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Date November 2013 Marks available 5 Reference code 13N.2.HL.TZ0.7
Level Higher level Paper Paper 2 Time zone Time zone 0
Command term Determine and State Question number 7 Adapted from N/A

Question

Part 2 The Doppler effect and optical resolution

The Doppler effect can be used to deduce that a particular star X is moving towards Earth.

Describe what is meant by the Doppler effect.

[2]
f.

One of the lines in the spectrum of atomic hydrogen has a frequency of 4.6×1016Hz as measured in the laboratory. The same line in the spectrum of star X is observed on Earth to be shifted by 1.3×1012Hz.

(i) State the direction of the observed frequency shift.

(ii) Determine the speed at which X is moving towards Earth stating any assumption that you have made.

[4]
g.

The star X has a companion star Y. The distance from Earth to the stars is 1.0×1018m. The images of X and Y are just resolved according to the Rayleigh criterion by a telescope on Earth with a circular eyepiece lens of diameter 5.0×10–2m.

(i) State what is meant by the statement “just resolved according to the Rayleigh criterion”.

(ii) The average wavelength of the light emitted by the stars is 4.8×10–7m. Determine the separation of X and Y.

[5]
h.

Markscheme

the observed change in frequency/wavelength of a wave;
emitted by a source moving away from or towards/relative to the observer;

f.

(i) a blue-shift / towards the blue end of the spectrum / to a higher frequency / OWTTE;

(ii) \(v = \left( {\frac{{c\Delta f}}{f} = } \right)\frac{{3 \times {{10}^8} \times 1.3 \times {{10}^{12}}}}{{4.6 \times {{10}^{16}}}}\);
8.5×103ms-1;
assume that the speed is very much less than speed of light;

g.

(i) the two stars are (just) seen as separate images;
if the central maximum of the diffraction image of one star coincides with the first minimum of the diffraction image of the other star / OWTTE;
Accept an appropriate diagram for second marking point.

(ii) \(\theta  = \left( {\frac{{1.22\lambda }}{b} = } \right)\frac{{1.22 \times 4.8 \times {{10}^{ - 7}}}}{{5.0 \times {{10}^{ - 2}}}}\) or 1.17×10-5rad;
\(\theta  = \frac{d}{{1.0 \times {{10}^{18}}}}\);
(d=)1.2×1013m;
Award [2 max] if 1.22 is missing, giving an answer of 0.98×1013.

h.

Examiners report

[N/A]
f.
[N/A]
g.
[N/A]
h.

Syllabus sections

Additional higher level (AHL) » Topic 9: Wave phenomena » 9.4 – Resolution
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