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Substitute the grating equation into the preceding result:Moving across the illuminated grating by one groove spacing changes the incident-plus-outgoing optical path length by $d(\sin \alpha +\sin \beta )=m\lambda$. Across length $W$, the total path difference is therefore $L$. Thus $L$ is the optical path difference between contributions from the two ends of the illuminated grating.

The number of coherent phase cycles across it is $L/\lambda =mN$, which is the intrinsic diffraction-limited spectral resolving power of the diffraction grating under the usual first-minimum criterion. In the slit-limited regime, $R\theta D=L$ also states how much sky angle can be accepted at a given resolution and aperture. Holding slit angle and resolution fixed while increasing telescope diameter requires a larger optical path span. The geometry gives $|L|\le 2W$; large incidence and diffraction angles increase resolution per unit grating length, though grazing beams become impractical.

The slit equations do not imply unlimited resolution when $\theta$ tends to zero. Once $\theta D$ is comparable to $\lambda$, finite-aperture diffraction matters and the actual resolving power is bounded by about $mN$.

An integral field spectrograph obtains spectra throughout a two-dimensional field instead of along one slit alone. Its reduced spectral data cube is $I(x,y,\lambda )$: two coordinates locate a spatial sampling element and the third labels wavelength. One slice at fixed wavelength is an image; one column at fixed spatial position is an optical spectrum. The third axis is spectral, not a third spatial direction.

Six ways to obtain such a cube illustrate the distinction between field reformatting and scanning:

The first four are simultaneous spatially multiplexed grating arrangements. The last two deliver equivalent cube coordinates by scanning; they are imaging spectrometers rather than simultaneous grating integral-field units, and variability during the scan can corrupt the cube. Detector packing, sampling, calibration and throughput determine the practical tradeoffs.

Atmospheric extinction changes the amplitude of the incoming light through absorption and scattering. Its wavelength dependence alters the measured optical spectrum and colors; clouds and aerosols introduce additional time dependence. Scattered moonlight and atmospheric emission increase sky brightness.

Atmospheric refraction changes the apparent position of the source. Its wavelength dependence, atmospheric dispersion, spreads a broadband image toward the zenith. The mean refractive index gradient therefore affects direction even in the absence of small-scale turbulence.

Atmospheric turbulence produces rapidly varying optical path lengths, distorting the phase and curvature of a nominally plane wavefront. Different pupil regions acquire different phase delays, producing astronomical seeing, image motion, and short-exposure speckle patterns. Different lines of sight sample different fluctuations, giving anisoplanatism.

Propagation through the fluctuating medium also changes the intensity through atmospheric scintillation, the familiar twinkling of stars. A phase-only adaptive optics correction can reduce atmospheric phase errors but does not remove absorption, all scintillation, or the atmospheric emission background.

First integral. If $F$ has no explicit $x$ dependence, the chain rule givesThe Euler-Lagrange equation makes this zero without requiring division by $y^{\prime }$. Hence the Beltrami identity is

Optical paths. By Fermat principle, a ray makes the optical path length stationary. Within the strip $0<y\le 1$, its functional for a graph $y(x)$ isThe Beltrami identity givesFor a nonvertical segment $k>0$, write $R=1/k$. Then $(y^{\prime }{)}^2=(R^2-y^2)/y^2$, soThus the optical rays in a reciprocal-height refractive index satisfyrestricted to the part in the strip. They are arcs of circles centered on $y=0$. Vertical lines are the additional solutions, obtained as a limiting case or directly from the constant horizontal optical momentum. The same curves are geodesics of the Poincaré half-plane model, whose length element is $ds/y$.

Horizontal displacement. Let the ray enter at $(x_0,1)$, travelling to the right and downward at angle $0<\theta <\pi /2$ to the vertical. The refractive index is continuous at $y=1$, so there is no instantaneous angular jump at entry. The tangent has $(dx/ds,dy/ds)=(\sin \theta ,-\cos \theta )$; hence $y^{\prime }(x_0)=-\cot \theta$. The first integral gives $k=\sin \theta$, or $R=\csc \theta$. The circle derivative $y^{\prime }=-(x-c)/y$ then fixes $c=x_0-\cot \theta$.

As $y\downarrow 0$, the rightward ray approaches $x=c+R$. The total limiting horizontal distance is thereforeFor vertical incidence, $D=0$ by continuity. For a leftward ray the displacement has the opposite sign, with the same magnitude. Since $\tan (\theta /2)$ increases on $[0,\pi /2]$,If entry is required to have a strictly downward component, this is a supremum rather than an attained maximum. A horizontal tangent at entry gives the limiting unit-radius arc, which immediately bends downward in the medium.

The lower boundary is a limit, rather than a point reached in finite optical travel time. On a nonvertical circular ray, $ds/dy=R/\sqrt{R^2-y^2}\to 1$ as $y\downarrow 0$. Thus the optical path length contains a divergent integral proportional to $\int dy/y$. The horizontal distance stays finite even though propagation toward $y=0$ takes infinite time.

