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Physics Tutorial 11
Superposition
SELF-CHECK QUESTIONS:
S1.
What is the principle of superposition?
S2.
How is a stationary wave formed?
S3.
What is the phase difference between particles located between the same two
successive nodes of a stationary wave? How about their amplitudes?
S4.
What is the longest possible wavelength of a stationary wave set up in a string fixed at
both ends? What about that of a stationary sound wave set up in a pipe with (i) one
end closed and (ii) with both ends open?
S5.
What is diffraction? Is the effect of diffraction more pronounced when the size of the
aperture is reduced or when it is enlarged?
S6.
Why is diffraction of light not observed easily (it was thought that light travels in a
straight line)?
S7.
Explain the term coherence and constructive and destructive interference.
S8.
In a double-slit experiment, what is the relation between the fringe separation x and
the wavelength , the slit separation a and the slit to screen distance D?
S9.
What is the grating formula?
S10. How do you calculate the maximum number of orders that can be observed when a
beam of monochromatic light passes through a diffraction grating?
PRACTICE QUESTIONS:
1.
An organ pipe of effective length 0.60 m is closed at one end. Given that the speed of
sound in air is 300 m s1, the two lowest resonant frequencies are
Ans: 125 Hz, 375 Hz
2.
A string fixed at both ends and of length L is plucked at its midpoint and emits its
fundamental note of frequency f1. When the string is plucked at a different point, the
first overtone frequency f2 is also produced. What is the ratio f2/f1 ? What is the speed of
the transverse waves in the string?
Ans: 2:1, v = 2 f1L or f2L
3.
A microwave transmitter emits waves which are reflected from a metal plate. The
reflected wave and the incident wave superpose to form a standing wave. When a
microwave detector was traversed along the standing wave, successive points at which
the detector detected zero intensity were located 1.5 cm apart. What is the frequency of
the waves?
Ans: 1.0  1010 Hz
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In a Young’s double-slit experiment using red light of wavelength 6.0  10-7 m, the slits
were 0.40 mm apart and the distance of the slits to the screen was 1.20 m.
(a) What is the separation of the fringes?
(b) What is the effect on the appearance of the fringes if
(i) the separation of the slits is decreased?
(ii) the screen is moved closer to the slits?
Ans: (a) 1.8 mm (b) fringe separation increases, fringe separation decreases.
5.
A diffraction grating has 400 lines per mm and is illuminated normally by
monochromatic light of wavelength 600 nm. Calculate
(a)
the grating spacing,
(b)
the angle to the normal at which the first order maximum is seen,
(c)
the number of diffraction maxima obtained.
Ans: (a) 2.5  106 m (b) 13.9 (c) 9
DISCUSSION QUESTIONS:
Stationary Waves
1.
A suspension bridge is to be built across a valley where it is known that the wind can
gust at 5 s intervals. It is estimated that the speed of transverse waves along the span
of the bridge would be 400 m s-1. Calculate the length of the span of the bridge when
there is danger of resonant motions at its fundamental frequency.
Ans: 1000 m
2.
One end of a horizontal string is attached to an oscillator and the other end passes over
a pulley. The string is kept under tension by means of a weight as shown below:
Pulley
Oscillator
Fig. 2
The frequency of oscillation of the oscillator is increased and at certain frequencies,
stationary waves are produced on the string.
(a) Draw the stationary wave on the string when the frequency is such that the
distance between the oscillator and the pulley corresponds to two wavelengths of
the wave on the string.
(b) On your diagram, label the position of a node on the string.
(c) Explain why a stationary wave is observed on the string only at particular
frequencies of vibration of the oscillator.
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A uniform tube, 60.0 cm long, stands vertically with its lower end dipping into water.
When the length above water is 14.8 cm, and again when it is 48.0 cm, the tube
resounds to a vibrating tuning-fork of frequency 512 Hz. Find the lowest frequency to
which the tube will resound when it is open at both ends.
Ans: 267 Hz
Interference
4.
In a double-slit arrangement, the distance between slits is 0.50 mm, and the slits are
1.0 m from the screen. Two interference patterns can be seen on the screen, one due
to light of 480 nm and the other 600 nm. What is the separation on the screen between
the third-order interference fringes of the different patterns?
Ans: 7.2  104 m
5.
A student sets up the apparatus illustrated in the Fig. 5 in order to observe two-source
interference fringes.
red
light
double slit
screen
Fig. 5
(a)
State a suitable separation for the two slits in the double slit.
(b)
State and explain what change, if any, occurs in the separation of the fringes and
in the contrast between bright and dark fringes observed on the screen, when
each of the following changes is made separately.
(i)
increasing the intensity of the red light incident on the double slit.
(ii) increasing the distance between the double slit and the screen.
(iii) reducing the intensity of light incident on one slit of the double slit.
6.
A source S of continuous waves a distance h from a plane reflector R produces regions
of high intensity such as C, C’ and C’’ as shown in Fig. 6. Account for this.
When the frequency of S is changed slowly, the regions C, C’ and C’’ move in the
direction D as shown. Account for this, and deduce whether the frequency has been
increased or decreased.
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R
h
C
S
C’
C’’
Fig. 6
D
In Appleton’s experiment, S was a radio transmitter on the Earth’s surface, and R was
the Heaviside layer – a reflecting layer in the atmosphere 80 km above the ground.
When the wavelength transmitted slowly changed from 200 m to 180 m, a receiver on
the ground 120 km away from S observed fluctuations in the received signal strength.
Calculate the number of signal strength maxima observed during this change of
frequency.
Ans: 44
7.
In Fig. 7, S1 and S2 are two coherent light sources in a Young’s two-slit experiment
separated by a distance 0.50 mm and O is a point equidistant from S1 and S2. O is on a
screen A which is 0.80 m from the slits.
When a thin parallel-sided piece of glass G of thickness 3.6 × 10−6 m is placed near S1
as shown, the centre of the fringe system moves from O to a point P. Calculate OP if
the wavelength of the monochromatic light from the two slits is 6.0 × 10−7 m and the
refractive index of the glass is 1.5.
Ans: 2.88 × 10−3 m
S1
G
P
0.50 mm
0.80 m
O
S2
Fig. 7
Diffraction Grating
8.
In the spectrum of white light obtained by using a certain diffraction grating, the second
and third orders partially overlap. What wavelength in the third-order spectrum will
appear at the angle corresponding to a wavelength of 650 nm in the second-order
spectrum?
Ans: 433 nm
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When a cadmium light is viewed through a diffraction grating having 500 lines per mm,
the first order spectral lines were observed at the stated angles.
Angle / 
Colour
18.78
red
14.74
green
13.89
light blue
13.53
dark blue
(a)
Find the wavelengths of each of these lines.
(b)
What is the maximum order of each colour?
Ans: 644nm, 509 nm, 480 nm, 468 nm, 3, 3, 4, 4
10.
A beam of red light from a laser is shone normally on to a diffraction grating. Bright light
is seen emerging at certain angles as shown in Fig. 10(a). Use the principle of
superposition to suggest a qualitative explanation of this effect.
1st order red
1st order violet
15.8
Light from gas
Red light from
laser
11.8
11.8 15.8
1st order violet
1st order red
Diffraction
grating
Diffraction
grating
Fig. 10(a)
Fig. 10(b)
A diffraction grating with a grating spacing of 2.20  106 m is used to examine the light
from a flowing gas. It is found that the first order violet light emerges at an angle 11.8
and the first order red light at an angle 15.8 as shown in Fig. 10 (b)
(a)
Calculate the wavelengths of these two colours.
(b)
Describe and explain what will be observed at an angle of 54.8.
(c)
Without making any further calculations, draw a sketch similar to Fig. 10(b)
showing the whole pattern observed.
Ans: 4.50  107 m, 5.99  107 m
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Challenging Question
11.
In Fig. 11, S1 and S2 are two equal point sources a distance d apart.
The sources emit waves of wavelength  in phase. P is a point on the line passing
through the midpoint of S1S2 and making an angle  with the centre line. The distances
of P from S1 and S2 are r1 and r2 respectively. The equation for the displacement y1 of
the wave arriving at P from S1 is
y1  yo sin{(2 /  )r1  t }
(1)
P
S1
d

