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-3
mm /3.5 = 1.0 mm = 1.0 x10 m.
= ya/D = 7.0 x10 m
-7
 Or: y3.5 = 3.6 mm, = y3.5a/3.5D = 7.0
x10 m
-7

 If the statement means the spacing between
4 dark fringes: i.e. 3 spaces
 The spacing between 4 dark fringes = 3y =
3.6 mm
-3
y = 3.6 mm /3 = 1.2 mm or 1.2 x10 m.
= ya/D = 8.4 x10 m
-7
i.e. 3 possible answers:
= 6.3x10 m, 8.4 x10 m, 7.0 x10 m
-7
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2c(ii) “the intensity variation of
fringes from the central
maximum” is NOT on the
syllabus.

st
5a(ii) Comparing the response of the 1 property:
human ear to the dB scale is  The Human ear can perceive/respond to a
NOT on the syllabus. very wide range of sound intensities. OR
SO 3.1 “discuss the response The response of the Human ear to sound
of the ear to incoming sound intensity is [roughly] logarithmic.
waves, in terms of sensitivity,  The dB scale is Logarithmic [It therefore
frequency response and matches the response of the ear]
intensity”. 2 property??:
nd
SO 3.3 “use the equation  The Human ear’s perception of loudness
Intensity level (in dB) = 10 log does not increase linearly with intensity but
⁄ ” is logarithmic OR
0
 Equal changes in Intensity are not perceived
by the Human ear as equal changes in
Loudness OR
 Change of loudness perceived by the Human
ear = changes in intensity/original intensity
= ∆/ [i.e. fractional changes in Intensity]

 On the dB scale, Intensity level = 10 log
I/I0
5b(ii)  For a standing wave: There is Possible Solutions:
a Node at the wall – No
vibration of particles  The student KNOWS that there is a
 There is an Antinode at the Node at the wall and an Antinode at the
speaker – Maximum vibration speaker. [as in the resonance tube]

of particles.

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