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Date May 2014 Marks available 2 Reference code 14M.3.SL.TZ2.19
Level Standard level Paper Paper 3 Time zone Time zone 2
Command term Outline Question number 19 Adapted from N/A

Question

This question is about interference.

Light from a laser is incident on two identical parallel slits. The light from the two slits produces a fringe pattern on a screen.

M14/4/PHYSI/SP3/ENG/TZ2/19

A central bright fringe is produced at C. The next bright fringe is produced at A. There is a dark fringe at B.

The light from the laser is coherent and monochromatic.

The distance from the two slits to the screen is 1.5 m. The distance BC is 1.8 mm and the distance between the slits is 0.30 mm.

Outline what is meant by the term

(i)     coherent.

(ii)     monochromatic.

[2]
a.

State the phase difference between the light waves from the two slits that meet at B.

[1]
b.

(i)     Show that the laser produces light of wavelength equal to 720 nm.

(ii)     State the path difference, in metres, between the waves that meet at B.

[4]
c.

Determine the number of lines per metre of the diffraction grating.

[3]
d.

Markscheme

(i)     constant/zero phase difference (between the light waves);

(ii)     single/same wavelength/frequency; (allow “narrow band” OWTTE)

Do not allow “single colour”.

a.

\(180^\circ /\pi {\text{rad}}\);

Do not accept \(\frac{\lambda }{2}\).

b.

(i)     use of \(\lambda  = \frac{{sd}}{D}\);

\(s = \underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{2}  \times 1.8{\text{ (mm)}}\); (award this mark for any evidence for the factor of 2)

\(\lambda  = \frac{{{\text{2}} \times {\text{1.8}} \times {\text{1}}{{\text{0}}^{ - 3}} \times {\text{0.30}} \times {\text{1}}{{\text{0}}^{ - 3}}}}{{{\text{1.5}}}}\) / OWTTE;

\({\text{(}} = 7.2 \times {10^{ - 7}}{\text{ m)}}\)

Exact answer is given, award marks for correct working.

(ii)     \({\text{3.6}} \times {\text{1}}{{\text{0}}^{ - 7}}{\text{ m}}\)\(\,\,\,\)or\(\,\,\,\)360 nm;

Allow ECF from (c)(i).

c.

use of \(d = \frac{{n\lambda }}{{\sin \theta }}\);

\(d = \frac{{3 \times 720 \times {{10}^{ - 9}}}}{{\sin 39^\circ }} = 3.4 \times {10^{ - 6}}{\text{ (m)}}\);

\(N = \left( {\frac{1}{{3.4 \times {{10}^{ - 6}}}} = } \right){\text{ }}2.9 \times {10^5}\);

Award [3] for a bald correct answer.

ECF examples:

Award [2 max] if n = 2 is used (gives 4.4 \( \times \) 105).

Award [2 max] if 78° is used (gives 4.5 \( \times \) 105).

Award [2 max] if 13° and n = 1 used (gives 3.1 \( \times \) 105).

d.

Examiners report

[N/A]
a.
[N/A]
b.
[N/A]
c.
[N/A]
d.

Syllabus sections

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