Here $E_y^2+E_z^2=9A^2$, so the tip of the electric field traces a circle at constant magnitude: this is circular polarization. At fixed $x=0$, it starts at $+y$ when $t=0$ and initially rotates toward $+z$. This explicit rotation specifies the sense without relying on competing handedness conventions.

Using $\mathbf{B}=v_p^{-1}\hat{\mathbf{x}}\times \mathbf{E}$, with $v_p=\omega /k$, givesIn vacuum replace $v_p$ by $c$. The magnetic flux density also has constant magnitude and is perpendicular to $\mathbf{E}$ at every instant.

Write $\psi =kx-\omega t$ and $v_p=\omega /k$. The electric field isIt has linear polarization, along the fixed line $E_z=6E_y$, at angle ${\tan }^{-1}6\simeq 80.54^{\circ }$ from $+y$ toward $+z$. For the associated plane electromagnetic wave, Faraday's law gives $\mathbf{B}=v_p^{-1}\hat{\mathbf{x}}\times \mathbf{E}$. HenceIn vacuum $v_p=c$. Both fields are transverse, and $\mathbf{E}\times \mathbf{B}$ points along $+x$.

In wave optics, a finite lens aperture produces diffraction. Contributions interfere constructively near the geometric focus but form a finite point spread function, rather than a point of infinite intensity. For a uniformly illuminated circular aperture of diameter $D$, the paraxial focal pattern is an Airy disk, with first dark-ring radius approximately $1.22\lambda f/D$. Aberrations can broaden or distort it further; even an ideal lens retains the finite diffraction scale.

For nonzero small $d$, stationarity of the travel time givesTo focus a bundle, this must hold for neighbouring off-axis rays, not merely for the automatically stationary axial ray $d=0$. Comparing with the thin lens equation, $1/a+1/b=1/f$, givesFor parallel incident rays $a\to \infty$, the image is at $b=f$. A centre-thick lens with $\alpha >0$ and $n>1$ is converging.

In the thin-lens approximation, the geometric path length isThe glass replaces approximately length $s(d)$ of air, rather than adding a further geometric segment. Its excess optical path length is therefore $(n-1)s(d)$. With $s(d)=s_0-\alpha d^2$,The approximations retain terms quadratic in the small ray height and neglect products of lens thickness with the squared ray slope. The distinction between geometric and optical path length explains why the glass contribution contains $n-1$, not $n$.

Fermat's principle states that the actual ray makes the travel time stationary under small variations of its path with endpoints fixed. Equivalently, for a stationary medium it makes the optical path length $\int nds$ stationary. “Stationary” allows minima, maxima, or other stationary paths; it does not require a strict minimum in every optical system.

For near-normal viewing through the flat top surface, the refracted ray appears to originate at an apparent depth $h_{\mathrm{app}}$. A ray at horizontal offset $x$ has $x=h\tan {\theta }_w=h_{\mathrm{app}}\tan {\theta }_a$, and Snell's law gives $n\sin {\theta }_w=\sin {\theta }_a$. In the paraxial limit,At an oblique viewing angle the exact ray-based apparent depth is $h\tan {\theta }_w/\tan {\theta }_a$; the simple ratio presumes the usual near-normal top view.

The result need not be the same. Optical dispersion changes the refractive index with wavelength. Under normal dispersion, the blue-light index is larger, changing the refracted rays and generally reducing the visible fraction in the one-interface model. The blue indices of both water and glass are not specified; they may also differ from each other, introducing refraction at their interface. Thus no unique numerical blue-light fraction follows from the red-light value alone.

Take a horizontal cross-section in units of $R$. The eye is at $E=(\sqrt{2},0)$, and the tadpole swims along the perpendicular diameter $T=(0,y)$ shown in the source. Since glass and water have the same red-light index, their internal interface causes no red-light refraction. Write an exit point on the circular wall as $P=(\cos \varphi ,\sin \varphi )$. An unobstructed outward ray to the eye has $|\varphi |\le \pi /4$.

For $\varphi \ge 0$, let $e$ be the magnitude of the emerging ray's angle to the outward normal and $i$ its internal angle. Geometry and Snell's law giveThe outgoing internal ray makes angle $\beta =\varphi -i$ to $+x$. Its backward continuation intersects the swim diameter atThe directly visible positive endpoint is the largest such intercept, including all single-pass refracted rays. Maximizing this one-variable expression gives ${\varphi }_{\ast }\simeq 38.07^{\circ }$ and $y_{\ast }\simeq 0.75744$. By symmetry the visible length is $2R{y}_{\ast }$, soChecking only the grazing exit at $\varphi =45^{\circ }$ gives $\sqrt{2}/[1+\sqrt{1.33^2-1}]\simeq 0.75350$, a close estimate but a slightly smaller fraction. The exact limiting ray occurs before that grazing ray, where two refracted paths merge at a caustic. This is a geometrical, single-pass visibility boundary; finite pupil size and detection thresholds would smooth the transition.

Fermat's principle says that the travel time, or equivalently the optical path length, is stationary under small path variations at fixed endpoints. For a flat interface, let the ray cross at horizontal coordinate $x$, with source and detector at heights $h_1,h_2$ and horizontal separation $D$. Apart from $1/c$, its travel time isSetting $d\mathcal{L}/dx=0$ givesThe angles are measured to the normal. This is Snell's law; “least time” is more generally a stationarity condition.

