| Date | May 2015 | Marks available | 2 | Reference code | 15M.3.SL.TZ2.21 |
| Level | Standard level | Paper | Paper 3 | Time zone | Time zone 2 |
| Command term | Explain | Question number | 21 | Adapted from | N/A |
Question
This question is about a double-slit experiment.
Coherent monochromatic light is incident on two narrow rectangular slits. The diagram shows the fringes produced on a screen that is some distance from the slits. M is the middle of the central bright fringe and P is the middle of the third bright fringe.
Explain why an interference pattern is produced on the screen.
The two slits are separated by 2.2 mm and the distance from the slits to the screen is 1.8 m. The wavelength of the light is 650 nm. Calculate the distance MP.
Markscheme
reference to:
diffraction at slits / slits are coherent sources;
path/phase difference;
constructive and destructive interference;
Do not reward just “interference” as this is mentioned in the question.
for single fringe: \(s = \frac{{650 \times {{10}^{ - 9}} \times 1.8}}{{2.2 \times {{10}^{ - 3}}}}\left( { = 5.3 \times {{10}^{ - 4}}\left( {\rm{m}} \right)} \right)\); } (also award this mark if the factor of 3 is seen in the numerator)
distance \(MP = \left( {5.3 \times {{10}^{ - 4}} \times 3 = } \right)1.6 \times {10^{ - 3}}\left( {\rm{m}} \right)\);
Allow ECF from first marking point.
Award [2] for a bald correct answer.
Examiners report
In part (a) diffraction at each slit, followed by a path difference and subsequent constructive or destructive interference was very often given, but sometimes in a clumsy fashion. It is evident that not all candidates take 30s to plan the order in which they are going to present the steps in their argument.
Part (b) was not difficult, but many lost 1 mark for not using n = 3. Highlight this fact in the stem and these kind of careless errors can be avoided.
