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In a transparent, stationary vacuum with Euclidean geometry and observers at rest in the same frame, the radiative transfer equation has zero absorption and emission, so $d{I}_{\nu }/ds=0$ along a light ray. The same specific intensity reaches the observer without geometric dilution. More generally, $I_{\nu }/{\nu }^3$ is a Lorentz invariant; collisionless propagation preserves this ratio along a ray even with gravitational or cosmological redshift, rather than preserving $I_{\nu }$ at shifted frequencies. ^[ref].

The elementary geometry is instructive. Two facing surface elements separated by distance $s$ exchange spectral radiant fluxThe receiver subtends $d{\Omega }_2=\cos {\theta }_{2}d{A}_2/s^2$ at the source. Describing that same traveling beam at the receiver instead uses $d{\Omega }_1=\cos {\theta }_{1}d{A}_1/s^2$. Both descriptions have identical geometric factors, so conservation of its radiant flux gives equal specific intensities at the two ends. This compares the same propagation direction at both ends, not emission in the opposite direction.

A uniform source remains equally bright per resolved direction under these assumptions, but occupies a smaller apparent solid angle farther away. For a small source of projected area $A_{\perp }$ at distance $d$ and a small receiver facing it, irradiance is approximately $E=L{\Omega }_{\mathrm{source}}\simeq L{A}_{\perp }/d^2$. The radiant flux intercepted by receiver area $A_d$ is approximately $E{A}_d$. With $L=10\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}$, $A_{\perp }=0.020{\mathrm{m}}^2$ and $A_d=10^{-4}{\mathrm{m}}^2$, distances $10$ and $20\mathrm{m}$ give $E\simeq 0.002$ and $0.0005\mathrm{W}{\mathrm{m}}^{-2}$, and received radiant fluxes of $0.20$ and $0.050\mu \mathrm{W}$. The inverse-square law concerns collected radiation from the whole source; $L$ remains $10$ at either distance.

A virtual image is the apparent origin of an outgoing diverging bundle: the backward extensions of its light rays intersect at the image point. Those extensions are geometric constructions, not backward-travelling light. A screen placed there cannot intercept the outgoing bundle to display that virtual image. An eye or a second optical lens receiving the actual outgoing light rays can nevertheless form a subsequent real image. “Virtual” describes this imaging stage, not whether the image can be seen or photographed. ^[ref].

For a converging thin lens with $0<u<f$, the same equations give $v<0$ and $m=f/(f-u)>1$: the virtual image is upright and enlarged. For the illustrative choice $f=10\mathrm{cm}$, $u=7.5\mathrm{cm}$, $h_o=0.30\mathrm{cm}$,The image appears farther from the optical lens than the object. A diverging thin lens with a real object instead has $v<0$ and $0<m<1$, so a virtual image need not be enlarged. A plane mirror provides another virtual image, located behind the reflecting surface by backward extensions of the reflected light rays. ^[ref].

For a direct viewing test, bring the target to $u=7.5\mathrm{cm}$, search for a sharp screen image on the outgoing side, then remove the screen and look through the optical lens toward the target. Expect an upright enlarged view if the eye receives the bundle and can focus. Record the eye position: $5\mathrm{cm}$ beyond the optical lens gives a predicted apparent distance $35\mathrm{cm}$. This adapts the Northern Virginia laboratory's comparison. ^[ref].

For a position measurement, use parallax localization of a virtual image. Place a movable pencil tip or narrow comparison marker beside the apparent image, and view that marker directly beside or above the optical lens, while viewing the image through it. Move the head a little sideways and move the marker along the bench until relative lateral motion vanishes. Read its coordinate $x_m$; the estimate is $v=x_m<0$. Deliberately place the marker nearer and farther to check that relative motion reverses on opposite sides of the match. This follows Orhan Aytür's lens laboratory, which uses a pencil to locate a virtual image. ^[ref].

Keep target and optical lens fixed, use one eye and the central image detail, and ensure the marker itself is not viewed through the optical lens. A marker blocked by the target, or a head movement that loses the bundle through the aperture, invalidates that trial. Repeat the match from both directions and record its depth range. This measures apparent depth; comparing apparent sizes at different depths does not directly measure transverse magnification. Near $u=f$, apparent depth becomes very large and the match is poorly constrained. A failed screen search alone therefore needs the distance check and the viewing test.

A wavefront connects points with the same wave phase at one specified time. It is a surface, rather than the trajectory of a particular piece of light. For a scalar monochromatic wave written as $u(\mathbf{r},t)=A(\mathbf{r})\cos \Phi (\mathbf{r},t)$, a wavefront is a connected level set of the phase function $\Phi$; different surfaces can share the same wave phase modulo $2\pi$. A plane wave has planar wavefronts. An outward wave from an ideal point source in a uniform medium has spheres as wavefronts, while an optical lens can turn nearly planar wavefronts into converging ones. The wavefront describes wave phase, not wave amplitude: drawing it does not say how bright the beam is. ^[ref]

The relation to light rays becomes precise in a stationary medium with isotropy, negligible loss and a real positive scalar refractive index $n(\mathbf{r},\omega )$. In the geometrical optics limit, spatial changes in the refractive index and wave amplitude are slow on the local wavelength scale. For a single smooth propagation branch, writeHere $\omega >0$ is the angular frequency, $c$ is the vacuum speed of light, $S$ has units of length, and the last relation is the optical eikonal equation. The local wavevector is $\mathbf{k}=\nabla \Phi =k_0\nabla S$. Its direction is normal to a smooth wavefront, since a gradient is normal to its level sets. A forward light ray has unit tangent vector $d\mathbf{r}/ds=\nabla S/n$, so it follows this normal under the stated assumptions. Along that light ray, $dS/ds=\nabla S\cdot (d\mathbf{r}/ds)=n$: the change in $S$ is the optical path length. This is the short geometric argument linking wavefronts, light rays and optical path length. ^[ref]

The diagram shows incident and transmitted branches separately, with the reflected branch omitted. Its stipulated refractive indices are $n_1=1$ and $n_2=3/2$, and its incident angle is $30^{\circ }$ to the interface normal. At a stationary interface, the frequency and tangential wavevector agree on the two sides: Snell's law gives $\sin {\theta }_2=1/3$, hence ${\theta }_2\approx 19.4712^{\circ }$. The wavefronts turn while their normal spacing decreases from ${\lambda }_0$ to $2{\lambda }_0/3$, where ${\lambda }_0$ is the vacuum wavelength. More explicitly, with $x$ normal to the interface and $y$ tangential, take $S_1=\sqrt{3}x/2+y/2$ for $x<0$ and $S_2=\sqrt{2}x+y/2$ for $x>0$. These functions agree at $x=0$; the drawn wavefronts are $S_j=m{\lambda }_0$ at $t=0$ with ${\varphi }_0=0$. This is an exact piecewise plane wave phase construction for the stipulated transparent media, not measured material data. ^[ref]

The normal-to-wavefront rule requires its assumptions. In a medium with anisotropy, the wavevector remains normal to a smooth wavefront, but the group velocity and energy flux need not follow it. ^[ref] Near a wave caustic, where multiple light rays meet, one smooth phase function may no longer describe the field; diffraction and the full wave amplitude matter. A wavefront is also not generally a surface of equal pulse arrival time. In a medium with optical dispersion, a monochromatic wave phase moves with the phase velocity, while a sufficiently narrow wave packet follows the group velocity. Even identical optical path lengths at one frequency can give different group delays. The usual equal-travel-time picture needs a common launch reference and a nondispersive ray model; an arbitrary broadband disturbance has no single monochromatic wavefront.^[ref]