Date | November 2013 | Marks available | 4 | 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.
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.
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.
Markscheme
the observed change in frequency/wavelength of a wave;
emitted by a source moving away from or towards/relative to the observer;
(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;
(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.