Use the $e^{i\omega t}$ phasor convention and interpret impedance matching as conjugate matching for the maximum power transfer theorem. The load electrical impedance of the series circuit isFor $Z=A+i\omega B$, the condition $Z_{RC}=Z^{\ast }$ equates the real parts and imaginary parts:These formulas require $B>0$, $C>0$ and a passive realization with $A\ge 0$, evaluated at the matching angular frequency; if $A,B$ depend on $\omega$, the equation for $\omega$ is implicit.

There is a convention ambiguity in the word “matched”. Literal equality $Z_{RC}=Z$, as required for equal bridge arms in part (b), instead gives $R=A$ and ${\omega }^{2}BC=-1$, requiring $B<0$. The inductive cases below use conjugate matching; literal equality is impossible for either passive inductive configuration at a finite positive angular frequency because its electrical reactance is positive while that of the series RC impedance is negative.

For viewing near the nadir, the required angular resolution isUsing the Rayleigh criterion with the first minimum of the single-slit diffraction pattern gives $\delta \theta \simeq \lambda /w$. Thus the minimum collecting width in that model isThe numerical coefficient depends on the aperture shape and the chosen resolution criterion. If the mirror has an unobstructed circular aperture, the Airy disk version of the Rayleigh criterion is $\delta \theta =1.22\lambda /D$, giving $D_{\min }=0.12298\mathrm{m}\simeq 12.3\mathrm{cm}$. The question does not specify mirror shape; $10.1\mathrm{cm}$ follows the slit-width model of part (i).

Take coordinate $y$ across the slit and observation angle $\theta$ from its normal vector. Under Fraunhofer diffraction, the relative optical path difference from $y$ is $y\sin \theta$. Uniform coherent illumination makes the complex amplitude proportional to the aperture integral:The light intensity is proportional to $|\mathcal{E}{|}^2$, soThe central value is the continuous limit of the sinc function. The first minima satisfy $\sin \theta =\pm \lambda /w$, if $\lambda \le w$. Infinite slit height removes spreading in the other transverse direction; this expression describes the slit-width direction.

Strictly, distance $D\gg w$ alone does not ensure Fraunhofer diffraction: one also needs $w^2/(\lambda D)\ll 1$, so that quadratic aperture-dependent wave phase is negligible, or observation in a suitable focal plane. For example, $w=1\mathrm{mm}$, $\lambda =500\mathrm{nm}$ and $D=1\mathrm{m}$ have $D/w=1000$ but $w^2/(\lambda D)=2$; the Fresnel diffraction correction is then substantial. The displayed result uses the intended far-field regime of the question and the usual scalar aperture approximation.

Approximate the atmosphere by concentric spherical layers of refractive index. At every boundary, Snell's law conserves $nr\sin \psi$, where $\psi$ is the angle between the light ray and the radius. Straight propagation between boundaries conserves the same quantity because $r\sin \psi$ is the distance of that line from the origin. In the smooth limit this is the spherical ray invariantAt the surface, a light ray apparently on the horizon is tangent, so $p=n_{s}R$. In vacuum far from the Earth, $n=1$, and the straight incident line has distance $p$ from the centre. Removing the atmosphere while retaining that incident line gives minimum altitudeThe detailed variation of refractive index with altitude changes the local curvature of the light ray, but not the conserved $p$. This conclusion assumes a spherically stratified atmosphere and a light ray which actually connects the surface to space; nonspherical structure or a trapping layer would require separate analysis.

The distance from the origin to the incident straight light ray is $d_a$ (taking $d_a\ge 0$). At its intersection with the sphere, let $i$ and $t$ be the angle of incidence and angle of refraction, measured from the radial normal vector. Elementary right triangle geometry gives $d_a=r_b\sin i$. By Snell's law,The refracted straight light ray inside the sphere has distance $d_b=r_b\sin t$ from the origin. ThereforeThe angle ${\theta }_b$ describes the normal vector of the new line; it need not equal the propagation angle. The existence of a crossing requires $|d_a|<r_b$. A signed convention for $d_a$ gives the same line equation after choosing the orientation of its normal vector consistently.

Using $\nabla \times \mathbf{E}=-{\partial }_t\mathbf{B}$ directly with the specified plane wave givesThe electric field, magnetic flux density and propagation direction form a right-handed triad, and for a nonzero wave amplitudeThe electric field and magnetic field have the same wave phase in vacuum.