centre line
S2
12
Fig. 11
(a)
Write down an equation, similar to equation (1), for the displacement y2 of the
waves arriving at P from S2.
(b)
Hence, or otherwise, write down an expression for the phase difference 
between the waves arriving at P from S1 and S2.
(c)
When P is a very great distance from the sources, S1P is approximately parallel to
S2P. Making this approximation, find an expression for  in terms of , d and .
(d)
Hence obtain the condition for P to be a point at which the amplitude of the
resultant disturbance is a maximum.
An oil film (refractive index n = 1.45) floating in water is illuminated by white light at
normal incidence. The film is 280 nm thick. By considering the interference of the light
rays reflected and oil–water and oil-air boundary, find
(a)
dominant observed colour in the reflected light and
(b)
The dominant colour in the transmitted light.
Note: when light enters from a less dense medium into a denser medium at
normal incidence, it will undergo a 180 phase jump. No such phase jumps
occurs when light enters an optically less than medium from a denser one.
Ans: (a) 541 nm (b) 406 nm
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Tutorial 11 Solutions:
S1
The principle of superposition states that when two waves of the same kind meet at a
point in space, their combined effect is given by the vector sum of the displacements
due to each of the waves at the point.
S2
A stationary wave is formed when two waves of the same frequency, same amplitude
and same wavelength propagating in opposite directions meet.
S3
These particles are in phase (i.e. phase difference = 0) and they have different
amplitudes. Their amplitudes may vary from 0 at the nodes to 2A at the antinodes.
S4
Twice the length of the string, four times the length of the pipe, twice the length of the
pipe.
S5
Diffraction is the bending and spreading out of light waves when they propagate
through a small aperture. It is more pronounced when the aperture size is reduced.
S6
Diffraction effects are only significant when the aperture size is comparable to the
wavelength of the waves. The wavelengths of visible Light are very short (10−7 m) as
compared the size of ‘ordinary’ openings such as doors and windows.
S7
Coherence refers to the situation in which two sources emit waves that have a constant
phase relationship.
Constructive (Destructive) interference occurs when two waves meet in phase (antiphase) producing a resultant wave of amplitude equal to the sum (difference) of the
amplitudes of the two waves.
D
a
S8
x 
S9
dsin = n
S10
Use
d
n
 sin   1  n 

