A documented experiment uses curved illumination rather than the uniform Fresnel slit propagation model. Laboratory Exercise No. 4: Diffraction, MIT 6.237, Fall 2024, printed pp. 5–6, gives a helium–neon laser at $632.8\mathrm{nm}$, a microscope objective optical lens, a circular aperture and a screen 5–40 cm from the object. Preparation includes comparing central brightness in Fresnel diffraction and Fraunhofer diffraction. A 40× objective produces a nonplanar incident wavefront; the opening is placed less than 1 cm from it. With objective-to-screen distance fixed, students move the opening, sketch patterns, and record positions of consecutive central nulls to estimate aperture size. The Fraunhofer diffraction pattern supplies a regime comparison. A separate spherical-obstacle task observes a bright centre inside the shadow. The handout supplies procedures, rather than measured distances or profiles.^[ref]

The circular-opening test probes cancellation of complex amplitudes at the centre while the source wavefront curvature changes. To interpret it, include that curvature in the aperture field $U_0$ and use a circular aperture, rather than substituting the plane-illuminated Fresnel slit propagation profile. Useful additional controls are to retain the same laser and opening, centre the illumination, repeat position readings and check background with the beam blocked. If a camera replaces the visual screen readout, avoid saturation and calibrate its position scale. Alignment, imperfect aperture edges, illumination variation and limited contrast near a null constrain the inferred size. A bright central obstacle spot illustrates why the sharp shadows of geometrical optics can fail; it does not imply that every Fresnel diffraction pattern has a bright centre.

Consider a uniform plane wave of unit complex amplitude transmitted by an ideal optical slit of full width $D$. Treat the field as invariant along the optical slit height: this is one-dimensional transverse Fresnel propagation, so the prefactor is $1/\sqrt{i\lambda z}$, rather than the two-dimensional prefactor $1/(i\lambda z)$. Removing the carrier $e^{ikz}$ givesHere $I_{\mathrm{inc}}$ is incident irradiance. This infinitely long slit model describes a transverse cut away from the ends of a sufficiently long physical slit; its finite conserved quantity is transmitted power per unit optical slit height.

Put $t=\xi /D$ and $q=xD/(\lambda z)$. Substitution reduces the integral toThis form separates the pattern's shape from its physical scale. Reflection $t\mapsto -t$ proves $A(-q,N_D)=A(q,N_D)$, so the light intensity is an even function of $x$. At $N_D=0$, direct integration gives $A(q,0)=\sin (\pi q)/(\pi q)$, with limit one at $q=0$. Hence the Fraunhofer diffraction comparison is the squared sinc function. In this paragraph the unnormalized sinc function takes argument $\pi q$.

For an original example, choose wavelength $\lambda =633\mathrm{nm}$ and full width $D=0.400\mathrm{mm}$. At $z=0.0200\mathrm{m}$, $N_D\simeq 12.6382$, $N_a\simeq 3.15956$ and the edge-to-centre wave phase change is $9.92604$ radians. At $z=2.00\mathrm{m}$, $N_D\simeq 0.126382$, $N_a\simeq 0.0315956$ and that wave phase change is only $0.0992604$ radians. Even the nearer screen is fifty slit widths away; that geometric comparison alone misses its strong Fresnel diffraction.

These are numerical predictions from the displayed scalar integral, using the composite trapezoidal rule, rather than measured profiles. Each curve is divided by its own peak light intensity; the two horizontal scales differ. At 2 cm the computed central irradiance is about $1.14812I_{\mathrm{inc}}$, with brighter off-axis maxima. At 2 m it is about $0.126272I_{\mathrm{inc}}$ and is the peak. The dashed Fraunhofer diffraction curve has its own central normalization; its first zeros are at $x=\pm \lambda z/D=\pm 3.165\mathrm{mm}$. The finite-distance Fresnel diffraction curve remains slightly nonzero there.

