Superposition 2008 prelim_solutions

1. B
2. C
d sin Ө = 2 λx = 3 λy λx/ λy = 3/2
3. A
Intensity is proportional to square of amplitude
4. a) It means that when two or more waves of the same type superpose or meet at a point in space, the resultant
amplitude at the point is the vector sum of all the individual wave amplitudes at that point.
b) (i) Sound waves from the two sources will meet or superpose at the point D. If they meet with constructive
interference, a loud sound (maxima) will be detected. This will occur when the path difference of the two
sound waves at D is an integral multiple of the wavelength.
If they meet with destructive interference, a soft or no sound (minima) will be detected. This will occur
when the path difference is (n + ½ ) times of the wavelength where n is any integer.
For the arrangement above, the path difference is fixed at the point D. When the frequency of the sound
is changed, the wavelength will change accordingly. At certain values of frequency, ( hence its
wavelength), when it fulfills the conditions for maxima or minima as stated earlier, maxima and minima
will be observed at the point D.
(ii)
(iii) The path difference is always zero for all frequencies. Hence maximum is always observed.
(iv) f = 510 Hz, λ = 340/510 = 0.667 m
using two-source interference equation, λ = ax/D,
maxima separation x = 0.667 x 40/9 = 2.963 m
zero order maxima is 9/2 =4.5 m below D.
first order maxima is 4.5 – 2.963 = 1.537 = 1.54 m below point D.
5. D
Vector addition of displacement is possible only for the same quantities
6. B
Minimum when two waves arrive in anti phase and destructive interference occurs
Source are in phase, hence path difference = odd number of ½ wavelength
S1 p – S2 P = ½ λ, 3( ½ λ) , 5( ½ λ)
Least possible distance S1P = 12.0 + 1/.2 (0.40) = 12.20m
7. C
d sin θ = n λ
let θ = 90 � (1 / 2 x 105) sin 90 = n ( 600 x 10 –9)
8. (a) (i) Adjust y-sensitivity to get the required height
Adjust time base to get a few cycles
(ii) amplitude of the resultant waves
(b) (i) Change in height alternate between maximum to minimum.
Constructive interference occurs at maxima where the two waves arrive in phase and the microphone
detects maximum loudness.
Destructive interference occurs at minima where the two waves arrive in antiphase and the microphone
detects minimum loudness.
(ii) λ = v / f = 340 / 3400 = 0.1 m
PQ = ½ Δ x = ½ (λ D / a) = ½ (0.1) ( 9) / 1.5 = 0.3 m
The equation is accurate only if the distance between XY and the loudspeakers is much greater than the
distance between the loudspeaker.
(c) Alternate loud and soft sound are detected. As the frequency is increased, wavelength decreases.
When path difference = ½λ 3/2λ 5/3λ , destructive interference occurs and soft sound is heard. When path
difference = λ, 2λ, 3λ, constructive interference occurs and loud sound is heard. Hence point Q passes through a
series of alternate constructive and destructive interference.
Alternate explanation.
The distance between adjacent maximum and minimum decreases as wavelength decreases gradually,
according to Δ x = λ D / a. Hence the positions of the maxima and minima will be shifted towards the central axis.
Hence Q will passes through alternate maxima and minima.
9. C

The fringe separation x is given by
.
Since the wavelength l and the distance D between the screen and the slits are unchanged, reducing the slit separation to
d/2 will cause the fringe separation to double. The slit width does not affect the interference fringe separation. However,
the intensity of the fringes will be reduced if the slit width is narrowed.
10. C
The relation for the diffraction grating is given by d sinθ = n l where d = slit separation
slit separation,
The question requires complete sets of the whole visible spectrum. As red light has the longest wavelength and the
greatest deviation, we need to find the principal maximum order for it.
Therefore, the maximum principal order, n = 3 (need to truncate the decimal value)
11. (a) (i) Diffraction refers refers to the bending of waves around an object or spreading of waves through an
aperture. [B1]
(ii) 1. The wavelength of sound is very much longer than the width of the corners that causes
diffraction (spreading) of sound waves around them. However, the wavelength of light is very
much shorter and little diffraction occurs. [B1]
2. The two light sources are not coherent. They do not have a constant phase relation. [B1]
(iii) The intensity of the central maximum decreases [B1]
and its width becomes broader (distance between the two first minimum increases) [B1]
(b) (i) Path difference = d sin θ1 – d sin θ2 [B1]
= nλ [B1]
(ii) Path difference = [M1]
= 4.3 m [M1]
Since the “sources” are now 180o out of phase,
1. For constructive interference path difference = ½λ
λ = 2 ( 4.26) = 8.5 m [A1]
.2 For destructive interference path difference = λ
λ = 4.3 m [A1]
(c) (i) d = 0.250 × 10
-3
m; λ = 546.1× 10
-9
m; D = 1.20 m
The fringe separation,
1. Distance between the first order maximum (bright line) on both sides
= 2 x [M1]
= 5.24 × 10
-3
m [A1]
2. Distance between the first order maxima on one side and second order bands
= 2.5 x [M1]
= 6.55 × 10
-3
m [A1]
(ii) Increase the velocity (kinetic energy) of electron by an accelerating potential difference. [B1]
Replace the parallel slits with crystal with a lattice structure. [B1]
The matter wavelength associated with the electron must be of the same order as the lattice
spacing. [B1]
(iii) The fringes will become more intense [B1]
and sharper. [B1]
12. (a) (i) 1.
2.

3. Destructive interference, resultant amplitude = 0 mm.
(ii) 1.
2.
3. Constructive interference, resultant amplitude = 40 mm.
(b) (i) 1. Coherent – constant phase difference
2. Monochromatic – single wavelength or frequency
3. Interference – the superposition of two of more waves to form a resultant wave
Note:
Superposition refers to individual wave particles
Interference refers to entire waves
4. Diffraction –the spreading (change in direction and intensities) of waves around an obstacle or
through an aperture (of same order of magnitude as wavelength) (without any change in
velocity)
(ii) Pass the laser beam through a polariser.
As the polariser is rotated in a plane perpendicular to the beam, the intensity of the emerging beam
should vary from a maximum to zero or vice versa.
(iii) 1. Yes. Red or Orange
Accept intermediate colours i.e. reddish orange
2.
3.
4. Adv – larger angle can be calculated ⇒ % error is smaller
Disadv – intensity of 3rd order diffracted light is lower than 1st order
13. B
14. (a) (i) ………………… C1
path difference = 0.631 m ………………… A1
(ii) wavelength = path difference as first constructive interference away from central max
= 0.631 m ………………… A1
(use of formula for Young’s Double Slit not accepted, as only approximates the situation)
(iii) ………………… A1
(b) (i) for formation of stationary wave in a pipe, the lowest frequency (fundamental mode) would be obtained
when Length of pipe = ½ (wavelength). ………………… B1
(or similar and correct relationship given)

Thus for a given frequency (or wavelength) of sound source, length of pipe needs to be adjusted to meet
the above relation, or vice versa. ………………… B1
(ii) Fund. Mode
(iii)
(iv)
15. C
Any two points either side of the centre have a phase difference of 180° (or π).
16.
17.
18.
19.
20
21.
22.
23.
24.
25. C
26. A
27. C

Superposition 2008 prelim_solutions

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Superposition 2008 prelim_solutions