光的世界︰矩陣光學六己

事物可『相連』未必能有『關係』的乎?俗話說︰事不關己,關心則亂!此事與矩陣光學何干耶??也許下面文字可說明一二吧!!

或許有讀者會問『哲學』與『 pyDatalog 物件導向設計』有關嗎?值得花時間討論這種問題嘛!其實西方『理性主義』中之思辨精神也就是一種『□□方法學』。然而『方法學』不只一種,『思潮』也不只有一次,如是可以了解,下面這篇文章的『題旨』︰

DDandOO

或將知道『理念本身並不講話』,不同『詮釋』下也許『內容』可大不同。在面對『雲端大數據』新潮流時,勇闖新世界之先,能多作點準備罷了。假使真可以『比較哲學』,也許讀一段莊子,大概就能體會東方思辨以『生生取向』之大不同的吧︰

莊子》‧《大宗師

死 生,命也,其有夜旦之常,天也。人之有所不得與,皆物之情也 。彼特以天為父,而身猶愛之,而況其卓乎!人特以有君為愈乎己 ,而身猶死之,而況其真乎!泉涸,魚相與處於陸,相呴以溼,相濡以沫,不如相忘於江湖。與其譽堯而非桀,不如兩忘而化其道。夫大塊載我以形,勞我以生,佚 我以老,息我以死。故善吾生者,乃所以善吾死也。夫藏舟於壑,藏山於澤,謂之固矣。然而夜半有力者負之而走,昧者不知也。藏大小有宜,猶有所遯。若夫藏天 下於天下,而不得所遯,是恆物之大情也。特犯人之形而猶喜之,若人之形者,萬化而未始有極也,其為樂可勝計邪!故聖人將遊於物之所不得遯而皆存。善妖善 老,善始善終,人猶效之,又況萬物之所係,而一化之所待乎!

夫 道,有情有信,無為無形;可傳而不可受,可得而不可見;自本自根,未有天地,自古以固存;神鬼神帝,生天生地;在太極之先而不為高,在六極之下而不為深; 先天地生而不為久,長於上古而不為老。豨韋氏得之,以挈天地;伏犧氏得之,以襲氣母;維斗得之,終古不忒;日月得之,終古不息;堪坏得之,以襲崑崙;馮夷 得之,以遊大川;肩吾得之,以處太山;黃帝得之,以登雲天;顓頊得之,以處玄宮;禺強得之,立乎北極;西王母得之,坐乎少廣 ,莫知其始,莫知其終;彭祖得之,上及有虞,下及五伯;傅說得之,以相武丁,奄有天下,乘東維,騎箕尾,而比於列星。

── 摘自《勇闖新世界︰ 《 pyDatalog 》【專題】之物件導向設計 ‧三

 

上篇『等效薄透鏡』說法,實則美矣!不過並未提及『節點』 nodal point 與『節面』 nodal plane 在哪哩?事實上它和『主平面』以及『主點』重合,可以驗以通過『節點』【※主點】之任意光 (0, \theta) ,離開角度 \theta 不變而得知︰

\left( \begin{array}{cc} 1 & 0 \\  - \frac{1}{f_{eff}} & 1 \end{array} \right) \left( \begin{array}{c} 0  \\  \theta \end{array} \right) = \left( \begin{array}{c} 0 \\ \theta \end{array} \right)

如是對一個『厚透鏡』來說

  \left( \begin{array}{cc} 1 - \alpha &  \beta  \\ - \frac{1}{f} & 1 + \gamma \end{array} \right)

,此處

\alpha = \frac{(n - 1) d}{n R_1}\beta = \frac{d}{n}

\frac{1}{f} = (n-1) \left( \frac{1}{R_1} - \frac{1}{R_2} + \frac{(n - 1) d}{n R_1 R_2} \right)\gamma = \frac{(n - 1) d}{n R_2}

主平面一

,這些點、面落在

p_1 = \frac{1 - D}{C} = \gamma f

p_2 = \frac{1 - A}{C} = - \alpha f

位置。

比較與『頂點』 vertex 度量對應之『前焦距』為 FFL = (1 + \gamma) f ,『後焦距』是 BFL = (1 - \alpha) f 的了。因此 f 就是 f_{eff} 的矣! !然而實務上即使知道了一光學系統之『前焦距』與『後焦距』,由於尚有 \alpha\gammaf 三個參數,且無法決定 f_{eff} 的 呦??