In charge-free, current-free vacuum, Maxwell's equations areTaking the curl of the third equation and using the fourth, the curl of the curl identity, and the zero divergence of the electric field givesThe magnetic flux density obeys the same wave equation. For the plane wave factor $e^{i(\omega t-kx)}$, the spatial Laplacian contributes $-k^2$ and the second time derivative contributes $-{\omega }^2$. Thusfor propagation in the positive $x$ direction. The zero-divergence conditions require $E_{0x}=B_{0x}=0$, while the curl equations require ${\mathbf{B}}_0=(k/\omega )\hat{\mathbf{x}}\times {\mathbf{E}}_0$. These conditions show that the proposed plane waves satisfy all four Maxwell's equations, rather than only their derived wave equations. Physical electric fields and magnetic fields are the real parts of the complex expressions.

Write the axial current density as $j(r)=Kr$. Normalize it using the total electric current:The enclosed electric current at radius $r$ is $I_{\mathrm{enc}}(r)=2\pi K{r}^3/3=I(r/a{)}^3$. The magnetostatic Ampère law now yieldsBoth profiles join the exterior magnetic flux density ${\mu }_{0}I/(2\pi a)$ at the surface, but the second vanishes quadratically at the axis because less electric current is concentrated near it.

Assume the electrical conductor is nonmagnetic, with magnetic permeability ${\mu }_0$, and neglect end effects. Rotational symmetry and translational symmetry make the magnetic flux density azimuthal and constant along a coaxial circle. The magnetostatic Ampère law therefore givesUniform current density is $j=I/(\pi a^2)$, so $I_{\mathrm{enc}}(r)=j\pi r^2=I{r}^2/a^2$. ThusThe direction follows the right-hand rule around the electric current. For a uniform linear magnetic material, replace ${\mu }_0$ by its magnetic permeability.

The non-inverting input is grounded, so the finite voltage gain gives $V_{\mathrm{OUT}}=-A{V}^{-}$, or $V^{-}=-V_{\mathrm{OUT}}/A$. Infinite input impedance makes Kirchhoff's current law at the inverting inputSubstitute $V^{-}=-V_{\mathrm{OUT}}/A$ and multiply by $R_1{R}_2$:Hence the inverting operational amplifier hasAs $A\to \infty$, the voltage gain tends to $-R_2/R_1$. Setting $V^{-}=0$ before taking that limit would discard precisely the correction being calculated.

An ideal operational amplifier has infinite input impedance, so neither input draws electric current, and zero output impedance, so its prescribed output voltage is independent of the output electric current within the ideal model. These are the two additional assumptions needed here.

For the convex vacuum void, use internal refractive index $1$, external refractive index $n$, and the original $\alpha >0$ in the immersed thin lens formula:A concave glass thin lens of the same curvature magnitude in vacuum has ${\alpha }_{\mathrm{concave}}=-\alpha$ and thereforeBoth elements diverge, but their physical focal lengths differ by a factor $n$ for $n>1$. For example, $n=1.5$ makes the void's focal length magnitude $1.5$ times that of the concave glass element. Their reduced optical powers $n_{\mathrm{outside}}/f$ are equal; physical focal length also depends on the surrounding refractive index.

In surrounding refractive index $n^{\prime }$, the geometrical segments contribute $n^{\prime }[\sqrt{a^2+d^2}+\sqrt{b^2+d^2}]$ to the optical path length, while replacement by the lens adds $(n-n^{\prime })\ell (d)$. Fermat's principle therefore givesThe physical focal length measured in the surrounding medium isFor $\alpha >0$ and $n>1$, increasing $n^{\prime }$ from $1$ towards $n$ makes the positive focal length grow monotonically from its vacuum value to infinity: $df/d{n}^{\prime }=n/[2\alpha (n-n^{\prime }{)}^2]>0$. At $n^{\prime }=n$, there is no refraction at the boundary, so the element has zero optical power and no finite optical focus. If $n^{\prime }>n$, its optical power changes sign and the same shape becomes diverging.

Fermat's principle makes the optical path length, or transit time in a stationary medium, stationary under small variations of the light ray with fixed endpoints. It need not be a global minimum. In the result of part (a), the coefficient of $d^2$ isFor a family of paraxial light rays from the object to its image, stationarity with respect to $d$ requires this coefficient to vanish. Therefore the thin lens equation isFor a distant object, $a\to \infty$, and the image position $b=f$ defines the focal length. HenceThe axial member $d=0$ is stationary by symmetry for arbitrary $b$; requiring the neighbouring non-axial light rays to meet the same image is what determines $b$.

