Use nonrelativistic dynamics, take $B>0$ along $+y$, and neglect radiation. The Lorentz force $q\mathbf{v}\times \mathbf{B}$ initially points along $+z$. It does no work, so the speed stays $v$ and the trajectory inside the field is circular, withStarting from the entry point, the orbit can be parametrized as $x=r\sin \theta$, $z=r(1-\cos \theta )$, where $\theta =\Omega t$. Its largest possible $x$ before turning back is $r$. Therefore it enters the strict region $x>b$ exactly whenFor that case, the first boundary crossing satisfies $\sin {\theta }_b=b/r$. The velocity components are $v_x=v\cos {\theta }_b$ and $v_z=v\sin {\theta }_b$, so the outgoing angle above the $+x$ direction isThe zero-field limit is straight motion. At $r=b$ the particle only touches the far boundary tangentially and does not enter $x>b$; for $r<b$ it turns around before reaching that boundary and eventually returns through the entrance side. The equality case must therefore not be included in a crossing condition.

Write ${\lambda }_0$ for the vacuum wavelength. Initially the identical air-filled boxes add equal optical paths. When the refractive index in one box falls from $n_{\mathrm{air}}$ to $n(P)$, the extra path in the box that remains full isIdentical fixed window contributions cancel. At the screen centre the geometric slit-path difference is zero, so, neglecting unequal transmission changes,Each extra optical path of one wavelength produces one full intensity period and one new maximum. Taking the stated count in the usual sense of 20 completed fringe periods during evacuation, excluding the initial maximum, and using $n(0)=1$ for vacuum givesTaking $N=20$, vacuum wavelength ${\lambda }_0\simeq 460\mathrm{nm}$ and $l=30\mathrm{mm}$,If the quoted $460\mathrm{nm}$ instead denotes the wavelength ${\lambda }_{\mathrm{air}}$ in ambient air, use ${\lambda }_0=n_{\mathrm{air}}{\lambda }_{\mathrm{air}}$ and the exact relation $n_{\mathrm{air}}=[1-N{\lambda }_{\mathrm{air}}/l{]}^{-1}$. It gives $1.00030676$, differing by less than $10^{-7}$ from the vacuum-wavelength estimate, well below one-fringe counting resolution.

The count convention matters: 20 maxima including both the initial and final maxima span only 19 periods, giving approximately $1.000291$. If the final fractional fringe is not measured, an integer count alone has a resolution of order ${\lambda }_0/l\simeq 1.53\times 10^{-5}$ in refractive index. The standard 20-period estimate should therefore not be interpreted as an arbitrarily precise index measurement.

An ideal focusing lens maps propagation angle to position in its back focal plane, with $z\simeq f\theta$ for paraxial light rays. The two-slit relative phase remains $2\pi s\sin \theta /\lambda$, so the focal-plane interference pattern has the same form with propagation distance replaced by focal length:The new peak intensity $I_{\max ,f}$ accounts for focusing and need not equal the earlier screen peak. The lens forms the angular Fraunhofer pattern at a finite focal-plane distance, so a long free propagation distance is no longer needed to obtain that angular pattern. The displayed position formula assumes the same narrow-slit and paraxial approximations.

For two identical narrow slits under coherent normal illumination, with $\lambda$ the wavelength in the propagation medium, let $I_s$ be the intensity contributed by either slit alone in the relevant angular range. At far-field angle $\theta$, the path difference is $\Delta =s\sin \theta$ and the phase difference is $2\pi \Delta /\lambda$. Adding the two complex amplitudes givesFor a screen perpendicular to the optical axis, $\tan \theta =z/d$. Thus the angular fringe factor uses $\sin \theta =z/\sqrt{d^2+z^2}$; in the paraxial region $|z|\ll d$ it reduces toThe supplied large-distance condition gives the Fraunhofer diffraction regime for the slit separation; the further replacement $\sin \theta \simeq z/d$ is a separate small-angle approximation. An unspecified finite slit width would add a single-slit envelope, and $I_s$ can then vary with angle. The formula above gives the interference factor in the ideal narrow-slit, equal-amplitude model; no absolute source intensity is supplied.

The signed normal-field profile is constant and negative below $z=-d$, rises linearly through the slab, jumps down at the negative sheet, then remains constant and positive above. More explicitly,There is no surface charge at the slab's lower face, so the field is continuous there; only its slope changes. At the ideal sheet, the field jump isexactly the negative-sheet boundary condition. The ideal sheet itself has distinct one-sided fields rather than one common field value.

The graph uses the illustrative ratio $\sigma /(\rho d)=1/2$, clearly separated from the symbolic general result. The zero occurs inside the lower half of the slab, and the two outside fields have equal magnitude and opposite sign.

The differential vacuum Gauss's law isIntegrate over a volume $\Omega$ with closed boundary $S$. The divergence theorem converts the left side into outward electric flux:This is the integral form. Surface charge can be treated by a thin-layer limit or by the corresponding jump in the normal field; the enclosed charge then includes the charged sheet. The surface orientation must be outward, and the field entering the integral is the total field, not merely the field of the enclosed charges alone.

