. . . possible. ¹

By thin film interference. If the wing is coated with a film of higher index of refraction than the wing material and a thickness equal to one fourth of the wavelength of blue light (at that index of refraction) then blue light will be preferentially reflected relative to other colours. If the film has a lower index than the wing material, the same effect can be achieved if its thickness is half the wavelength, because in that case both reflected rays have the same phase reversal so the net path length difference needs to be a full wavelength. (I gave full credit for the first sentence of this answer.)

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. . . contribution? ²

Maxwell proposed the modification of Ampère's Law by adding the ``displacement current'' (rate of change of the electric flux through the loop) to the charge-flow current linking the loop.

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. . . represent? ³

The speed of light, $c = 1/\sqrt{\mu_0 \, \epsilon_0} = 2.99792458 \times 10^{8}$ m/s. You can work this out from the given values of the two constants, but it's a lot quicker to just remember this important fact.

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. . . answers] ⁴

The second and fourth answers are correct.

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. . . Explain. ⁵

They can never be equal. Both forces have the same dependence on distance, so the ratio of their strengths will always be the same. (This was a trick question designed to catch people who start calculating before they think.)

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. . . given. ⁶

If the plates carry a charge Q then the surface charge density is $\sigma = Q/A$. The field between the plates would thus be $E = \sigma/\epsilon_\circ = Q/\epsilon_\circ A$ if the gap were empty. The effect of the dielectric is to reduce E by a factor $\epsilon_\circ / \epsilon$. Thus in region 1 the field is $E_1 = Q/\epsilon_1 A$ and in region 2 the field is $E_2 = Q/\epsilon_2 A$. Moving from one plate to the other we travel a distance d/2 along E₁ and another distance d/2 along E₂ for a net potential change of $V = E_1 d/2 + E_2 d/2 = (Qd/2A)(1/\epsilon_1 + 1/\epsilon_2)$. Since C is defined by Q = CV or C = Q/V, we have .

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. . . coil? ⁷

The resonant frequency of the LC circuit is $1/\sqrt{LC} = \omega = 2 \pi f = 10^3$ s^-1 so LC = 10^-6 s². The time constant of the RC circuit is $\tau = RC = 0.1$ s for $R = 10^3 \; \Omega$ so 10³ C =0.1 or C = 10^-4 F [Farads]. Thus 10^-4 L = 10^-6 or [Henries].

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. . . identical. ⁸

The third circuit is simple because the array of resistors can be replaced by a single equivalent resistor. The fourth (rightmost) circuit is obviously not simple because it has unlike components in parallel; this can also be seen by picturing what it will ``do'' when the circuit is initially completed. The second circuit looks topologically suspicious, and would not be ``simple'' if the elements of the same type (resistors or capacitors) were not identical; but since both capacitors must have the same charge at all times and since the discharge path for each is also identical, a simple RC circuit with one equivalent capacitor and one equivalent resistor would have the same behaviour. The same is true of the first (leftmost) circuit for the same reasons, although it might be tedious to calculate the values of the equivalent R and C; in this case one must have a good eye for symmetry to see that each corner has three identical resistors in parallel, each of which collects charge from two identical capacitors. Again, if the symmetry were broken (e.g. by having one capacitor or resistor different from the others) the circuit would no longer be simple.

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. . . $e = 2.718\cdots$) ⁹

Boltzmann showed that the probability a system being in a state of energy $\varepsilon$ while in thermal equilibrium with a heat reservoir at temperature $\tau$ is proportional to ${\displaystyle e^{-\varepsilon/\tau} }$. For the state with zero energy this is just 1, so the relative probability of being in the other state is just ${\displaystyle e^{-\varepsilon/\tau} }$. If we want this to equal e^-1 we must have $\tau = \varepsilon = 0.01$ eV $= 0.01 \times (k_{\rm B} \times 11,600$ K) $= k_{\rm B} T$ or T = 116 K.

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. . . awake? ¹⁰

Following the same argument as before, we are looking for ${\displaystyle e^{-\varepsilon/\tau} = 1}$. This is only true if $\varepsilon/\tau = 0$, which for finite $\varepsilon$ is only true when $\tau \to \infty$. You don't really have to do the mathematics to realize this, though, if you think about what the Boltzmann distribution describes. Anyone who got stuck on this part ought to have gotten a good retroactive hint from the graphs in the next part!

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. . . $\varepsilon$. ¹¹

The probability at $\tau = 0$ is zero; it stays ``pretty small'' until it shoots up around $\tau \approx \varepsilon$ and then slowly approaches 1/2 as $\tau \to \infty$. Don't forget that factor of 2! The best you can do it to make it equally probable to be either asleep or awake. (See previous question.)

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. . . $\tau$. ¹²

This is just an exponential decay from 1/2 at $\varepsilon = 0$ to zero as $\varepsilon \to \infty$, with a value of ${1 \over 1+e}$ at $\varepsilon = \tau$.

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. . . axis. ¹³

Imagine a cylindrical Gaussian surface of length L centred on the same axis, with radius r < R. The electric field is normal to the curved surface and has a constant magnitude everywhere on that surface (by symmetry) so the surface integral in Gauss' Law has the value $2 \pi r L E$. The enclosed charge is $\pi r^2 L \rho$ so Gauss' Law gives  . QED.

