JPS6385501A - Fresnel zone plate - Google Patents

Abstract

PURPOSE:To increase the utilizing efficiency of X-rays and to shorten exposing time by providing multiple stages of Fresnel zones which condense light with diffractions of the higher order nearer the outer side from the center. CONSTITUTION:The n-th ring radius of the Fresnel zone plate where the m-th order diffracted light condenses is as expressed by the equation I if the radius of the n-th ring of the Fresnel zone plate 2 and the width of the aperture part thereof are designated respectively as rn, dn; the distance between an X-ray source 1 and the Fresnel zone plate 2 is designated as (a) and the distance between side plate 2 and a condensing point 3 as (b). In the equation, (m) denotes the order of diffractions. The equation II is obtd. by selecting the width dn<(m)> in the aperture of the n-th ring of the m-th order diffraction Fresnel zone plate in such a manner that the optical path difference between the X-ray passing the X-ray source 1-diffraction point 4-condensing point 3 and the X-ray passing the X-ray source 1-bottom end 5 of the n-th ring-condensing point 3 attains mlambdax/2. The multistage Fresnel zone plate is attained by taking the number (n) (radius rn<(m)>) of the zones until the min. manufacturable ring width is attained and increasing the order of diffractions for the larger aperture diameter than said aperture diameter.

Description

DETAILED DESCRIPTION OF THE INVENTION Field of the Invention The present invention relates to Fresnel zone plates used as active elements in the field of X-ray radiation.

P] Conventional technology Because X-rays are difficult to use for all refraction groups.

A Fresnel zone plate that makes full use of diffraction is used as an X-ray imaging element.

Fresnel zone plates are currently manufactured using three methods: mechanical marking using a ruling engine, electron beam linear drawing, and holographic exposure.

Fresnel zone plates require a ring width that is narrower toward the outside, and the outermost ring width is from λX/2 to λx/4 (
λX: used wavelength). Therefore, the diameter of a Fresnel zone plate for X-rays with a short wavelength (several to several hundred amperes) is determined by the minimum ring width that can be manufactured.

However, the minimum ring width that can be manufactured by any of the above methods is about 0.2 μm, and therefore only a Fresnel zone plate with a small diameter of about 0.1×1 could be manufactured. Due to the small diameter, the amount of X-rays that can be used is limited and the exposure time is long.

This has caused the inconvenience of deteriorating the flake.

The purpose of the present invention is to fully realize a large diameter Fresnel zone plate while keeping the ring width that can be manufactured at present.

2) First, we will explain the case where it is possible to accurately mark a ring at the wavelength λX of X-rays by using mechanical marking or electron beam direct writing.

In FIG. 1, n'tr of Fresnel zone plate 2
The radius of the eye ring and the width of the opening are rn and d, respectively.
n, X-ray ffJ 1 and Fresnel zone plate 2
Let the distance between the plate 2 and the condensing point (the position of the aberration-free imaging point of X-rays) 3 be b.

Generally, if this is a Fresnel zone plate of m-order diffraction, X-ray # ] −' Above the 8th ring @ (diffraction point)
4 - The optical path difference between the X-rays passing through the condensing point 3 and the X-rays passing through the X-ray source 1 - the upper end of the (n-1)th ring 6 - the condensing point 3t is mλX0 5 ears, a "+9-J Confucian Society 7-5 Ward+b" = mλ・(n
, m is an integer) ・・・・・・−・・・・・・・−・
・(1) However, o=O.

Equation (1) holds true for n from 1 to r+1, so these n
If we line up these equations and take the sum of each, we get the 9th order.

r+r −a −b = nrnlx −(2) From this.

This represents the entire radius of the n-th ring of the Fresnel zone plate that focuses the 9 mm order light, and 1 represents the diffraction order.

X-ray source l - Diffraction point 4 - Xi passing through focal point 3 and X-ray source] -
Width of the opening of the nth ring of the 1m-order diffraction Fresnel zone plate dnfp>
).

becomes.

A multistage 7-Renel zone plate is constructed by increasing the number of zones n (radius,
Increasing the aperture beyond n(m) and υ is attempted to be achieved by increasing the number of diffraction orders.

Therefore, if the ring can be scored accurately, and the minimum manufacturable ring width dsis is given, the fabrication of the multi-step Fresnel zone plate is as follows.

tzu, 1st order diffraction Frenet ~ zone play) (m = 1) and radius r. (1), Calculate the ring width d n (11 and increase ni, bH1''>'Exceed> dn I +'
? Find the number of zones nI such that '. In the first-order diffraction, the radius ".!(1)" is the production limit, so for the ninth order of diffraction m f 2, dn2(2
) >'xiyr>dn2-J'4'. Find 12 k such that a diffraction)
, radial force r. , (1) Color r-U (2J size and 9 Shio times R-(2)≠S-melon, 3rd order diffraction up to r13(3)...- Fresnel zone is expressed by equation (3) and (4)
The radius r as given by Eq. ←) and ring width dn (m
), it is possible to increase the diameter of the Fresnel zone plate using higher-order diffraction.

