Optics 066 · Diffraction, Fourier optics, and computational imaging

Schlieren Flow Viewer

A Toepler lens-type Schlieren bench turns a known transparent index field into directional grayscale: trace a point, rotate the opaque blade, change source size and expose saturation.

Selected object-point deflection

\theta_x,\theta_y

$0.50$

Directional focal shift / raw point intensity

q,I_p

$50\%$

Empty-field background / saturated sample fraction

I_{\rm bg},S_{\rm clip}

$0.00\pi$

Model regime $\text{valid model regime}$

Validated scientific apparatus

Schlieren Flow Viewer

Synthetic index field (named buttons below)

o

0

Signed refractive-index anomaly amplitude

\Delta n_0

-1\times10^{-5}

Uniform line-of-sight field thickness

L

40\,\mathrm{mm}

Transverse field-width scale

w

4\,\mathrm{mm}

Common focal length (moves conjugate hardware)

f

1000\,\mathrm{mm}

Source-image radius at knife plane

a

60\,\mu\mathrm m

Knife normal offset (positive darkens background)

b

0\,\mu\mathrm m

Knife sensitive normal (clockwise from object right)

\alpha

0^{\circ}

Knife: 0 removed, 1 inserted

k

1

Probe original object horizontal position

x_p

4\,\mathrm{mm}

Probe original object vertical position (positive down)

y_p

0\,\mathrm{mm}

Remove the knife: the transparent test field disappears. Insert it and invisible index gradients become bright and dark edges. Rotate the blade to reveal another direction; choose an annular field to turn its apparent shading without moving the field. Click the receiving image to follow one object point into the enlarged source-image cutoff. The blade is opaque and rejected light stops there. Source and cutoff positions, raw images and readouts share one geometric model; no automatic image normalization.

Prescribed static index fields, uniform thickness and straight-path geometric thin-screen approximation; uniform incoherent circular source and ideal unit-inverting relay. Not CFD, actual shocks, shadowgraph, BOS or density recovery. No diffraction, noise or expert certification; apparatus scales differ.

Start with the essentials

Focus question: How does a razor reveal invisible air—and why does rotating it erase one direction?
One-sentence intuition: Only the deflection component normal to the blade changes transmission: $q=f\boldsymbol\theta\cdot\hat{\mathbf e}_{\alpha}$ . Rotating the knife changes shading without changing density. A uniform index offset produces no contrast.

Core mathematical model

Straight-path index integration

\begin{aligned}n(x,y,z)&=n_0+\Delta n_0F(x,y)\\\boldsymbol\theta&\approx\frac{L\Delta n_0}{n_0}\nabla F\end{aligned}

Coordinates are millimetres: object right and down, propagation through the thickness. The small perturbation is collapsed to a thin screen; rays are not iteratively bent through a full volume.

Finite disk, exact geometric cutoff

\begin{aligned}u&=(q-b)/a\\I/I_0&=\frac12+\frac{\arcsin u+u\sqrt{1-u^2}}{\pi}\\&\qquad(-1<u<1)\end{aligned}

Integrate the accepted area of a uniform incoherent source image of radius a. Outside the interval transmission is zero or one. The blade blocks the lower normal-coordinate half-plane; positive knife offset reduces empty-field brightness.

Sensitivity and saturation

\begin{aligned}b=0:\quad I/I_0&\approx\frac12+\frac{2q}{\pi a}\\|q|&\ll a\\\left|q-b\right|&\ge a\Rightarrow I/I_0\in\{0,1\}\end{aligned}

The linear expression applies only near the centre, not as a global clipped ramp. Exact disk-area integration supplies the record at every setting; black or white regions no longer encode a unique gradient.

Two conjugacies, one relay

\begin{aligned}\mathbf r_K&=f(\boldsymbol\eta+\boldsymbol\theta)\\\mathbf r_i&=-\mathbf r_o,\qquad m=-1\\z_K&=2f,\quad z_i=4f\end{aligned}

The test plane is zero; collector and relay lenses are at one and three focal lengths. A source coordinate becomes illumination angle eta. Changing the common focal length moves the hardware while maintaining unit-inverted imaging. Lens shapes are schematic, not an aberration model.

Common difficulties

A bright edge is not a hot spot or a velocity arrow

Typical misconceptionBrightness directly measures temperature or flow speed; a darker image always means a larger gradient.

Better mental modelThe record depends on directional integrated index gradients, source size and knife position. Once the source disk is wholly accepted or rejected, intensity saturates and loses gradient magnitude. A single view does not recover a three-dimensional density field or velocity.

Run the experiment

01
Scene 1: Invisible plume made visible
Remove then reinsert the knife. Click a bright or dark pixel and compare the source disk with the absorbing blade.
What to observe: Knife removed means unit intensity for this ideal in-focus thin screen; it is not a simulated shadowgraph.
02
Scene 2: Annular field: rotate the shading
Choose the annular field and rotate sensitivity by a quarter turn. Compare the same raw profiles, not automatically rescaled images.
What to observe: The angular source plane and image plane are different conjugates. The camera inverts both transverse object coordinates.
03
Scene 3: Smooth front: saturation is lost information
Select the smooth front; make the source small and the focal length long. Look for clipped black or white areas, then rotate the normal perpendicular to the front gradient.
What to observe: Smaller source radius gives greater small-signal sensitivity but a narrower unsaturated gradient range.