Write $T$ for transit time and $\ell (y)=t_0-\alpha y^2$ for lens thickness to avoid using $t$ for both. In the thin lens approximation, the two geometrical distances to the lens plane are $\sqrt{a^2+d^2}$ and $\sqrt{b^2+d^2}$. Their sum counts the lens segment as vacuum. Replacing that segment by material of refractive index $n$ adds $(n-1)\ell (d)$ to the optical path length. Thus, to the retained thin-element accuracy,Using the Taylor expansion $\sqrt{1+u}=1+u/2+O(u^2)$ for $d\ll a,b$ givesThe omitted geometrical terms include $O(d^4/a^3+d^4/b^3)$ in optical path length, together with finite-thickness corrections discarded by the thin lens and paraxial approximations. The addition is $n-1$, because the underlying distance already includes propagation across the thickness once in vacuum.

Define positive $\alpha$ by the right-hand rule about $+x$, starting from the loop in the $x$-$y$ plane at $\alpha =0$. Its magnetic dipole moment becomesThe corresponding $x$ component of magnetic dipole torque is ${\tau }_x=-m(B_z\sin \alpha +B_y\cos \alpha )$. Integrating ${\tau }_x=-dE/d\alpha$ gives the mechanical magnetic dipole potential energywhere the arbitrary additive constant is chosen as zero. Set $B_{\perp }=\sqrt{B_y^2+B_z^2}$. For $B_{\perp }>0$, a positive-amplitude form isEquivalently, ${\alpha }_0=atan2(B_y,-B_z)$ modulo $2\pi$. With this positive choice, minimum potential energy occurs at $\alpha ={\alpha }_0+\pi$, where the rotating magnetic dipole moment aligns with the projection of $\mathbf{B}$ into the $y$-$z$ plane. If $B_y=B_z=0$, the allowed torque vanishes, $E_0=0$ and ${\alpha }_0$ is arbitrary. Holding electric current constant includes exchange with its supply; differentiating only the isolated loop's stored magnetic energy would not give the appropriate mechanical potential energy.

The elemental torque about the origin is $d\mathbf{\tau }=\mathbf{r}\times d\mathbf{F}$. Substitution of part (a) givesOnly its $x$ component drives the allowed rotation about the $x$ axis. Using ${\int }_0^{2\pi }{\sin }^2\theta d\theta =\pi$ and ${\int }_0^{2\pi }\sin \theta \cos \theta d\theta =0$, the projected torque isThe magnetic dipole moment is $\mathbf{m}=I\pi a^2\hat{\mathbf{z}}$, so the full electromagnetic torque isIts $x$ component agrees exactly. The $y$ component is a real electromagnetic torque, but the bearing constraining motion to the $x$ axis supplies a cancelling reaction torque. For an allowed infinitesimal rotation $d\mathbf{\alpha }=\hat{\mathbf{x}}d\alpha$, only $\mathbf{\tau }\cdot d\mathbf{\alpha }={\tau }_{x}d\alpha$ does mechanical work.

Use the source orientation, anticlockwise viewed from $+z$. The position and oriented line element areThe Lorentz force on the electric current element is $d\mathbf{F}=Id\mathbf{l}\times \mathbf{B}$. Evaluating the cross product givesEach component integrates to zero around $0\le \theta <2\pi$, so ${\mathbf{F}}_{\mathrm{net}}=0$. Equivalently, uniform magnetic flux density permits $I\oint d\mathbf{l}\times \mathbf{B}=I(\oint d\mathbf{l})\times \mathbf{B}=0$. Uniformity is essential; a nonuniform magnetic field can exert a net force on a magnetic dipole.

Put $s=e^{-R\Delta t/L}$ and $J=V/R$, with $0<s<1$. The first interval leaves electric current $I_1=s{I}_0$. Position B reconnects the battery, so the driven RL circuit obeys $L\dot{I}+RI=V$. Multiplication by the integrating factor $e^{Rt/L}$, or subtraction of the constant solution $J$, givesConsequentlyThe same two intervals give the recurrence relation $I_{2n+2}=s^2{I}_{2n}+J(1-s)$. Its fixed point is $I_{\ast }=J/(1+s)$. Subtracting this fixed point makes the recurrence relation homogeneous:This is the upper endpoint of the eventual periodically switched RL circuit response, immediately before switching from B to A. The lower endpoint is $sJ/(1+s)$. Their sum is $J$, consistent with equal charging and discharging intervals.

Position A disconnects the battery and closes the resistor-inductor loop. Taking clockwise electric current as positive, Kirchhoff's voltage law for this RL circuit givesThe electric current through an ideal inductor remains continuous across a switch with no impulsive voltage. Thus $I(0^{+})=I_0$, and separation of variables givesIf the switch remains at A, the same result holds for every $t>0$. The time constant is $L/R$; the stored magnetic energy is dissipated by Joule heating.