With $D=5\mathrm{mm}$, the circular-aperture diffraction limit is ${\theta }_R=244\lambda$ when $\lambda$ is in metres. No wavelength is specified, so a unique numerical diffraction angle cannot be deduced. Taking representative visible light at $\lambda =550\mathrm{nm}$ givesFor a roughly square receptor arrangement with surface density $n=10^5{\mathrm{mm}}^{-2}$, the characteristic receptor pitch is $s\simeq n^{-1/2}=3.16\mathrm{micrometres}$. With the retina $f_{\mathrm{eye}}=17\mathrm{mm}$ behind the lens, that pitch subtendsThis is an order-of-magnitude estimate of the point-separation scale from retinal sampling. A periodic intensity pattern needs at least two samples per period under the Nyquist–Shannon sampling theorem, giving a corresponding period angle of about $3.72\times 10^{-4}\mathrm{rad}$ or $1.28\mathrm{arcmin}$. A different receptor packing changes the pitch by an order-one geometric factor; density alone does not specify a full sampling lattice or a contrast threshold.

At a $5\mathrm{mm}$ pupil and the chosen visible wavelength, the optical diffraction limit is somewhat finer than the receptor pitch. The supplied model is therefore not purely diffraction limited: retinal sampling is comparable to, and slightly more restrictive than, ideal aperture diffraction. Real optical aberrations and neural processing can further reduce resolution. Different wavelengths or a smaller pupil can change which effect dominates.

Under the Rayleigh criterion, two equally bright, mutually incoherent point sources are just resolved when the central maximum of one diffraction pattern lies at the first minimum of the other. Their intensities, rather than their electric-field amplitudes, add. For an ideal circular aperture of diameter $D$, the first dark ring of the Airy disk givesThe angle is in radians in the small-angle limit. This is a conventional resolution criterion for a specified aperture and wavelength; it is not an absolute prohibition on estimating a source separation by fitting noisy data under additional assumptions.

The magnetic dipole moment is $\mathbf{\mu }=IA\hat{\mathbf{n}}$ and its magnetic torque is ${\mathbf{\tau }}_{\mathrm{mag}}=\mathbf{\mu }\times \mathbf{B}$. Its component along the rotation axis opposes the imposed rotation. For positive $\omega$, the external driving torque needed to keep the angular velocity constant isAveraging over a complete rotation, with $⟨{\sin }^2\omega t⟩=1/2$, givesAs an independent sign and dimension check, $\omega ⟨{\tau }_{\mathrm{drive}}⟩=R⟨I^2⟩$. Thus the mechanical input power equals the mean Joule heating; a passive loop cannot supply a sustained driving torque. This average assumes the same negligible-self-inductance approximation as the current expression.

Let $A=\pi r^2$ and choose the rotating area normal to coincide initially with $+z$. At time $t$ its angle to the applied magnetic field is $\omega t$, so the external magnetic flux is ${\Phi }_B=BA\cos \omega t$. Faraday's law therefore givesIn the high-electrical resistance approximation, neglecting the loop's self-inductance, Ohm's law givesThe displayed signs specify positive circulation by the chosen normal; reversing the reference circulation reverses both signs. The high-electrical resistance assumption is quantitatively $\omega L\ll R$ if the loop has self-inductance $L$. Otherwise the current lags the induced electromotive force and must be obtained from $L\dot{I}+RI=\mathcal{E}$.

Define the requested electric potential difference as $\Delta V=V(b)-V(-a)$. Since $E_x=-dV/dx$,soGeometrically, the integral is the area of two triangles under the signed electric field profile. For $C>0$, the right-hand plane is at higher electric potential. Choosing $V(-a)=0$, a continuous interior electric potential is $V=C(x+a{)}^2/(2{ϵ}_{0}a)$ in the first slab and $V=Ca/(2{ϵ}_0)+C(x-x^2/(2b))/{ϵ}_0$ in the second. These expressions also verify the boundary values directly.

Translational and reflection symmetry parallel to the slabs imply $\mathbf{E}=E_x(x)\hat{\mathbf{x}}$. The differential Gauss's law is $d{E}_x/dx=\rho /{ϵ}_0$. Starting at the left boundary with $E_x(-a)=0$ givesThere is no singular surface charge at $x=0$, so the normal electric field is continuous there: $E_x(0)=-C/{ϵ}_0$. Integrating through the positive slab givesThe second expression indeed returns to zero at $x=b$, because the two total charges per unit area are $-C$ and $+C$. For $C>0$, the electric field throughout the interior points from the positive slab towards the negative slab, along $-x$; reversing $C$ reverses the directions. The piecewise linear profile is a useful consequence of the uniform volume charge density.

