IPhO 2006 · Theory problem 3

Five order-of-magnitude estimates: Digital Camera, Hard-boiled Egg, Lightning, Capillary Vessels, Skyscraper

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This question consists of five independent parts. Each of them asks for an estimate of an order of magnitude only, not for a precise answer. Enter all your answers into the Answer Script.

Digital Camera Consider a digital camera with a square CCD chip with linear dimension \(L=35\ \mathrm{mm}\) having \(N_p=5\ \mathrm{Mpix}\) (\(1\ \mathrm{Mpix}=10^6\) pixels). The lens of this camera has a focal length of \(f=38\ \mathrm{mm}\). The well known sequence of numbers (2, 2.8, 4, 5.6, 8, 11, 16, 22) that appear on the lens refer to the so called F-number, which is denoted by \(F\#\) and defined as the ratio of the focal length and the diameter \(D\) of the aperture of the lens, \(F\#=f/D\).

3.1 (1.0) Find the best possible spatial resolution \(\Delta x_{\min}\), at the chip, of the camera as limited by the lens. Express your result in terms of the wavelength \(\lambda\) and the F-number \(F\#\) and give the numerical value for \(\lambda=500\ \mathrm{nm}\).

3.2 (0.5) Find the necessary number \(N\) of Mpix that the CCD chip should possess in order to match this optimal resolution.

3.3 (0.5) Sometimes, photographers try to use a camera at the smallest practical aperture. Suppose that we now have a camera of \(N_0=16\ \mathrm{Mpix}\), with the chip size and focal length as given above. Which value is to be chosen for \(F\#\) such that the image quality is not limited by the optics?

3.4 (0.5) Knowing that the human eye has an approximate angular resolution of \(\phi=2\ \mathrm{arcmin}\) and that a typical photo printer will print a minimum of 300 dpi (dots per inch), at what minimal distance \(z\) should you hold the printed page from your eyes so that you do not see the individual dots?

Data

1 inch \(=25.4\ \mathrm{mm}\)
1 arcmin \(=2.91\times10^{-4}\ \mathrm{rad}\)

Hard-boiled egg An egg, taken directly from the fridge at temperature \(T_0=4^\circ\mathrm{C}\), is dropped into a pot with water that is kept boiling at temperature \(T_1\).

3.5 (0.5) How large is the amount of energy \(U\) that is needed to get the egg coagulated?

3.6 (0.5) How large is the heat flow \(J\) that is flowing into the egg?

3.7 (0.5) How large is the heat power \(P\) transferred to the egg?

3.8 (0.5) For how long do you need to cook the egg so that it is hard-boiled?

Hint You may use the simplified form of Fourier’s Law \(J=\kappa\Delta T/\Delta r\), where \(\Delta T\) is the temperature difference associated with \(\Delta r\), the typical length scale of the problem. The heat flow \(J\) is in units of \(\mathrm{W\,m^{-2}}\).

Data

Mass density of the egg: \(\mu=10^3\ \mathrm{kg\,m^{-3}}\)
Specific heat capacity of the egg: \(C=4.2\ \mathrm{J\,K^{-1}\,g^{-1}}\)
Radius of the egg: \(R=2.5\ \mathrm{cm}\)
Coagulation temperature of albumen (egg protein): \(T_c=65^\circ\mathrm{C}\)
Heat transport coefficient: \(\kappa=0.64\ \mathrm{W\,K^{-1}\,m^{-1}}\) (assumed to be the same for liquid and solid albumen)

Lightning An oversimplified model of lightning is presented. Lightning is caused by the build-up of electrostatic charge in clouds. As a consequence, the bottom of the cloud usually gets positively charged and the top gets negatively charged, and the ground below the cloud gets negatively charged. When the corresponding electric field exceeds the breakdown strength value of air, a disruptive discharge occurs: this is lightning.

Figure 1

Idealized current pulse flowing between the cloud and the ground during lightning.

Answer the following questions with the aid of this simplified curve for the current as a function of time and these data:

Distance between the bottom of the cloud and the ground: \(h=1\ \mathrm{km}\);
Breakdown electric field of humid air: \(E_0=300\ \mathrm{kV\,m^{-1}}\);
Total number of lightning striking Earth per year: \(32\times10^6\);
Total human population: \(6.5\times10^9\) people.

3.9 (0.5) What is the total charge \(Q\) released by lightning?

3.10 (0.5) What is the average current \(I\) flowing between the bottom of the cloud and the ground during lightning?

3.11 (1.0) Imagine that the energy of all storms of one year is collected and equally shared among all people. For how long could you continuously light up a 100 W light bulb for your share?

