WO2022195805A1 - 顕微鏡システム - Google Patents
顕微鏡システム Download PDFInfo
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- WO2022195805A1 WO2022195805A1 PCT/JP2021/011097 JP2021011097W WO2022195805A1 WO 2022195805 A1 WO2022195805 A1 WO 2022195805A1 JP 2021011097 W JP2021011097 W JP 2021011097W WO 2022195805 A1 WO2022195805 A1 WO 2022195805A1
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- 238000001514 detection method Methods 0.000 claims abstract description 149
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- 238000003384 imaging method Methods 0.000 claims abstract description 64
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- 238000002834 transmittance Methods 0.000 description 97
- 238000010586 diagram Methods 0.000 description 71
- 238000004088 simulation Methods 0.000 description 33
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- G—PHYSICS
- G02—OPTICS
- G02B—OPTICAL ELEMENTS, SYSTEMS OR APPARATUS
- G02B21/00—Microscopes
- G02B21/06—Means for illuminating specimens
- G02B21/08—Condensers
- G02B21/082—Condensers for incident illumination only
- G02B21/084—Condensers for incident illumination only having annular illumination around the objective
-
- G—PHYSICS
- G02—OPTICS
- G02B—OPTICAL ELEMENTS, SYSTEMS OR APPARATUS
- G02B21/00—Microscopes
-
- G—PHYSICS
- G02—OPTICS
- G02B—OPTICAL ELEMENTS, SYSTEMS OR APPARATUS
- G02B21/00—Microscopes
- G02B21/0004—Microscopes specially adapted for specific applications
- G02B21/002—Scanning microscopes
- G02B21/0024—Confocal scanning microscopes (CSOMs) or confocal "macroscopes"; Accessories which are not restricted to use with CSOMs, e.g. sample holders
- G02B21/0032—Optical details of illumination, e.g. light-sources, pinholes, beam splitters, slits, fibers
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- G—PHYSICS
- G02—OPTICS
- G02B—OPTICAL ELEMENTS, SYSTEMS OR APPARATUS
- G02B21/00—Microscopes
- G02B21/06—Means for illuminating specimens
-
- G—PHYSICS
- G02—OPTICS
- G02B—OPTICAL ELEMENTS, SYSTEMS OR APPARATUS
- G02B21/00—Microscopes
- G02B21/36—Microscopes arranged for photographic purposes or projection purposes or digital imaging or video purposes including associated control and data processing arrangements
- G02B21/365—Control or image processing arrangements for digital or video microscopes
Definitions
- the present invention relates to microscope systems.
- the image of the object is acquired with the measurement optical system.
- An image calculation method is used to calculate what kind of image the object model image will be. Therefore, in this reconstruction method, two things, the measurement optical system and the image calculation method, are important.
- the optical system of a microscope can be used as the measurement optical system.
- a microscope optical system uses a halogen lamp or an LED to acquire an image of a specimen.
- Halogen lamps and LEDs are incoherent light sources.
- Illumination using an incoherent light source is classified into incoherent illumination, coherent illumination, and partial coherent illumination depending on the lighting conditions. These lights will be explained.
- Koehler illumination is used in microscopes.
- the light source is placed on the focal plane of the condenser lens, or an image of the light source is formed on the focal plane of the condenser lens.
- Light emitted from each point of the light source is converted into parallel rays by a condenser lens.
- the specimen is thus illuminated with a parallel beam.
- Changing the size of the light source changes the spatial coherence of the illumination light on the specimen plane. Imaging characteristics change as the spatial coherence of the illumination light changes.
- the light source can be regarded as a point light source. Illumination that irradiates a specimen with light from a point light source is called coherent illumination.
- Image computation methods can be classified into linear computation and nonlinear computation.
- Linear arithmetic considers only one scatter within the sample.
- Non-linear arithmetic takes into account not only single scattering but also multiple scattering.
- the linear operation uses the first-order Born approximation.
- Born approximation even if more than one scattering occurs in the specimen, the more than two scatterings are ignored.
- sample information and output information are determined in a one-to-one relationship. Therefore, the output information can be calculated analytically.
- the output information is, for example, an image of the specimen.
- the image of the object model reconstructed using linear arithmetic is obtained by deconvolving the measured image of the specimen with the point spread intensity distribution.
- Nonlinear calculation is a calculation method that takes into account the fact that multiple scattering occurs in a sample.
- One of the nonlinear operations is the beam propagation method.
- the beam propagation method replaces the object model with multiple thin layers. Then, the image of the object model is calculated by sequentially calculating the wavefront change when the light passes through each layer.
- the beam propagation method can calculate the image of the object model more accurately than the linear operation.
- Non-Patent Document 1 discloses a method for restoring the refractive index distribution of a specimen by optimization calculation. This technique utilizes the beam propagation method. Spatial coherent illumination is also used in image acquisition.
- a three-dimensional optical property is, for example, the refractive index distribution of an object.
- Non-Patent Document 1 illumination is performed with one luminous flux. In this case, it is necessary to change the illumination angle in order to illuminate the sample from multiple directions. Therefore, it takes time to acquire an image. As a result, it takes a long time from the start of image acquisition to the completion of object model estimation.
- the present invention has been made in view of such problems, and an object of the present invention is to provide a microscope system in which the time from the start of image acquisition to the completion of object model estimation is short.
- a microscope system includes: Equipped with an incoherent light source, a detection optical system, and an imaging device,
- An incoherent light source is a light source that emits light that is not temporally coherent
- the detection optical system is an optical system that forms an optical image of the specimen
- the imaging element receives an optical image of the specimen formed by the detection optical system.
- Coherent illumination is illumination with light that is spatially coherent
- the direction in which the specimen is irradiated with the light beam differs for each coherent illumination
- the light beams of the coherent illumination pass through different first regions, Each of the first regions satisfies the following condition (1), At least one of the intervals between two adjacent first regions is characterized by satisfying the following condition (2).
- LS is the area of the first region (unit: mm 2 ); PS is the area of the pupil of the detection optical system (unit: mm 2 ); d is the distance between two adjacent first regions (unit: mm); T is the diameter of the pupil of the detection optical system (unit: mm); is.
- a microscope system comprises: An incoherent light source, an illumination optical system, a detection optical system, and an imaging device
- An incoherent light source is a light source that emits light that is not temporally coherent
- the detection optical system is an optical system that forms an optical image of the specimen
- the imaging element receives an optical image of the specimen formed by the detection optical system
- Coherent illumination is illumination with light that is spatially coherent
- the direction in which the specimen is irradiated with the light beam differs for each coherent illumination
- the light beams of the coherent illumination are positioned in different second regions, Each of the second regions satisfies the following condition (6), At least one of the intervals between two adjacent second regions is characterized by satisfying the following condition (7).
- LS' is the area of the second region (unit: mm 2 ); PS′ is the pupil area of the illumination optical system (unit: mm 2 ); d' is the distance between two adjacent second regions (unit: mm); T′ is the diameter of the pupil of the illumination optical system (unit: mm); is.
- FIG. 1 shows the pupil plane of the detection optical system of a 1st example. It is a figure which shows the pupil plane of the detection optical system of a 2nd example. It is a figure which shows the pupil plane of the detection optical system of a 3rd example. It is a figure which shows the pupil plane of the detection optical system of a 4th example. It is a figure which shows the pupil plane of the detection optical system of a 5th example. It is a figure which shows the pupil plane of the detection optical system of a 6th example.
- FIG. 1st example It is a figure which shows the pupil plane of the detection optical system of a 2nd example. It is a figure which shows the pupil plane of the detection optical system of a 3rd example. It is a figure which shows the pupil plane of the detection optical system of a 4th example. It is a figure which shows the pupil plane of the detection optical system of a 5th example. It is a figure which shows the pupil plane of the detection optical system of a 6th example.
- FIG. 11 is a diagram showing a pupil plane of a detection optical system of a seventh example; It is a figure which shows the pupil plane of the detection optical system of the 8th example.
- FIG. 21 is a diagram showing a pupil plane of a detection optical system of a ninth example; It is a figure which shows the microscope system of this embodiment. It is a figure which shows the pupil plane of an illumination optical system. It is a figure which shows the microscope system of this embodiment.
- FIG. 10 is a diagram showing an aperture member; It is a figure which shows the microscope system of this embodiment. 4 is a flow chart of a first simulation; It is a figure which shows the optical system used by simulation.
- FIG. 10 is a diagram showing an image of an estimated sample; FIG.
- FIG. 4 is a diagram illustrating correction of a wavefront
- FIG. 10 is a diagram showing the gradient of a sample
- FIG. 10 is a diagram showing the gradient of a sample
- It is a figure which shows the microscope system of this embodiment. It is a figure which shows a picked-up image.
- 10 is a flow chart of a second simulation
- 10 is a flow chart of a second simulation
- It is a figure which shows the optical system used by simulation.
- FIG. 4 is a diagram showing a wavefront at an acquisition position of a photographed image and a wavefront at an imaging plane
- FIG. 10 is a diagram showing an image of an estimated sample
- FIG. 4 is a diagram illustrating correction of a wavefront
- FIG. 2 shows the sample gradient and wavefront propagation
- Fig. 2 shows the sample gradient and wavefront propagation
- FIG. 10 is a diagram showing the gradient of a sample
- FIG. 10 is a diagram showing the gradient of a sample
- FIG. 11 shows an aperture member and a reconstructed putative sample
- FIG. 10 is a diagram showing images of the aperture member and the specimen in the measurement of the second example; It is a figure which shows the estimation sample of a 2nd example.
- a thin specimen is called a "thin specimen”
- a thick specimen is called a "thick specimen”.
- the microscope system of this embodiment includes an incoherent light source, a detection optical system, and an imaging device.
- An incoherent light source is a light source that emits light that is not temporally coherent.
- the detection optical system is an optical system that forms an optical image of the sample, and the imaging element receives the optical image of the sample formed by the detection optical system.
- Multiple coherent illuminations are simultaneously provided at the specimen by light emitted from an incoherent light source, where coherent illumination is illumination with light that is spatially coherent. The direction in which the specimen is irradiated with the light flux differs for each coherent illumination. On the pupil plane of the detection optical system, the coherent illumination light beams pass through different first regions.
- the first area is an area of the light flux passing through the pupil plane of the detection optical system, each of the first areas satisfies the following condition (1), and at least one of the intervals between the two adjacent first areas is The interval satisfies the following condition (2).
- LS is the area of the first region (unit: mm 2 ); PS is the area of the pupil of the detection optical system (unit: mm 2 ); d is the distance between two adjacent first regions (unit: mm); T is the diameter of the pupil of the detection optical system (unit: mm); is.
- FIG. 1 is a diagram showing a microscope system and captured images of this embodiment.
- FIG. 1(a) is a diagram showing a microscope system.
- FIG.1(b) is a figure which shows a picked-up image.
- the microscope system 1 includes an incoherent light source 2, a detection optical system 3, and an imaging device 4.
- the incoherent light source 2 is a light source that emits light that is temporally incoherent. Coherent illumination is performed on the sample 5 by light emitted from an incoherent light source. Coherent illumination is illumination with light that is spatially coherent.
- each of the plurality of light beams is irradiated onto the sample 5 from different directions.
- the luminous flux L1 and the luminous flux L2 are illustrated.
- a luminous flux L1 and a luminous flux L2 are luminous fluxes emitted from the incoherent light source 2 to irradiate the sample 5 .
- the luminous flux L1 and the luminous flux L2 are mutually independent luminous fluxes, and the sample 5 is irradiated with them at the same time.
- the direction of the light flux L1 with respect to the specimen 5 and the direction of the light flux L2 with respect to the specimen 5 are different.
- the luminous flux L1 and the luminous flux L2 are respectively coherent illumination light fluxes.
- Specimen 5 is a thin specimen.
- the focal position Fo of the detection optical system 3 is positioned inside the sample 5 .
- the detection optical system 3 is an optical system that forms an optical image 5' of the specimen 5. Light emitted from the specimen 5 is condensed on the imaging plane IP by the detection optical system 3 . An optical image 5' is formed on the imaging plane IP.
- the imaging plane of the imaging device 4 is positioned on the imaging plane IP.
- the imaging element 4 receives an optical image 5 ′ of the specimen 5 formed by the detection optical system 3 .
- An image of the optical image 5 ′ is acquired by the imaging device 4 .
- the photographed image I mea (r) shown in FIG. 1B is obtained. r indicates the two-dimensional coordinates of (x, y).
- the specimen 5 is a thin specimen, one photographed image is acquired. Therefore, the detection optical system 3 and the imaging element 4 do not move in the optical axis direction. Also, the sample 5 does not move in the optical axis direction.
- the luminous flux emitted from the sample 5 reaches the pupil position Pu of the detection optical system 3 .
- the sample 5 is simultaneously irradiated with a plurality of light beams. Therefore, a plurality of light beams reach the pupil position Pu at the same time. Moreover, each of the plurality of light beams is irradiated onto the sample 5 from different directions. Therefore, at the pupil position Pu, a plurality of light beams pass through different regions.
- FIG. 2 is a diagram showing the pupil plane of the detection optical system.
- the first area is the area of the light flux passing through the pupil plane of the detection optical system.
- the number of first regions on the pupil plane 10 of the detection optical system 3 is eight.
- the first area 11 is an area through which the luminous flux L1 passes through the pupil plane 10 .
- the first area 12 is an area through which the luminous flux L2 passes through the pupil plane 10 .
- each of the first regions satisfies the following condition (1).
- LS is the area of the first region (unit: mm 2 );
- PS is the area of the pupil of the detection optical system (unit: mm 2 ); is.
- Each of the multiple luminous fluxes is an independent luminous flux. Therefore, the emission positions of the light beams in the light source are different for each light beam.
- a light beam emitted from the light source is applied to the sample. At this time, it is preferable that the sample is irradiated with one wavefront in the irradiation of the light flux emitted from one emission position.
- the microscope system of the present embodiment can acquire a captured image I mea (r).
- the captured image I mea (r) can be used, for example, to estimate the refractive index distribution. Estimation of the refractive index distribution will be described later.
- a partially coherent imaging beam propagation method is used to compute a computed image from the estimated sample.
- the parameter (refractive index distribution) of the estimated sample is changed by the gradient descent method or the like so that the calculated image approaches the captured image I mea (r).
