Structured Questions
? WRONG ASSUMPTION
2. For wavelength 589.0 nm, we have:
2.50 10 m sin6
4 589.0 10 m
9
For wavelength 589.6 nm, we have:
2.5 10 m sin ' 4 589.6 10 m6
9
calculation of angle for both wavelengths [1]
We thus have:
angularseparation 70.624 70.459 0.165 0.10
This is greater than minimum required , therefore students will be able to observe the wavelengths as separate images .
calculating the difference between the 2 angles and correct conclusion [1]
[Total: 12]
29. [RI/H2/Prelims 2010/P3/6]
a. Any two [B1] × 2
The waves must be coherent .
The waves must have approximately the same amplitude .
The waves must be unpolarised or polarised in the same plane (for transverse waves).
The waves must interfere to give regions of maxima (constructive interference) and minima (destructive interference).
b. Intensity required at 12 km away:
i. Shorter wavelengths means the anti-nodal lines will be closer to one another. [B1]
Hence, aircrafts may “lock on” to the wrong line of maxima / difficult to identify the central line of maxima / difficult to differentiate the lines of maxima. [B1]
ii. Since the two radio waves are in phase [B1], along centre-line, path difference is always zero / phase difference is always zero / P & Q are equidistant from any point on the centre-line. Hence constructive interference occurs. [B1]
d.
i. P is nearer to the aircraft , hence intensity (or amplitude) of signal should be higher . [B1]
ii. From Figure 24, phase difference = 2 rad
. [B1]
iii. Since phase difference = 2 rad
Distance from P to plane 4800 m 180 m 480 m 4827.30 m Distance from Q to plane 4800 m 230 m 480 m 4829.42 m
i. If aircraft is on the central anti-nodal line , it should detect maximum signals from both frequencies / the maximum signal will be stronger . [B1]
OR
If the aircraft is on the wrong anti-nodal line , only one of the frequencies will show a strong signal . [B1]
ii. If the ratio is an integer ratio, higher orders of maxima will still coincide/overlap [B1].
Hence the aircraft could still detect maximum signals from both frequencies even though it is not on the central anti-nodal line . [B1]
f. Advantage: Can still work under low visibility conditions / Use of detector to align aircraft is more accurate than using visual inspection . [B1]
Disadvantage: Possible interference of signals from other sources (e.g. radio stations, telecommunication base stations, etc) / It is more costly to install the emitters and receivers on every airplane . [B1]
[Total: 20]
30. [RVHS/H2/Prelims 2010/P3/4]
a. The distances of a point on XY from the two sources are different.
When the difference in distance is integer multiples of wavelength apart , the wave meet in phase and constructive interference occurs. This results in maximum intensity.
When the difference in distance is half-integer multiples of wavelength apart , the wave meet out-of-phase and destructive interference occurs. This results in minimum intensity. [2]
b.
i. We have:
2
2BR 12 m 5.5 m 13.200 m 13.2 m (3 s.f.) (shown) [1]
ii. Since a maximum is detected at P and next at R, the path difference from A and B at R is 1
. [1]
Thus,
BR AR 13.2 m 12.5 m 0.7 m (1 d.p.)
[1]
iii. We have:
470 Hz
0.7 m 329 m s
1 330 m s (2 s.f.)1v f
[2]
c. Distances to Q from A and B are different.
Since intensity at a position from a point source is inversely proportional to the square of the distance between them, the intensities of the waves arriving at Q from A and B are different .
Since intensity is proportional to the square of the amplitude , the amplitude of the waves arriving from A and B will be different .
Q is a position with destructive interference without complete cancellation of the waves occurs. Hence the intensity at Q is not zero. [3]
[Total: 10]
31. [TJC/H2/Prelims 2010/P3/6]
a. It is a phenomenon in which waves from two or more coherent waves superpose with one another producing a resultant wave whose amplitude is given by the Principle of Superposition . [1]
b.
i. Sound from the 2 sources undergo interference. As the source S1 moves, the path difference changes . [1]
When the path difference between S1P and S2P is an integral number of the wavelength , constructive interference occurs at P and a maxima is detected. [1]
When the path difference between S1P and S2P is an odd integral number of half wavelength , destructive interference occurs at P and a minima is detected. [1]
ii. At destructive interference, path difference 0.082 m
2
iii. At this new frequency, since constructive interference is detected at P,
S X1 0.082 m
i. Waves travel down the tube and get reflected . [1]
The incident and the reflected waves, both having the same amplitude , frequency and speed travelling in opposite directions superpose to form standing wave . [1]
ii. Air column in tube has a natural frequency of vibration . [1]
When the fork’s frequency is equal to the natural frequency of the air column, resonance occurs ; there is maximum energy transfer and so maximum amplitude of vibrations occurs , leading to maximum loudness . [1]
When the fork’s frequency is not equal to the natural frequency of the air column, no resonance occurs and loudness drops . [1]
iii. Sketch: Antinode at top, node at surface of water, 1 antinode and 1 node in between [1]
Therefore, antinode is 0.50 cm above the top of the tube / 16.2 cm above water surface . [1]
[Total: 20]
32. [YJC/H2/Prelims 2010/P3/6 (part)]
a. Slit separation on the diffraction grating:
6
0.40 m
It is before the second-order fringe of the 750-nm light at an angle of 64.16°.
Hence overlapping occurs since the minimum angle of the third-order spectrum is larger than the maximum angle of the second-order spectrum . [3]
c. No overlapping occurs. [1]
The spectrum is brighter . [1]
d. A white , bright spot/fringe. [1]
[Total: 10]
33. [VJC/H2/Prelims 2010/P3/2]
a.
i. Incident waves travelling along the wire to the ends are reflected .
The incident and reflected waves travelling in opposite directions have the same frequency and amplitude . They superpose to form standing waves . [2]
ii.
0.40 m
b.
i. When the support is shifted, the natural frequency of the wire is changed and no longer matches the driving frequency of the periodic force produced by the
alternating current in the wire and magnetic field.
The system no longer resonates and hence its amplitude decreases .
The frequency of the a.c. source must therefore be the natural frequency of the wire which is 50 Hz. [3]
ii. The length l can be reduced to 0.20 m to double the fundamental frequency . The weights attached to the wire can be increased to increase the speed of the waves to 80 m s-1. [2]
[Total: 11]