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A submerged object looks shallower because its emerging light rays bend away from the surface normal vector. Their backward extensions give an approximate virtual image nearer the surface. For an object at perpendicular depth $d$ in a medium of refractive index $n_{\mathrm{obj}}$, viewed through a flat boundary from a medium of refractive index $n_{\mathrm{obs}}$, the near-normal paraxial approximation gives $d_{\mathrm{par}}=(n_{\mathrm{obs}}/n_{\mathrm{obj}})d$. Thus the model water depth $12\mathrm{cm}$ appears approximately $9\mathrm{cm}$ from air. This is a bundle-of-light rays statement, not a deflection of the axial light ray.^[ref]

For general transmitted light rays, specify what “depth” means. Put the object at $(0,-d)$ below a horizontal boundary $z=0$. A light ray reaches the boundary at $(x,0)$, with object-side angle ${\theta }_{\mathrm{obj}}$ and observer-side angle ${\theta }_{\mathrm{obs}}$, both measured from the upward normal vector. For $x>0$, its backward extension meets the object's normal line $x=0$ at $z=-d_{\mathrm{axis}}$, whereThis axis-intercept depth is exact for that one transmitted light ray in the stated model. It tends to $d_{\mathrm{par}}$ as the angles tend to zero; at exactly zero the quotient is interpreted as a limit. These elementary tangent, sine and cosine relations show precisely where the small-angle approximation enters.

The original calculations below use $d=12\mathrm{cm}$, $n_{\mathrm{obj}}=4/3$ and $n_{\mathrm{obs}}=1$:

At appreciable obliquity these depths differ, so the full outgoing bundle has no single exact virtual image on the object's normal line. An oblique eye samples a narrow bundle; its local apparent position also depends on that bundle and cannot be inferred by declaring $d_{\mathrm{axis}}$ to be a universal image depth. A plane refracting boundary gives approximate image formation near normal incidence, with departures at larger angles. The construction assumes a flat surface, uniform transparent media with isotropy, and a propagating transmitted light ray; it does not extend to total internal reflection.

The requested source-image relation is the thin lens equation, using positive real-object distance $u$ and signed image distance $v$:Here $f>0$ for a converging lens, and $v<0$ denotes a virtual image. This is usually distinguished from Lensmaker's equation, $1/f=(n-1)(1/R_1-1/R_2)$ for a thin lens in air with the usual signed surface radii. The latter relates focal length to the lens material and curvature, rather than to object and image positions.^[ref]

For a magnifying glass held close to the eye, compare the angle $h/u$ with the unaided near-point angle $h/D_n$, where $D_n=25\mathrm{cm}$. Maximum usable angular magnification occurs with the virtual image at the near point, so $v=-D_n$. ThusThe object lies inside the front focal length. The emerging light rays diverge, while their backward extensions meet at an upright enlarged virtual image on the object's side. With the image at infinity instead, the relaxed-eye angular magnification would be $D_n/f=6.25$, which is smaller. A finite eye-to-lens distance $s$ changes the near-point result to $1+(D_n-s)/f$; the quoted maximum uses the conventional $s\simeq 0$ limit.^[ref]

For the adjacent reflected minima of an air-film-air layer in thin-film interference, reflection at the front surface acquires a phase reversal, while reflection at the rear does not. Near normal incidence the reflected minima therefore obey $2nt=m\lambda$ for integral m. A single intermediate maximum means that the two stated minima are consecutive orders:Thus $2nt=3000\mathrm{nm}$, with orders6 and5, andThe intermediate reflected maximum is at $\lambda =3000/5.5\simeq 545\mathrm{nm}$. This assumes negligible dispersion of the supplied refractive index and an air environment on both sides. ^[ref]

The angular separation is $\Delta \theta =10^{11}/(2\times 10^{17})=5.00\times 10^{-7}\mathrm{rad}$. Each star's angular diameter is below $5\times 10^{-9}\mathrm{rad}$, so treating it as a point source is justified at these diffraction scales.

At300nm the Airy disk first-zero radius is $1.22\lambda /D=1.525\times 10^{-7}\mathrm{rad}$. The separation is3.28 times this radius: two well-separated bright peaks occur. The stars are mutually incoherent, so their point spread functions add as intensities, not amplitudes:The value at u=0 is defined by continuity. Faint diffraction rings remain on either side of each principal peak. ^[ref]

At1000nm the first-zero radius is $5.083\times 10^{-7}\mathrm{rad}$, slightly larger than the $5.00\times 10^{-7}\mathrm{rad}$ separation. The images therefore overlap strongly and just fail the conventional Rayleigh criterion, rather than forming two clean isolated peaks.