- the computation time is proportional to the number of wavefronts irradiated to the specimen. If the upper limit of condition (1) is exceeded, a plurality of wavefronts will be applied to the specimen by irradiation of the light flux emitted from one emission position. Therefore, the calculation time becomes too long.
- At least one of the intervals between the two adjacent first regions satisfies the following condition (2).
- d is the distance between two adjacent first regions (unit: mm);
- T is the diameter of the pupil of the detection optical system (unit: mm); is.
- An optical image is formed by the light flux in one first area, and an optical image is formed by the light flux in the other first area. Since the angles of incidence on the sample are different for the two light beams, the two optical images are also different.
- half of the first area preferably satisfies condition (2).
- condition (2) When half of the first region satisfies condition (2), mutually different information can be obtained from two adjacent optical images, and a large number of mutually different information can be obtained.
- condition (2) When estimating the refractive index distribution, it can be estimated with relatively high accuracy.
- the microscope system of this embodiment preferably satisfies the following condition (3).
- LSi is the area of the i-th first region (unit: mm 2 );
- PS is the area of the pupil of the detection optical system (unit: mm 2 );
- n is the number of first regions; is.
- condition (3) is the total area obtained by adding the respective areas of the first region.
- some of the first regions are located within the first annular region, and the first annular region is an area having a radius of 50% or more of the pupil region of the detection optical system.
- the first annular region is an area having a radius of 50% or more of the pupil region of the detection optical system.
- FIG. 3 is a diagram showing the pupil plane of the detection optical system of the first example.
- FIG. 3 shows a first region in the pupil plane 10 of the detection optics.
- the first area 20 and the first area 21 are positioned on the circumference.
- the first area 20 is located outside the circumference where the first area 21 is located. Either one of the first area 20 and the first area 21 is sufficient.
- one circle is drawn with a dashed line.
- the radius of the circle is 50% of the radius of pupil 10 .
- the area outside the circle is the first annular area 30 .
- the first annular region 30 is a region with a radius of 50% or more.
- the first region 20 and the first region 21 are located in the first annular region 30 .
- the first example satisfies the conditions (1), (2) and (3).
- relatively fine structures can be estimated.
- condition (2) The more first regions that satisfy condition (2), the shorter the calculation time. Even if some of the first regions do not satisfy the condition (2), the computation time is shortened. It is desirable that more than half of the first regions satisfy condition (2).
- the first regions are arranged so as to form a double circle within the first annular region.
- a double circle is formed in the first annular region 30 by the first region 20 and the first region 21 . Therefore, a large number of mutually different information can be acquired.
- relatively fine structures can be estimated.
- some of the first regions are located within the second annular region, and the second annular region has a radius of 70% to 90% of the pupil region of the detection optical system.
- a region is preferred.
- FIG. 4 is a diagram showing the pupil plane of the detection optical system of the second example.
- FIG. 4 shows a first region in the pupil plane 10 of the detection optics.
- the first area 20 and the first area 21 are positioned on the circumference.
- the first area 20 is located outside the circumference where the first area 21 is located. Only the first region 20 may be present and the first region 21 may be absent.
- the radius of the inner circle is 70% of the radius of pupil 10 .
- the radius of the outer circle is 90% of the radius of pupil 10 .
- a region sandwiched between the two circles is the second annular region 31 .
- the second annular region 31 is a region with a radius of 70% to 90%.
- the first region 20 is located in the second annular region 31.
- the second example satisfies the conditions (1), (2) and (3).
- relatively fine structures can be estimated. Some of the first regions do not have to satisfy condition (2). It is desirable that more than half of the first regions satisfy condition (2).
- some of the first regions are located within the third annular region, and the third annular region has a radius of 50% to 70% of the pupil region of the detection optical system. A region is preferred.
- FIG. 5 is a diagram showing the pupil plane of the detection optical system of the third example.
- FIG. 5 shows a first region in the pupil plane 10 of the detection optics.
- the first area 20 and the first area 21 are positioned on the circumference.
- the first area 20 is located outside the circumference where the first area 21 is located. Only the first region 21 may be present and the first region 20 may be absent.
- the radius of the inner circle is 50% of the radius of pupil 10 .
- the radius of the outer circle is 70% of the radius of pupil 10 .
- the area sandwiched between the two circles is the third annular area 32 .
- the third annular region 32 is a region with a radius of 50% to 70%.
- the first region 21 is located in the third annular region 32.
- the third example satisfies the conditions (1), (2) and (3).
- a relatively fine structure can be estimated. Some of the first regions do not have to satisfy condition (2). It is desirable that more than half of the first regions satisfy condition (2).
- some of the first regions are located within the first circular region, and the first circular region is the central region of the pupil region of the detection optical system relative to the first annular region. is preferably
- FIG. 6 is a diagram showing the pupil plane of the detection optical system of the fourth example.
- FIG. 6 shows a first region in the pupil plane 10 of the detection optics.
- the first area 20 and the first area 40 are positioned on the circumference.
- the first area 20 is positioned outside the circumference of the circumference where the first area 40 is positioned.
- one circle is drawn with a dashed line.
- the area outside the circle is the first annular area 30 .
- the area inside the circle is the first circular area 50 .
- the first circular area 50 is an area closer to the center than the first annular area.
- the first area 40 is located in the first circular area 50.
- the fourth example satisfies the conditions (1), (2) and (3).
- the fourth example not only relatively fine structures but also relatively coarse structures can be estimated.
- Some of the first regions do not have to satisfy condition (2). It is desirable that more than half of the first regions satisfy condition (2).
- the first regions are arranged so as to form a circle within the first circular region.
- a circle is formed in the first circular area 50 by the first area 40 .
- relatively coarse structures can be estimated.
- the first regions are located within the second circular region, and the second circular region is a region having a radius of 50% or less of the pupil region of the detection optical system. is preferred.
- FIG. 7 is a diagram showing the pupil plane of the detection optical system of the fifth example.
- FIG. 7 shows a first region in the pupil plane 10 of the detection optics.
- the first area 40 and the first area 41 are positioned on the circumference.
- the first area 41 is positioned outside the circumference of the circumference where the first area 40 is positioned. Either one of the first area 40 and the first area 41 is sufficient.
- one circle is drawn with a dashed line.
- the radius of the circle is 50% of the radius of pupil 10 .
- the area inside the circle is the second circular area 51 .
- the second circular area 51 is an area with a radius of 50% or less.
- the first area 40 and the first area 41 are located in the second circular area 51 .
- the fifth example satisfies the conditions (1), (2) and (3).
- a relatively rough structure can be estimated. Some of the first regions do not have to satisfy condition (2). It is desirable that more than half of the first regions satisfy condition (2).
- the first regions are arranged so as to form a circle within the second circular region.
- a circle is formed in the second circular area 51 by the first area 40 and the first area 41 . Therefore, a large number of mutually different information can be obtained.
- relatively coarse structures can be estimated.
- some of the first regions are located within the fourth annular region, and the fourth annular region has a radius of 30% or more to 50% of the pupil region of the detection optical system. is preferably in the region of
- FIG. 8 is a diagram showing the pupil plane of the detection optical system of the sixth example.
- FIG. 8 shows a first region in the pupil plane 10 of the detection optics.
- the first area 40 and the first area 41 are positioned on the circumference.
- the first area 41 is positioned outside the circumference of the circumference where the first area 40 is positioned. Only the first region 41 may be present, and the first region 40 may be absent.
- the radius of the inner circle is 30% of the radius of pupil 10 .
- the radius of the outer circle is 50% of the pupil 10 radius.
- the area between the two circles is the fourth annular area 52 .
- the fourth annular region 52 is a region with a radius of 30% to 50%.
- the first region 41 is located in the fourth annular region 52.
- the sixth example satisfies the conditions (1), (2) and (3).
- a relatively rough structure can be estimated. Some of the first regions do not have to satisfy condition (2). It is desirable that more than half of the first regions satisfy condition (2).
- the microscope system of this embodiment some of the first regions are located within the third circular region, and the third circular region is a region having a radius of 30% or less of the pupil region of the detection optical system. is preferred.
- FIG. 9 is a diagram showing the pupil plane of the detection optical system of the seventh example.
- FIG. 9 shows the first region in the pupil plane 10 of the detection optics.
- the first area 40 and the first area 41 are positioned on the circumference.
- the first area 41 is positioned outside the circumference of the circumference where the first area 40 is positioned. Only the first region 40 may be present, and the first region 41 may be absent.
- one circle is drawn with a dashed line.
- the radius of the circle is 30% of the radius of pupil 10 .
- the area inside the circle is the third circular area 53 .
- the third circular area 53 is an area with a radius of 30% or less.
- the first area 40 is located in the third circular area 53.
- the seventh example satisfies the conditions (1), (2) and (3).
- a relatively rough structure can be estimated. Some of the first regions do not have to satisfy condition (2). It is desirable that more than half of the first regions satisfy condition (2).
- the pupil of the detection optical system is divided into four fan shapes with the same central angle, it is preferable that one of the first regions is located in all four fan shapes.
- FIG. 10 is a diagram showing the pupil plane of the detection optical system of the eighth example.
- FIG. 10 shows the first region in the pupil plane 10 of the detection optics.
- the first regions 60 are arranged in a grid pattern.
- FIG. 10 two orthogonal straight lines are drawn.
- the two straight lines divide the pupil plane 10 into a first fan-shaped region 70 , a second fan-shaped region 71 , a third fan-shaped region 72 and a fourth fan-shaped region 74 .
- the central angle in one fan-shaped region is equal to the central angle in the other fan-shaped region.
- the first regions 60 are located in all of the first fan-shaped region 70, the second fan-shaped region 71, the third fan-shaped region 72, and the fourth fan-shaped region 74. .
- the eighth example satisfies the conditions (1), (2) and (3).
- the refractive index distribution in the eighth example, not only relatively fine structures but also relatively coarse structures can be estimated.
- Some of the first regions do not have to satisfy condition (2). It is desirable that more than half of the first regions satisfy condition (2).
- the microscope system of this embodiment it is preferable that some of the first regions are paired across the center of the pupil of the detection optical system.
- FIG. 11 is a diagram showing the pupil plane of the detection optical system of the ninth example.
- FIG. 11 shows the first area on the pupil plane 10 of the detection optical system.
- the first area 80 and the first area 81 are positioned on the circumference.
- the first area 80 and the first area 81 form a pair across the center C of the pupil of the detection optical system.
- the ninth example satisfies the conditions (1), (2) and (3).
- a relatively rough structure can be estimated. Some of the first regions do not have to satisfy condition (2). It is desirable that more than half of the first regions satisfy condition (2).
- the number of first regions is smaller than in the seventh example.
- it can be estimated in a shorter time than the estimation time in the seventh example.
- each of the first regions preferably satisfies the following condition (4).
- the optical image becomes dark.
- the SN in the captured image I mea (r) deteriorates.
- the captured image I mea (r) can be used, for example, to estimate the refractive index distribution. If the SN in the captured image I mea (r) deteriorates, the estimation accuracy deteriorates.
- At least one of the intervals between the two adjacent first regions preferably satisfies the following condition (5).
- d is the distance between two adjacent first regions (unit: mm);
- T is the pupil diameter of the detection optical system (unit: mm) is.
- condition (2) It is desirable that the first region that satisfies condition (2) also satisfies condition (5).
- the microscope system of this embodiment preferably satisfies the following condition (A). 4 ⁇ n ⁇ 100 (A) here, n is the number of first regions; is.
- the lower limit of condition (A) If the lower limit of condition (A) is not reached, the number of first regions will decrease. Therefore, the optical image becomes dark. If the upper limit of condition (A) is exceeded, for example, it takes a long time to estimate the refractive index profile.
- the microscope system of this embodiment includes an incoherent light source, an illumination optical system, a detection optical system, and an imaging device.
- An incoherent light source is a light source that emits light that is not temporally coherent.
- the detection optical system is an optical system that forms an optical image of the sample, and the imaging element receives the optical image of the sample formed by the detection optical system.
- Multiple coherent illuminations are simultaneously provided at the specimen by light emitted from an incoherent light source, where coherent illumination is illumination with light that is spatially coherent.
- the directions in which the specimen is irradiated with the light beams are different for each of the coherent illuminations, and the light beams of the coherent illuminations are positioned in mutually different second regions on the pupil plane of the illumination optical system.
- Each of the second regions satisfies condition (6) below, and at least one interval between two adjacent second regions satisfies condition (7) below.
- FIG. 12 is a diagram showing the microscope system of this embodiment. The same numbers are assigned to the same configurations as in FIG. 1, and the description thereof is omitted.
- a microscope system 90 includes an incoherent light source 2 , an illumination optical system 91 , a detection optical system 3 and an imaging device 4 .
- the incoherent light source 2 When the incoherent light source 2 is arranged at a position away from the pupil plane PI of the illumination optical system 91, the light emitted from the incoherent light source 2 passes through the pupil plane PI. When the incoherent light source 2 is arranged at the position of the pupil plane PI, light is emitted from the pupil plane PI. The light emitted from the incoherent light source 2 passes through the illumination optical system 91 and irradiates the specimen 5 .
- FIG. 13 is a diagram showing the pupil plane of the illumination optical system.
- the number of second regions on the pupil plane 100 of the illumination optical system 91 is eight.
- the pupil plane 100 multiple lights are located in different regions. Each position of the second region is different from each other.
- a second area indicates the position of each of the plurality of lights.
- the second area 101 is an area where the light flux L1 passes through the pupil plane 100, or an area where the light flux L1 is generated.
- the second area 102 is an area through which the light flux L2 passes through the pupil plane 100, or an area where the light flux L2 is generated.
- the luminous flux L1 and the luminous flux L2 are respectively coherent illumination light fluxes.
- each of the second regions satisfies the following condition (6).
- LS' ⁇ PS' ⁇ 10 ⁇ 3 (6) here, LS' is the area of the second region (unit: mm 2 ); PS′ is the pupil area of the illumination optical system (unit: mm 2 ); is.
- condition (6) is the same as that of condition (1).
- At least one of the intervals between two adjacent second regions satisfies the following condition (7).
- d' is the distance between two adjacent second regions (unit: mm);
- T′ is the pupil diameter of the illumination optical system (unit: mm), l is.
- condition (7) is the same as that of condition (2).