但思『焦、焦』面『參考系』︰

牛頓成像公式

 

可用牛頓成像公式︰

x \cdot x' = FFL * BFL

假使考察『薄透鏡』之『焦、焦』面矩陣形制︰

\left( \begin{array}{cc} 1 & f \\  0 & 1 \end{array} \right) \left( \begin{array}{cc} 1 & 0 \\  - \frac{1}{f} & 1 \end{array} \right) \left( \begin{array}{cc} 1 & f \\  0 & 1 \end{array} \right) = \left( \begin{array}{cc} 0 & f \\ - \frac{1}{f} & 0 \end{array} \right)

成像條件可以推導為︰

\left( \begin{array}{cc} 1 & x' \\  0 & 1 \end{array} \right) \left( \begin{array}{cc} 0 & f \\  - \frac{1}{f} & 0 \end{array} \right) \left( \begin{array}{cc} 1 & x  \\  0 & 1 \end{array} \right) = \left( \begin{array}{cc} - \frac{x'}{f} & f - \frac{x' x}{f} \\ - \frac{1}{f} & - \frac{x}{f}  \end{array} \right)

\therefore x x' = f^2

那麼當 x = x' = f 時,放大率等於1 ,如此當可確定 f 的吧 !!??

不知『厚透鏡』的『焦、焦』面矩陣形制是否也如此的呢???

pi@raspberrypi:~ ipython3 Python 3.4.2 (default, Oct 19 2014, 13:31:11)  Type "copyright", "credits" or "license" for more information.  IPython 2.3.0 -- An enhanced Interactive Python. ?         -> Introduction and overview of IPython's features. %quickref -> Quick reference. help      -> Python's own help system. object?   -> Details about 'object', use 'object??' for extra details.  In [1]: from sympy import *  In [2]: from sympy.physics.optics import FreeSpace, FlatRefraction, ThinLens, GeometricRay, CurvedRefraction, RayTransferMatrix  In [3]: init_printing()  In [4]: f, z, α, β, γ = symbols('f, z, α, β, γ')  In [5]: 厚透鏡 = RayTransferMatrix(1 - α, β, -1/f, 1 +γ)  In [6]: 厚透鏡 Out[6]:  ⎡-α + 1    β  ⎤ ⎢             ⎥ ⎢ -1          ⎥ ⎢ ───    γ + 1⎥ ⎣  f          ⎦  In [7]: 前焦平面 = RayTransferMatrix(1, (1 + γ) * f, 0, 1)  In [8]: 前焦平面 Out[8]:  ⎡1  f⋅(γ + 1)⎤ ⎢            ⎥ ⎣0      1    ⎦  In [9]: 後焦平面 = RayTransferMatrix(1, (1 - α) * f, 0, 1)  In [10]: 後焦平面 Out[10]:  ⎡1  f⋅(-α + 1)⎤ ⎢             ⎥ ⎣0      1     ⎦  In [11]: 厚透鏡 * 前焦平面 Out[11]:  ⎡-α + 1  f⋅(-α + 1)⋅(γ + 1) + β⎤ ⎢                              ⎥ ⎢ -1                           ⎥ ⎢ ───              0           ⎥ ⎣  f                           ⎦  In [12]: 後焦平面 * 厚透鏡 * 前焦平面 Out[12]:  ⎡ 0   f⋅(-α + 1)⋅(γ + 1) + β⎤ ⎢                           ⎥ ⎢-1                         ⎥ ⎢───            0           ⎥ ⎣ f                         ⎦  In [13]:  </pre>    果真f \cdot (1 - \alpha) \cdot (1 + \gamma) + \beta = f的嗎??!!  若知\frac{1}{f}可以表示成︰\frac{1}{f} = \frac{\alpha \gamma + \alpha - \gamma}{\beta}於是f \cdot (1 - \alpha) \cdot (1 + \gamma) + \beta= f \cdot \left[ (1 - \alpha) \cdot (1 + \gamma)  + \frac{\beta}{f} \right]= f \cdot  \left[ (1 - \alpha) \cdot (1 + \gamma)  +  \alpha \gamma + \alpha - \gamma \right]= f \cdot 1$

 

誠神奇焉!!!