Capillary Vessels Let us regard blood as an incompressible viscous fluid with mass density \(\mu\) similar to that of water and dynamic viscosity \(\eta=4.5\ \mathrm{g\,m^{-1}\,s^{-1}}\). We model blood vessels as circular straight pipes with radius \(r\) and length \(L\) and describe the blood flow by Poiseuille’s law,

\[ \Delta p=RD, \]

the Fluid Dynamics analog of Ohm’s law in Electricity. Here \(\Delta p\) is the pressure difference between the entrance and the exit of the blood vessel, \(D=S\upsilon\) is the volume flow through the cross-sectional area \(S\) of the blood vessel and \(\upsilon\) is the blood velocity. The hydraulic resistance \(R\) is given by

\[ R=\frac{8\eta L}{\pi r^4}. \]

For the systemic blood circulation (the one flowing from the left ventricle to the right auricle of the heart), the blood flow is \(D\approx100\ \mathrm{cm^3\,s^{-1}}\) for a man at rest. Answer the following questions under the assumption that all capillary vessels are connected in parallel and that each of them has radius \(r=4\ \mu\mathrm{m}\) and length \(L=1\ \mathrm{mm}\) and operates under a pressure difference \(\Delta p=1\ \mathrm{kPa}\).

3.12 (1.0) How many capillary vessels are in the human body?

3.13 (0.5) How large is the velocity \(\upsilon\) with which blood is flowing through a capillary vessel?

Skyscraper At the bottom of a 1000 m high skyscraper, the outside temperature is \(T_{\mathrm{bot}}=30^\circ\mathrm{C}\). The objective is to estimate the outside temperature \(T_{\mathrm{top}}\) at the top. Consider a thin slab of air (ideal nitrogen gas with adiabatic coefficient \(\gamma=7/5\)) rising slowly to height \(z\) where the pressure is lower, and assume that this slab expands adiabatically so that its temperature drops to the temperature of the surrounding air.

3.14 (0.5) How is the fractional change in temperature \(dT/T\) related to \(dp/p\), the fractional change in pressure?

3.15 (0.5) Express the pressure difference \(dp\) in terms of \(dz\), the change in height.

3.16 (1.0) What is the resulting temperature at the top of the building?

Data

Boltzmann constant: \(k=1.38\times10^{-23}\ \mathrm{J\,K^{-1}}\)
Mass of a nitrogen molecule: \(m=4.65\times10^{-26}\ \mathrm{kg}\)
Gravitational acceleration: \(g=9.80\ \mathrm{m\,s^{-2}}\)

Official English text of the problem.

Answers

The main estimates use the diffraction scale \(\lambda F\#\), boiling water at \(100^\circ\mathrm{C}\), and \(E_0h\) as the characteristic lightning voltage.

  • 3.1: \(\Delta x_{\min}\sim\lambda F\#\). At the largest available aperture, \(F\#=2\), this is \(\boxed{1.0\ \mu\mathrm{m}}\). The circular-aperture Rayleigh value is \(1.22\ \mu\mathrm{m}\).
  • 3.2: \(\boxed{N\sim1.2\times10^3\ \mathrm{Mpix}}\), or \(8.2\times10^2\ \mathrm{Mpix}\) with the Rayleigh coefficient.
  • 3.3: \(\boxed{F\#=16}\) using \(\Delta x\sim\lambda F\#\). A strict Rayleigh comparison gives \(\boxed{F\#=11}\).
  • 3.4: \(\boxed{z_{\min}\approx0.15\ \mathrm{m}=15\ \mathrm{cm}}\).
  • 3.5: \(\boxed{U\approx1.7\times10^4\ \mathrm{J}}\).
  • 3.6: \(\boxed{J\sim2.5\times10^3\ \mathrm{W\,m^{-2}}}\).
  • 3.7: \(\boxed{P\sim19\ \mathrm{W}}\).
  • 3.8: \(\boxed{t\sim8.7\times10^2\ \mathrm{s}\approx15\ \mathrm{min}}\).
  • 3.9: \(\boxed{Q=5\ \mathrm{C}}\).
  • 3.10: \(\boxed{I=5.0\times10^4\ \mathrm{A}=50\ \mathrm{kA}}\).
  • 3.11: \(\boxed{t\sim7.4\times10^4\ \mathrm{s}\approx20\ \mathrm{h}}\). A completely discharging capacitor model gives approximately \(10\ \mathrm{h}\).
  • 3.12: \(\boxed{N_{\mathrm{cap}}\approx4.5\times10^9}\) capillary vessels.
  • 3.13: \(\boxed{\upsilon\approx4.4\times10^{-4}\ \mathrm{m\,s^{-1}}=0.44\ \mathrm{mm\,s^{-1}}}\).
  • 3.14: \(\boxed{\displaystyle\frac{dT}{T}=\frac{\gamma-1}{\gamma}\frac{dp}{p}=\frac27\frac{dp}{p}}\).
  • 3.15: \(\boxed{\displaystyle dp=-\rho g\,dz=-\frac{mpg}{kT}\,dz}\), where \(\rho\) is the local air density.
  • 3.16: \(\boxed{T_{\mathrm{top}}\approx294\ \mathrm{K}\approx21^\circ\mathrm{C}}\).

