 579 nm is incident normally to the surface.

M14/4/PHYSI/HP3/ENG/TZ2/14

Determine the minimum thickness of the oil layer that gives rise to the least amount of light being reflected.

[3]
a.

Describe the change in the intensity of the reflected light as the thickness of the oil layer in (a) is gradually increased.

[2]
b.

Markscheme

use of \(m = 1\);

\(2 \times 1.51 \times t = 1 \times 579 \times {10^{ - 9}}\);

\(t = 1.92 \times {10^{ - 7}}{\text{ (m)}}\);

Award [3] for a bald correct answer.

Award [2 max] for use of \(m = \frac{1}{2}\) giving 9.6 \( \times \) 10–8 (m).

Award [2 max] for answer of \(\frac{\lambda }{2}\) for air (2.9 \( \times \) 10–7 (m)).

a.

intensity increases;

intensity then decreases and increases repeatedly;

when thickness becomes very large the intensity becomes constant;

b.

Examiners report

In (a) the correct thin film formula from the data booklet was sometimes chosen, but many candidates were able to answer the question using first principles. A common mistake was was to give the half wavelength value in air rather than oil or to forget the phase change on the hard reflection.

a.

Few could give a well reasoned answer to (b) and were often lucky to score one mark.

b.

Syllabus sections

Additional higher level (AHL) » Topic 9: Wave phenomena » 9.3 – Interference
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