- the detection optical system includes an objective lens and an imaging lens
- the illumination optical system includes a condenser lens
- the area of the second region is expressed by the following equation (8).
- the pupil diameter of the illumination optical system is preferably represented by the following equation (9).
- PS′ (FLcd ⁇ NA) 2 ⁇ (8)
- T′ FLcd ⁇ NA (9) here, FLcd is the focal length of the condenser lens (unit: mm), NA is the numerical aperture of the objective lens; is.
- the microscope system of this embodiment preferably further includes an aperture member.
- Each of the plurality of luminous fluxes is emitted from each of the plurality of independent regions on the predetermined surface.
- the predetermined surface is a surface perpendicular to the optical axis of the detection optical system and located on the opposite side of the specimen from the detection optical system.
- the aperture member is arranged on a predetermined plane and has a plurality of independent transmission areas.
- the transmissive regions are regions that transmit light, and each of the transmissive regions corresponds to each of the second regions.
- FIG. 14 is a diagram showing the microscope system of this embodiment. The same numbers are assigned to the same configurations as in FIG. 12, and the description thereof is omitted.
- the microscope system 110 includes an incoherent light source 2, an illumination optical system 91, an aperture member 111, a detection optical system 3, and an imaging element 4.
- each of the plurality of light beams is emitted from each of the plurality of regions on the predetermined plane.
- Each of the plurality of regions are independent of each other.
- the predetermined surface is a surface perpendicular to the optical axis of the detection optical system 3 and located on the opposite side of the specimen 5 from the detection optical system 3 .
- the pupil plane PI of the illumination optical system 91 is the predetermined plane.
- the opening member 111 is arranged on a predetermined surface.
- the aperture member 111 is arranged on the pupil plane PI.
- the microscope system 110 may include a processor 112. By providing the processor 112, for example, the refractive index profile of the sample 5 can be estimated.
- FIG. 15 is a diagram showing an opening member.
- the aperture member 111 has a plurality of independent transmissive regions 112 .
- Each of the multiple transmissive regions 112 is independent of each other.
- a light-shielding region 113 surrounds the transmissive region 112 .
- the transmissive area 112 is an area that transmits light. As the light passes through the transmissive region 112, the specimen 5 is irradiated with the light flux L1. As the light passes through another transmission region 112, the specimen 5 is irradiated with the light flux L2.
- the transmission area 112 is the area of light on the pupil plane PI.
- Each of the transmissive regions 112 corresponds to each of the second regions.
- the transmissive regions 112 are positioned on the circumference of the four circles.
- the four circles are the first circle, the second circle, the third circle, and the fourth circle from the outer edge of the opening member toward the center.
- a transmissive region 112a is located on the circumference of the first circle.
- a transmissive region 112b is located on the circumference of the second circle.
- a transmissive region 112c is positioned on the circumference of the third circle.
- a transmissive region 112d is positioned on the circumference of the fourth circle. If the refractive index distribution is estimated using the aperture member 111, it can be estimated with high accuracy.
- a transmission region is provided in the center of the aperture member 111 . Positioning of the opening member 111 can be facilitated by providing the transmission region in the center. However, a central transmissive region is not absolutely necessary.
- the second region is located on the pupil plane PI and the first region is located on the pupil plane Pu.
- the pupil plane PI can be conjugated with the pupil plane Pu.
- the second region is conjugated with the first region. Since the second area is the transmissive area of the aperture member, the first area can be regarded as the transmissive area of the aperture member.
- FIGS. 3-11 can be viewed as showing specific examples of aperture members.
- FIG. 3 shows a first example of the opening member.
- a first region 20 and a first region 21 represent the transmissive regions of the aperture member.
- Some of the aperture members are located in a region with a radius of 50% or more of the pupil region of the illumination optical system.
- Some of the transmissive regions of the aperture member are arranged to form double circles within the region with a radius of 50% or more.
- the first example satisfies the conditions (6) and (7).
- relatively fine structures can be estimated.
- FIG. 4 shows a second example of the opening member.
- some of the transmissive regions of the aperture member are located in a region with a radius of 70% to 90% of the pupil region of the illumination optical system.
- the second example satisfies conditions (6) and (7).
- relatively fine structures can be estimated.
- FIG. 5 shows a third example of the opening member.
- some of the transmissive regions of the aperture member are located in a region with a radius of 50% to 70% of the pupil region of the illumination optical system.
- the third example satisfies conditions (6) and (7).
- a relatively fine structure can be estimated.
- FIG. 6 shows a fourth example of the opening member.
- some of the transmission areas of the aperture member are located in the area closer to the center than the area with a radius of 50% or more in the pupil area of the illumination optical system.
- the fourth example satisfies the conditions (6) and (7).
- the refractive index distribution in the fourth example, not only relatively fine structures but also relatively coarse structures can be estimated.
- FIG. 7 shows a fifth example of the opening member.
- some of the transmissive regions of the aperture member are located in a region with a radius of 50% or less of the pupil region of the illumination optical system.
- the fifth example satisfies the conditions (6) and (7).
- a relatively rough structure can be estimated.
- FIG. 8 shows a sixth example of the opening member.
- some of the transmission regions of the aperture member are located in a region with a radius of 30% or more to 50% of the pupil region of the illumination optical system.
- the sixth example satisfies the conditions (6) and (7).
- a relatively rough structure can be estimated.
- FIG. 9 shows a seventh example of the opening member.
- some of the transmissive regions of the aperture member are located in regions with a radius of 30% or less in the pupil region of the illumination optical system.
- the seventh example satisfies the conditions (6) and (7).
- a relatively rough structure can be estimated.
- FIG. 10 shows an eighth example of the opening member.
- the pupil of the illumination optical system is divided into four fan shapes with the same central angle, one of the transmission regions of the aperture member is positioned in all four fan shapes.
- the eighth example satisfies the conditions (6) and (7).
- the eighth example not only relatively fine structures but also relatively coarse structures can be estimated.
- FIG. 11 shows a ninth example of the opening member.
- some of the transmissive regions of the aperture member are paired across the center of the pupil of the illumination optical system.
- the ninth example satisfies the conditions (6) and (7).
- a relatively rough structure can be estimated.
- the number of transmission regions of the opening member is smaller than that in the seventh example.
- it can be estimated in a shorter time than the estimation time in the seventh example.
- each of the plurality of light beams is preferably emitted from each of the plurality of independent regions on the predetermined plane.
- the predetermined surface is a surface perpendicular to the optical axis of the detection optical system and located on the opposite side of the specimen from the detection optical system.
- a plurality of incoherent light sources are arranged on a predetermined surface, and each of the incoherent light sources corresponds to each of the second regions.
- FIG. 16 is a diagram showing the microscope system of this embodiment. The same numbers are assigned to the same configurations as in FIG. 12, and the description thereof is omitted.
- the microscope system 120 includes an incoherent light source 121, an illumination optical system 122, a detection optical system 3, and an imaging device 4.
- each of the plurality of light beams is emitted from each of the plurality of regions on the predetermined plane.
- Each of the plurality of regions are independent of each other.
- the predetermined surface is a surface perpendicular to the optical axis of the detection optical system 3 and located on the opposite side of the specimen 5 from the detection optical system 3 .
- the pupil plane PI of the illumination optical system 122 is the predetermined plane.
- the incoherent light source 121 is arranged on a predetermined plane. In the microscope system 120, an incoherent light source 121 is placed in the pupil plane PI.
- the incoherent light source 121 has a plurality of independent light emitting regions 121a. Each of the plurality of light emitting regions 121a is independent of each other.
- the light emitting area 121a is an area that emits light.
- the specimen 5 is irradiated with the light beam L1 by the light emitted from the light emitting region 121a.
- the specimen 5 is irradiated with the light beam L2 by the light emitted from the other light emitting region 121a.
- the light emitting region 121a is a region of light on the pupil plane PI. Each of the light emitting regions 121a corresponds to each of the second regions.
- FIG. 15 can be regarded as a diagram showing an incoherent light source. 3-11 can also be viewed as examples of incoherent light sources. Therefore, the effects produced by the aperture members of the first to ninth examples are also produced when an incoherent light source is used.
- the microscope system of this embodiment preferably further includes a processor.
- the processor obtains the wavefront passing through the estimated specimen modeled on the specimen by forward propagation calculation for each luminous flux, calculates the intensity distribution at the imaging position of the detection optical system corresponding to the wavefront for each luminous flux, A calculated image is generated by summing the intensity distributions, and optimization processing is performed to reduce the difference between the calculated image and the measured image output from the imaging device, thereby reconstructing the estimated sample.
- the microscope system of this embodiment includes a processor. By providing the processor, the microscope system of this embodiment can reconstruct an estimated specimen. In reconstructing the estimated specimen, for example, the refractive index profile of the specimen is estimated.
- the wavefront passing through the estimated sample that models the sample is obtained by forward propagation calculation for each luminous flux.
- the intensity distribution at the imaging position of the detection optical system corresponding to the wavefront is calculated for each light beam.
- a calculated image is generated by summing the intensity distribution of each light flux. Optimization processing is performed to reduce the difference between the calculated image and the measured image output from the imaging device.
- the specimen 5 is positioned between the illumination optical system 91 and the detection optical system 3.
- the sample 5 is positioned between the illumination optical system 122 and the detection optical system 3.
- a measurement optical system is formed by the illumination optical system and the detection optical system.
- Light beams are simultaneously incident on the specimen 5 from a plurality of directions.
- a sample 5 is illuminated by light rays incident simultaneously from a plurality of directions.
- the illumination by each light beam is coherent illumination.
- An optical image 5 ′ of the specimen 5 is formed by the detection optical system 3 .
- the captured image I mea (r) shown in FIG. 1B is obtained.
- the captured image I mea (r) is input to the processor.
- the processor reconstructs the estimated sample using the captured image I mea (r). In reconstruction, a simulation is performed.
- the sample is a thin sample.
- the specimen is a thick specimen.
- the optical system used in the first simulation is the measurement optical system of the microscope system 110 shown in FIG.
- specimen 5 is a thin specimen.
- a focal position Fo of the detection optical system 3 is located inside the specimen 5 .
- the distance between the focal position Fo and the surface 5a of the sample 5 is ⁇ z1.
- the specimen 5 is a thin specimen, one photographed image is acquired. Therefore, the detection optical system 3 and the imaging element 4 do not move in the optical axis direction. Also, the sample 5 does not move in the optical axis direction.
- FIG. 17 is a flowchart of the first simulation. Before describing the flow chart, the estimated sample and wavefront will be described.
- FIG. 18 is a diagram showing an optical system used in the simulation.
- the optical system used in the simulation is the same as the measurement optical system that acquired the captured image I mea (r). In the simulation, instead of sample 5, estimated sample 130 is used.
- FIG. 18 shows the estimated sample 130, the wavefront f in m (r), the amplitude transmittance T s (r), the wave front g out m (r), the wave front u m (r) at the captured image acquisition position, and the imaging
- the wavefront u img m (r) in plane is shown.
- the aperture member 111 is arranged on the pupil plane PI.
- the second area is the transmission area of the opening member 111 .
- the second area can be regarded as a light source, since light is emitted from the transmissive area.
- FIG. 18 shows the second area from the first light source to the NLSth light source.
- the second area can be arranged at the pupil position of the illumination optical system 91 .
- the simulation includes the steps of estimating the estimated specimen, calculating the image of the estimated specimen, optimizing the refractive index distribution of the estimated specimen, updating the estimated specimen, and reconstructing the structure of the estimated specimen. and outputting.
- step S10 the number N_LS of the second regions is set.
- the transmission area of the aperture member is positioned in the pupil plane of the illumination optical system.
- the transmissive area is the second area. Therefore, in step S10 , the number of transmissive regions is set in NLS.
- Step S20 is a step of estimating an estimated sample. For the sample 5, one photographed image is acquired. Since the putative sample 130 is a thin sample, it can be considered as one thin layer. Therefore, the initial value of the amplitude transmittance is set once.
- step S20 an initial value is set for the amplitude transmittance T s (r) in the estimated sample 130.
- the estimated sample 130 is a modeled sample of the sample 5 . Therefore, it is preferable that the refractive index distribution of the specimen 5 can be used as the refractive index distribution of the estimated specimen 130 .
- the refractive index profile of the sample 5 cannot be accurately obtained from the captured image I mea (r). Therefore, the refractive index distribution of the estimated sample 130 must be estimated.
- the refractive index profile n s (r) of the putative sample 130 can be converted to amplitude transmittance T s (r). Therefore, in step S20, the initial value of the amplitude transmittance T s (r) in the estimated sample 130 is set.
- k 0 is 2 ⁇ / ⁇ with respect to the wavelength ⁇ of the illumination light
- n 0 is the refractive index of the medium dz is the thickness of the specimen, is.
- step S30 the value of variable m is initialized. Steps S41, S42, S43, S44, and S45, which will be described later, are executed for all light sources.
- the variable m represents the number of times these steps have been performed.
- Steps S40 and S50 are steps for calculating the image of the estimated specimen.
- the number of estimated sample images is the same as the number of captured images. Since the number of captured images is one, the number of images of the estimated sample is also one.
- Step S40 includes steps S41, S42, S43, S44, S45, S46, and S47.
- step S41 the wavefront f in m (r) incident on the estimated sample 130 is calculated.
- f in m (r) represents the wavefront of light emitted from the first to NLS -th light sources.
- the first to NLS -th light sources are positioned, and the first wavefront is the wavefront emitted from each light source. represents the wavefront.
- each of the second regions can be regarded as a point light source.
- illumination light Lm is emitted from the m-th light source.
- Illumination light Lm is incident on the estimated specimen 130 .
- the wavefront f in m (r) is represented by Equations (11) and (12).
- k is 2 ⁇ n 0 /r
- n 0 is the refractive index of the medium
- ⁇ x,m and ⁇ y,m are incident angles to the estimated sample, is.
- step S42 the wavefront g out m (r) emitted from the estimated sample 130 is calculated.
- the wavefront g out m (r) is given by equation (13).
- T s (r) is the amplitude transmittance of the estimated sample; is.
- Wavefront g out m (r) is the wave front after wave front f in m (r) has passed through estimated sample 130 . Since wavefront f in m (r) represents the first wavefront, wavefront g out m (r) represents the second wavefront.
- the wavefront g out m (r) can be directly calculated from the wave front f in m (r) as shown in equation (13).