d
PRACTICE QUESTIONS:
1.
Since this is a closed pipe, the two lowest frequencies will be the first and the third
harmonics. The wavlengths are 4L and 4L/3 respectively.
Hence, the frequencies are:
First Harmonic: f1 = 300  2.4 = 125 Hz
Third Harmonic: f3 = 300  0.8 = 375 Hz
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Since the speed of the wave propagating along the string is a constant,
f2 1 2


f1 2 1
The speed is f   = f1  2 L = 2 f1L
3.
The nodes are separated 1.5 cm apart. Hence the wavelength is 3.0 cm.
By v = f   ,
3  108 = f  0.03

4.
(a)
(b)
f = 1.0  1010 Hz
Separation of fringes x 
D
a

6.0  107  1.2
 1.8  103 mm
4  104
(i) If a is decreased, then x, the fringe separation increases.
(ii) If D is decreased, then x, the fringe separation decreases.
5.
(a)
400 lines per mm = 400 000 lines per metre.
Grating spacing d = 1/N = 1 400 000 = 2.5  106 m
d sin   n
(b)

2.5  10 6 sin   1 6  10 7

  13.9
(c)
sin 
n
d

n
1
d
 n
d


2.5  106
 4.17
6  107
Hence, n = 4. Adding the maxima of the ‘other side’ and that in the straightthrough direction, the total number of maxima is 4 + 4 + 1 = 9
DISCUSSION QUESTIONS:
1.
Frequency of wind = 0.2 Hz. Hence, resonance occurs when the fundamental
frequency of the bridge is 0.2 Hz too. The resonant wavelength is:

v 400

 2000 m
f
0.2
If this is the fundamental mode, then  = 2L so that L = 1000 m
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(a)
Pulley
N
N
N
Oscillator
(c)
3.
To get resonance, the frequency of the oscillator must be equal to the various
harmonics of the string.
Let the speed of sound be v.
c
c
14.8 cm
48 cm
60 cm
We form two equations using the two resonant length:

(1)
 14.8  c
4
3
(2)
 48  c
4
Subtract (1) from (2), we get  = 66.4 cm  v = 0.664  512 = 340 m s1
Substituting  = 66.4 into (1), we get c = 1.8 cm.
Let the lowest frequency to which the tube will resound when it is open at both ends be
f. The corresponding wavelength is  = 2  (60 + 2c) = 127.2 cm. Hence,
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f  1.272  340