The summed Airy profile can still have a shallow central dip and two local maxima: Rayleigh's criterion is a resolution convention, not the exact condition for a profile to become single-peaked. The two-panel original plot in part(a) shows the correct intensity sums at both wavelengths. ^[ref]

With a positive charge magnitude e, the relaxation-time equation for electrons is $m_e^{\ast }\dot{\mathbf{v}}=-e\mathbf{E}-m_e^{\ast }\mathbf{v}/\tau$. Steady drift is opposite the field, ${\mathbf{v}}_d=-e\tau \mathbf{E}/m_e^{\ast }$, so the positive electron mobility and electrical conductivity areThis is the weak-field Drude model; the mobility is a magnitude, not the signed velocity-to-field ratio.

For an intrinsic nondegenerate semiconductor, $n_i=p_i=\sqrt{N_c{N}_v}{e}^{-E_g/(2k_{B}T)}$. Equal electron and hole masses give $N_c=N_v$ and intrinsic chemical potential midway between the band edges. At300K, the exponential is $e^{-19.34}=3.98\times 10^{-9}$. Using the later supplied mass $0.015m_e$ givesThis is many orders below a typical metal's carrier density. Since ${\sigma }_i=e{n}_i({\mu }_e+{\mu }_h)$, ordinary finite carrier mobilities leave its intrinsic conduction negligible on a metallic scale; the gap alone does not specify an exact conductivity without the mobilities.

Under the explicitly stipulated full donor ionization, $n\simeq N_d=10^{24}{\mathrm{m}}^{-3}$ and the supplied mobility is $0.0100{\mathrm{m}}^2{\mathrm{V}}^{-1}{\mathrm{s}}^{-1}$. ThereforeThe band diagram places the intrinsic chemical potential at midgap and a shallow isolated donor level just below the conduction edge. The doped chemical potential moves upward. Quantitatively, the degeneracy test for a fully ionized light-mass donor gas gives $N_d/N_c=21.7$, so a nondegenerate logarithmic chemical-potential formula is inconsistent. The parabolic-band zero-temperature estimate is $E_{F,d}-E_c={\hslash }^2(3{\pi }^2{N}_d{)}^{2/3}/(2m^{\ast })=0.243\mathrm{eV}$; its300K value is about0.241eV. The diagram accordingly shows the doped chemical potential inside the conduction band, not between an isolated donor level and the edge.

The hydrogenic donor state scaling using the supplied hydrogen values isThe binding is much smaller than $k_{B}T=25.9\mathrm{meV}$, indicating a shallow donor on the isolated-state estimate. However, the mean donor spacing is only $N_d^{-1/3}=10\mathrm{nm}$, well below the orbit radius. The donor-state overlap estimate invalidates an independent isolated-level ionization model at this concentration. Shallow binding supports abundant mobile carriers, but the full ionization versus isolated donor occupation distinction means full ionization cannot be justified solely by comparing binding with temperature while simultaneously retaining isolated levels and a degenerate chemical potential. The conductivity above follows the question's carrier-density assumption; an overlapping impurity/conduction-band description is needed for a self-consistent microscopic account. ^[ref]

For the stated standard configuration, the Lorentz transformation isDifferentiate along the moving object's worldline and divide $d{x}^{\prime }$ by $d{t}^{\prime }$ to obtain the velocity-addition formula:For a receding source, two emissions separated by proper time $1/f^{\prime }$ occur $\gamma /f^{\prime }$ apart in S. The second signal travels an additional distance $v\gamma /f^{\prime }$, so its reception interval is $\gamma (1+v/c)/f^{\prime }$. Hence the longitudinal relativistic Doppler effect isThe inverse velocity transformation for the forward throw givesThe ball approaches the stationary receiver, so its forward light is blueshifted. The received wavelength is $400\sqrt{(1-3/5)/(1+3/5)}=\boxed{200\mathrm{nm}}$, in the ultraviolet. The stated monochromatic emission is outside the visible band. Other reflected or emitted frequencies have not been specified. ^[ref]

Online text: openstax.org/books/university-physics-volume-3/pages/2-8-microscopes-and-telescopes

Section of University Physics Volume 3 by Samuel J. Ling and others. The existing book record supplies the whole-book authors and 2016 publication year; no separately credited section author or date is inferred.

Cited passages: Microscopes, Figure 2.38; equations 2.35–2.37 and near-point versus relaxed viewing.