SCIENCE · INVESTIGATE AND EXPLAIN

Derive the aperture transform and its central limit

Fraunhofer diffraction adds contributions from every point of an aperture with position-dependent phase. For uniform monochromatic illumination and a scalar far-field approximation, a point x across a slit contributes exp(i2πux), where u=sinθ/λ. This spatial frequency has inverse-length units; it is not automatically an angle.

Learning outcome: Integrate cosine across the aperture independently, verify the central limit and sign of successive amplitude lobes, then compare width changes without confusing normalized shape with power.

Starting recap

Retain complex amplitude before intensity, definite-integral boundaries, normalization, calibration covariance and unused evidence.

All lessons are open directly. Your saved experiments remain when you leave and return.

Try

At w=2 and u=0.25, πwu=π/2. Field amplitude is 2/π≈0.636620 and intensity 4/π²≈0.405285. The first zero lies at u=0.5 and the central lobe spans width 1 in inverse reference length.

Open the experiment

Predict

Why is the central intensity finite even though the quotient contains u in its denominator?

Your prediction and reason

Experiment

Integrate cosine across the aperture independently, verify the central limit and sign of successive amplitude lobes, then compare width changes without confusing normalized shape with power.

Integrate a rectangular aperture amplitude

Worked teaching model · synthetic values · Uniform scalar aperture, normal monochromatic illumination, Fraunhofer observation. Normalized field is (1/w)∫ exp(i2πu x)dx over −w/2..w/2. Central normalization removes width-dependent throughput; it is not total collected power.

Slit width: 2 reference length Spatial frequency sinθ/λ: 0.25 inverse reference length

Reset keeps your prediction, captures and practice history.

Integrate the field across the aperture before taking its squared magnitude; the central limit is exactly one.

Uniform scalar aperture, normal monochromatic illumination, Fraunhofer observation. Normalized field is (1/w)∫ exp(i2πu x)dx over −w/2..w/2. Central normalization removes width-dependent throughput; it is not total collected power.

The vertical axis is rescaled to this case’s actual range; equal display height does not imply equal optical power. The finite curve is illustrative; the ledger preserves the exact case values and stated numerical quadrature.

Original conditional optical ledger
Quantity	Value	Unit
Aperture phase half-span	1.57080	radians
Normalized signed real amplitude	0.636620	ratio
Normalized intensity	0.405285	ratio
First positive zero	0.500000	inverse reference length
Central lobe zero-to-zero width	1	inverse reference length

Virtual original records only. No laser, intense-light source, instrument alignment, physical apparatus operation or completed practical competence is supplied.

Integrate the field across the aperture before taking its squared magnitude; the central limit is exactly one.

Observation to attach to a capture

Captures preserve recipes and replay their calculated results. They are synthetic model evidence, not physical measurements.

Understand

1. Establish the optical mechanism

Normalize the field by the aperture width w: A(u)=(1/w)∫ from −w/2 to w/2 exp(i2πux)dx. The odd sine contributions cancel, leaving sin(πwu)/(πwu). The intensity normalized to its own central value is A². Taking intensity before integrating would discard interference between different aperture locations.

At u=0, the defining integral is exactly one. The formula’s apparent 0/0 is removable because sin z/z tends to one. Nonzero zeros satisfy wu=m for integer m, so the first positive zero is 1/w and the central zero-to-zero width is 2/w. Negative field lobes represent a π phase change, not negative intensity.

This central normalization hides the change in throughput and central absolute intensity when width changes. Comparing two normalized curves does not establish equal transmitted power. The far-field model additionally requires a suitable observation distance or Fourier plane and an aperture sufficiently large relative to wavelength for scalar approximations; the dimensionless controls are not instrument setup instructions.

2. Work the original evidence

At w=2 and u=0.25, πwu=π/2. Field amplitude is 2/π≈0.636620 and intensity 4/π²≈0.405285. The first zero lies at u=0.5 and the central lobe spans width 1 in inverse reference length.

3. Predict and investigate

Why is the central intensity finite even though the quotient contains u in its denominator?

Integrate cosine across the aperture independently, verify the central limit and sign of successive amplitude lobes, then compare width changes without confusing normalized shape with power.

Transfer the derivation to a changed case

Use w=3 and u=0.5. Find amplitude, intensity and first positive zero.

Retain your prediction, intermediate reasoning and assumptions before opening public feedback.

Compare a worked response

The amplitude is −2/(3π)≈−0.212207, intensity 4/(9π²)≈0.0450316 and first zero 1/3. Squaring removes the field sign, not its physical phase meaning.

Primary mechanism context. Checked 31 October 2026. Original prose, records and diagrams; no copied source exercises or physical light-source operation. Retain first reasoning and public-feedback exposure in the course evidence workspace when registered.

Apply

Public practice checks a specific claim. It does not establish independent mastery.

Use w=3 and u=0.5. Find amplitude, intensity and first positive zero.

Your explanation and limitations

Continue

Completion requires a saved prediction, at least one capture and a reflection. Presence checks do not grade your reasoning.

Skipping retains experiments and awards no mastery.

Previous: Sum interface phases and the complete reflected series Next: Separate width from separation using two spatial scales

Save a science backup or import one

Science work is stored locally in this browser, separately from mathematics and AI. It does not sync automatically. Use this science-only backup; this checkout's older all-subject export does not include science.

A legacy file is available only when the complete snapshot fits the 36 MB limit. For larger snapshots, prepare a complete file set below; never discard notebooks to make a smaller file.

Choose a science backup (maximum 36 MB)

Matching imported notebook drafts replace their current drafts. Omitted lessons and known first attempts remain. Export your current work first if you want to retain both drafts.

Complete file sets and course backups

A file set can hold more work than one science file. Keep its manifest and every part together. Each file is limited to 36 MB. This format needs this version of LearnAI or a compatible newer version; older versions can use only the legacy single-file backup above.

Export scope

The set includes every saved science course and global continuation at this moment. Later edits need a new export. A conflicting tab exports its own visible work, which may differ from saved work in another tab.

Choose the manifest and all parts together

0 file(s) selected. File order and filenames do not determine identity. Missing, changed or conflicting parts block the entire restore; no good part is applied by itself.

Science storage: loading. No account or remote computation is required.