4.
f = 267 Hz
By x 
D
a
,
fringe separation for 480 nm 
4.8  107  1
5  104
 9.6  104 m
fringe separation for 600 nm 
6.0  107  1
5  104
 1.2  10 3 m

Hence, separation between third-order interference fringes  3 1.2  103  9.6  104
 7.2  10
5.
4

m
(a)
0.5 mm
(b)
(i) The contrast is improved because the bright fringes become brighter.
(ii) The spacing increases (use x = D/a)
(iii) Contrast decreases because the dark fringes are not completely dark.
6.
In regions C, C’ and C’’, the signal arriving directly from the source and the signals
reflected from the reflector R interfere constructively and result in a high intensity signal.
As the frequency is changed, the wavelength changes correspondingly. Since
constructive interference occurs when the path difference between the direct signals
and the reflected signals is equal to a multiple of the wavelength, C, C’ and C’’ must
shift in order the condition is satisfied.
Since the path difference increases, the frequency has been decreased.
Path length 1  120 km
Path length 2  2  602  802  200 km
Hence, path difference = 80 km
When  = 200 m, this corresponds to 80 000 ÷ 200 = 400 wavelengths
When  = 180 m, this corresponds to 80 000 ÷ 180 = 444.4 wavelengths
Hence, the signal strength will go through 44 maxima.
7.
Extra optical path introduced to S1 = 1.5 ×(3.6 ×10−6) − 3.6 ×10−6 = 1.8 ×10−6 m
This means that physically, S1P must be 1.8 ×10−6 m (or 3 wavelengths) less than S2P.
Hence, OP = 3 ×x = 3 × 6.0 × 10−7 × 0.80 ÷ 5.0 × 10−4 = 2.88 × 10−3 m
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If two wavelengths 1 and 2 are diffracted to the same angle, then


9.
10.
d sin  n11  n2 2
3  1  2  650
1  433 nm
d = 1/500 = 0.002 mm = 2 ×10−6 m
Angle / 
  d sin / nm
Max. Order = d/
18.78
644
3
14.74
509
3
13.89
480
4
13.53
468
4
When light passes through the slits of the grating, they are diffracted in various angles.
At some angles, the path difference between light rays emanating from adjacent slits is
equal to a multiple of the wavelength of the light. This results in constructive
interference and bright light is observed in these angles. In any other directions, the
light rays have path difference different from a multiple of the wavelength and the rays
will sum up to a near-zero resultant.
(a)
(b)
 = 11.8

 = d sin11.8 = 4.50  107 m
 = 15.8

 = d sin15.8 = 5.99  107 m
At  = 54.8, d sin = 1.80  106 m. This is equal to 4  4.50  107 m or 3  5.99
 107 m. Hence, the 4th order of the violet light will overlaps with the 3rd order of
the red light at this angle.
(c)
Light from gas
11.
2008
(a)
y 2  yo sin{(2 /  )r2  t }
(b)

(c)
2

 r2  r1 
In this case, r2  r1  d sin  . Hence,

2

d sin 
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(d)
For maximum disturbance,   2n , where n 
2

2008
0
.i.e.
d sin   2n
d sin   n
12.
Light is reflected at both the air-oil interface and the oilwater interface. At the air-oil interface, light is reflected
when it is entering a denser medium. Hence, there is a
180 phase change. No such phase change occurs for
the reflection at the oil-water surface since refractive
index of water is only 1.3.
The dominant colour of the reflected light corresponds
to the wavelength that experiences constructive
interference.
(a)
Hence, condition for constructive interference is:
o.p.d.  n  2d   2m  1

2
1.45  2  2.8  107   2m  1

2
1.624  106
 
2m  1
6
When m = 0,  = 1.624  10 m (infra-red)
m = 1,  = 5.41  107 m (green)
m = 2,  = is in the uv region.
Hence, green light is strongly reflected.
(b)
For the transmitted light, there is no phase jump. Hence, for constructive
interference,
o.p.d.  n  2d  m
1.45  2  2.8  107  m
8.12  107
m
7
When m = 1,  = 8.12  10 m (infra-red)
m = 2,  = 4.06  107 m (violet)


Hence, violet light is strongly reflected.
12
oil
water