Normalization hides the brightness change, but not the change in shape. The Fourier transform propagation multiplier has unit complex modulus, so Plancherel theorem givesThis is conservation of transmitted power per unit height in the ideal scalar model. An integral over only the displayed screen interval misses the tails; a peak-normalized profile is not a unit-integral probability density. Spreading and decreasing brightness together preserve the transmitted power.

A Fresnel number compares squared aperture size with the product of wavelength and propagation distance. For an optical slit of full width $D=2a$, explicitly distinguish the full-width and half-width conventions:The full-width convention is used throughout the Fresnel slit propagation example below. The definition $D^2/(\lambda z)$ also appears in the Delft treatment.^[ref]

For uniform incident wave phase, $N_D\ll 1$ makes the omitted aperture wave phase small and gives the Fraunhofer diffraction regime. When $N_D$ is of order one or larger, retain it and use Fresnel diffraction, provided the paraxial approximation still holds. These are continuous changes in accuracy, rather than a physical boundary at one selected number. For a rectangular aperture, check both transverse dimensions. Incident wavefront curvature contributes its own wave phase through $U_0$; the aperture-to-screen Fresnel number alone does not describe a spherical source's illumination.

The scaling is revealing: doubling the optical slit width at fixed wavelength and distance multiplies the Fresnel number by four. Maintaining the same Fresnel number then requires four times the distance. Two systems can consequently have the same dimensionless pattern despite very different physical lengths.

The quadratic wave phase follows from an ordinary distance calculation. Write ${\rho }^2=(x-\xi {)}^2+(y-\eta {)}^2$. The Taylor expansion of the distance to a screen point isIn the slowly varying propagation prefactor, $r\simeq z$ suffices. In the complex exponential $e^{ikr}$, however, a small distance error is multiplied by the large wavenumber $k$. Retaining the quadratic term while dropping the quartic term requires both $\rho /z\ll 1$ and $k{\rho }^4/(8z^3)\ll 1$ over the contributing aperture and the screen region being considered. The second condition is a wave phase accuracy requirement; the first is the geometric paraxial approximation.

Expanding the retained quadratic term separates three effects:The outside quadratic factor changes the wave phase at each screen point but has unit complex modulus, so it leaves that screen's light intensity unchanged. The quadratic factor inside the integral changes relative wave phases between contributions from different aperture points, so it changes their wave interference. The remaining linear factor supplies a Fourier transform. Fraunhofer diffraction drops the aperture quadratic factor when it is nearly constant; it does not require dropping the screen quadratic factor when predicting a later coherent field.

There is also a useful proof from the Helmholtz equation. Substituting $U=e^{ikz}u$ gives $u_{zz}+2ik{u}_z+u_{xx}+u_{yy}=0$. Neglecting $u_{zz}$ yields the parabolic wave equation. A transverse Fourier transform evolves each spatial frequency $\mathbf{\nu }$ by $\exp [-iz|\mathbf{\nu }{|}^2/(2k)]$, using the transform kernel $e^{-i\mathbf{\nu }\cdot \mathbf{x}}$. Its inverse Fourier transform is the quadratic Fresnel propagator kernel above. Thus quadratic wave phase in space and quadratic wave phase in spatial frequency are two views of the same propagation.

For an axial point source a distance $p$ before a square aperture and an observer a distance $q$ beyond it, source and observation wave phase curvatures add: $R_{\mathrm{eff}}^{-1}=p^{-1}+q^{-1}$. With aperture side $d$, put $w=(d/2)\sqrt{2/(\lambda R_{\mathrm{eff}})}$. Separation into two one-dimensional Fresnel integrals gives the field normalized to its unobscured value,At an off-axis observer $(X,Y)$, centre the dimensionless coordinates on $x_0=pX/(p+q)$ and $y_0=pY/(p+q)$. If $D_x,D_y$ are the directed Cornu spiral chords between the corresponding aperture-edge parameters, $\psi /{\psi }_0=D_x{D}_y/(2i)$. This is a product of complex interval factors, rather than a sum of two intensities. The unobscured field is evaluated at the same observation point and includes its common propagation wave phase.

