Physics in Everyday Life
Prof George Lafferty, 12 Nov 2014
Notes for Lectures 13 and 14
1
1.1
THE EYE AS AN OPTICAL INSTRUMENT
... continued
Resolving power of the eye
Light passing through any aperture undergoes diffraction, which results in blurring
at the edges of the image. Light passing through a circular aperture, such as an
iris, will produce a circular pattern as shown in figure 1. The central bright spot
contains about 84% of the light and is known as the Airy disk. Further diffraction
rings can be seen outside of this disk.
Figure 1: Diffraction of light by a circular apperture. (Taken from the internet and
to be acknowledged.)
Figure 2 shows the diffraction pattern for light of wavelength 500 nm from
two point sources passing through an aperture of diameter 6 mm. The y-axis shows
the intensity while the x-axis represents the angular separation of the sources, in
radians, subtended at the aperture. It can be seen that when the sources are 0.4 mrad
(±0.2 mrad) apart, the image diffraction patterns are well separated, and so would
be discernable as separate sources by an eye. As the angle between the two sources
decreases the diffraction patterns begin to overlap and when the separation is at
about 0.1 mrad, the patterns are effectively merged into one, and the two sources
would be perceived as a single light.
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The Rayleigh criterion for resolving two such sources of light (or indeed any
type of electromagnetic radiation) is based on the peak of one pattern coinciding with
the first minimum of the other. If the patterns are closer together than this, then
they cannot be resolved (at least according to Rayleigh’s criterion). This condition
requires
1.22λ
θmin =
D
where θmin is the smallest resolvable angle between the sources, λ is the wavelength
of the light and D is the diameter of the aperture. Figure 3 illustrates the definitions
of the quantities for a human eye.
If we assume an iris of diameter 6 mm and a wavelength of 550 nm, we see that
θmin = 10−4 rad = 0.1 mrad. In fact the resolving power of the eye is not quite as
good as this theoretical value, but is limited by the size of the light-receptor cells on
the retina, which are about 2 µm in diameter. In order for two points of light to be
perceived as two separate sources, there must be at least one inactive cell between
two excited cells on the iris. Therefore the excited cells need to be at least about
4 µm apart, and this limits the resolving power to θmin ≈ 3 × 10−4 rad. While some
people have eyes that can get close to this, in most cases imperfections in the iris
and the lens degrade the resolving power to about 5 × 10−4 rad.
Experiments have shown that about 60 photons of light need to reach the
cornea to produce a perceptible flash of light. Of these, about half are absorbed
by the eye before they can reach the retina, while about 5 of the remaining 30 are
actually absorbed by the photo-receptor cells on the retina. These initiate chemical
changes that result in electrical signals being sent along the optic nerve to the
brain, where the wonderful process of interpretation of the signals as an external
visual image all takes place.
2
2.1
SOUND AND HEARING
Sound as a longitudinal wave
Sound is in essence a wave phenomenon, whereby energy is transmitted through
a medium as waves or pulses. The waves are longitudinal, which means that the
associated disturbance is in the same direction as the wave propagation direction.
This may be contrasted with transverse waves, such as electromagnetic radiation,
surface water waves, waves on a stretched string of a musical instrument etc., where
the disturbance is at right angles (i.e. transverse) the the direction of propagation
of the wave.
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As in a transverse wave, the transfer of energy in a sound wave does not involve bulk translational motion of the the medium carrying the disturbance. For
example, for sound in air, the air molecules typically have a harmonic, backwards
and forwards, motion superimposed on their random thermal motion. The sound
propagates through the medium by “layers” of air oscillating and successively setting in motion subsequent layers in the direction of propagation. (Similarly, in a
transverse wave on a string a piece of string moves from side to side while the wave
itself moves along the string.) In speech and hearing, the disturbance starts with
motion of the vocal chords of the speaker producing fluctuations in air pressure that
propagate out to a listener whose ear drums respond to these pressure fluctuations,
converting the signal to an electrical one that is send for interpretation by the brain.
Since a sound signal in a gas is passed on by motion of the molecules of the
medium through which is passes, it cannot travel faster than the typical speed of a
molecule. In air at STP, a typical molecular speed is about 500 m s−1 . In fact the
speed of sound is given more precisely by the elastic properties of medium:
s
B
v=
ρ
where B is the bulk modulus and ρ is the density. The bulk modulus of a gas is
rather analagous to the Young’s modulus for a wire. In the case of a gas, the bulk
modulus is defined as bulk stress over bulk strain:
B=
∆p
∆V /Vo
where the strain ∆p is the change in pressure and the stress is the fractional change
in volume ∆V /V0 .
In p
a solid with Young’s modulus Y and density ρ, the speed of sound is given
by v = Y /ρ. In this case, Y represents the ratio of tensile stress (force per unit
area applied to the solid) to tensile strain (fractional change in length).