The source's variable inductor and variable resistor are in parallel. Let their combined complex impedance beUse the left and right source terminals as the drive nodes. The upper divider has arms $Z_V$ and $i\omega L_1$; the lower divider has arms $R_1$ and $R+i\omega L$. At bridge balance, the detector draws no electric current and the divider midpoint electric potential differences agree, soSubstitution and cancellation at a nonzero drive frequency giveEquating real and imaginary parts yieldsThe first expression has units of resistance and the second units of inductance. Both are independent of the drive amplitude and, for ideal frequency-independent components, of the chosen nonzero drive frequency. At zero frequency the ideal inductors short their arms and the cancellation would not be valid as a measurement of $L$.

Two applications are precision comparison of an unknown electrical resistance with calibrated standards in a Wheatstone bridge, and detecting small resistance changes in a strain gauge or resistance thermometer. Alternating-current bridges additionally compare capacitance, inductance and dielectric losses; this particular bridge determines the two parameters of a non-ideal inductor from a null measurement.

Choose the phasor convention $V(t)=\mathrm{Re}(\tilde{V}{e}^{i\omega t})$ and similarly for the electric current. In a linear sinusoidal steady state, differentiation becomes multiplication by $i\omega$, and complex impedance is $Z=\tilde{V}/\tilde{I}$. Kirchhoff's current law and Kirchhoff's voltage law then become algebraic equations that retain both amplitude and phase information.

For a resistor, $V=RI$, so $Z_R=R$. For a capacitor, $I=C\dot{V}$, while an inductor obeys $V=L\dot{I}$. ThereforeThe resistor's electric potential difference and current are in phase. The inductor's electric potential difference leads its current by $\pi /2$; the capacitor's current leads its electric potential difference by $\pi /2$. Replacing $e^{i\omega t}$ by $e^{-i\omega t}$ conjugates every impedance but leaves these physical lead-and-lag statements unchanged. The phasor method describes the steady state; switch-on transients must be added separately when initial conditions are relevant.

For a lens between an object and a screen, $u+v=D$ and the thin lens relation requires $uv=fD$. Eliminating $v$ gives the quadratic equationFor $D>4f$, both roots lie between $0$ and $D$, and they exchange the object and image distances. Both therefore focus the image on the same screen, with reciprocal absolute magnifications. The distance between lens positions isThe existence qualification is essential: for $D=4f$ the two roots coincide and there is only one distinct position; for $D<4f$ there is none. For example, $D=3f$ makes the discriminant negative, providing a concrete counterexample to an unconditional assertion of two positions. This is the basis of the Bessel lens displacement method, with $f=(D^2-l^2)/(4D)$ when two distinct positions can be measured.

A real image is a point where the actual outgoing rays converge; a screen placed there can intercept the focused light. A virtual image is a point from which the outgoing rays appear to diverge: only their backward extensions intersect, so a screen at that apparent point does not receive a focused real image from the same rays.

For a converging thin lens with $f>0$ and a real object a distance $u>0$ away, the lens relation gives $v=fu/(u-f)$. If $u>f$, then $v>0$: the image is real, on the opposite side, and inverted. If $0<u<f$, then $v<0$: the image is virtual, on the same side as the object, upright and enlarged, with $m=-v/u=f/(f-u)>1$.

At $u=f$, the emergent rays are parallel and the image is at infinity; there is no finite focused image. For an object very far away, $v\to f$ from above. These are conditions for a real object in the stated lens convention, rather than claims about all systems with virtual incident objects.

For a real image, $u>f$ and $v=fu/(u-f)>0$. The object-to-image separation isIts excess above $4f$ factors into a nonnegative expression:ThereforeAt this minimum the image is inverted and has unit absolute magnification. The restriction to a real object and a real image matters: the positive separation used here is not the same quantity as a signed distance to a virtual image.

Adopt positive object and real-image distances $u$ and $v$ measured from a converging thin lens, with the object on the left. A ray through the lens centre is undeviated in the thin-lens approximation. For object height $h$ and signed image height $h^{\prime }$, similar triangles give $h^{\prime }/h=-v/u$.