A Fresnel half-period zone is a region of an aperture whose optical path to an observation point differs from the axial path by an interval of half a wavelength. For an axial optical point source a distance $s$ before the aperture and an axial observer a distance $r$ beyond it, the paraxial path excess at aperture radius $u$ is $u^2(s^{-1}+r^{-1})/2$. The outer radius of zone $n$ is thereforeSuccessive zones contribute approximately opposite phases to the Huygens-Fresnel principle integral. The finite source distance matters: using the plane-wave formula $u_n^2=n\lambda r$ requires $s\to \infty$.

For a circular aperture of radius $a$, paraxial plane-wave illumination and propagation distance $z$, define $N_F=a^2/(\lambda z)$. On axis, the radial Fresnel diffraction integral normalized by the unobstructed field givesFor a finite axial point source a distance $s$ before the aperture and detector a distance $r$ after it, replace $z$ by $sr/(s+r)$ in the phase. The formula assumes illumination amplitudes and obliquity factors approximately constant across the aperture. Odd and even numbers of Fresnel half-period zones give bright and dark axial extrema; large local intensity is compatible with global conservation of energy.

Under the same scalar paraxial assumptions as Fresnel diffraction by a circular aperture, Babinet's principle gives the complementary opaque-disk field $U_d/U_0=e^{i\pi N_F}$ on axis. Its axial intensity equals the unobstructed intensity, producing the Arago spot inside the geometric shadow. The off-axis pattern depends on the full Fresnel integral; it is not obtained by subtracting aperture intensities from the incident intensity. Finite source size, misalignment and imperfect circular edges reduce the ideal spot contrast.

The Arago spot is the bright axial point in the shadow of a circular obstruction under suitably coherent illumination. In ideal scalar diffraction theory, secondary fields from the circular edge arrive in phase on axis. Fresnel diffraction by an opaque disk gives an axial intensity equal to the unobstructed intensity in its paraxial model; this is not a claim that the whole geometric shadow is bright.

At the geometric shadow boundary of a straight opaque half-plane, the transmitted Fresnel integral is exactly half the unobstructed complex amplitude for ideal point illumination. The Cornu spiral chord is from its origin to one limiting endpoint, rather than between both limiting endpoints. Therefore $U/U_0=1/2$ and $I/I_0=1/4$. Half the unobstructed intensity would confuse a field fraction with its square. Edge thickness, source extent and alignment can modify the ideal result.

A spectrogram displays a signal's changing local frequency spectrum, usually by color proportional to squared short-time Fourier transform magnitude. With window function $w$, one convention isThe window selects a neighborhood of time $t$ before taking its Fourier transform. A sinusoid gives a persistent frequency ridge, while a brief pulse spreads across frequencies and occupies a short time range.

A longer window separates nearby persistent frequencies more clearly but mixes events across a longer interval. A shorter window localizes transients but broadens their spectral features. Changing the window function changes this tradeoff and the spectral leakage. Zero padding can sample the displayed frequency curve more finely; it does not add observed data or restore the resolution lost through a short measurement interval. The phase is discarded in the plotted squared magnitude, so the spectrogram should not be confused with a complete set of complex transform coefficients.

For four identical long slits of width $a$ centred at $x=\pm D/2,\pm 3D/2$, the aperture Fourier transform givesHere $\mathrm{sinc}x=\sin x/x$. Normalized to the on-axis light intensity,Principal array directions have $qD=2\pi m$; a single-slit diffraction envelope zero cancels such an order when $ma/D$ is a nonzero integer. The closest central array zeros satisfy $qD=\pm \pi /2$. Finite slit width generally shifts the actual maxima of the complete product slightly away from the nominal principal array directions.