Solution

3.1. For an ideal lens, diffraction at the aperture limits the angular separation of distinguishable objects to the scale

\[ \theta_{\min}\sim\frac{\lambda}{D}. \]

For distant objects, an angular separation \(\theta\) corresponds to a separation \(f\theta\) in the focal plane. Thus

\[ \boxed{\Delta x_{\min}\sim f\frac{\lambda}{D}=\lambda F\#.} \]

At \(\lambda=500\ \mathrm{nm}\),

\[ \Delta x_{\min}\sim0.50(F\#)\ \mu\mathrm{m}. \]

The best resolution occurs at the largest aperture, corresponding to the smallest listed F-number, \(F\#=2\). Its diameter is \(D=f/2=19\ \mathrm{mm}\), giving

\[ \boxed{\Delta x_{\min}\sim1.0\ \mu\mathrm{m}.} \]

For a circular aperture, the Rayleigh criterion gives the more specific expression

\[ \Delta x_{\mathrm{Rayleigh}}=1.22\lambda F\#, \]

and hence \(1.22\ \mu\mathrm{m}\) at \(F\#=2\). The coefficient is not essential to the requested order-of-magnitude estimate.

3.2. For a square chip with square pixels of pitch \(a\), the total number of pixels is approximately \((L/a)^2\). Matching the pixel pitch to the optical resolution scale gives

\[ N_{\mathrm{pixels}}\sim \left(\frac{L}{\Delta x_{\min}}\right)^2. \]

Therefore,

\[ N_{\mathrm{pixels}}\sim \left(\frac{35\times10^{-3}}{1.0\times10^{-6}}\right)^2 =1.225\times10^9. \]

In megapixels,

\[ \boxed{N\sim1.2\times10^3\ \mathrm{Mpix}.} \]

Using the Rayleigh value instead gives

\[ N_{\mathrm{Rayleigh}} =\frac{1}{10^6} \left(\frac{35\times10^{-3}}{1.22\times10^{-6}}\right)^2 \approx8.2\times10^2\ \mathrm{Mpix}. \]

Both estimates require about \(10^3\ \mathrm{Mpix}\).

3.3. The pixel pitch of the 16 Mpix chip is

\[ a=\frac{L}{\sqrt{16\times10^6}} =\frac{35\ \mathrm{mm}}{4000} =8.75\ \mu\mathrm{m}. \]

For the optics not to set a coarser resolution than the pixels, require

\[ \lambda F\#\lesssim a. \]

Thus

\[ F\#\lesssim\frac{8.75\ \mu\mathrm{m}}{0.500\ \mu\mathrm{m}} =17.5. \]

The smallest aperture corresponds to the largest allowed F-number. Among the available settings, this gives

\[ \boxed{F\#=16.} \]

At this setting the estimated diffraction scale is \(8.0\ \mu\mathrm{m}\); at \(F\#=22\), it would be \(11\ \mu\mathrm{m}\).

If the Rayleigh coefficient is retained and the inequality is enforced strictly,

\[ 1.22\lambda F\#\leq a \quad\Longrightarrow\quad F\#\leq14.3. \]

The largest listed setting then becomes \(F\#=11\). This discrete choice depends on the resolution convention.

3.4. At 300 dpi, adjacent printed dots have separation

\[ d=\frac{25.4\ \mathrm{mm}}{300} =8.47\times10^{-5}\ \mathrm{m}. \]

At viewing distance \(z\), their angular separation is approximately \(d/z\). They cease to be individually resolved when

\[ \frac{d}{z}\lesssim\phi. \]

Using

\[ \phi=2(2.91\times10^{-4}) =5.82\times10^{-4}\ \mathrm{rad}, \]

we obtain

\[ \boxed{ z_{\min}\approx\frac{d}{\phi} =\frac{8.47\times10^{-5}}{5.82\times10^{-4}} \approx0.15\ \mathrm{m}. } \]

3.5. Treat the egg as a sphere with uniform density and specific heat capacity. Neglect the shell’s thermal effects and any additional enthalpy of coagulation, for which no data are supplied.