反想從『焦、焦』面倒推『主平面』與『主點』位置,終究是一樣的吧☆

 

 

 

 

 

 

 

 

 

 

 

 

 

光的世界︰矩陣光學六戊

在《光的世界︰矩陣光學六丙》文章中,我們談過厚透鏡

  \left( \begin{array}{cc} 1 - \alpha &  \beta  \\ - \frac{1}{f} & 1 + \gamma \end{array} \right)

,此處

\alpha = \frac{(n - 1) d}{n R_1}\beta = \frac{d}{n}

\frac{1}{f} = (n-1) \left( \frac{1}{R_1} - \frac{1}{R_2} + \frac{(n - 1) d}{n R_1 R_2} \right)\gamma = \frac{(n - 1) d}{n R_2}

 

。如薄透鏡 d \to 0 一般,都具有前、後聚焦的性質,然而它的數學表達式太過麻煩,並不方便論理與應用。假使一個任意複雜之光學系統竟能夠『等效』於一片『薄透鏡』豈非太美妙耶!且聽聽 Justin Peatross 和 Michael Ware 先生們之大哉論也

主平面一

主平面二

主平面三

主平面四

※ 註︰ 若一光學系統光線進、出介質折射率相同,則

det  \left( \begin{array}{cc} A &  B  \\ C & D \end{array}  \right)  = 1

顯然 C 不等於零。

 

這將人們帶入『主平面』和『主點』的世界

Cardinal point (optics)

In Gaussian optics, the cardinal points consist of three pairs of points located on the optical axis of a rotationally symmetric, focal, optical system. These are the focal points, the principal points, and the nodal points.[1] For ideal systems, the basic imaging properties such as image size, location, and orientation are completely determined by the locations of the cardinal points; in fact only four points are necessary: the focal points and either the principal or nodal points. The only ideal system that has been achieved in practice is the plane mirror,[2] however the cardinal points are widely used to approximate the behavior of real optical systems. Cardinal points provide a way to analytically simplify a system with many components, allowing the imaging characteristics of the system to be approximately determined with simple calculations.

250px-Cardinal-points-1.svg

The cardinal points of a thick lens in air.
F, F’ front and rear focal points,
P, P’ front and rear principal points,
V, V’ front and rear surface vertices.

Explanation

The cardinal points lie on the optical axis of the optical system. Each point is defined by the effect the optical system has on rays that pass through that point, in the paraxial approximation. The paraxial approximation assumes that rays travel at shallow angles with respect to the optical axis, so that  \sin\theta\approx\theta and  \cos\theta\approx 1.[3] Aperture effects are ignored: rays that do not pass through the aperture stop of the system are not considered in the discussion below.

Focal planes

The front focal point of an optical system, by definition, has the property that any ray that passes through it will emerge from the system parallel to the optical axis. The rear (or back) focal point of the system has the reverse property: rays that enter the system parallel to the optical axis are focused such that they pass through the rear focal point.

BackFocalPlane.svg

Rays that leave the object with the same angle cross at the back focal plane.

The front and rear (or back) focal planes are defined as the planes, perpendicular to the optic axis, which pass through the front and rear focal points. An object infinitely far from the optical system forms an image at the rear focal plane. For objects a finite distance away, the image is formed at a different location, but rays that leave the object parallel to one another cross at the rear focal plane.

A diaphragm or “stop” at the rear focal plane can be used to filter rays by angle, since:

  1. It only allows rays to pass that are emitted at an angle (relative to the optical axis) that is sufficiently small. (An infinitely small aperture would only allow rays that are emitted along the optical axis to pass.)
  2. No matter where on the object the ray comes from, the ray will pass through the aperture as long as the angle at which it is emitted from the object is small enough.

Note that the aperture must be centered on the optical axis for this to work as indicated. Using a sufficiently small aperture in the focal plane will make the lens telecentric.