Fermat principle states that the optical path length $\mathcal{S}=\int nds$, or equivalently travel time $\mathcal{S}/c_0$, is stationary under small variations of the ray with fixed endpoints. Stationarity need not mean a global minimum.

For a plane ray written as $y(x)$, use $\mathcal{S}[y]=\int F(x,y,y^{\prime })dx$, with $F=n(x,y)\sqrt{1+y^{\prime 2}}$. Its stationary paths obey the Euler-Lagrange equationIt follows by taking the first variation, integrating the term in $\delta y^{\prime }$ by parts, and using zero endpoint variations.

The Fermat principle makes the optical path length stationary. For a graph $y=\xi (x)$ the variational integrand isThe Euler-Lagrange equation isUse the chain rule $dn(x,\xi (x))/dx=n_x+n_y{\xi }^{\prime }$ and differentiate ${\xi }^{\prime }/\sqrt{1+{\xi }^{\prime 2}}$ to obtainMultiplying through and combining the $n_y$ terms gives the required second-order ordinary differential equation with no derivatives in denominators:All partial derivatives of the refractive index are evaluated along the ray.

The Fermat principle states that the actual path of a light ray makes its travel time stationary under small admissible variations with fixed endpoints. Equivalently, its optical path length is stationary:where $v$ is the local light speed, $n$ is the refractive index, and the two integrals differ by the constant vacuum light speed. In a uniform medium this reduces to stationarity of geometric path length. The principle is a stationarity statement; for the planar reflection considered next the stationary path is also the unique minimum.

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.

The slide adds an optical path length $\Delta =(n-1)t=50\mu \mathrm{m}$ in the covered path. If the covered slit is at positive transverse position, its geometric path to $y>0$ is shorter, giving total path difference $\Delta -by/D$. The zero-difference fringe therefore moves toward that slit:This is about $94$ fringe periods. The leading grating-angle correction, $b\sin \theta =\Delta$ and $y=D\tan \theta$, gives $50.06$ mm, consistent with the paraxial result. In a strictly monochromatic periodic pattern, identifying the continuously tracked central order is essential: the final intensity pattern determines displacement only modulo one fringe spacing.

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.

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 principle states that the optical path length $\int nds$, equivalently the travel time in the nondispersive stationary isotropic ray model, is stationary for the actual light ray between fixed endpoints. It need not be a minimum in every optical system.

For two homogeneous media separated by $y=0$, place the fixed endpoints at $(0,h_1)$ and $(D,-h_2)$, with positive heights. If the ray crosses at $(x,0)$, its optical path isStationarity with respect to the crossing point givesThe two fractions are $\sin {\theta }_1$ and $\sin {\theta }_2$, with angles measured from the interface normal. Thuswhich is Snell's law. Dividing $\mathcal{L}$ by the vacuum light speed gives the corresponding travel time; minimizing geometric distance without refractive-index weighting would incorrectly predict a straight ray through unlike media.

Monochromatic light has one frequency, or a bandwidth sufficiently narrow for the intended measurement. A lamp can illuminate the experiment through a narrow spectral filter or monochromator, followed by a small source aperture at the focus of a collimating lens. The aperture supplies spatial coherence across the slits; the lens supplies approximately parallel incident wavefronts. Two independently illuminated incoherent slits would not provide a stationary fringe pattern.

In the Fraunhofer approximation, the optical path difference at observation angle $\theta$ is $s\sin \theta$. For equal single-slit intensities ${\mathcal{I}}_s$, adding the two wave amplitudes givesThe thin-slit idealization suppresses the individual diffraction envelope. If additionally $|Y|\ll D$, then $\sin \theta \simeq Y/D$ and $\mathcal{I}(Y)\simeq 4{\mathcal{I}}_s{\cos }^2(\pi sY/(\lambda D))$. The far-field condition does not by itself impose this small observation angle.

For order $m=10$ at $Y=0.05\mathrm{m}$,The paraxial estimate is $sY/(10D)=5.88\times 10^{-7}\mathrm{m}$, consistent to the precision of the data.

Let the upper slit contain the glass plate and take positive $Y$ towards that slit. The extra optical path length is $\Delta =(n_g-n_a)t=5.0\times 10^{-5}\mathrm{m}$. Its total upper-minus-lower path difference is $\Delta -s\sin \theta$. The order-zero fringe therefore satisfiesIt moves towards the covered slit. Using $Y_0\simeq D\Delta /s=0.425\mathrm{m}$ here is a poor small-angle approximation: $30^{\circ }$ is not small. The order-zero fringe is being tracked as the optical delay is introduced; because the pattern is periodic in phase, many other bright fringes still lie near the original centre. Neglecting unequal glass transmission changes visibility but not the zero-path-difference condition.

Fermat's principle states that the travel time, or equivalently the optical path length $\int nds$ in a stationary medium, is stationary under small variations of the path with endpoints held fixed. It need not be a global minimum.