The egg’s mass is

\[ M=\frac43\pi R^3\mu =\frac43\pi(2.5\times10^{-2})^3(10^3) =6.54\times10^{-2}\ \mathrm{kg}. \]

Convert the specific heat capacity:

\[ C=4.2\times10^3\ \mathrm{J\,kg^{-1}\,K^{-1}}. \]

The energy needed to raise the egg from \(T_0\) to the coagulation temperature is estimated by

\[ U\approx MC(T_c-T_0). \]

Hence

\[ U\approx (6.54\times10^{-2})(4.2\times10^3)(65-4) =1.68\times10^4\ \mathrm{J}, \]

so

\[ \boxed{U\approx1.7\times10^4\ \mathrm{J}.} \]

3.6. Assume ordinary atmospheric boiling, so \(T_1\approx100^\circ\mathrm{C}\). For a characteristic heat flux, take the conduction length to be the egg radius and the temperature contrast to be the initial water-to-egg contrast:

\[ \Delta r\sim R,\qquad \Delta T\sim T_1-T_0=96\ \mathrm{K}. \]

Fourier’s law then gives the inward heat-flux magnitude

\[ J\sim\kappa\frac{T_1-T_0}{R} =0.64\frac{96}{2.5\times10^{-2}}. \]

Therefore,

\[ \boxed{J\sim2.5\times10^3\ \mathrm{W\,m^{-2}}.} \]

The actual temperature gradient changes during cooking; this is its characteristic scale.

3.7. Heat enters through the egg’s surface, whose area is

\[ A=4\pi R^2 =7.85\times10^{-3}\ \mathrm{m^2}. \]

Consequently,

\[ P\sim JA =\left(2.46\times10^3\right) \left(7.85\times10^{-3}\right) \approx19.3\ \mathrm{W}. \]

Thus

\[ \boxed{P\sim19\ \mathrm{W}.} \]

3.8. Divide the required heating energy by the characteristic heat-transfer power:

\[ t\sim\frac{U}{P}. \]

Combining the previous expressions also gives

\[ t\sim \frac{\frac43\pi R^3\mu C(T_c-T_0)} {4\pi R^2\kappa(T_1-T_0)/R} = \frac{\mu C R^2}{3\kappa} \frac{T_c-T_0}{T_1-T_0}. \]

Numerically,

\[ \boxed{ t\sim\frac{1.68\times10^4}{19.3} \approx8.7\times10^2\ \mathrm{s} \approx15\ \mathrm{min}. } \]

This estimate has the expected thermal-diffusion scaling \(t\propto R^2\).

3.9. The transferred charge is the area under the current–time graph:

\[ Q=\int_0^\tau I(t)\,dt. \]

The pulse is triangular, so its area is independent of when the peak occurs:

\[ Q=\frac12 I_{\max}\tau. \]

With \(I_{\max}=100\ \mathrm{kA}=10^5\ \mathrm{A}\) and \(\tau=0.1\ \mathrm{ms}=10^{-4}\ \mathrm{s}\),

\[ \boxed{Q=\frac12(10^5)(10^{-4})=5\ \mathrm{C}.} \]

3.10. The average current over the duration of the pulse is

\[ I=\frac{Q}{\tau} =\frac{I_{\max}}{2}. \]

Therefore,

\[ \boxed{I=5.0\times10^4\ \mathrm{A}=50\ \mathrm{kA}.} \]

3.11. Approximating the electric field between cloud and ground as uniform, the voltage at breakdown is

\[ V_0\approx E_0h =(300\times10^3)(10^3) =3.0\times10^8\ \mathrm{V}. \]

The electrical energy released is

\[ W=\int V(t)I(t)\,dt. \]

Using \(V_0\) as its characteristic voltage gives

\[ W\sim V_0Q =(3.0\times10^8)(5) =1.5\times10^9\ \mathrm{J} \]

per lightning strike.

The annual energy share per person is therefore

\[ W_{\mathrm{person}} \sim \frac{32\times10^6}{6.5\times10^9} (1.5\times10^9) \approx7.4\times10^6\ \mathrm{J}. \]

A \(100\ \mathrm{W}\) bulb consumes \(100\ \mathrm{J}\) each second, so

\[ \boxed{ t\sim\frac{W_{\mathrm{person}}}{100} \approx7.4\times10^4\ \mathrm{s} \approx20\ \mathrm{h}. } \]

For a cloud–ground system modeled as a capacitor that discharges completely, the voltage falls with the remaining charge. Its released energy is then

\[ W=\frac12 QV_0, \]

giving approximately \(3.7\times10^4\ \mathrm{s}\), or \(10\ \mathrm{h}\), per person. Both models give a lighting time of order one day.