- step S43 the wavefront u m (r) at the acquisition position of the captured image is calculated.
- the acquisition position of the photographed image is the focal position Fo of the detection optical system 3 on the specimen side when the photographed image is acquired.
- Wavefront u m (r) is represented by Equation (14).
- ⁇ z1 is the distance from the surface of the estimated specimen to the acquisition position of the photographed image
- ⁇ is the wavelength
- u is a two-dimensional notation of pupil plane coordinates ( ⁇ , ⁇ )
- F 2D is the two-dimensional Fourier transform
- F 2D ⁇ 1 is the two-dimensional inverse Fourier transform
- step S60 which will be described later, the residual is calculated.
- the captured image and the image of the estimated sample are used to calculate the residual.
- the distance between the focal position Fo and the surface 5a is ⁇ z1. Assuming that the sign of the distance measured in the traveling direction of light is positive, the captured image acquisition position is - ⁇ z1 away from the surface 5a.
- the captured image acquisition position is - ⁇ z1 away from the surface 130a of the estimated specimen 130.
- the wavefront at the acquisition position of the captured image is the wavefront at the position separated from the surface 130a by - ⁇ z1.
- the wavefront u m (r) in Equation (14) is a wavefront propagated by ⁇ z1 in the direction opposite to the traveling direction of light from the wavefront g out m (r). This wavefront is the wavefront at a position - ⁇ z1 away from the surface 130a. Therefore, the wavefront u m (r) in Equation (14) represents the wavefront at the captured image acquisition position.
- the acquisition position of the photographed image and the position of the surface 5a are different.
- sample 5 is a thin sample, the value of ⁇ z1 is very small. Therefore, the acquisition position of the captured image and the position of the surface 5a can be regarded as substantially the same.
- the estimated sample 130 is also a thin sample. Therefore, the position of the surface 130a and the position separated from the surface 130a by - ⁇ z1 can be regarded as substantially the same. That is, the position of the wavefront g out m (r) and the position of the wave front u m (r) can be regarded as substantially the same. In this case, instead of the wavefront u m (r), the wavefront g out m (r) can also be used.
- step S44 the wavefront u img m (r) on the imaging plane is calculated.
- the wavefront u m (r) is propagated to the imaging plane IP. At this time, it passes through the detection optical system 3 .
- the detection optical system 3 forms a Fourier optical system. Therefore, as shown in Equation (15), the wavefront u img m (r) on the imaging plane IP can be calculated using the wavefront u m (r) and the pupil function P(u) of the detection optical system. .
- step S45 the wavefront u img m (r) is squared.
- the wavefront u img m (r) represents the amplitude of the light. Therefore, the light intensity is calculated by squaring the wavefront u img m (r).
- 2 represents the light intensity distribution on the imaging plane IP. Assuming that the first intensity distribution is the light intensity distribution at the imaging position of the detection optical system,
- Wavefront f in m (r), wave front g out m (r), wave front u m (r), and wave front u img m (r) are the illumination light emitted from the m-th light source, that is, from one light source It represents the wavefront generated by the emitted illumination light.
- An estimated specimen image I est (r) is generated by illumination light emitted from all light sources. Therefore, the wavefronts f in m (r), g out m (r), u m (r), and u img m (r) must be determined for all light sources.
- step S46 it is determined whether or not the value of the variable m matches the number NLS of the second regions. If the determination result is NO, step S47 is executed. If the determination result is YES, step S50 is executed.
- step S47 If the judgment result is NO: m ⁇ N LS , If the determination result is NO, 1 is added to the value of the variable m in step S47. After step S47 ends, the process returns to step S41.
- step S47 the value of variable m is increased by one. Therefore, for another light source, the wavefront f in m (r) is calculated in step S41, the wavefront g out m (r) is calculated in step S42, the wavefront u m (r) is calculated in step S43, and the wavefront u m (r) is calculated in step S44. , and
- Steps S41, S42, S43, S44, and S45 are repeated until
- FIG. 19 is a diagram showing an image of an estimated specimen.
- the estimated sample image I est (r) is an image when the wavefront u img m (r) is obtained for all light sources. As shown in FIG. 19, the wavefront u img m (r) is calculated for each light source, and
- a residual is represented by Formula (17). As shown in Equation (17), the residual is calculated from the captured image I mea (r) and the estimated sample image I est (r).
- Equation (17) expresses the norm of the matrix.
- a norm is represented by Formula (18).
- step S70 the residual is compared with the threshold. If the determination result is NO, step S80 is executed. If the determination result is YES, step S110 is executed.
- step S80 the value of variable m is initialized. Steps S91 and S92, which will be described later, are executed for all light sources. The variable m represents the number of times these steps have been performed.
- Step S90 is a step of optimizing the refractive index distribution of the estimated sample.
- Step S90 includes steps S91, S92, S93, and S94.
- step S91 the wavefront u' m (r) is calculated.
- the captured image I mea (r) and the estimated sample image I est (r) are used.
- a wavefront u′ m (r) is the wavefront at the acquisition position of the captured image.
- Wavefront u' m (r) is represented by Equation (19).
- FIG. 20 is a diagram showing wavefront correction.
- FIG. 20(a) is a diagram showing a wavefront before correction emitted from an estimated sample.
- FIG. 20B is a diagram showing the wavefront before correction at the acquisition position of the captured image.
- FIG. 20(c) is a diagram showing the corrected wavefront at the acquisition position of the captured image.
- FIG. 20(d) is a diagram showing a corrected wavefront emitted from the estimated sample.
- the estimated specimen image I est (r) is calculated based on the wavefront u img m (r). Further, as shown in FIGS. 19 and 20(b), the wavefront u img m (r) is calculated based on the wave front u m (r).
- the amplitude transmittance T s (r) is used to calculate the wavefront u m (r).
- Amplitude transmittance T s (r) is the estimated amplitude transmittance. This amplitude transmittance T s (r) is different from the amplitude transmittance of the sample 5 at the first execution of step S90.
- the difference between the amplitude transmittance T s (r) and the amplitude transmittance of the sample 5 increases, the difference between the estimated sample image I est (r) and the captured image I mea (r) also increases. Therefore, the difference between the estimated specimen image I est (r) and the captured image I mea (r) can be regarded as reflecting the difference between the amplitude transmittance T s (r) and the amplitude transmittance of the specimen 5. can be done.
- Equation (11) the estimated specimen image I est (r) and the captured image I mea (r) are used to correct the wavefront u m (r).
- the corrected wavefront that is, the wavefront u' m (r) is obtained.
- a new amplitude transmittance T s (r) can be calculated by using the wavefront u′ m (r). Wavefront u′ m (r) is different from wavefront u m (r). Therefore, the new amplitude transmittance T s (r) differs from the amplitude transmittance when the wavefront u m (r) was calculated.
- the wavefront u′ m (r) can be used to calculate the amplitude transmittance T s (r).
- the wavefront g out m (r) is required to calculate the amplitude transmittance T s (r).
- the position of the wavefront u′ m (r) and the position of the wavefront g out m (r) are different. Therefore, in order to calculate the amplitude transmittance T s (r), the wavefront g′ out m (r) is required as shown in FIG. 20(d).
- Wavefront g′ out m (r) is represented by Equation (20). Since the wavefront u′ m (r) is the wavefront after correction, the wavefront g′ out m (r) is also the wavefront after correction.
- the acquisition position of the photographed image is a position away from the surface 130a by - ⁇ z1.
- the position of the surface 130a is ⁇ z1 away from the captured image acquisition position. Therefore, the wavefront at the position of the surface 130a becomes the wavefront at a position separated by ⁇ z1 from the captured image acquisition position.
- the wavefront g′ out m (r) in Equation (20) is a wavefront obtained by propagating the wavefront u′ m (r) by ⁇ z1 in the traveling direction of light. This wavefront is the wavefront at a position separated by ⁇ z1 from the shadow image acquisition position. Therefore, the wavefront g′ out m (r) in equation (20) represents the wavefront at the position of surface 130a.
- the wavefront at surface 130 a is the wavefront after f in m (r) has passed through putative sample 130 .
- f in m (r) represents the first wavefront.
- the wavefront g′ out m (r) represents the second wavefront.
- the value of ⁇ z1 is very small.
- the estimated sample 130 is a thin sample. Therefore, the acquisition position of the captured image and the position separated by ⁇ z1 from the acquisition position of the captured image can be regarded as substantially the same. That is, the position of the wavefront u′ m (r) and the position of the wavefront g out m (r) can be regarded as substantially the same. In this case, instead of the wavefront g'outm (r), the wavefront u'm ( r) can also be used.
- step S92 the sample gradient ⁇ T s m (r) is calculated.
- the sample gradient ⁇ T s m is expressed by Equation (21).
- Gradient descent for example, can be used to calculate the sample gradient ⁇ T s m (r).
- f * is the complex conjugate of f
- ⁇ is a normalization constant to prevent division by zero, is.
- the amplitude transmittance T s (r) is used to calculate the wavefront g out m (r).
- Amplitude transmittance T s (r) is the estimated amplitude transmittance. Therefore, this amplitude transmittance T s (r) differs from the amplitude transmittance of the sample 5 .
- the difference between the wavefront g out m (r) and the wave front g′ out m (r) can be regarded as reflecting the difference between the amplitude transmittance T s (r) and the amplitude transmittance of the sample 5.
- the wavefront f in m (r), the amplitude transmittance T s (r), the wavefront g out m (r), and the wave front g′ out m (r) are known. Therefore, as shown in Equation (21), using wavefront f in m (r), amplitude transmittance T s (r), wavefront g out m (r), and wave front g′ out m (r), The slope ⁇ T s m (r) can be calculated.
- FIG. 21 is a diagram showing the gradient of the sample.
- the sample gradient ⁇ T s m (r) obtained in step S92 represents the sample gradient in the illumination light emitted from one light source.
- the sample gradient ⁇ T s m (r) is determined by the illumination light emitted from all light sources. Therefore, the sample gradient ⁇ T s m (r) must be determined for all sources.
- step S93 it is determined whether or not the value of the variable m matches the number of light sources NLS . If the determination result is NO, step S94 is executed. If the determination result is YES, step S100 is executed.
- step S94 If the judgment result is NO: m ⁇ N LS , If the determination result is NO, 1 is added to the value of the variable m in step S94. After step S94 ends, the process returns to step S91.
- step S94 the value of variable m is incremented by one. Therefore, for another light source, the wavefront u′ m (r) is calculated in step S91, and the sample gradient ⁇ T s m (r) is calculated in step S92.
- Steps S91 and S92 are repeated until the sample gradients ⁇ T s m (r) are obtained for all light sources.
- FIG. 22 is a diagram showing sample gradients.
- sample gradients ⁇ T s m (r) are determined for all light sources.
- Step S100 is a step of updating the estimated samples.
- Equation (22) The updated amplitude transmittance T s (r) is represented by Equation (22). here, ⁇ is the correction factor for the slope of the sample, is.
- equation (23) can be used to further update the amplitude transmittance T s (r).
- step S100 ends, the process returns to step S30.
- Steps S30 to S100 are executed with the updated amplitude transmittance T s (r).
- the updated amplitude transmittance T s (r) gradually approaches the amplitude transmittance of the sample 5 . That is, the residual becomes smaller. Over time, the residual will be less than the threshold.
- the refractive index profile of the putative sample is calculated.
- the amplitude transmittance T s (r) obtained is identical or nearly identical to the amplitude transmittance of the sample 5 .
- a refractive index distribution n(r) can be obtained from the obtained amplitude transmittance T s (r) and Equation (1).
- the structure of the estimated specimen can be reconstructed.
- the reconstructed estimated sample structure can be output to, for example, a display device.
- the estimated sample 130 is a thin sample. In the first simulation, the structure of the thin specimen can be reconstructed.
- the amplitude transmittance T s (r) obtained in step S110 is the same as or substantially the same as the amplitude transmittance of the sample 5 .
- the refractive index distribution n(r) can also be regarded as the same as or substantially the same as the refractive index distribution of the specimen 5 . Therefore, the structure of the reconstructed putative sample 130 can be considered identical or nearly identical to the structure of the sample 5 .
- steps S40, S50, and S90 are repeatedly performed. As a result, the amplitude transmittance T s (r) is updated. As described above, steps S40 and S50 are steps for calculating the image of the estimated sample. Step S90 is a step of optimizing the refractive index profile of the estimated sample.
- the amplitude transmittance T s (r) represents the estimated sample. Therefore, the estimated sample is updated by repeatedly executing the step of calculating the image of the estimated sample and the step of optimizing the refractive index distribution of the estimated sample.
- the optical system used in the second simulation is the measurement optical system of microscope system 120 shown in FIG.
- FIG. 23 is a diagram showing the microscope system of this embodiment. The same numbers are assigned to the same configurations as in FIG. 16, and the description thereof is omitted.
- the specimen 140 is a thick specimen. Light beams are simultaneously incident on the specimen 140 from a plurality of directions.
- FIG. 23 shows the luminous flux L1 and the luminous flux L2.
- the light emitted from the specimen 140 is condensed on the imaging plane IP by the detection optical system 3 .
- An optical image 140' is formed on the imaging plane IP.
- Optical image 140 ′ is an optical image of specimen 140 .
- the microscope system 120 has a moving stage 141.
- the moving stage 141 moves in the direction of the optical axis AX.
- the captured image is used to optimize the refractive index distribution of the estimated specimen. Since the specimen 140 is a thick specimen, multiple captured images are acquired. In order to acquire a plurality of captured images, the sample 140 is fixed and the focal position of the detection optical system 3 is moved by the moving stage 141 .
- the detection optical system 3 has, for example, an infinity correction objective lens and an imaging lens.
- the focal position of the detection optical system 3 can be moved by moving the objective lens.
- the detection optical system 3 and the imaging element 4 may be fixed, and the sample 140 may be moved.
- FIG. 24 is a diagram showing a captured image.
- FIG. 24(a) is a diagram showing a photographed image at the first position.
- FIG. 24(b) is a diagram showing a photographed image at the second position.
- FIG. 24(c) is a diagram showing a photographed image at the third position.
- FIG. 24(d) is a diagram showing a photographed image at the fourth position.
- Photographed image I mea1 (r) An image at a distance of 3 ⁇ z from the surface 140a.