Let a light ray join a point at height $h_1$ above a plane interface to a point at depth $h_2$ below it, with horizontal separation $D$. If the crossing point is horizontally $\xi$ from the first endpoint, its optical path length isSetting its derivative to zero givesThe two geometric ratios are the sines of the incident and refracted angles, measured from the interface normal vector. Hence Snell's law is

The paraxial approximation restricts light rays to small slopes relative to the optical axis, so $\sin \beta \simeq \tan \beta \simeq \beta$ and geometric lengths can be expanded to second order in height. Treat the source's optical lens as a thin lens located at $x=0$, with $|y|\ll u,|v|$ and small surface slopes.

For positive real conjugate distances $u,v$, the collapsed-plane geometric path has lengthTo thin-element accuracy, a segment $s(y)=s_0-\alpha y^2$ of this path lies inside the optical lens, replacing the same length of air. The geometric air path is $S-s$, the material path is $s$, and the optical path length and travel time are thereforeThis distinguishes geometric length from optical path length; adding $ns$ to $S$ without subtracting the replaced air would count that segment twice.

For a nonaxial light ray, Fermat's principle gives $d\mathcal{L}/dy=0$ and henceThe focal length is obtained with the incident object at infinity:The constant thickness $s_0$ supplies a common delay but does not change the leading focal length. Since $\alpha$ has units of inverse length, this formula has the correct dimensions. These are thin lens results; paraxial approximation alone does not remove finite-thickness corrections in a general thick lens.

The slide adds optical path length $\Delta =(n-1)t=0.5(0.1\mathrm{mm})=50\mu \mathrm{m}$ to the covered beam. Put the covered slit at transverse coordinate $+s/2$. For positive screen coordinate $x$ its geometrical path is shorter by approximately $sx/d$, so the total covered-minus-uncovered optical path difference isThe zero-order fringe follows equal total optical path lengths, ${\Delta }_{\mathrm{tot}}=0$. HenceThe covered beam needs a shorter geometric path to compensate its extra optical path length. This is the paraxial estimate; exact point-to-point distances give $x_0\simeq 5.006\mathrm{cm}$. “Central” here follows the zero-order fringe continuously as the slide is introduced. At one fixed wavelength, the final monochromatic wave interference pattern is periodic, so intensity alone identifies a fringe position only modulo the fringe spacing. Slide reflections and absorption have been neglected.

Write $T$ for transit time and $\ell (y)=t_0-\alpha y^2$ for lens thickness to avoid using $t$ for both. In the thin lens approximation, the two geometrical distances to the lens plane are $\sqrt{a^2+d^2}$ and $\sqrt{b^2+d^2}$. Their sum counts the lens segment as vacuum. Replacing that segment by material of refractive index $n$ adds $(n-1)\ell (d)$ to the optical path length. Thus, to the retained thin-element accuracy,Using the Taylor expansion $\sqrt{1+u}=1+u/2+O(u^2)$ for $d\ll a,b$ givesThe omitted geometrical terms include $O(d^4/a^3+d^4/b^3)$ in optical path length, together with finite-thickness corrections discarded by the thin lens and paraxial approximations. The addition is $n-1$, because the underlying distance already includes propagation across the thickness once in vacuum.

Fermat's principle makes the optical path length, or transit time in a stationary medium, stationary under small variations of the light ray with fixed endpoints. It need not be a global minimum. In the result of part (a), the coefficient of $d^2$ isFor a family of paraxial light rays from the object to its image, stationarity with respect to $d$ requires this coefficient to vanish. Therefore the thin lens equation isFor a distant object, $a\to \infty$, and the image position $b=f$ defines the focal length. HenceThe axial member $d=0$ is stationary by symmetry for arbitrary $b$; requiring the neighbouring non-axial light rays to meet the same image is what determines $b$.

In surrounding refractive index $n^{\prime }$, the geometrical segments contribute $n^{\prime }[\sqrt{a^2+d^2}+\sqrt{b^2+d^2}]$ to the optical path length, while replacement by the lens adds $(n-n^{\prime })\ell (d)$. Fermat's principle therefore givesThe physical focal length measured in the surrounding medium isFor $\alpha >0$ and $n>1$, increasing $n^{\prime }$ from $1$ towards $n$ makes the positive focal length grow monotonically from its vacuum value to infinity: $df/d{n}^{\prime }=n/[2\alpha (n-n^{\prime }{)}^2]>0$. At $n^{\prime }=n$, there is no refraction at the boundary, so the element has zero optical power and no finite optical focus. If $n^{\prime }>n$, its optical power changes sign and the same shape becomes diverging.

Compare a directly transmitted ray emerging at $A$ with a ray that makes one extra round trip inside the Fabry–Pérot interferometer and emerges at $C$. Both emerge parallel at angle $\theta$ to the plate normal. The extra internal distance is $2d/\cos \theta$, and their emergence points differ laterally by $AC=2d\tan \theta$. The first ray has already travelled $AC\sin \theta$ towards a common outgoing wavefront when the second reaches $C$. The air-gap optical path difference in a Fabry–Pérot interferometer follows because optical path length equals geometric path length in the air-gap approximation:The subtraction is essential: comparing distances only inside the plates would compare different outgoing wavefronts. The corresponding propagation phase difference is $2\pi \Delta /\lambda$. Coating reflection phases can add a separate phase offset; they do not change this geometric optical path difference.