Date | November 2011 | Marks available | 5 | Reference code | 11N.2.HL.TZ0.9 |
Level | Higher level | Paper | Paper 2 | Time zone | Time zone 0 |
Command term | Calculate and Determine | Question number | 9 | Adapted from | N/A |
Question
Part 2 Resolution
(i) State the wave phenomenon that limits the resolution of the eye.
(ii) State the Rayleigh criterion for determining if the images of two objects are just resolved.
An advertising sign contains two straight vertical sections that emit light.
The vertical sections are separated by a horizontal distance of 0.13 m. An observer views them from a distance of 720 m. The wavelength of the emitted light is 510 nm and the diameter of the aperture of the observer’s eye is 3.0 mm.
(i) Determine if the images formed on the retina of the observer will be resolved.
(ii) One of the vertical sections is switched off. The observer looks at the illuminated vertical section. The diameter of the aperture of the observer’s eye is now 2.5 mm.
Calculate the angular width of the central maximum of the diffraction pattern formed on the observer’s retina.
Markscheme
(i) diffraction;
(ii) the first minimum of one diffraction pattern;
falls on central maximum of other diffraction pattern;
(i) \(\theta = \left( {1.22 \times \frac{{5.1 \times {{10}^{ - 7}}}}{{3.0 \times {{10}^{ - 3}}}} = } \right)2.1 \times {10^{ - 4}}\left( {{\rm{rad}}} \right)\);
angular separation of sections\( = \left( {\frac{{0.13}}{{720}} = } \right)1.8 \times {10^{ - 4}}\left( {{\rm{rad}}} \right)\)/ required minimum
separation for resolution\( = \left( {2.1 \times {{10}^{ - 4}} \times 720 = } \right)0.15{\rm{m}}\);
they cannot be resolved;
Ignore omission of 1.22 (gives θ=1.7×10–4 and (ECF) are just resolved).
Award [0] for a bald correct answer.
(ii) \(\theta = \left( {1.22 \times \frac{{510 \times {{10}^{ - 9}}}}{{3.5 \times {{10}^{ - 3}}}} = } \right)1.78 \times {10^{ - 4}}\left( {{\rm{rad}}} \right)\);
\(3.6 \times {10^{ - 4}}\left( {{\rm{rad}}} \right)\);
Ignore omission of 1.22 (gives 2.9×10–4 (rad)).