- Photographed image I mea2 (r) An image at a position at a distance of 2 ⁇ z from the surface 140a.
- Photographed image I mea3 (r) An image at a position ⁇ z from the surface 140a.
- Photographed image I mea4 (r) Image of surface 140a.
- the captured image I mea1 (r), the captured image I mea2 (r), the captured image I mea3 (r), and the captured image I mea4 (r) are input to the processor.
- the processor reconstructs the estimated sample using the four captured images. In reconstruction, a simulation is performed.
- 25 and 26 are flowcharts of the second simulation. The same processing as the processing in the first flow chart is given the same number, and the explanation is omitted. Before describing the flow chart, the estimated sample and wavefront will be described.
- FIG. 27 is a diagram showing an optical system used in the simulation. The same numbers are assigned to the same configurations as in FIG. 16, and the description thereof is omitted.
- the optical system used in the simulation is the same measurement optical system that acquired the captured image I mea1 (r), the captured image I mea2 (r), the captured image I mea3 (r), and the captured image I mea4 (r).
- Estimated samples 150 are used instead of samples 140 in the simulation.
- FIG. 27 illustrates the estimated sample 150, wavefront f in m (r), amplitude transmittance T z (r), and wavefront g out m (r).
- the wavefront g out m (r) can be directly calculated from the wave front f in m (r) as shown in equation (13).
- the estimated sample is a thick sample, it is difficult to directly calculate the wavefront goutm (r) from the wavefront finm ( r).
- the estimated sample 150 is a thick sample. Therefore, the estimated sample 150 is replaced with a plurality of thin layers along the optical axis direction. Then, for each thin layer, the wavefronts on both sides of the layer are calculated.
- FIG. 28 is a diagram showing wavefronts in each layer. Calculation of the wavefront will be described later.
- the number of layers can be the same as the number of acquired images. However, the number of layers may be greater than the number of acquired images. In FIG. 28, the number of layers is the same as the number of acquired images.
- step S10 the number N_LS of the second regions is set.
- an incoherent light source is located in the pupil plane of the illumination optics. Therefore, in step S10, the number of light sources is set to NLS .
- step S200 the number of layers N IM is set.
- the estimated sample 150 is a thick sample.
- the number of layers N IM represents the number of thin layers.
- Photographed images are acquired at a plurality of positions on the specimen 140 .
- the number of layers NIM can be set freely.
- the number of layers N IM can be greater than the number of locations from which shot images were acquired.
- the number of thin layers is seven.
- images of seven putative specimens are calculated.
- the simulation uses the captured image and the image of the estimated specimen in the thin layer, as described below. Therefore, the seven positions for which the images of the estimated specimen are calculated include the four positions for which the photographed images are obtained.
- the relationship between the seven positions and the captured image can be as follows.
- the number 1 represents the position of the first layer. At this position, the captured image I mea1 (r) is acquired. Also, at this position, the image of the estimated specimen in the first layer is calculated. Therefore, the image of the estimated specimen in the first layer and the captured image I mea1 (r) are used in the steps described later.
- the number 2 represents the position of the second layer. There is no captured image acquired at this position.
- the number 3 represents the position of the third layer. At this position, the captured image I mea2 (r) is acquired. Also, at this position, the image of the estimated sample in the third layer is calculated. Therefore, the image of the estimated specimen in the third layer and the captured image I mea2 (r) are used in the steps described later.
- the number 4 represents the position of the 4th layer. There is no captured image acquired at this position.
- the number 5 represents the position of the fifth layer. At this position, the captured image I mea3 (r) is acquired. Also, at this position, the image of the estimated sample in the fifth layer is calculated. Therefore, the image of the estimated specimen in the fifth layer and the captured image I mea3 (r) are used in the steps described later.
- the number 6 represents the position of the 6th layer. There is no captured image acquired at this position.
- the number 7 represents the position of the seventh layer. At this position, the captured image I mea4 (r) is acquired. Also, at this position, the image of the estimated sample in the seventh layer is calculated. Therefore, the image of the estimated specimen in the seventh layer and the captured image I mea4 (r) are used in the steps described later.
- step S210 the number of corrections N-- CR is set.
- step S220 the value of variable z is initialized.
- Step S231, which will be described later, is executed for all acquisition positions.
- a variable z represents the number of times step S231 has been executed.
- Step S230 is a step of estimating an estimated sample.
- Four captured images are acquired for the specimen 140 .
- the estimated sample 150 has been replaced with four thin layers. Therefore, the initial value of the amplitude transmittance is set four times.
- Step S230 has steps S231, S232, and S233.
- step S231 an initial value is set for the amplitude transmittance T z (r) in the estimated sample 150.
- An intensity transfer equation may be used to set the initial values.
- the intensity transport equation is disclosed, for example, in the following references. M. R. Teague, “Deterministic phase retrieval: a Greens function solution,” J. Opt. Soc. Am. 73, 1434-1441 (1983)
- Equation (24) The intensity transport equation at the focal position Z0 is represented by Equation (24).
- ⁇ 2 is the second-order Laplacian
- k is the wavenumber
- ⁇ Z0 (r) is the sample phase distribution on the imaging plane
- I Z0 is the average light intensity of the optical image
- ⁇ I meaZ0 (r)/ ⁇ Z is the differential image of two defocused images separated by ⁇ z from the imaging plane, is.
- Equation (24) the phase distribution ⁇ Z0 (r) of the specimen can be easily obtained from the focused image and the two defocused images.
- the phase distribution ⁇ Z0 (r) is calculated from the focused image and the two defocused images.
- a focused image for example, is obtained by moving an objective lens at regular intervals in the optical axis direction. In this case, a plurality of focus images are acquired discretely along the optical axis. Therefore, two defocused images are also acquired discretely.
- phase distribution ⁇ Z0 (r) represented by Equation (24) is the phase distribution in the plane perpendicular to the optical axis. Since the focused image and the two defocused images are obtained discretely, the plane representing the phase distribution ⁇ Z0 (r) is also discretely positioned along the optical axis.
- the phase distribution ⁇ z (r) can be converted to amplitude transmittance T s (r). In this way, an initial value can be set for the amplitude transmittance T z (r).
- phase distribution ⁇ Z0 obtained by the intensity transport equation can be used for the phase distribution ⁇ z (r).
- step S232 it is determined whether or not the value of the variable z matches the number of acquisition positions NIM . If the determination result is NO, step S233 is executed. If the determination result is YES, step S30 is executed.
- step S233 If the judgment result is NO: z ⁇ N IM ) If the determination result is NO, 1 is added to the value of the variable z in step S233. After step S233 ends, the process returns to step S231.
- step S233 the value of variable z is incremented by one. Therefore, an initial value is set to the amplitude transmittance T z (r) in step S231 for another acquisition position.
- Step S231 is repeated until initial values are set for all acquisition positions.
- step S30 the value of variable m is initialized. Steps S240, S41, S42, S251, and S260, which will be described later, are performed for all light sources. The variable m represents the number of times these steps have been performed.
- I estz (r) represents the image of the estimated specimen 30 .
- the image of putative specimen 150 has been replaced by four thin layers. I estz (r) thus represents the image of the thin layer.
- Steps S250 and S270 are steps for calculating the image of the estimated specimen.
- the number of estimated sample images is the same as the number of captured images. Since the number of captured images is four, the number of images of the estimated sample is also four.
- Step S250 includes steps S41, S42, S251, S252, S253, and S260.
- step S41 the wavefront f in m (r) incident on the estimated sample 150 is calculated.
- the wavefront f in m (r) is represented by the above equations (11) and (12).
- step S42 the wavefront g out m (r) emitted from the estimated sample 150 is calculated.
- the wavefront g out m (r) is calculated based on the wave front f in m (r).
- the estimated sample 150 has been replaced by four thin layers. Therefore, the wavefront is calculated at each of the thin layers.
- the four thin layers are evenly spaced.
- the distance between two adjacent layers is ⁇ z.
- a wavefront propagates between the two layers. Therefore, ⁇ z represents the propagation distance.
- Equation (26) The wavefront f 1 m (r) in the first layer is represented by Equations (26) and (12).
- the position of the first layer matches the position of the surface 150b of the estimated sample 150.
- FIG. A wavefront f in m (r) is incident on the surface 150b.
- the wavefront f 1 m (r) represents the wavefront f in m (r).
- wavefront f in m (r) is shown.
- a wavefront g 1 m (r) in the first layer is represented by Equation (27).
- T 1 (r) is the amplitude transmittance in the first layer; is.
- the wavefront f 2 m (r) in the second layer is the wavefront when the wavefront g 1 m (r) propagates by ⁇ z.
- Wavefront f 2 m (r) is represented by equation (28) ⁇ z is the distance between two adjacent layers; ⁇ is the wavelength, u is a two-dimensional notation of pupil plane coordinates ( ⁇ , ⁇ ), F 2D is the two-dimensional Fourier transform; F 2D ⁇ 1 is the two-dimensional inverse Fourier transform, is.
- a wavefront g 2 m (r) in the second layer is represented by Equation (29).
- T 2 (r) is the amplitude transmittance in the second layer; is.
- the wavefront f 3 m (r) in the third layer is the wavefront when the wavefront g 2 m (r) propagates by ⁇ z.
- a wavefront f 3 m (r) in the third layer is represented by Equation (30).
- a wavefront g 3 m (r) in the third layer is represented by Equation (31). here, T 3 (r) is the amplitude transmittance in the third layer; is.
- the wavefront f 4 m (r) in the fourth layer is the wavefront when the wavefront g 3 m (r) propagates by ⁇ z.
- a wavefront g 4 m (r) in the fourth layer is represented by Equation (33).
- T 4 (r) is the amplitude transmittance in the fourth layer; is.
- the position of the fourth layer matches the position of the surface 150a of the estimated sample 150.
- FIG. A wavefront g out m (r) emerges from the surface 150a.
- the wavefront g4m (r) represents the wavefront goutm ( r).
- the wavefront goutm ( r) is shown instead of the wavefront g4m ( r).
- the wavefront g out m (r) can be calculated by replacing it with a plurality of thin layers and obtaining the wavefront propagating between the two layers.
- step S251 the value of variable z is initialized. Steps S261, S262, and S263, which will be described later, are executed for all acquisition positions. The variable z represents the number of times these steps have been performed.
- Step S260 includes steps S261, S262, S263, S264, and S265.
- step S261 the wavefront u z m (r) at the acquisition position of the captured image is calculated.
- the wavefront u z m (r) is represented by Equation (34). here, ⁇ D is the distance from the surface of the estimated specimen to the thin layer, is.
- step S262 the wavefront u imgz m (r) on the imaging plane is calculated.
- the wavefront u imgz m (r) is represented by Equation (35).
- step S263 the wavefront u imgz m (r) is squared.
- the wavefront u imgz m (r) represents the amplitude of the light. Therefore, the light intensity is calculated by squaring the wavefront u imgz m (r).
- 2 represents the light intensity distribution on the imaging plane IP. Assuming that the first intensity distribution is the light intensity distribution at the imaging position of the detection optical system,
- step S264 it is determined whether or not the value of the variable z matches the number of acquisition positions N_IM . If the determination result is NO, step S265 is executed. If the determination result is YES, step S252 is executed.
- step S265 If the judgment result is NO: z ⁇ N IM , If the determination result is NO, 1 is added to the value of the variable z in step S265. After step S265 ends, the process returns to step S261.
- step S265 the value of the variable z is incremented by one. Therefore, steps S261, S262, and S263 are executed for another acquisition position.
- Steps S261, S262, and S263 are repeated until initial values are set for all acquisition positions.
- step S250 will be explained using the first layer and the fourth layer.
- the second and third layers may be considered in the same way as the first layer.
- FIG. 29 is a diagram showing the wavefront at the acquisition position of the captured image and the wavefront at the imaging plane.
- FIG. 29(a) is a diagram showing two wavefronts in the first layer.
- FIG. 29(b) is a diagram showing two wavefronts in the fourth layer.
- the photographed image I mea1 (r) is an image at a position at a distance of 3 ⁇ z from the surface 140a.
- the first layer is 3 ⁇ z away from the surface 150a. Therefore, the position of the first layer corresponds to the acquisition position of the captured image I mea1 (r).
- the exit position of wavefront g out m (r) coincides with surface 150a. As shown in FIG. 29(a), the exit position of the wavefront g out m (r) is different from the position of the first layer. The first layer is 3 ⁇ z away from the exit location of the wavefront g out m (r).
- the wavefront u img1 m (r) on the imaging plane is calculated from Equation (35) in step S262.
- step S263 the light intensity
- the captured image I mea2 (r) is an image at a position with a distance of 2 ⁇ z from the surface 140a.
- the second layer is 2 ⁇ z away from surface 150a. Therefore, the position of the second layer corresponds to the acquisition position of the captured image I mea2 (r).
- the exit position of the wavefront g out m (r) is different from the position of the second layer.
- the second layer is 2 ⁇ z away from the exit location of the wavefront g out m (r).
- step S262 After the wavefront u 2 m (r) is calculated, the wavefront u img2 m (r) on the imaging plane is calculated in step S262.
- step S263 the light intensity
- the captured image I mea3 (r) is an image at a position with a distance of ⁇ z from the surface 140a.
- the third layer is separated from surface 150a by ⁇ z. Therefore, the position of the third layer corresponds to the acquisition position of the captured image I mea3 (r).
- the exit position of the wavefront g out m (r) is different from the position of the third layer.
- the third layer is ⁇ z away from the exit position of the wavefront g out m (r).
- step S262 After the wavefront u 3 m (r) is calculated, the wavefront u img3 m (r) on the imaging plane is calculated in step S262.
- step S263 the light intensity
- the captured image I mea4 (r) is an image of the surface 140a.
- the fourth layer conforms to surface 150a. Therefore, the position of the fourth layer corresponds to the acquisition position of the captured image I mea4 (r).
- the exit position of wavefront g out m (r) is surface 150a.
- the emission position of the wavefront g out m (r) is the same as the position of the fourth layer.
- the wavefront u 4 m (r) in the fourth layer is the same as the wavefront g out m (r).
- Wavefront g out m (r) can be replaced by wave front u 4 m (r).
- the exit position of wavefront g out m (r) is surface 150a. As shown in FIG. 16(b), the emission position of the wavefront g out m (r) is the same as the position of the fourth layer.