Only the ideal inductor stores energy in this circuit; ideal resistors dissipate energy as heat. With $I_1(\infty )=V/(4R)$,The question excludes the battery's internal energy, and the continuously generated resistor heat is not stored electromagnetic energy.

Put $e_t=e^{-t/\tau }$. From $u=V/2-R{I}_1$ and $I_0=I_1+I_2$,At $t=0^{+}$ the three values are $I_1=0$ and $I_2=I_0=V/(4R)$. At late times they are $I_1=V/(4R)$, $I_2=V/(8R)$, and $I_0=3V/(8R)$. Thus $I_1$ and $I_0$ increase monotonically while $I_2$ decreases; every transient has the same time constant. By $10\tau$, the exponential correction is only $e^{-10}\simeq 4.54\times 10^{-5}$ of its initial amplitude.

The constant particular solution is $A=V/(4R)$, and the homogeneous equation has decay rate $2R/L$. Imposing $I_1(0)=0$ givesAs a separate check, the Thévenin equivalent seen by $L$ has open-circuit voltage $V/2$ and resistance $R+(2R\parallel 2R)=2R$, again giving $\tau =L/(2R)$.

The source diagram has $2R$ in series with two parallel branches: one containing $R$ then $L$, and the other containing $2R$. Let their common voltage be $u(t)$, measured from the upper node to the lower return. Kirchhoff's current law and the branch laws giveUsing $I_2=u/(2R)$ in the source equation yields $u=V/2-R{I}_1$. Substitution into the coil branch givesThe switch was open for a long time, so $I_1(0^{-})=0$. Inductor current continuity gives $I_1(0^{+})=0$ for a finite applied voltage.

At $(x,0)$ the distances from the two centres are equal:By the electrostatic shell theorem, a uniformly charged spherical shell gives zero field inside and its point-charge field outside. For $x<x_1$ the point is inside both shells. For $x_1<x<x_2$ only the small, negatively charged shell contributes. For $x>x_2$ both contribute, soThe requested magnitude isIt jumps from zero to $k_{e}q/a^2$ at $x_1$, decreases to $k_{e}q/(4a^2)$ just before $x_2$, jumps down to $k_{e}q/(8a^2)$ just after $x_2$, and then falls as $x^{-3}$. The ideal surface charge makes the field discontinuous at each crossing, so the one-sided limits, rather than a unique value on the sheet, specify the graph. The far field is the equatorial field of an electric dipole with electric dipole moment $\mathbf{p}=-qa\hat{\mathbf{y}}$.

Put charges $+Q$ and $-Q$ on the concentric conductors. Gauss's law gives $E(r)=Q/(4\pi {ϵ}_0{r}^2)$ between them, $a<r<2a$. The potential difference isThe capacitance of this vacuum spherical capacitor is thereforeThe conductors are equipotentials; a uniform dielectric filling would replace ${ϵ}_0$ by its permittivity.

In matter, Gauss's law for free electric charge isThe area normal points out of $\mathcal{V}$. In vacuum $\mathbf{D}={ϵ}_0\mathbf{E}$, so these become $\nabla \cdot \mathbf{E}=\rho /{ϵ}_0$ and $\oint \mathbf{E}\cdot d\mathbf{S}=Q_{\mathrm{enclosed}}/{ϵ}_0$.

The band has mean wavelength $\bar{\lambda }\simeq 530\mathrm{nm}$ and width $\Delta \lambda =20\mathrm{nm}$. At order $m$, the positions of the extreme-wavelength maxima differ by about $|m|D\Delta \lambda /b$. Comparing this spread with a fringe spacing $D\bar{\lambda }/b$ givesEquivalently, the coherence length scale is ${\ell }_c\sim {\bar{\lambda }}^2/\Delta \lambda \simeq 14\mu \mathrm{m}$. Thus the usual bandwidth estimate isThis is a washout scale, not a sharp definition of “clearly”. For a spectrum uniform in inverse wavelength $\sigma =1/\lambda$, the interference visibility is $|\mathrm{sinc}(\pi \Delta \sigma \delta )|$, with $\mathrm{sinc}u=\sin u/u$; its first zero occurs near order $26.5$. Fringes well inside that limit have stronger contrast. A half-fringe-smear criterion would instead retain about $13$ orders each side. The precise visible number needs the spectral distribution and a contrast threshold, neither supplied. If the slide remains in place and dispersion is neglected, the envelope is centred on the shifted zero-path-difference position and this order-count scale is unchanged.

Define the phase difference as current phase minus voltage phase. Since $\tilde{I}=\tilde{V}/Z$,It decreases continuously from $+\pi /2$ at low frequency, through zero at ${\omega }_0=1/\sqrt{LC}$, to $-\pi /2$ at high frequency. Thus the current leads in the capacitive limit and lags in the inductive limit. Defining voltage phase minus current phase would reverse the graph's sign.