Next, a case will be described in which a multi-stage Fresnel zone plate is manufactured by holographic exposure. When accurate marking lines are possible, there is no aberration in principle, but in holographic exposure, the exposure wavelength λ■ and the used wavelength λX are different, so
The resulting aberrations must also be taken into account. The factors that determine the maximum radius of the Fresnel zone plate in m-order diffraction include the minimum ring width that can be manufactured, aberration at the focal point, and conservation of the diffraction order m.

First, as shown in Fig. 2, when the maximum radius γ0 that can be manufactured in the previous diffraction order (m-1) is given, Lower end of n@th ring 6-
Unless the optical path difference at the focal point 3 becomes +71λX, find the radius r! of the first fringe of the next (m-th order iPI Lite).

σ+ 7.2 Tonishi β♀r, "-q-Rikishoji) 7-~λ・
Yoji・・・・・・・・・・・・-(5) Next, when making this ring with the entire exposure wavelength λH.

io and f+ correspond to the (n-1) i-th and n-th radii I n -+ + I n at λ11, respectively, and the convergence point 7 - diffraction point 4 - convergence point 8 of the two spherical waves at the exposure wavelength
The condensing positions ξ and η of the spherical waves are determined so that the optical path difference between the convergence point 7, the lower end of the ring 6, and the convergence point 8 becomes λH.

F ear) depression π + national vote −ξ−η−(・−1)λ・Je”;−
'r, "+Pconn" -e -77= nλ, °=
-(6) It is sufficient to determine n, ξ, and η so as to satisfy this equation (6), but these cannot be determined uniquely.

For now, let's consider only the case of ζ2η as a first approximation.

become that way. However, (m) is the 0th order representing the diffraction order, and the maximum radius anvil in which the given diffraction order m is preserved.

As shown in Figure 3, the 7'C (11-1)th and n
th ring (radius rn-+.

The imaging positions (focus points) of the m-th and (m+1) next new party of X-ray λX by rn) are 9 and 10, respectively (the distance from the Fresnel zone plate 2 is bm and bm+1 respectively)
If so.

However, cx=p −az+rn−+” (>O) δ=5
7τ7;7. Now, the diffraction order m of the convergent light closest to the aberration-free condensing point 3 is bm++ (b (bm
==-・If defined by (9), then equation (8),
From equation (9), β = F7 is a Gaussian symbol.The outermost ring (radius r, ) made by the exposure wavelength λtl and the exposure distance ξ(m) If the total number of diffraction orders of the X-rays focused on is mN, then in order for this to be equal to the initially given diffraction order mVC, + rHnor rs+12<< a2
．． When approximated by b2.

・－・－・・・・・・－(111 becomes.

2 The radius of the n-th ring when exposed on the horn sphere a (ξ=η) is given by, where N is the number of rings ax when r is determined, and N is r N < rm a x < j N+1 When defined as .

becomes. Diffraction order m initially given from formulas 03 and 12
Find the maximum radius r9 that preserves.

is given by

Also, as shown in Figure 3, the aberration CN at the imaging plane is expressed as (-N = (
1-b””-m) rH・・・・・・・・・・・・・・・・
...Defined by Tomoe.

The maximum radius of the Fresnel zone plate in the m-th order of diffraction is determined by the equation +1) and the equation 02 from the conservation of diffraction orders.

Minimum swing width that can be suppressed by D type and further manufactured
Given in and the allowable aberration C at the imaging point, dmi
It is restricted so that n < dN, C > CN.

(e) Using 5.4 people as the wavelength λX used in the example, a=15Qjll
Assume that b=680 layers〇 First, the calculation result of the maximum radius t when accurate marking lines are possible is shown in FIG. However, the minimum line width that can be manufactured is dmin
was set to 0.2 μm. Next, the results of holographic exposure are shown in FIG. The exposure wavelength λH is He −(:d
4416 of lasers, using + dmin = 0
．． It was set to 14 μm and C=0.18M.

In addition, in the above-mentioned holographic exposure method, for the sake of simplicity, we considered the simplest case, but it is not necessary to use a spherical wave with ξ = η. Alternatively, the exposure wavefront etc. may be optimized.

(f) Effects According to the present invention, it is possible to increase the diameter of the Fresnel zone plate in the X-ray region.

This increases the utilization efficiency of X-rays, and can be expected to have effects such as shortening exposure time and improving μm swiftness.

[Brief explanation of the drawing]

Figure 1 is a diagram completely explaining the general m-order diffraction Fresnel p-zone plate, Figure 2 is a diagram explaining the m-order fresnel zone plate made by small holographic exposure, and Figure 3 is a diagram explaining the m-order fresnel zone plate made by holographic exposure. A diagram explaining image formation using a Fresnel zone plate, Figure 4 is an example of calculating the maximum radius that can be manufactured when accurate marking lines are possible, and Figure 5 is a calculation of the maximum radius when manufacturing using the holographic exposure method. This is an example. 1. Upper end of the ring 7.8 - Convergence point of two spherical waves at the exposure wavelength 9 - m due to the nth zone and the (row 1)th zone
Next time the new party's image location

Claims (1)

[Claims] A Fresnel zone plate characterized by being composed of multi-stage Fresnel zones that converge light with higher order diffraction from the center outward.