- the wavefront u 4 m (r) in the fourth layer is the same as the wavefront g out m (r). Wavefront g out m (r) can be replaced by wave front u 4 m (r).
- step S262 After the wavefront u 4 m (r) is calculated, the wavefront u img4 m (r) on the imaging plane is calculated in step S262.
- step S263 the light intensity
- a wavefront uzm(r) and a wavefront uimgzm ( r ) represent wavefronts generated by illumination light emitted from the m -th light source, that is, illumination light emitted from one light source.
- An estimated specimen image I estz (r) is generated at the acquisition position by illumination light emitted from all light sources. Therefore, wavefronts must be obtained for all light sources.
- Step S242 is executed.
- Wavefront f in m (r), wave front g out m (r), wave front u z m (r), and wave front u imgz m (r) are the illumination light emitted from the m-th light source, that is, one light source represents the wavefront generated by the illumination light emitted from
- An estimated sample image I estz (r) is generated by illumination light emitted from all light sources. Therefore, the wavefronts f in m (r), g out m (r), u z m (r), and u imgz m (r) must be determined for all light sources.
- step S252 it is determined whether or not the value of variable m matches the number of light sources NLS . If the determination result is NO, step S253 is executed. If the determination result is YES, step S270 is executed.
- step S253 If the judgment result is NO: m ⁇ N LS , If the determination result is NO, 1 is added to the value of the variable m in step S253. After step S253 ends, the process returns to step S41.
- step S253 the value of variable m is increased by one. Therefore, for another light source, the wavefront f in m (r) is calculated in step S41, the wavefront g out m (r) is calculated in step S42, the wavefront u z m (r) is calculated in step S261, and the wavefront u z m (r) is calculated in step S261. Wavefront u imgz m (r) is calculated in S262, and
- Steps S41, S42, S251, and S260 are repeated until
- FIG. 30 is a diagram showing an image of an estimated specimen.
- FIG. 30(a) is a diagram showing an image of an estimated sample in the first layer.
- FIG. 30(b) is a diagram showing an image of an estimated sample in the fourth layer.
- the estimated sample image I est1 (r) is an image when the wavefront u img1 m (r) is obtained for all light sources.
- the estimated sample image I est4 (r) is an image when the wavefront u img4 m (r) is obtained for all light sources.
- the wavefront u img1 m (r) is calculated for each light source,
- the estimated sample image I est1 (r) in the first layer is calculated.
- the wavefront u img4 m (r) is calculated for each light source,
- the estimated sample image I est4 (r) in the fourth layer is calculated.
- a residual is represented by Formula (37). As shown in Equation (37), the residual is calculated from the captured image I meaz (r) and the estimated sample image I estz (r).
- the number of captured images is four, and the number of estimated sample images is also four. Therefore, the residual in the first layer is calculated from I mea1 (r) and I est1 (r). The residual in the second layer is calculated from I mea2 (r) and I est2 (r). The residual in the third layer is calculated from I mea3 (r) and I est3 (r). The residual in the fourth layer is calculated from I mea4 (r) and I est4 (r).
- the residual used in step 70 is calculated from the residual in the first layer, the residual in the second layer, the residual in the third layer, and the residual in the fourth layer.
- step S70 the residual is compared with the threshold. If the determination result is NO, step S290 is executed. If the determination result is YES, step S110 is executed.
- step S290 the value of variable L is initialized. Steps S301, S302, S303, S304, and S310, which will be described later, are executed the number of times set in step S210.
- the variable L represents the number of times these steps have been performed.
- Step S300 includes steps S301, S302, S303, S304, S305, S306, and S310.
- step S301 one is randomly selected from 1 to N IM .
- step S311 which will be described later, the corrected wavefront is calculated. One photographed image and one estimated sample image are used to calculate the corrected wavefront.
- step S270 images of a plurality of estimated specimens are calculated.
- One image of the estimated sample is used to calculate the wavefront after correction. Therefore, the estimated sample image used for calculating the wavefront after correction is selected from among a plurality of estimated sample images.
- the selected number is 1, the number 1 represents the first layer.
- the captured image at the first acquisition position corresponds to the estimated specimen image in the first layer. Therefore, the captured image at the first acquisition position and the image of the estimated sample in the first layer are used to calculate the corrected wavefront.
- the selected number represents the fourth layer.
- the captured image at the fourth acquisition position corresponds to the estimated specimen image in the fourth layer. Therefore, the captured image at the fourth acquisition position and the image of the estimated sample in the fourth layer are used to calculate the corrected wavefront.
- step S302 the value selected at step S301 is input to the variable zL.
- one number is randomly selected from the numbers 1 to NIM . For example, if the selected number is 1, 1 will be input to the variable zL in step S302.
- step S303 the value of variable m is initialized. Steps S311, S312, and S313, which will be described later, are executed for all light sources. The variable m represents the number of times these steps have been performed.
- Step S310 is a step of optimizing the refractive index distribution of the estimated sample.
- Step S310 includes steps S311, S312, S313, S314, and S315.
- step S311 the wavefront u' zL m (r) is calculated.
- the wavefront u′ zL m (r) is the wavefront at the layer location indicated by the value of the variable zL.
- the captured image I meazL (r) and the estimated sample image I estzL (r) are used to calculate the wavefront u′ zL m (r).
- the captured image I meazL (r) is a captured image at a position indicated by the value of the variable zL in the captured image I meaz .
- the estimated specimen image I estzL (r) is the estimated specimen image at the position indicated by the value of the variable zL in the estimated specimen image I estz .
- Wavefront u' zL m (r) is represented by Equation (38).
- FIG. 31 is a diagram showing wavefront correction.
- FIG. 31(a) is a diagram showing a wavefront before correction emitted from an estimated sample.
- FIG. 31(b) is a diagram showing the wavefront before correction at the acquisition position of the captured image.
- FIG. 31(c) is a diagram showing the corrected wavefront at the acquired position of the captured image.
- FIG. 31(d) is a diagram showing a corrected wavefront emitted from the estimated sample.
- the estimated sample image I est1 (r) is calculated based on the wavefront u img1 m (r). Further, as shown in FIGS. 30(a) and 31(b), the wavefront u img1 m (r) is calculated based on the wave front u 1 m (r).
- the amplitude transmittance Tz ( r ) is used to calculate the wavefront u1m (r).
- Amplitude transmittance T z (r) is the estimated amplitude transmittance.
- this amplitude transmittance T z (r) is different from the amplitude transmittance of sample 140 .
- the difference between the estimated specimen image I estz (r) and the captured image I meaz (r) increases. Therefore, the difference between the estimated specimen image I estz (r) and the captured image I meaz (r) can be regarded as reflecting the difference between the amplitude transmittance T z (r) and the amplitude transmittance of the specimen 140. can be done.
- Equation (38) the estimated specimen image I est1 (r) and the captured image I mea1 (r) are used to correct the wavefront u 1 m (r). As a result, a corrected wavefront, ie, the wavefront u′ 1 m (r) is obtained as shown in FIG. 31(c).
- a new amplitude transmittance can be calculated by using the wavefront u′ 1 m (r). Wavefront u′ 1 m (r) is different from wavefront u 1 m (r). Therefore, the new amplitude transmittance is different from the amplitude transmittance when the wavefront u 1 m (r) was calculated.
- step S312 the corrected wavefront g′ out m,zL (r) is calculated.
- a wavefront g′ out m,zL (r) is a wavefront when the wavefront u′ zL m (r) propagates by ⁇ D.
- the wavefront g′ out m,zL (r) is represented by Equation (39).
- the wavefront u′ 1 m (r) can be used to calculate the amplitude transmittance T z (r).
- the wavefront at the position of the wavefront g out m (r) is required to calculate the amplitude transmittance T z (r).
- the position of the wavefront u′ 1 m (r) and the position of the wavefront g out m (r) are different. Therefore, in order to calculate the amplitude transmittance T z (r), the wavefront g′ out m,1 (r) is required as shown in FIG. 31(d).
- a wavefront g′ out m,1 (r) is a wavefront when the wavefront u′ 1 m (r) propagates by 3 ⁇ z.
- step S313 the sample gradient ⁇ T z m,zL (r) is calculated.
- ⁇ T z m,zL (r) is the sample gradient when illuminated by the m-th light source and corrected with the captured image and the estimated sample image at the layer position indicated by the value of the variable zL.
- the sample gradient ⁇ T z m,zL is expressed by Equation (40).
- Gradient descent for example, can be used to calculate the sample gradient ⁇ T z m,zL (r).
- f * is the complex conjugate of f ⁇ is a normalization constant to prevent division by zero; is.
- the putative sample 150 has been replaced with multiple thin layers. Therefore, it is necessary to calculate the sample gradient ⁇ T z m,zL (r) for each thin layer.
- FIG. 32 is a diagram showing sample gradient and wavefront propagation.
- FIG. 32(a) is a diagram showing the gradient of the sample.
- FIG. 32(b) is a diagram showing the propagation of wavefronts.
- the amplitude transmittance T 4 (r) is used to calculate the wavefront g out m (r). Amplitude transmittance T 4 (r) is the estimated amplitude transmittance. Therefore, this amplitude transmittance T 4 (r) differs from the amplitude transmittance of sample 140 .
- the difference between the wavefront g out m (r) and the wave front g′ out m,1 (r) is considered to reflect the difference between the amplitude transmittance T 4 (r) and the amplitude transmittance of the sample 140. be able to.
- the wavefront goutm ( r ) can be used instead of g4m ( r).
- g' 4 m,1 (r) is the same as g' out m,1 ( r), so if g' 4 m,1 (r) is used instead of g' 4 m,1 (r), good.
- the sample gradient ⁇ T 3 m,1 (r) is calculated.
- Calculation of the sample gradient ⁇ T 3 m,1 (r) requires the wavefront at the position of the wavefront g 3 m (r).
- the wavefront f′ 4 m,1 (r) is required as shown in FIG. 32(b).
- FIG. 33 is a diagram showing the gradient of the sample and the propagation of the wavefront.
- FIG. 33(a) is a diagram showing the propagation of the wavefront and the gradient of the sample.
- FIG. 33(b) is a diagram showing the propagation of wavefronts.
- the wavefront g′ 3 m,1 (r) is the wavefront when the wavefront f′ 4 m,1 (r) propagates by ⁇ z. This is the propagation of the wavefront from the 4th layer to the 3rd layer.
- the gradient of the sample can be calculated in the same way as for the third layer.
- FIG. 34 is a diagram showing sample gradients.
- the sample gradient ⁇ T 1 m,1 (r) in the first layer the sample gradient ⁇ T 2 m,1 (r) in the second layer, and the sample gradient ⁇ T 3 m,1 (r ), and the sample gradient ⁇ T 4 m,1 (r) in the fourth layer.
- the sample gradient ⁇ T z m,1 (r) obtained in step S313 is illuminated by the m-th light source and corrected by the photographed image at the position of the first layer and the estimated sample image at the position of the first layer. is the slope of the time sample.
- the sample gradient ⁇ T z m,1 (r) is determined by the illumination light emitted from all light sources. Therefore, the sample gradient ⁇ T z m,1 (r) must be determined for all light sources.
- step S314 it is determined whether or not the value of the variable m matches the number of light sources NLS . If the determination result is NO, step S315 is executed. If the determination result is YES, step S304 is executed.
- step S315 If the judgment result is NO: m ⁇ N LS , If the determination result is NO, 1 is added to the value of the variable m in step S315. After step S315 ends, the process returns to step S311.
- step S315 the value of variable m is incremented by one. Therefore, for another light source, the wavefront u z ′ m,1 (r) is calculated in step S311, the wavefront g out ′ m,1 (r) is calculated in step S312, and the sample gradient ⁇ T z m , 1 (r) are calculated.
- Steps S311, S312, and S313 are repeated until the sample gradients ⁇ T z m,1 (r) are obtained for all light sources.
- FIG. 35 is a diagram showing sample gradients.
- the sample gradient ⁇ T z m. 1 (r) is sought.
- Step S304 is a step of updating the estimated samples.
- Equation (42) The updated amplitude transmittance T z (r) is represented by Equation (42). here, ⁇ is the correction factor for the slope of the sample, is.
- step S305 it is determined whether or not the value of the variable L matches the number of corrections NCR . If the determination result is NO, step S306 is executed. If the determination result is YES, step S30 is executed.
- step S306 If the judgment result is NO: m ⁇ N CR ) If the determination result is NO, 1 is added to the value of the variable L in step S306. After step S306 ends, the process returns to step S301.
- step S301 one is randomly selected from 1 to N IM . Based on the selected numbers, the image and acquisition position of the estimated sample to be used for correction are determined.
- step S311 the wavefront u z ′ m,1 (r) is calculated, in step S312, the wavefront g out ′ m,1 (r) is calculated, and in step S313, the sample gradient ⁇ T z m,1 (r) is calculated, and the amplitude transmittance T z (r) is updated in step S304.
- Steps S301, S302, S303, S304, and S310 are repeated until the set number of corrections is completed.
- Step S30 are executed with the updated amplitude transmittance T z (r).
- the updated amplitude transmittance T s (r) gradually approaches the amplitude transmittance of the sample 140 . That is, the residual becomes smaller. Over time, the residual will be less than the threshold.
- step S110 the refractive index distribution of the estimated sample is calculated.
- the resulting amplitude transmittance T z (r) is identical or nearly identical to the amplitude transmittance of sample 140 .
- a refractive index distribution n z (r) can be obtained from the obtained amplitude transmittance T z (r) and Equation (1).
- the structure of the estimated sample can be reconstructed using the refractive index distribution n z (r) obtained in step S110.
- the reconstructed estimated sample structure can be output to, for example, a display device.
- the estimated sample 150 is a thick sample. In a second simulation, the structure of the thick specimen can reconstruct the 3D configuration of the putative specimen.
- the amplitude transmittance T z (r) obtained in step S110 is the same or substantially the same as the amplitude transmittance of sample 140 .
- the refractive index profile n z (r) can also be considered identical or substantially identical to the refractive index profile of the sample 140 . Therefore, the structure of the reconstructed putative sample 150 can be considered identical or nearly identical to the structure of the sample 6 .
- steps S250, S270, and S310 are repeatedly performed. As a result, the amplitude transmittance T z (r) is updated. As described above, steps S250 and S270 are steps for calculating the image of the estimated sample. Step S310 is a step of optimizing the refractive index profile of the estimated sample.