For the modified circuit, $I_i=V_i/R_i$ and ${\dot{V}}_o=-V_i/(R_i{C}_f)$. With $V_i=V_0\cos \omega t$,The sinusoidal component has amplitude $V_0/(\omega R_i{C}_f)$ and leads the cosine input by $90^{\circ }$; the constant is set by the initial capacitor charge. This assumes the output remains within the amplifier's linear range.

Replace the feedback resistor by a capacitor $C_f$, retaining the input resistor $R_i$. The virtual ground makes the input charge flow onto the feedback capacitor rather than into the ideal amplifier. Integrating its current givesIf the calorimeter delivers $Q_{\mathrm{in}}=\kappa \mathcal{E}$ for deposited energy $\mathcal{E}$, then $\Delta V_o=-\kappa \mathcal{E}/C_f$. A reset establishes the initial output; a practical large resistor across $C_f$ limits long-time drift, but is omitted from the ideal integration interval.

With $R_f=R_i$, the output and input voltage magnitudes are equal. The signal power supplied at the input is $P_i=V_{i,\mathrm{rms}}^2/R_i$, while the load receives $P_L=V_{o,\mathrm{rms}}^2/R_L$. HenceThis is $10$ dB of power gain into the specified load. Output energy comes from the amplifier's supply rails; unity voltage-gain magnitude does not imply unity power gain. Dissipation in the feedback resistor is additional to the specified load power.

The input source supplies $I_i=V_i/R_i$ into the virtual ground, so the complete circuit's input resistance is $R_i$. The ideal output is a zero-resistance controlled voltage source:These are properties of the closed-loop circuit, distinct from the infinite input resistance of the amplifier terminals themselves.

Ground the non-inverting input and take the ideal operational amplifier in stable, unsaturated negative feedback. Its inverting input is a virtual ground and draws no current. Kirchhoff's current law there givesThe minus sign is the phase inversion of the inverting amplifier.

In uniform fields, a rigid neutral electric dipole with both charges moving at the selected common velocity has zero net translational force. To leading nonrelativistic order its torque is $\mathbf{\tau }=\mathbf{p}\times (\mathbf{E}+\mathbf{v}\times \mathbf{B})$, which also vanishes at the velocity-selector condition. Thus it continues undeflected and, initially nonrotating, does not start rotating in the ideal balanced fields. An unbalanced effective field could exert a torque; the dipole orientation is not otherwise specified.

The upper positive plate gives a downward field $E=V/a$, while $\mathbf{B}$ points into the page. For a rightward electron, the electric force is upward and the magnetic force downward. Its initial speed isThese are equal to the precision intended by the rounded plate spacing: the arrangement is a velocity selector. Accordingly,It is not appropriate to calculate a large electric deflection and ignore magnetic bending. For completeness, if all printed values are treated literally, let $v_d=E/B$, $u=v_0-v_d$, and ${\omega }_c=eB/m_e$. The exact nonrelativistic crossed-field motion iswith $v_x=v_d+u\cos {\omega }_{c}t$ and $v_y=-u\sin {\omega }_{c}t$. At the exit solve $x(t_e)=0.50$ m, then use $\theta =atan2(v_y,v_x)$. The small rounding mismatch produces nanometre-scale oscillations, with $|\theta |\le {\sin }^{-1}(|u|/v_d)\simeq 6.32\times 10^{-4}$ rad. Its precise oscillatory exit angle is not robust to the supplied rounding.

For the centred screen, $|y|\le 0.30$ m. Retaining the screen-angle geometry givesThere are about $\bar{\sigma }{\Delta }_{\max }=10.045$ carrier orders on each side. With equal spectral intensities, direct extrema of the intensity formula giveTheir contrast declines toward the edges: the first equal-intensity visibility zero occurs at $\Delta =1/(2\Delta \sigma )=5.25\mu \mathrm{m}$, just beyond the screen's $5.145\mu \mathrm{m}$ limit. Edge visibility is only about $0.031$, so fewer fringes may be discernible at a practical contrast threshold.

The screen reaches $|y|/D=0.6$, where a paraxial conversion of its full width is inaccurate. Also, wavelength-independent detector sensitivity does not itself establish equal source intensities. The boxed count specifies the usual equal-intensity idealization; without spectral weights and a visibility threshold, a unique experimental count is not fixed.

Let $J_1,J_2$ be the single-slit intensities at the two wavelengths. At a screen position $y$, the path difference is approximately $\Delta =d\sin \theta$, with $\sin \theta =y/\sqrt{D^2+y^2}$. Fields from the two slits interfere at each wavelength, while a detector averaging over the optical beat period adds the intensities of the two distinct wavelengths:For equal single-slit intensities $J_1=J_2=J$, define $\bar{\sigma }=(1/{\lambda }_1+1/{\lambda }_2)/2$ and $\Delta \sigma =1/{\lambda }_2-1/{\lambda }_1$. ThenThe first cosine gives the interference visibility envelope. Near the centre one may further replace $\Delta$ by $dy/D$. The formula neglects unspecified single-slit envelopes and the slow change of single-slit intensity across the screen.

For a point charge $q_j$, Coulomb's law gives ${\mathbf{E}}_j=q_j\hat{\mathbf{r}}/(4\pi {ϵ}_0{r}^2)$. The flux through an oriented surface element isHere $d\Omega$ is its signed solid angle seen from the charge. A closed surface subtends total oriented solid angle $4\pi$ for a point inside and zero for a point outside. Thus the flux of each inside charge is $q_j/{ϵ}_0$ and each outside charge has zero net flux. By superposition,This is Gauss's law; a surface passing through a point-charge singularity is excluded.