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. . . hole. ¹⁴

The electric field points away from the axis, so the force on a (negatively charged) electron points back toward the axis and has a magnitude proportional to the distance from the axis. This constitutes a linear restoring force and the electron has mass (inertia), so simple harmonic motion is automatic. QED

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. . . Explain. ¹⁵

Let r be the radius of the solenoid and d be the width of the wire. The length of the solenoid is $\ell = N d$ where N is the number of turns. $B = \mu_0 n I$ where n = turns/unit length = $N/\ell = 1/d$, so for a given I, B is independent of the diameter 2r, as long as there is enough wire to make the solenoid ``long'' ( $\ell \gg r$). Otherwise the Ampère's Law formula for B will be an overestimate and the true B will be smaller. Thus a strictly correct answer would be, ``All other things being equal, smaller is slightly better.''

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. . . Explain. ¹⁶

$I = {\cal E}/R$ and $R = \rho \ell/d^2$, so $I = {\cal E} d^2/\rho N d = {\cal E} d/\rho N$. [1] But for a given V = L d² (where L is the total length of the wire) the number of turns is $N = L/2\pi r$ or $N = V/2\pi r d^2$. Put this back into [1] to get ${\displaystyle I = {{\cal E} \over \rho} \cdot {2\pi r d^3 \over V} }$ and so ${\displaystyle B = \mu_0 I {N \over \ell} = \mu_0 {I \over d} = \left(2\pi \mu_0 {\cal E} r \over \rho V \right) d^2 }$ implying that short and thick is lots better, until you start to have $\ell$ comparable to r, at which point you eventually reach the point of diminishing returns. Also, of course, you can't make $\ell < 2\pi r$ or you won't even be able to make up a single turn! There is another practical limit to this strategy: it is actually difficult to produce enormous currents at moderate voltage because any actual power supply has a resistance of its own, usually higher than that of a putative single turn of copper bar. The would-be Electrical Engineer may want to perform a calculation of the optimum conductor dimensions to generate the maximum B with the least possible power consumption, given a finite resistance in the power supply itself . . . .

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. . . constants.) ¹⁷

As the bar drops at a speed v, the magnetic flux through the loop formed by the bar, rails and resistor decreases at a rate $d\Phi_M/dt = \ell v B$, giving an induced EMF V of the same magnitude. By Lenz' Law this EMF will ``try'' to replace the flux into the page: a current will be induced in such a direction as to make its own magnetic field into the page, namely clockwise. The resulting current will be $I = V/R = \ell v B / R$. A current of this size in a perpendicular magnetic field B produces a force $F = I B \ell$ on the bar. Checking with the right hand rule confirms that the force will be up, thus counteracting the downward force of gravity. The magnetic force increases as the velocity increases, until it exactly balances mg, at which velocity (v_t) there is no net force and terminal velocity will have been reached. Thus $(\ell v_t B/R)(B \ell) = mg$ or  .

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. . . side. ¹⁸

The wide diffraction envelope from the narrow slit width has a first minimum at $\sin \vartheta_1^D = \lambda/a = 0.5/25 = 0.02$ whereas the $m^{\rm th}$ principal maximum of the interference pattern from the 3 slits is at $\sin \vartheta_m^I = m \lambda/d = m(0.5/100) = 0.005 m$. The position x on a flat screen 1 m away is given by $x \hbox{\rm [m]} = \tan \vartheta$. All the angles are small out to at least $\vartheta_1^D$, so we can use the small angle approximations $\tan \vartheta \approx \sin \vartheta \approx \vartheta$ to get x₁^D = 0.02 m (2 cm) and x_m^I = 0.005 m m (0.5 m cm). Note that x₁^D is right on top of x₄^I and the principal maxima are evenly spaced. The only other thing you need to remember to get the full pattern is that for N = 3 slits there are N-1 = 2 minima and N-2 = 1 secondary maximum between the principal maxima. The pattern therefore looks like this:

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. . . maximum. ¹⁹

For the central maximum (or any principal maximum) the phasor diagram is three vectors of the same length added tip-to-tail all in the same direction, so the net amplitude is three times that of each. When the three individual phasors are perpendicular as shown, the net amplitude is just the missing side of the square, or 1/3 as big as for the central maximum. Since the intensity is the square of the amplitude, this gives 1/9 of the central intensity.

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. . . maximum? ²⁰

Rayleigh's criterion says that two lines are just resolvable if the principal maximum of one falls right on top of the first minimum beyond that principal maximum of the other. A grating gives larger dispersion (change of angle with wavelength) for higher order principal maxima, but here we are using m=1. Thus the criterion is that $\vartheta_1^I(\lambda + \Delta \lambda)$, the angle of the first principal maximum for the second wavelength, is equal to the angle $\vartheta_{1+1/N}^I(\lambda)$ of the first local minimum beyond the first principal maximum for the original wavelength. That this occurs 1/N = 1/3 of the way to the second principal maximum can be seen by picturing the phasor diagram for that first minimum: an equilateral triangle with each phasor $\delta = 2\pi(1 + 1/3) = 360^\circ + 120^\circ$ out of phase with the next. Since $\delta/2\pi = (d/\lambda) \sin \vartheta \approx d \vartheta/\lambda$, this value of $\delta$ corresponds to $\vartheta = (1 + 1/3) \lambda/d$. At the same angle we have the first principal maximum ( $\delta' = 2\pi$) for $\lambda + \Delta \lambda$, that is, $\vartheta = (\lambda + \Delta \lambda)/d$. Combining these results gives $(1 + 1/3) \lambda = \lambda + \Delta \lambda$ or  . Note that you could get the same answer by just plugging into the formula for the resolving power of a grating at the $m^{\rm th}$ principal maximum, but that wouldn't be nearly as much fun!

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