- Amplitude transmittance T z (r) represents an estimated sample. Therefore, the estimated sample is updated by repeatedly executing the step of calculating the image of the estimated sample and the step of optimizing the refractive index distribution of the estimated sample.
- FIG. 36 is a diagram showing the simulation results of the first example.
- FIG. 36(a) shows a specimen.
- FIG. 36(b) shows an aperture member.
- FIG. 36(c) shows an image of the specimen.
- FIG. 36(d) shows the reconstructed estimated samples.
- the specimen is a photonic crystal fiber (hereinafter referred to as "PCF").
- PCF a photonic crystal fiber
- a plurality of through holes are formed inside the clad.
- a through-hole is called a core.
- the PCF is immersed in liquid.
- the core is thus filled with liquid.
- Numerical values of various parameters are as follows. Specimen outer diameter 230 ⁇ m Core diameter 6 ⁇ m Core refractive index 1.466 Clad refractive index 1.462 Refractive index of liquid 1.466 Objective lens numerical aperture 1.4 Objective lens magnification 60 times Objective lens focal length 3 mm Numerical aperture of the second region 0.3, 1.25 Focal length of condenser lens 7mm Diameter of transmission region of aperture member 0.2 mm Wavelength of illumination light 0.7 ⁇ m
- the numerical aperture of the second region is represented by the numerical aperture of the condenser lens. As shown in FIG. 36(b), the second area is positioned on the circumference of the two circles.
- the luminous flux emitted from the second area located on the inner circumference can be regarded as a luminous flux corresponding to a numerical aperture of 0.3.
- the luminous flux emitted from the second area located on the outer circumference can be regarded as a luminous flux corresponding to a numerical aperture of 1.25.
- the second area is positioned on the circumference of the two circles.
- the radius to the inner second region is the radius of the inner circle.
- FIG. 37 is a diagram showing an aperture member and a reconstructed estimated sample.
- Figures 37(a), 37(b), 37(c) and 37(d) show the opening member.
- Figures 37(e), 37(f), 37(g) and 37(h) show reconstructed estimated samples.
- the estimated sample is reconstructed with the numerical aperture of the objective lens set to 1.4 and the wavelength of the illumination light set to 0.7 ⁇ m.
- the numerical apertures of the second region are 0.3 and 0.5.
- the core extends in the Z-axis direction.
- the numerical apertures of the second region are 1.05 and 1.25.
- artifacts are present in the reconstructed estimated sample. Thin images of the core are present on the left and right.
- the numerical apertures of the second region are 0.3, 0.5, 1.05, and 1.25.
- artifacts are present in the reconstructed estimated sample. However, compared to FIG. 37(f), artifacts are reduced.
- the numerical apertures of the second region are 0.3, 0.5, 1.05, and 1.25. However, two opening members are used.
- artifacts are present in the reconstructed estimated sample. However, the artifact is of the same degree as in FIG. 37(g).
- the reconstruction can adjust the proportion of relatively coarse structure estimation and relatively fine structure estimation. By performing such adjustment, the reconstruction time can be shortened.
- FIG. 38 is a diagram showing images of the aperture member and the specimen in the measurement of the second example.
- FIG. 38(a) shows an aperture member.
- FIGS. 38(b) and 38(c) show images of the specimen.
- the specimen is a grid-like structure.
- the specimen is immersed in oil with a refractive index of 1.518. Therefore, the space enclosed by the lattice is filled with oil.
- the numerical aperture of the second region is represented by the numerical aperture of the condenser lens. As shown in FIG. 37(a), the second area is positioned on the circumference of the two circles.
- the luminous flux emitted from the second area located on the inner circumference can be regarded as a luminous flux corresponding to a numerical aperture of 0.3.
- the luminous flux emitted from the second area located on the outer circumference can be regarded as a luminous flux corresponding to a numerical aperture of 0.45.
- the second area is located on the circumference of the two circles.
- the radius to the inner second region is the radius of the inner circle.
- FIG. 39 is a diagram showing the results of the second example.
- FIGS. 39(a) and 39(b) show estimated samples with initial values set.
- Figures 39(c), 39(d) and 39(e) show reconstructed estimated samples.
- initial values are set so that the outline of the estimated sample can be determined.
- initial values are set using the intensity transport equation.
- FIGS. 37(c) and 37(e) A comparison between FIGS. 37(c) and 37(e) and a comparison between FIGS. 37(d) and 37(e) reveals that setting the initial values can reconstruct the estimated sample with higher accuracy. can.
- the microscope system 1 shown in FIG. 1(a) can be equipped with an illumination optical system.
- the microscope system 1 preferably satisfies the conditions (6) and (7).
- the microscope system 1 can also comprise an aperture member.
- the present invention is suitable for microscope systems in which the time from the start of image acquisition to the completion of object model estimation is short.
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Abstract
Description
I=PSF*O
ここで、*は、コンボリューションを表す。
インコヒーレント光源と、検出光学系と、撮像素子と、を備え、
インコヒーレント光源は、時間的にコヒーレントでない光を射出する光源であり、
検出光学系は、標本の光学像を形成する光学系であり、
撮像素子は、検出光学系により形成された標本の光学像を受光し、
標本では、インコヒーレント光源から射出した光によって複数のコヒーレント照明が同時に行われ、
コヒーレント照明は、空間的にコヒーレントである光による照明であり、
標本に対して光束が照射される方向は、コヒーレント照明それぞれで異なり、
検出光学系の瞳面では、コヒーレント照明それぞれの光束が互いに異なる第1領域を通過し、
第1領域のそれぞれは、以下の条件(1)を満たし、
隣接する2つの第1領域の間隔のうち少なくとも1つの間隔は、以下の条件(2)を満たすことを特徴とする。
LS<PS×10-3 (1)
0.05×T<d (2)
ここで、
LSは、第1領域の面積(単位はmm2)、
PSは、検出光学系の瞳の面積(単位はmm2)、
dは、隣接する2つの第1領域の間隔(単位はmm)、
Tは、検出光学系の瞳の直径(単位はmm)、
である。
インコヒーレント光源と、照明光学系と、検出光学系と、撮像素子と、を備え、
インコヒーレント光源は、時間的にコヒーレントでない光を射出する光源であり、
検出光学系は、標本の光学像を形成する光学系であり、
撮像素子は、検出光学系により形成された標本の光学像を受光し、
標本では、インコヒーレント光源から射出した光によって複数のコヒーレント照明が同時に行われ、
コヒーレント照明は、空間的にコヒーレントである光による照明であり、
標本に対して光束が照射される方向は、コヒーレント照明それぞれで異なり、
照明光学系の瞳面では、コヒーレント照明それぞれの光束が互いに異なる第2領域に位置し、
第2領域のそれぞれは、以下の条件(6)を満たし、
隣接する2つの第2領域の間隔のうち少なくとも1つの間隔は、以下の条件(7)を満たすことを特徴とする。
LS’<PS’×10-3 (6)
0.05×T’<d’ (7)
ここで、
LS’は、第2領域の面積(単位はmm2)、
PS’は、照明光学系の瞳の面積(単位はmm2)、
d’は、隣接する2つの第2領域の間隔(単位はmm)、
T’は、照明光学系の瞳の直径(単位はmm)、
である。
LS<PS×10-3 (1)
0.05×T<d (2)
ここで、
LSは、第1領域の面積(単位はmm2)、
PSは、検出光学系の瞳の面積(単位はmm2)、
dは、隣接する2つの第1領域の間隔(単位はmm)、
Tは、検出光学系の瞳の直径(単位はmm)、
である。
LS<PS×10-3 (1)
ここで、
LSは、第1領域の面積(単位はmm2)、
PSは、検出光学系の瞳の面積(単位はmm2)、
である。
0.05×T<d (2)
ここで、
dは、隣接する2つの第1領域の間隔(単位はmm)、
Tは、検出光学系の瞳の直径(単位はmm)、
である。
LSiは、i番目の第1領域の面積(単位はmm2)、
PSは、検出光学系の瞳の面積(単位はmm2)、
nは、第1領域の数、
である。
PS×10-6<LS (4)
ここで、
LSは、第1領域の面積(単位はmm2)、
PSは、検出光学系の瞳の面積(単位はmm2)、
である。
d<0.5×T (5)
ここで、
dは、隣接する2つの第1領域の間隔(単位はmm)、
Tは、検出光学系の瞳の直径(単位はmm)
である。
4≦n≦100 (A)
ここで、
nは、第1領域の数、
である。
LS’<PS’×10-3 (6)
0.05×T’<d’ (7)
ここで、
LS’は、第2領域の面積(単位はmm2)、
PS’は、照明光学系の瞳の面積(単位はmm2)、
d’は、隣接する2つの第2領域の間隔(単位はmm)、
T’は、照明光学系の瞳の直径(単位はmm)、
である。
LS’<PS’×10-3 (6)
ここで、
LS’は、第2領域の面積(単位はmm2)、
PS’は、照明光学系の瞳の面積(単位はmm2)、
である。
0.05×T’<d’ (7)
ここで、
d’は、隣接する2つの第2領域の間隔(単位はmm)、
T’は、照明光学系の瞳の直径(単位はmm)、l
である。
PS’=(FLcd×NA)2×π (8)
T’=FLcd×NA (9)
ここで、
FLcdは、コンデンサレンズの焦点距離(単位mm)、
NAは、対物レンズの開口数、
である。
k0は、照明光の波長λに対して2π/λ、
n0は、媒質の屈折率
dzは、標本の厚み、
である。
ここで、
Ts(r)は、推定標本の振幅透過率、
である。
ここで、
Δz1は、推定標本の表面から撮影画像の取得位置までの距離、
λは、波長、
uは、瞳面座標(ξ,η)の2次元表記、
F2Dは、2次元フーリエ変換、
F2D -1は、2次元フーリエ逆変換、
である。
判断結果がNOの場合、ステップS47で、変数mの値に1が加算される。ステップS47が終ると、ステップS41に戻る。
判断結果がYESの場合、ステップS50で、|uimg m(r)|2の足し合わせが行われる。その結果、推定標本の像Iest(r)が算出される。推定標本の像Iest(r)は、式(16)で表される。
ここで、
f*は、fの複素共役、
δは、ゼロ除算を防ぐための正規化定数、
である。
判断結果がNOの場合、ステップS94で、変数mの値に1が加算される。ステップS94が終ると、ステップS91に戻る。
判断結果がYESの場合、ステップS100で、振幅透過率Ts(r)が更新される。ステップS100は、推定標本を更新するステップするステップである。
ステップ110では、推定標本の屈折率分布が算出される。得られた振幅透過率Ts(r)は、標本5の振幅透過率と同一か、又は、略同一である。得られた振幅透過率Ts(r)と式(1)から、屈折率分布n(r)が求まる。
撮影画像Imea1(r):表面140aからの距離が3×Δzの位置の画像。
撮影画像Imea2(r):表面140aからの距離が2×Δzの位置の画像。
撮影画像Imea3(r):表面140aからの距離がΔzの位置の画像。
撮影画像Imea4(r):表面140aの画像。