A uniformly illuminated rectangle of width $w$ and height $h$ has field proportional to $\mathrm{sinc}(q_{x}w/2)\mathrm{sinc}(q_{y}h/2)$, with angular spatial wavenumbers $q_x,q_y$ measured in inverse length. The first envelope zeros give $w=2\pi /|q_{x,0}|$ and $h=2\pi /|q_{y,0}|$. Two identical rectangles with horizontal centre separation $s$ introduce an additional $\cos (q_{x}s/2)$ factor, so the interference period is $\Delta q_x=2\pi /s$.

The factors $2\pi$ depend on using angular spatial frequency; axes in cycles per length would use the reciprocal widths directly. Fourier light intensity alone cannot determine an overall aperture translation or distinguish a mask from its inverted copy. Extra assumptions about shape and coherent illumination are necessary for a unique geometrical reconstruction.

Extend $F_L(x)=x(L-|x|)$ from $[-L,L]$ to a periodic function of period $2L$, with $L>0$. It is odd and continuous, so its Fourier series contains only sine terms. Its Fourier coefficients areTwo applications of integration by parts prove the coefficient formula. The first has no endpoint contribution because $x(L-x)$ vanishes at both endpoints; the second leaves a multiple of $1-\cos (n\pi )$. Thus only odd frequencies survive, giving the displayed Fourier sine series. Its absolutely summable coefficients give uniform convergence, and Fourier orthogonality identifies its sum with $F_L$.

Taking $L=1$, ${\int }_{-1}^1{F}_1(x{)}^{2}dx=2{\int }_0^1{x}^2(1-x{)}^{2}dx=1/15$. Parseval's identity therefore gives

On $0\le x\le L$, the Fourier sine series of ${\sin }^2(\pi x/L)$ has zero even-index coefficients and odd-index coefficients $b_n=8/[\pi n(4-n^2)]$. Substitute $u=\pi x/L$ in the coefficient integral and use ${\sin }^{2}u\sin (nu)=\frac{1}{2}\sin (nu)-\frac{1}{4}\sin ((n+2)u)-\frac{1}{4}\sin ((n-2)u)$. For odd $n$, integrating gives $b_n=(2/\pi )[1/n-\frac{1}{2}(1/(n+2)+1/(n-2))]$; reflection about the midpoint makes the even coefficients zero. The $O(n^{-3})$ decay gives absolute and uniform convergence, and Fourier orthogonality identifies the sum. In the heat equation with zero endpoint temperatures, multiply each coefficient by $\exp [-\kappa (n\pi /L{)}^{2}t]$. With insulated endpoints, the same initial profile is instead just the constant and second cosine modes of a Fourier cosine series.

The period-$L$ extension of $f(x)=Kx/L$ on $-L/2<x<L/2$ has Fourier coefficients $a_0=a_n=0$ and $b_n=K(-1{)}^{n+1}/(\pi n)$. ConsequentlyThe Fourier series converges to the ramp away from its jumps and to zero at every jump, the average of the one-sided limits. For $K\ne 0$, Termwise differentiation of a Fourier series fails here as an ordinary convergent series: the differentiated summands have nondecaying amplitude $2K/L$. The classical derivative equals $K/L$ between jumps, whereas the periodic distributional derivative includes negative point masses of weight $K$ at the jumps.

On $(0,1)$, integration by parts gives the displayed Fourier sine series. For normalized Dirichlet eigenfunctions $y_i=\sqrt{2}\sin [(i+1)\pi x]$, indexed from $i=0$, it has the form $A[y_0+{\sum }_{i\ge 1}{c}_i{y}_i]$, with $A=4\sqrt{2}/{\pi }^3$ and $c_i=[1-(-1{)}^{i+1}]/[2(i+1{)}^3]$. Its Rayleigh quotient for $-d^2/d{x}^2$ is $10$, since ${\int }_0^1(u^{\prime }{)}^{2}dx=1/3$ and ${\int }_0^1{u}^{2}dx=1/30$. Reflection symmetry forces $c_1=0$, because that first excited eigenfunction is antisymmetric about $1/2$. Thus the first excited state's contribution to the error is exactly zero; the first nonzero correction is $c_2=1/27$.