A second ray leaves the object parallel to the axis, reaches the lens at height $h$ and then passes through the far focal point. Its height at distance $v$ is $h^{\prime }=h(1-v/f)$. Equating the two descriptions givesThe signed transverse magnification is $\boxed{m=h^{\prime }/h=-v/u}$. Since a real image has $v>0$ for $u>0$, $m<0$: it is inverted. Its size relative to the object is $|m|=v/u$.

The derivation assumes the paraxial approximation, a negligible lens thickness and a focal length defined in the surrounding medium. The ray diagrams show actual paths as solid lines and backward extrapolations as dashed lines.

By symmetry, the resultant Coulomb force on each charge points along the line from the centroid to its vertex. Their accelerations and subsequent speeds are equal, so the centroid remains fixed and the equilateral triangle expands without rotating. Each charge follows a straight radial trajectory; the separation grows and the repulsive force decreases.

For example, if the instantaneous side is $s$ and the centroid-to-charge distance is $r=s/\sqrt{3}$, the two pair forces have a $60^{\circ }$ angle between them. Their resultant is $\sqrt{3}{k}_e{q}^2/s^2=k_e{q}^2/(\sqrt{3}{r}^2)$, along the outward radius. This confirms that the symmetric shape persists.

Assuming release from rest, nonrelativistic motion and negligible radiation losses, energy conservation givesThe three velocity vectors sum to zero, consistent with momentum conservation. Their magnitudes approach the displayed value as the remaining electrostatic potential energy tends to zero.

Take the electrostatic potential energy to vanish for infinitely separated point charges, and write $k_e=1/(4\pi {ϵ}_0)$. The first charge costs no work to place. Bringing the second to distance $a$ costs $k_e{q}^2/a$; the third is distance $a$ from both existing charges and costs $2k_e{q}^2/a$. Thus the external quasistatic assembly work isEquivalently, sum over the three distinct pairs in the equilateral triangle. Counting both ordered pairs without a compensating factor of one half would double the answer. The positive result applies to either sign of $q$, since all three electric charges have the same sign.

Outside the wire, the Amperian loop encloses the entire electric current, soIts direction is again azimuthal, fixed by the right-hand rule. The inside and outside formulas join continuously at the maximum $B(a)={\mu }_{0}I/(2\pi a)$. The outer branch decays as $1/r$.

For a steady uniform current density in a nonmagnetic wire, the enclosed electric current at radius $r<a$ is $I_{\mathrm{enc}}=I{r}^2/a^2$. A circular Amperian loop and Ampère's law then giveThe magnetic field circles the wire according to the right-hand rule relative to the current direction. Its magnitude increases linearly from zero on the axis. The uniform-current assumption excludes skin effect.

Put $t_1=0.005\mathrm{s}$, $\tau =t-t_1$, and $I_1=2(1-e^{-1.875})$. When the switch reopens, the circuit again contains the battery and all three series elements. Inductor current continuity and capacitor voltage continuity giveThus $I^{\prime }(0^{+})=(V_0-R{I}_1)/L\simeq 115.0\mathrm{A}{\mathrm{s}}^{-1}$, not zero. Differentiating the loop equation gives the homogeneous current equationThe battery is still connected; its constant voltage disappears on differentiating, and the eventual capacitor voltage is $V_0$.

With $\alpha =R/(2L)=187.5{\mathrm{s}}^{-1}$ and ${\omega }_0=1/\sqrt{LC}$, the inequality $\alpha <{\omega }_0$ makes the circuit underdamped. The damped angular frequency and ordinary frequency areThe full current satisfying both initial conditions isIts amplitude envelope decays as $e^{-\alpha \tau }$. ThereforeThis is measured after reopening; on the clock starting with the first closure it is $t\simeq 0.02098\mathrm{s}$. An instantaneous current zero is not an amplitude decay measurement.

Initially the open switch leaves a series capacitor-resistor-inductor circuit driven by the battery. After a long time the electric current is zero and the capacitor voltage is $V_0$. Closing the switch bypasses and, under the stated idealization, instantaneously discharges the capacitor. The inductor current remains continuous at zero.

The remaining RL circuit obeys $L\dot{I}+RI=V_0$, soThe capacitor's impulsive discharge through the ideal short is distinct from the finite electric current through the coil. A finite switch resistance would resolve that separate fast transient.