| Date | May 2012 | Marks available | 5 | Reference code | 12M.2.HL.TZ2.12 |
| Level | Higher level | Paper | Paper 2 | Time zone | Time zone 2 |
| Command term | Determine and State | Question number | 12 | Adapted from | N/A |
Question
Part 2 Resolution and the Doppler effect
Radio telescopes can be used to locate distant galaxies. The ability of such telescopes to resolve the images of galaxies is increased by using two telescopes separated by a large distance D. The telescopes behave as a single radio telescope with a dish diameter equal to D.
The images of two distant galaxies G1 and G2 are just resolved by the two telescopes.
(i) State the phenomenon that limits the ability of radio telescopes to resolve images.
(ii) State the Rayleigh criterion for the images of G1 and G2 to be just resolved.
(iii) Determine, using the following data, the separation d of G1 and G2 .
Effective distance of G1 and G2 from Earth = 2.2 ×1025 m
Separation D = 4.0 ×103 m
Wavelength of radio waves received from G1 and G2= 0.14 m
Due to the Doppler effect, light from distant galaxies is often red-shifted.
(i) Describe, with reference to the Doppler effect, what is meant by red-shift.
(ii) The frequency of a particular spectral line as measured in the laboratory is 4.57 ×1014 Hz. The same line in the spectrum of a distant galaxy has a frequency that is lower than the laboratory value by 6.40 ×1011 Hz. Determine the speed with
which the galaxy is receding from Earth.
Markscheme
(i) diffraction;
(ii) the (central) maximum of the diffraction pattern of one image coincides with the first minimum of the diffraction pattern of the other image;
Allow mark for clear diagram.
(iii) angular separation of G1 and \({G_2} = \frac{d}{{2.2 \times {{10}^{25}}}}\);
\(\frac{d}{{2.2 \times {{10}^{25}}}} = \left( {1.22\frac{\lambda }{D} = } \right)\frac{{1.22 \times 0.14}}{{4.0 \times {{10}^3}}}\);
d9.41020 m;
(i) if there is relative motion between source/galaxy and observer/Earth;
the observed frequency/wavelength will differ from the source frequency/wavelength;
the observed frequency will be lower / the observed wavelength will be greater if the direction of relative motion is away from the source;
(to award the mark it must be clear in which direction a red-shift occurs)
Award [3] for a good description that mentions all of the above.
Award [3] for a clear annotated diagram that shows all of the above points.
(ii) \(\frac{{\Delta f}}{f} = \frac{v}{{3 \times {{10}^8}}} = \left( {\frac{{6.40 \times {{10}^{11}}}}{{4.57 \times {{10}^{14}}}} = } \right)1.40 \times {10^{ - 3}}\);
\(v = \left( {3 \times {{10}^8} \times 1.40 \times {{10}^{ - 3}} = } \right)4.20 \times {10^5}{\rm{m}}{{\rm{s}}^{ - 1}}\);
Examiners report
(i) Many recognized that the phenomenon at work is diffraction. Common incorrect responses included “Rayleigh criterion” and “interference”.
(ii) Candidates often failed to make clear that their answer referred to the diffraction patterns of G1 and G2, it was common to see candidates writing about the “first maximum of G1” etc.
(iii) Many were able to negotiate this two-step calculation with ease.
(i) Descriptions of red shift in terms of the Doppler effect were well done with many high marks seen.
(ii) Candidates frequently became confused in this calculation as to the meaning of the symbols quoted in the Data Booklet. Examiners often saw answers of about 3 x 108 m s–1 (or greater) with no realization by the candidate that an answer of this magnitude is implausible.
