Diffraction around an occulter Hale COLLAGE (CU ASTR-7500) “Topics in Solar Observation Techniques” Spring 2016, Part 1 of 3: Off-limb coronagraphy & spectroscopy Lecturer: Prof. Steven R. Cranmer APS Dept., CU Boulder steven.cranmer@colorado.edu http://lasp.colorado.edu/~cranmer/ Lecture 4: Diffraction around an occulter

Brief overview Goals of Lecture 4: Understand why we need to block out the bright solar disk. Derive basic physics of diffraction around an “occulter.”

Intensity above the limb Visible-light emission from the solar corona: K-corona due to Thomson- scattering of free electrons F-corona due to scattering of inner heliospheric dust

Intensity above the limb Visible-light emission from the solar corona: K-corona due to Thomson- scattering of free electrons F-corona due to scattering of inner heliospheric dust Compare with sky brightness: typical hazy sky clear day sky (mountain-top?) eclipse totality sky

F corona = Zodiacal light

Separating the K & F coronae Near the Sun, K corona is linearly polarized; F corona is unpolarized (more about this soon…) K corona follows magnetic structure; F corona is ~spherically symmetric Original meanings: F = Fraunhofer, K = kontinuierlich (“continuous”) F-corona: dust is cold; scatters full solar spectrum, absorption lines & all. K-corona: free electrons are hot; Doppler broadening smears out lines in the spectrum.

How do we go about observing the corona?

Off-limb geometry The Sun’s disk subtends solid angle Ω, angular radius = 959” = 0.267o Our goal: observe light coming from larger elongation angles

Off-limb geometry For observations not too far from the Sun, “observing rays” are nearly parallel: x = Line of sight (LOS) impact parameter in the “plane of sky” (POS) heliocentric radius of any point along the LOS

What can an “artificial eclipse” achieve? i.e., why do we need to block out the Sun? Even if every ray is separable from every other ray, the huge dynamic range of intensities is usually too much for a detector to handle. Of course, the rays get mixed up by diffraction. geometrical optics physical optics ! Trade-offs: The more we occult the Sun, the better we can beat down diffraction. More over-occulting → we can’t see regions nearest to the Sun.

Basic optical paths (linear cut) We want to block rays from the solar disk, while letting in coronal light:

Basic optical paths (linear cut) In reality, some fraction of the disk light is diffracted around the occulter edge and be reflected by the mirror, thus contaminating the coronal light.

Diffraction Born & Wolf: Principles of Optics, Hecht: Optics, ASTR-5550, PHYS-4510 Interference of electromagnetic waves when they interact with “obstacles”

The Huygens-Fresnel principle Point-sources of radiation emit spherical wave-fronts. Huygens: Every point on a spherical wave-front (surface of constant phase) acts as a point-source for secondary spherical waves. Fresnel: E-field at any point can be constructed as the superposition of all “incoming” wave-fronts. Kirchhoff: Above is a direct consequence of the wave equation derivable from Maxwell’s equations. In vacuum, the wave-fronts coalesce in predictable ways…

Diffraction: aperture geometry Goal: Given E(x,y,z,t) at source P0 , compute E(x,y,z,t) at observation point P. Note: distances r & s vary a lot for different points Q inside the aperture; distances r′ & s′ are known & fixed.

Follow the waves: P0 → Q → P A spherical wave starts at P0 The field at point Q in the aperture (at distance r from P0) is: Collect time-dependence into amplitude:

Follow the waves: P0 → Q → P Fresnel found that the differential amount of electric field that reaches point P after passing through a differential area (dA) of the aperture is: i.e., But what is the normalization constant K ? Units: 1 / Length It tells us the relative efficiency of the secondary-wave generation. Two ways to derive it: Forget Huygens-Fresnel, use Maxwell’s eqns! Assume dA covers all space… EP = unobstructed E

Follow the waves: P0 → Q → P In any case, χ = “scattering angle” (prevents secondary waves going back to the source) In telescopes, these angles are usually small; assume χ ≈ 0 . If the angles are all small, then that means the aperture diameter (call it D) is << the distances r, r′, s, s′ Thus, can we can replace r & s by r′ & s′ ?

Follow the waves: P0 → Q → P In any case, χ = “scattering angle” (prevents secondary waves going back to the source) In telescopes, these angles are usually small; assume χ ≈ 0 . If the angles are all small, then that means the aperture diameter (call it D) is << the distances r, r′, s, s′ Thus, can we can replace r & s by r′ & s′ ? In denominator: Yes In exponent: No!

Follow the waves: P0 → Q → P We need to specify distances in terms of a known coordinate system…

Follow the waves: P0 → Q → P (x,y) tell us the location of point P (the detector) (ξ,η) tell us the location of point Q (“inside the aperture”) Remember: We’ve already assumed that D (the maximum extent of ξ and η) is << r′ & s′ Thus, all ratios above are << 1

Follow the waves: P0 → Q → P We’re getting closer to useful expressions… The exp argument depends on the difference between actual & reference paths:

Simplify some things Assume P0 is so astronomically distant that waves at Q are plane waves. The reference distance s′ (from aperture to detector) is often called R D R x ≈ Rθ and thus… Thus, if original plane wave was unobstructed, it would reach detector with:

Fraunhofer vs. Fresnel diffraction Fraunhofer: for “small” apertures, quadratic terms in ξ and η can be neglected. → eikf integral becomes a spatial Fourier transform of the aperture → appropriate for slits, pinholes, and finite-sized optics in telescopes Fresnel: need to keep all terms.

Fresnel diffraction: rectangular aperture Sneaky trick: take point P along the z-axis, but move around the aperture: x y ξ1 ξ2 η1 η2 Thus, we care only about for which

Fresnel integrals

Fresnel integrals “Cornu spiral”

Fresnel diffraction: straight-edge Expand 3 of the 4 sides of the rectangular aperture to infinity… x y ξ1 ξ2 η1 η2 v1 < 0 : observer (at x=0) in direct sunlight v1 > 0 : observer in the shadow of occulter

Fresnel diffraction: straight-edge

Fresnel diffraction: straight-edge If the calculation is re-done with a circular occulter or a circular aperture…

Far inside the shadow… Let’s work it out for a simple idea for a solar coronagraph… Visible light (λ = 500 nm) Typical distance between occulter & primary mirror (R = 2 m) Try a range of θ corresponding to coronal “elongation angles”

Far inside the shadow…

Far inside the shadow… What are our options? We have to move the occulter much further away (like the distance of the Moon?), or we need to experiment with other shapes (other than a straight-edge), or we need to just deal with the diffracted light after it enters the telescope. λ = 500 nm R = 2 m

Next time How do real coronagraphs solve (or at least minimize) the problem of diffraction?