Chapter 16: Diffraction solutions
1.
[4 marks]
(VCE 2001 Q11)
When the wavelength is long the sound
waves can diffract around the head, so
the intensity for both ears will be roughly
the same. The sound takes a little of time
to travel to the other ear, the brain can
detect the time difference between the
compressions reaching the first ear then
the other. Using this time difference the
brain calculates where the sound is
coming from.
Short wavelengths
This does not work as well when the
wavelength is smaller, the time difference
is too small and the brain cannot detect
the difference as well. The way we
overcome this is that the smaller
wavelengths do not diffract around the
head as well, and therefore the intensity
on one side of the head is more intense
than the other. The brain uses this to
determine where the sound is coming
from.
2.
[2 marks]
3.
[2 marks]
(VCE 2003 Q7)
Use v = f
 340 = 6000  
340
=
6000
= 5.7cm (ANS)
(Make sure you round off correctly)
4.
[2 marks]
(VCE 2003 Q8)
The amount of diffraction depends upon
the wavelength of the sound compared with
the width of the speaker.

This is expressed as the ratio
where d is
w
the width of the speaker and diffraction is
the most when this ratio is > 1.
At 6000Hz the wave length is 5.7cm. This
means that not much diffraction will occur,
and so this sound will be very directional,
and only project where the actual speaker is
pointing. This means that Mustafa won’t
hear this frequency very well.
Rebus will hear all the frequencies being
produced, but only the lower frequencies
will diffract sufficiently for Mustafa to hear.
(VCE 2002 Q6)
The amount of diffraction depends upon
the wavelength of the sound compared with
the width of the door.

This is expressed as the ratio
where w is
w
the width of the door and diffraction is the
most when this ratio is > 1, ie when  > 1m.
Since v = f (the wave equation) and
v = 340m/s, the corresponding frequency is
340Hz.
An estimate of what frequencies will have a

reduced intensity is when the
ratio is
w
less than 1.
This will result in all frequencies above
340Hz diffracting less to some extent and so
won’t be heard as loudly by Peta.
 340 – 22000Hz (ANS)
5.
[3 marks]
(VCE 2003 Q9)
Sound can’t travel directly to the houses, so
it must get there through transmission,
diffraction and reflection.
Barrier B absorbed and reflected sound,
thus reducing the noise level at all houses.
Some sound would diffract over the barrier
to reach all houses, but more would reach
house 3 because it does not have to diffract
as much. Sound would also reflect from
barrier A to reach house 3.
(You needed to make the last comment to
get full marks for this question.)
10.
6.
[2 marks]
(VCE 2003 Q10)
The sound reflected by barrier A was now
scattered (diffuse reflection) by the rough
wall, so a lot less sound was reflected. Some
extra sound was also absorbed by the wall.
This will have much more effect on house 3,
as this as the one that had most of the
reflected sound getting to it.
7.
[4 marks]
(VCE 2004 Q7)
The graph of amplitude arises as a result of
diffraction of the sound passing through the
gap.
Diffraction means that the sound spreads
out after passing through the gap. The
amount of diffraction depends upon the
ratio of the wavelength to the gap size.
Since there is a lot of diffraction then

~ 1.
w
The gap size is 0.30 m and so wavelengths
greater than 0.30 m will show significant
diffraction. Given that the speed of sound is
340ms-1 then a wavelength of 0.30 m
corresponds to a frequency of 1133Hz.
Hence, the approximate frequency used was
about 1000 Hz.
(Answers in the range 500 – 1500 Hz were
accepted.)
[2 marks]
(VCE 2004 Q8)
The curve shows very little diffraction and
so the wavelength is considerably less than
0.30m. Correspondingly the frequency will
be considerably higher than 1000 Hz, as
calculated in question 549. Thus, the
frequency of 10 000 Hz as listed in A is best
fits with Lee’s prediction.
 A (ANS)
9.
[2 marks]
D (ANS)
(VCE 2004 Sample Q7)
(VCE 2004 Sample Q8)
The amount of diffraction depends upon
the wavelength of the sound compared with
the width of the speaker.

This is expressed as the ratio
where w is
w
the width of the speaker and diffraction is
the most when this ratio is > 1. The higher
frequencies with their smaller wavelengths
don’t diffract enough for Leo to hear them.
Louise will hear all the frequencies because
she is directly in front of the speaker. She
will hear all the frequencies produced by
the speaker, in the correct proportions.
11.
[4 marks]
(VCE 2004 Pilot Q10)
340
100
= 3.4 m (ANS)
a) Use λ =
b) The low frequency sound (100 Hz) with a
wavelength of 3.4m will diffract around the
screen because the screen is smaller than the
wavelength. These sounds will be heard
well.
The higher frequencies of the dialogue (3000
Hz) will have λ around 0.1 m and so will
not diffract around the screen. These sounds
will not be heard well
12.
8.
[3 marks]
[3 marks]
(VCE 2006 Q1)
Jamie is listening to the sound of an
orchestra through a small gap in a partly
open sliding door. When the sound wave
travels through the gap, [diffraction] occurs
and spreading of the wave results. High
pitched (frequency) instruments such as
flutes experience [less] spreading than
lower pitched instruments. As the size of
the gap decreases, the angle of spreading
will [increase].
13.
[3 marks]
60 = 10 log
(VCE 2007 Q8)
I
I0
where I0 = 1 x 10-12
 6.0 = log

I
1  10 -12
I
= 106.0
1  10 -12
 I = 106.0  1  10-12
 I = 10-6.0
I = 1  10-6 W m-2 (ANS)
14.
a)
[1 + 2 = 3 marks]
18.
(VCE 2007 Q9)
200 Hz (ANS)
b) To reach the point Y, the sound will
need to diffract. The amount of diffraction
depends upon the wavelength of the sound
compared with the width of the stage
opening (w)

This is expressed as the ratio
w
The higher frequencies with their smaller
wavelengths don’t diffract as much and so
will sound softer at the point Y.
15.
[2 marks]
(VCE 2007 Q10)
The sound level has dropped by 9dB. Each
drop of 3dB corresponds to a halving of the
intensity. Therefore the intensity has halved
three times, so the ratio is 8:1
 C (ANS)
16.
[2 marks]
(VCE 2007 Q11)
Doubling the distance will decrease the
sound intensity (Wm-2) by a factor of 4. This
leads to a 6dB drop in sound level.
 B (ANS)
17.
few rows will not hear the sound at the
same level as those people sitting towards
the middle of the room. Thus the sound will
be distorted for the people in the outer
edges of the first few rows due to diffraction
the lower frequencies will be louder to these
people compared to the higher frequencies.
 B (ANS)
[2 marks]
(VCE 2008 Q12)
The gap on the stage is approximately 3.0m,
sound with of more than 100 Hz will not
diffract as much as those below this value.
This means that the outer edges of the first
[2 marks]
(VCE 2008 Q13)
The sound will reflect off the walls in many
different directions and this will lead to
many paths for the sound to reach people
from the one speaker. This will then lead to
interference in some parts of the theatre.
 D (ANS)