3.12. The relevant SI quantities are

\[ \eta=4.5\times10^{-3}\ \mathrm{Pa\,s},\qquad r=4\times10^{-6}\ \mathrm{m},\qquad L=10^{-3}\ \mathrm{m}, \]

\[ \Delta p=10^3\ \mathrm{Pa},\qquad D_{\mathrm{total}}=100\ \mathrm{cm^3\,s^{-1}} =10^{-4}\ \mathrm{m^3\,s^{-1}}. \]

Poiseuille’s law gives the volume flow through one capillary:

\[ D_{\mathrm{cap}} =\frac{\Delta p}{R} =\frac{\pi r^4\Delta p}{8\eta L}. \]

Substitution yields

\[ D_{\mathrm{cap}} = \frac{\pi(4\times10^{-6})^4(10^3)} {8(4.5\times10^{-3})(10^{-3})} \approx2.23\times10^{-14}\ \mathrm{m^3\,s^{-1}}. \]

Since the identical capillaries are connected in parallel, each experiences the same pressure difference, and their volume flows add:

\[ D_{\mathrm{total}}=N_{\mathrm{cap}}D_{\mathrm{cap}}. \]

Thus

\[ \boxed{ N_{\mathrm{cap}} =\frac{10^{-4}}{2.23\times10^{-14}} \approx4.5\times10^9. } \]

3.13. The velocity appearing in \(D=S\upsilon\) is the mean velocity over the cross-section. Therefore,

\[ \upsilon=\frac{D_{\mathrm{cap}}}{\pi r^2} =\frac{\Delta p\,r^2}{8\eta L}. \]

Numerically,

\[ \upsilon= \frac{(10^3)(4\times10^{-6})^2} {8(4.5\times10^{-3})(10^{-3})}. \]

Hence

\[ \boxed{ \upsilon\approx4.4\times10^{-4}\ \mathrm{m\,s^{-1}} =0.44\ \mathrm{mm\,s^{-1}}. } \]

3.14. For a fixed quantity of ideal gas undergoing a quasistatic adiabatic expansion,

\[ pV^\gamma=\text{constant}. \]

Taking logarithmic differentials,

\[ \frac{dp}{p}+\gamma\frac{dV}{V}=0. \]

The ideal-gas law gives \(T\propto pV\), with \(T\) measured in kelvin, so

\[ \frac{dT}{T} =\frac{dp}{p}+\frac{dV}{V} =\left(1-\frac1\gamma\right)\frac{dp}{p}. \]

Thus

\[ \boxed{ \frac{dT}{T} =\frac{\gamma-1}{\gamma}\frac{dp}{p} =\frac27\frac{dp}{p}. } \]

Pressure and temperature both decrease as the slab rises.

3.15. Hydrostatic force balance on a thin horizontal layer gives

\[ dp=-\rho g\,dz, \]

where \(\rho\) is the local mass density.

For nitrogen molecules of mass \(m\), the ideal-gas law gives

\[ p=\frac{\rho}{m}kT, \qquad \rho=\frac{mp}{kT}. \]

Therefore,

\[ \boxed{ dp=-\rho g\,dz =-\frac{mpg}{kT}\,dz. } \]

3.16. Substitute the hydrostatic pressure gradient into the adiabatic temperature relation:

\[ \frac{dT}{T} = \frac{\gamma-1}{\gamma} \left(-\frac{mg}{kT}\,dz\right). \]

The temperature cancels from the right-hand side after multiplication by \(T\), giving a constant lapse rate:

\[ \frac{dT}{dz} =-\frac{\gamma-1}{\gamma}\frac{mg}{k}. \]

For the supplied constants,

\[ \frac{dT}{dz} = -\frac27 \frac{(4.65\times10^{-26})(9.80)} {1.38\times10^{-23}} \approx-9.43\times10^{-3}\ \mathrm{K\,m^{-1}}. \]

Over \(H=1000\ \mathrm{m}\), the temperature drop is approximately \(9.4\ \mathrm{K}\). Hence

\[ T_{\mathrm{top}} =T_{\mathrm{bot}} -\frac{\gamma-1}{\gamma}\frac{mgH}{k} \approx303.15-9.43 =293.72\ \mathrm{K}. \]

Therefore,

\[ \boxed{T_{\mathrm{top}}\approx294\ \mathrm{K}\approx21^\circ\mathrm{C}.} \]

The solution was checked against the official solution.