In air at STP, the value of the bulk modulus is B = 1.4 × 105 Pa, and the
speed of sound is vsound = 344 m s−1 . In everyday units, this speed is equivalent
to about one mile in five seconds. All campers should know how to estimate the
distance of a lightning strike: start counting seconds as soon as the flash is seen;
every five seconds represents one mile of distance. If successive strikes are getting
closer, then it is time to worry.
Consider a longitudinal wave propagating along the x direction. The displacement of any particle in the wave is parallel to the x axis, but we need to draw this
on a graph along the y axis. In general, the displacement of a particle in the wave
is a function of both position along the wave and of the time i.e. y = y(x, t). We
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stress again that y does not represent the direction transverse to the wave, but displacement in the direction of propagation of the wave. If we look at a snapshot of a
pure sound wave at time t = 0 (see figure 4), the displacement may be represented
as
2πx
= A cos kx,
y(x, t = 0) = A cos
λ
where k = 2π/λ is called the wavenumber, which has SI units of m−1 . A snapshot
at a fixed position along the wave, say x = 0, may be written as
y(x = 0, t) = A cos 2πνt = A cos ωt,
where ν is the frequency of the wave, and ω = 2πν is the angular frequency. The
period of the wave is T = 1/ν = 2π/ω. For a wave travelling to the right, the full
expression, which will be considered in detail in PHYS10132, is
y(x, t) = A cos(kx − ωt).
In a sound wave it is the fluctuations in pressure, rather than displacement,
that are deteced. In the ear, for example, pressure fluctuations on the ear drum generate the signal that goes to the brain. As shown in figure 4, the pressure fluctuates
about its average value and the wave is 90◦ out of phase with the displacement wave.
Where the average displacement is zero, particles are either moving into or moving
out of the region, creating high or low pressure. Where the displacement has its
maximum positive or negative value, the particles are moving in unison to positive
or negative values of x, so that the average pressure change at those positions is
zero.
Thus the pressure, like the displacement, varies sinusoidally about the average
value, and is given by
p(x, t) = BkA sin(kx − ωt),
where p represemts the “gauge pressure” i.e. the difference from the average pressure. The amplitude of the pressure wave is pmax = BkA = B(2π/λ)A.
A moderately loud sound wave may have pmax = 0.03 Pa, which is of order
10 times atmospheric pressure, and a frequency iof 1 kHz. We can calculate the
displacement amplitude using A = pmax /Bk. For B = 1.4 × 105 Pa, frequency
ν = 1 kHz and speed v = 344 m s−1 , the wavenumber is k = 2π/λ = 2πν/v and
A turns out to have a value of 1.2 × 10−8 m. We see that the amplitude of the
displacement wave is tiny, but the very small pressure fluctuations associated with
this are detectable by the human ear. In fact, at the threshold of hearing, the human
ear can detect pressure fluctuations as small as 3×10−10 times standard atmospheric
pressure.
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2.2
Energy in sound
The intensity, I, of a sound wave (or indeed any wave) is defined as the time averaged
rate of transport of energy per unit area, which has units of W m−2 . In our picture
of a longitudinal sound wave, pressure pulses from a source propagate through the
medium with a speed v. The energy in the wave can be related to the work done
by the force associated with the pressure. The work done is is force times distance
moved by the particles of the medium, so that the rate of work is given by the force
times the velocity of the particles. Since the pressure is force per unit area, then
the intensity, or rate of transport of energy per unit area, is given by the product of
the pressure and the velocity.
We already have the displacement of the particles, y(x, t) = A cos(kx − ωt), so
the particle velocity is
∂y
= ωA sin(kx − ωt).
vy =
∂t
This expression represents the forward-backward motion of the particles transmitting the wave, and is not the same as the speed of the wave itself. We’ve already written down an expression for the gauge pressure in the wave p(x, t) = BkA sin(kx−ωt)
and therefore the rate of energy transfer per unit area is pvy = BωkA2 sin2 (kx − ωt).
We see that for some fixed value of x, the energy transfer varies between zero
and a maximum value of BωkA2 . The intensity is the time-averaged rate, and so we
need to average over a full period of the wave to account for the variations during
a period. It is a generally useful result (exercise: show this) that the integral of a
squared cosine or sine function over a period of the oscillations is equal to 1/2. In
this case
Z t=2π/ω
1
sin2 (kx − ωt)dt = .
2
0
(This result appears in other areas of physics, such as in the power in an alternating
current circuit.)
Therefore the intensity in a sound wave is given by
1
I = BωkA2 .
2
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Figure 2: Diffraction patterns for light from two point sources of light of wavelength
500 nm passing through an aperture of diameter 6 mm. The diagrams from top to
bottom show the effect of decreasing angular separation between the two sources.
Figure 3: Illustrating the angle θ made by light from two sources a distance x apart
at a distance L from the eye, which has an iris diameter of D. The Raleigh criterion,
θmin = 1.22λ/D gives the minimum resolvable value of θ.
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Figure 4: Showing the correspondence between displacement and pressure in a sound
wave of constant frequency. (Taken from the internet and adapted. To be acknowledged.)
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