| Date | May 2015 | Marks available | 3 | Reference code | 15M.2.HL.TZ2.4 |
| Level | Higher level | Paper | Paper 2 | Time zone | Time zone 2 |
| Command term | Determine and Outline | Question number | 4 | Adapted from | N/A |
Question
This question is about the properties of waves.
Microwaves from a microwave transmitter are reflected from two parallel sheets, A and B. Sheet A partially reflects microwave energy while allowing some to pass through. All of the microwave energy incident on sheet B is reflected.
Sheet A is fixed and sheet B is moved towards it. While sheet B is moving, the intensity of the signal detected at the receiver goes through a series of maximum and minimum values.
Outline why a minimum in the intensity occurs for certain positions of sheet B.
The apparatus is arranged to demonstrate diffraction effects.
The microwaves emerge from the transmitter through an aperture that acts as a single slit.
(i) Outline what is meant by diffraction.
(ii) A maximum signal strength is observed at P. When the receiver is moved through an angle \(\theta \), a first minimum is observed. The width of the aperture of the transmitter is 60 mm. Estimate the value of \(\theta \).
Microwaves can be used to demonstrate polarization effects. Outline why an ultrasound receiver and transmitter cannot be used to demonstrate polarization.
Markscheme
mention of interference;
interference is between reflected waves from both reflectors;
minimum caused (by destructive interference) when crest meets trough/when path difference is \(\frac{\lambda }{2}\) / (completely) out of phase / phase difference of \(\pi \)/180º / OWTTE;
minimum occurs when twice the distance between plates is \(\left( {n + \frac{1}{2}} \right)\lambda \);
Ignore references to standing waves.
(i) spreading out of a wave; (do not allow “bending” even if context is obstacle)
when it meets an aperture/gap/slit/obstacle;
Allow credit for answers appearing on clear labelled diagram for both marks.
(ii) \(\left( {\theta = \frac{{32}}{{60}} = } \right)0.533\left( {{\rm{rad}}} \right)\) or 30.6(º);
Award [0] for calculation that uses 1.22 (0.65 rad).
Award [0] for 0.533º or 30.6 rad.
At least one centre is using the abbreviation c for rad. Please allow this.
sound waves (in air) are longitudinal;
longitudinal waves cannot be polarized / only transverse waves can be polarized;
Award [0] for any suggestion that ultrasound is an electromagnetic wave.