M. R. Teague, “Deterministic phase retrieval: a Greens function solution,” J. Opt. Soc. Am. 73, 1434-1441(1983)
∇2は、2次のラプラシアン、
kは、波数、
φZ0(r)は,結像面での標本の位相分布、
IZ0は,光学像の平均光強度、
δImeaZ0(r)/δZは、結像面から±△zだけ離れた2つのデフォーカス像の差分像、
である。
判断結果がNOの場合、ステップS233で、変数zの値に1が加算される。ステップS233が終ると、ステップS231に戻る。
ステップS30で、変数mの値が初期化される。後述のステップS240、ステップS41、ステップS42、ステップS251、及びステップS260は、全ての光源に対して実行される。変数mは、これらのステップが実行された回数を表している。
λは、波長、
uは、瞳面座標(ξ,η)の2次元表記、
F2Dは、2次元フーリエ変換、
F2D -1は、2次元フーリエ逆変換、
である。
判断結果がNOの場合、ステップS265で、変数zの値に1が加算される。ステップS265が終ると、ステップS261に戻る。
波面gout m(r)の射出位置は、表面150aである。図29(b)に示すように、波面gout m(r)の射出位置は、第4層の位置と同じである。
第4層における波面u4 m(r)は、波面gout m(r)同じである。波面gout m(r)を波面u4 m(r)に置き換えることができる。
波面u4 m(r)が算出されると、ステップS262で、結像面における波面uimg4 m
ステップS242が実行される。
判断結果がNOの場合、ステップS253で、変数mの値に1が加算される。ステップS253が終ると、ステップS41に戻る。
判断結果がYESの場合、ステップS270で、|uimgz m(r)|2の足し合わせが行われる。その結果、推定標本の像Iestz(r)が算出される。推定標本の像Iestz(r)は、式(36)で表される。
ステップS280で、残差が算出される。残差は、式(37)で表される。式(37)に示すように、残差は、撮影画像Imeaz(r)と推定標本の像Iestz(r)とから算出される。
ステップS290では、変数Lの値が初期化される。後述のステップS301、ステップS302、ステップS303、ステップS304、及びステップS310は、ステップS210で設定した回数だけ実行される。変数Lは、これらのステップが実行された回数を表している。
f*は、fの複素共役
δは、ゼロ除算を防ぐための正規化定数、
である。
波面f4 m(r)を、波面g’3 m,1(r)に置き換える。
波面g3 m(r)を、波面f’4 m,1(r)に置き換える。
ΔD=-Δzとする。
判断結果がNOの場合、ステップS315で、変数mの値に1が加算される。ステップS315が終ると、ステップS311に戻る。
判断結果がYESの場合、ステップS304で、振幅透過率Tz(r)が更新される。ステップS304は、推定標本を更新するステップするステップである。
判断結果がNOの場合、ステップS306で、変数Lの値に1が加算される。ステップS306が終ると、ステップS301に戻る。
判断結果がYESの場合、ステップS30に戻る。更新された振幅透過率Tz(r)で、ステップS30からステップS300までが実行される。
ステップS110では、推定標本の屈折率分布が算出される。得られた振幅透過率Tz(r)は、標本140の振幅透過率と同一か、又は、略同一である。得られた振幅透過率Tz(r)と式(1)から、屈折率分布nz(r)が求まる。
標本の外径 230μm
コアの直径 6μm
コアの屈折率 1.466
クラッドの屈折率 1.462
液体の屈折率 1.466
対物レンズの開口数 1.4
対物レンズの倍率 60倍
対物レンズの焦点距離 3mm
第2領域の開口数 0.3、1.25
コンデンサレンズの焦点距離 7mm
開口部材の透過領域の直径 0.2mm
照明光の波長 0.7μm
検出光学系の瞳の面積(PS’) 55.4mm2
第2領域の面積(LS’) 0.0058mm2
LS’/PS’ 1.0×10-4
検出光学系の瞳直径(T’) 8.4mm
隣接する2つの第2領域の間隔(d’) 0.47mm
d’/T’ 0.056
第2領域の面積 (0.086/2)2×3.14=0.0058
第2領域の直径 0.2/7×3=0.086
検出光学系の瞳直径 2×1.4×3=8.4
内側の第2領域までの半径 0.3×3=0.9
隣接する2つの第2領域の間隔
0.9×sin30/cos15=0.47
開口部材 再構成した推定標本
図37(a) 図37(e)
図37(b) 図37(f)
図37(c) 図37(g)
図37(d) 図37(g)
対物レンズの開口数 1.42
対物レンズの倍率 60倍
対物レンズの焦点距離 3mm
第2領域の開口数 0.3、0.45
コンデンサレンズの焦点距離 7.1mm
開口部材の透過領域の直径 0.2mm
検出光学系の瞳の面積(PS’) 57.0mm2
第2領域の面積(LS’) 0.0056mm2
LS’/PS’ 9.8×10-5
検出光学系の瞳直径(T’) 8.52mm
隣接する2つの第2領域の間隔(d’) 0.47mm
d’/T’ 0.055
第2領域の面積 (0.0845/2)2×3.14=0.0056
第2領域の直径 0.2/7.1×3=0.0845
検出光学系の瞳直径 2×1.42×3=8.52
内側の第2領域までの半径 0.3×3=0.9
隣接する2つの第2領域の間隔
0.9×sin30/cos15=0.47
初期値を設定したの推定標本 再構成した推定標本
図39(a) 図39(c)
図39(b) 図39(d)
なし 図39(e)
2 インコヒーレント光源
3 検出光学系
4 撮像素子
5 標本
5a 表面
5’ 光学像
10 検出光学系3の瞳面
11、12、20、21 第1領域
30 第1円環領域
31 第2円環領域
32 第3円環領域
40、41 第1領域
50 第1円領域
51 第2円領域
52 第4円環領域
53 第3円領域
60 第1領域
70 第1の扇型領域
71 第2の扇型領域
72 第3の扇型領域
74 第1の扇型領域
80、81 第1領域
90 顕微鏡システム
91 照明光学系
100 照明光学系の瞳面
101、102 第2領域
110 顕微鏡システム
111 開口部材
112 プロセッサ
113 遮光領域
112a、112b、112c、112d 透過領域
120 顕微鏡システム
121 インコヒーレント光源
122 照明光学系
130 推定標本
130a 表面
140 標本
140’ 光学像
141 移動ステージ
150 推定標本
150a 表面
L1、L2 光束
Fo 検出光学系の焦点位置
Pu 検出光学系の瞳位置
IP 結像面
C 検出光学系の瞳の中心
IP 結像面
PI 照明光学系の瞳面
Claims (25)
- インコヒーレント光源と、検出光学系と、撮像素子と、を備え、
前記インコヒーレント光源は、時間的にコヒーレントでない光を射出する光源であり、
前記検出光学系は、標本の光学像を形成する光学系であり、
前記撮像素子は、前記検出光学系により形成された前記標本の光学像を受光し、
前記標本では、前記インコヒーレント光源から出射した光によって複数のコヒーレント照明が同時に行われ、
前記コヒーレント照明は、空間的にコヒーレントである光による照明であり、
前記標本に対して光束が照射される方向は、前記コヒーレント照明それぞれで異なり、
前記検出光学系の瞳面では、前記コヒーレント照明それぞれの光束が互いに異なる第1領域を通過し、
前記第1領域のそれぞれは、以下の条件(1)を満たし、
隣接する2つの前記第1領域の間隔のうち少なくとも1つの間隔は、以下の条件(2)を満たすことを特徴とする顕微鏡システム。
LS<PS×10-3 (1)
0.05×T<d (2)
ここで、
LSは、前記第1領域の面積(単位はmm2)、
PSは、前記検出光学系の瞳の面積(単位はmm2)、
dは、隣接する2つの前記第1領域の間隔(単位はmm)、
Tは、前記検出光学系の瞳の直径(単位はmm)、
である。 - 前記第1領域の半数が、前記条件(2)を満たすことを特徴とする請求項1に記載の顕微鏡システム。
- 前記第1領域のいくつかは、第1円環領域内に位置し、
前記第1円環領域は、前記検出光学系の瞳領域のうち、半径が50%以上の領域であることを特徴とする請求項1に記載の顕微鏡システム。 - 前記第1領域のいくつかは、前記第1円環領域内において、2重の円を形成するように並んでいることを特徴とする請求項4に記載の顕微鏡システム。
- 前記第1領域のいくつかは、第2円環領域内に位置し、
前記第2円環領域は、前記検出光学系の瞳領域のうち、半径が70%から90%の領域であることを特徴とする請求項4に記載の顕微鏡システム。 - 前記第1領域のいくつかは、第3円環領域内に位置し、
前記第3円環領域は、前記検出光学系の瞳領域のうち、半径が50%から70%の領域であることを特徴とする請求項6に記載の顕微鏡システム。 - 前記第1領域のいくつかは、第1円領域内に位置し、
前記第1円領域は、前記検出光学系の瞳領域のうち、前記第1円環領域より中心側の領域であることを特徴とする請求項4に記載の顕微鏡システム。 - 前記第1領域のいくつかは、前記第1円領域内において、円を形成するように並んでいることを特徴とする請求項8に記載の顕微鏡システム。
- 前記第1領域のいくつかは、第2円領域内に位置し、
前記第2円領域は、前記検出光学系の瞳領域のうち、半径が50%以下の領域であることを特徴とする請求項1に記載の顕微鏡システム。 - 前記第1領域のいくつかは、前記第2円領域内において、円を形成するように並んでいることを特徴とする請求項10に記載の顕微鏡システム。
- 前記第1領域のいくつかは、第4円環領域内に位置し、
前記第4円環領域は、前記検出光学系の瞳領域のうち、半径が30%以上から50%の領域であることを特徴とする請求項10に記載の顕微鏡システム。 - 前記第1領域のいくつかは、、第3円領域内に位置し、
前記第3円領域は、前記検出光学系の瞳領域のうち、半径が30%以下の領域であることを特徴とする請求項12に記載の顕微鏡システム。 - 前記検出光学系の瞳を中心角が等しい4つの扇型に分けたとき、4つの前記扇型の全てに前記第1領域のいずれかが位置していることを特徴とする請求項1に記載の顕微鏡システム。
- 前記第1領域のいくつかは、前記検出光学系の瞳の中心を挟んで対をなしていることを特徴とする請求項1に記載の顕微鏡システム。
- 前記第1領域のそれぞれは、以下の条件(4)を満たすことを特徴とする請求項1に記載の顕微鏡システム。
PS×10-6<LS (4) - 隣接する2つの前記第1領域の間隔のうち少なくとも1つの間隔は、条件(2)及び以下の条件(5)を満たすことを特徴とする請求項1に記載の顕微鏡システム。
d<0.5×T (5) - 照明光学系を更に備え、
前記照明光学系の瞳面では、前記コヒーレント照明それぞれの光束が互いに異なる第2領域に位置し、
前記第2領域のそれぞれは、以下の条件(6)を満たし、
隣接する2つの前記第2領域の間隔のうち少なくとも1つの間隔は、条件(2)及び以下の条件(7)を満たすことを特徴とする請求項1に記載の顕微鏡システム。
LS’<PS’×10-3 (6)
0.05×T’<d’ (7)
ここで、
LS’は、前記第2領域の面積(単位はmm2)、
PS’は、前記照明光学系の瞳の面積(単位はmm2)、
d’は、隣接する2つの前記第2領域の間隔(単位はmm)、
T’は、前記照明光学系の瞳の直径(単位はmm)、
である。 - 開口部材を更に備え、
前記コヒーレント照明それぞれの光束は、所定面上の独立した複数の領域のそれぞれから射出され、
前記所定面は、前記検出光学系の光軸と直交する面であり、かつ、前記標本に対して前記検出光学系と反対側の位置にある面であり、
前記開口部材は、前記所定面に配置され、独立した複数の透過領域を備え、前記透過領域は、光を透過する領域であり、
前記透過領域のそれぞれは、前記第1領域のそれぞれに対応することを特徴とする請求項1に記載の顕微鏡システム。 - 前記コヒーレント照明それぞれの光束は、所定面上の独立した複数の領域のそれぞれから射出され、
前記所定面は、前記検出光学系の光軸と直交する面であり、かつ、前記標本に対して前記検出光学系と反対側の位置にある面であり、
前記インコヒーレント光源は、前記所定面に複数配置され、
前記インコヒーレント光源のそれぞれは、前記第1領域のそれぞれに対応することを特徴とする請求項1に記載の顕微鏡システム。 - インコヒーレント光源と、照明光学系と、検出光学系と、撮像素子と、を備え、
前記インコヒーレント光源は、時間的にコヒーレントでない光を射出する光源であり、
前記検出光学系は、標本の光学像を形成する光学系であり、
前記撮像素子は、前記検出光学系により形成された前記標本の光学像を受光し、
前記標本では、前記インコヒーレント光源から出射した光によって複数のコヒーレント照明が同時に行われ、
前記コヒーレント照明は、空間的にコヒーレントである光による照明であり、
前記標本に対して光束が照射される方向は、前記コヒーレント照明それぞれで異なり、
前記照明光学系の瞳面では、前記コヒーレント照明それぞれの光束が互いに異なる第2領域に位置し、
前記第2領域のそれぞれは、以下の条件(6)を満たし、
隣接する2つの前記第2領域の間隔のうち少なくとも1つの間隔は、以下の条件(7)を満たすことを特徴とする顕微鏡システム。
LS’<PS’×10-3 (6)
0.05×T’<d’ (7)
ここで、
LS’は、前記第2領域の面積(単位はmm2)、
PS’は、前記照明光学系の瞳の面積(単位はmm2)、
d’は、隣接する2つの前記第2領域の間隔(単位はmm)、
T’は、前記照明光学系の瞳の直径(単位はmm)、
である。 - 前記検出光学系は、対物レンズと、結像レンズと、を備え、
前記照明光学系は、コンデンサレンズを備え、
前記第2領域の面積は、以下の式(8)で表され、
前記照明光学系の瞳の直径は、以下の式(9)で表されることを特徴とする請求項21に記載の顕微鏡システム。
PS’=(FLcd×NA)2×π (8)
T’=FLcd×NA (9)
ここで、
FLcdは、前記コンデンサレンズの焦点距離(単位mm)、
NAは、前記対物レンズの開口数、
である。 - 開口部材を更に備え、
前記複数の光束のそれぞれは、所定面上の独立した複数の領域のそれぞれから射出され、
前記所定面は、前記検出光学系の光軸と直交する面であり、かつ、前記標本に対して前記検出光学系と反対側の位置にある面であり、
前記開口部材は、前記所定面に配置され、独立した複数の透過領域を備え、前記透過領域は、光を透過する領域であり、
前記透過領域のそれぞれは、前記第2領域のそれぞれに対応することを特徴とする請求項21に記載の顕微鏡システム。 - 前記複数の光束のそれぞれは、所定面上の独立した複数の領域のそれぞれから射出され、
前記所定面は、前記検出光学系の光軸と直交する面であり、かつ、前記標本に対して前記検出光学系と反対側の位置にある面であり、
前記インコヒーレント光源は、前記所定面に複数配置され、
前記インコヒーレント光源のそれぞれは、前記第2領域のそれぞれに対応することを特徴とする請求項21に記載の顕微鏡システム。 - プロセッサを更に備え、
前記プロセッサは、
前記標本をモデル化した推定標本を通過する波面を、前記光束毎に順伝搬演算で求め、
前記光束毎に、前記波面に対応する前記検出光学系の結像位置における強度分布を算出し、
前記光束毎の前記強度分布を足し合わせることで計算像を生成し、
前記計算像と、前記撮像素子から出力された測定像と、の差を小さくする最適化処理を行うことで、前記推定標本を再構築することを特徴とする請求項1または請求項21に記載の顕微鏡システム。
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JP2014063151A (ja) * | 2012-08-29 | 2014-04-10 | Canon Inc | 顕微鏡用照明光学系およびこれを用いた顕微鏡 |
JP2015179208A (ja) * | 2014-03-19 | 2015-10-08 | キヤノン株式会社 | 撮像装置および画像処理システム |
WO2016125281A1 (ja) * | 2015-02-05 | 2016-08-11 | 株式会社ニコン | 構造化照明顕微鏡、観察方法、及び制御プログラム |
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Publication number | Priority date | Publication date | Assignee | Title |
---|---|---|---|---|
JP2014044254A (ja) * | 2012-08-24 | 2014-03-13 | Canon Inc | 顕微鏡用照明光学系およびこれを用いた顕微鏡 |
JP2014063151A (ja) * | 2012-08-29 | 2014-04-10 | Canon Inc | 顕微鏡用照明光学系およびこれを用いた顕微鏡 |
JP2015179208A (ja) * | 2014-03-19 | 2015-10-08 | キヤノン株式会社 | 撮像装置および画像処理システム |
WO2016125281A1 (ja) * | 2015-02-05 | 2016-08-11 | 株式会社ニコン | 構造化照明顕微鏡、観察方法、及び制御プログラム |
Non-Patent Citations (2)
Title |
---|
"High-resolution 3D refractive index microscopy of multiple-scattering samples from intensity images", OPTICA, vol. 6, no. 9, 2019, pages 1211 - 1219 |
M. R. TEAGUE: "Deterministic phase retrieval: a Greens function solution", J. OPT. SOC. AM., vol. 73, 1983, pages 1434 - 1441, XP001026262, DOI: 10.1364/JOSA.73.001434 |
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