An unwanted ellipse is an experimental warning: a finite-amplitude pendulum can develop additional orbit-axis precession even without Earth's rotation. A symmetrical, firm support and careful launch reduce this effect.^[ref] Distinguish that nonlinear effect from the tiny ellipse already present in the Exact linear Foucault pendulum trajectory. The linear calculation establishes a controlled approximation, not the complete motion of an arbitrary large swing.

Support anisotropy creates a second competing mechanism. As an illustrative modification, let the two horizontal restoring angular frequencies be ${\omega }_x$ and ${\omega }_y$. Even with $\alpha =0$, a diagonal release from rest givesWhen ${\omega }_x\ne {\omega }_y$, the relative phase changes on the scale $1/|{\omega }_x-{\omega }_y|$: the trace changes shape without any terrestrial Coriolis force. It need not be uniform precession of a Foucault pendulum. To cleanly resolve the terrestrial effect over times of order $1/|\alpha |$, a useful design target is $|{\omega }_x-{\omega }_y|\ll |\alpha |$. Near the equator that target becomes particularly demanding. Compare fast periods along perpendicular launch directions to look for this frequency splitting; repeat at several launch orientations to detect preferred suspension axes.

Other controls address distinct alternatives: shelter the apparatus from air currents, track the support against the building, measure both orbit axes, and compare early and late fitted directions as the amplitude decays. A sustaining drive, if used in a different apparatus, must supply mechanical energy without imposing an azimuthal bias; an undriven run provides a useful comparison. A change in swing direction alone cannot exclude these effects. Agreement of the signed rate with independently known geographical latitude and Earth's rotation, together with stable repeat launches and small competing effects, is the inference the experiment is designed to support.

For release from rest relative to the building, take $x(0)=A>0$, $y(0)=\dot{x}(0)=\dot{y}(0)=0$. The exact solution of the stated coupled linear differential equations isIndeed $u(0)=A$ and $\dot{u}(0)=i\alpha A$, so $\dot{z}(0)=0$. In the slowly rotating coordinates, the ellipse has semiaxes $A$ and $A|\alpha |/\beta$. A bob released from rest therefore does not describe a perfectly straight line even in this ideal linear model. Replacing the bracket by $\cos {\omega }_{0}t$ gives the familiar leading approximation but introduces a spurious initial transverse velocity $-i\alpha A$. The exact expression keeps the initial condition and separates that tiny orbit width from the slow rotation of its long axis.

For the illustrative $\ell =20\mathrm{m}$ and $g=9.81m/{\mathrm{s}}^2$, the fast period $2\pi /\beta$ is approximately $8.9714\mathrm{s}$. At $45^{\circ }$ north, $|\alpha |/\beta \simeq 7.3624\times 10^{-5}$: the transverse semiaxis is too small to distinguish at the figure's scale. Nevertheless the long-axis orientation changes by about $-63.8138^{\circ }$ in six hours. Changing the sign of geographical latitude conjugates the entire solution, reflecting the trajectory across the eastward axis; setting $\alpha =0$ gives $z=A\cos {\omega }_{0}t$ exactly. These are stronger checks than merely drawing a rotating line.

The coupled equations give an additional independent constraint. Multiplying them by $\dot{x}$ and $\dot{y}$ and adding cancels the Coriolis force's contributions:This is mechanical energy per unit mass in the reduced model; precession does not require a loss of mechanical energy. Also, direct differentiation givesFor this release the constant is $\alpha A^2$. The exact trajectory satisfies both identities. They test the coupled dynamics separately from the visible slow rotation and express the fact that the Coriolis force does no work.