Similarly, the allowed range of angles on the output side of the lens can be filtered by putting an aperture at the front focal plane of the lens (or a lens group within the overall lens). This is important for DSLR cameras having CCD sensors. The pixels in these sensors are more sensitive to rays that hit them straight on than to those that strike at an angle. A lens that does not control the angle of incidence at the detector will produce pixel vignetting in the images.

BackFocalPlane_aperture.svg

Angle filtering with an aperture at the rear focal plane.

Principal planes and points

The two principal planes have the property that a ray emerging from the lens appears to have crossed the rear principal plane at the same distance from the axis that that ray appeared to cross the front principal plane, as viewed from the front of the lens. This means that the lens can be treated as if all of the refraction happened at the principal planes. The principal planes are crucial in defining the optical properties of the system, since it is the distance of the object and image from the front and rear principal planes that determines the magnification of the system. The principal points are the points where the principal planes cross the optical axis.

If the medium surrounding the optical system has a refractive index of 1 (e.g., air or vacuum), then the distance from the principal planes to their corresponding focal points is just the focal length of the system. In the more general case, the distance to the foci is the focal length multiplied by the index of refraction of the medium.

For a thin lens in air, the principal planes both lie at the location of the lens. The point where they cross the optical axis is sometimes misleadingly called the optical centre of the lens. Note, however, that for a real lens the principal planes do not necessarily pass through the centre of the lens, and in general may not lie inside the lens at all.

Lens_shapes.svg

Various lens shapes, and the location of the principal planes.

Nodal points

The front and rear nodal points have the property that a ray aimed at one of them will be refracted by the lens such that it appears to have come from the other, and with the same angle with respect to the optical axis. The nodal points therefore do for angles what the principal planes do for transverse distance. If the medium on both sides of the optical system is the same (e.g., air), then the front and rear nodal points coincide with the front and rear principal points, respectively.

The nodal points are widely misunderstood in photography, where it is commonly asserted that the light rays “intersect” at “the nodal point”, that the iris diaphragm of the lens is located there, and that this is the correct pivot point for panoramic photography, so as to avoid parallax error.[4][5][6] These claims generally arise from confusion about the optics of camera lenses, as well as confusion between the nodal points and the other cardinal points of the system. (A better choice of the point about which to pivot a camera for panoramic photography can be shown to be the centre of the system’s entrance pupil.[4][5][6] On the other hand, swing-lens cameras with fixed film position rotate the lens about the rear nodal point to stabilize the image on the film.[6][7])

Cardinal-points-2.svg

N, N’ The front and rear nodal points of a thick lens.

Surface vertices

The surface vertices are the points where each optical surface crosses the optical axis. They are important primarily because they are the physically measurable parameters for the position of the optical elements, and so the positions of the cardinal points must be known with respect to the vertices to describe the physical system.

In anatomy, the surface vertices of the eye’s lens are called the anterior and posterior poles of the lens.[8]

 

也可一睹『相對』 □ 面 ○ 點 『定位』之風采乎!!!

 

 

 

 

 

 

 

 

 

 

 

送別七月半

君不見久地長天一時盡?人情依依道

送別》! 李叔同

長亭外,古道邊
芳草碧連天
晚風拂柳笛聲殘
夕陽山外山

天之涯,地之角
知交半零落
一瓢濁酒盡餘歡
今宵別夢寒

草碧色,水綠波
南浦傷如何
人生難得是歡聚
唯有別離多

情千縷,酒一杯
聲聲離笛催
問君此去幾時來
來時莫徘徊

韶光逝,留無計
今日卻分訣
驪歌一曲送別離
相顧卻依依

聚雖好,別雖悲
世事堪玩味
來日後會相予期
去去莫遲疑

 

 

 

 

 

 

 

 

 

 

 

光的世界︰矩陣光學六丁

Justin Peatross 和 Michael Ware 兩位先生解釋『成像條件

成像條件一

成像條件二

 

宛如順手捻來不費力氣!甚至三言兩語說明了經典的『幾何光學』三條線!!

幾何光學三條線一

幾何光學三條線二

 

此處僅補之以『幾何』的推導︰

GeoGebra(在 raspberrypi)_105

 

上圖且將『物』 y_1、『像』- y_2 多畫幾次。同時參照 Figure 9.14 ,依相似三角形可得︰

\frac{f}{d_o} =  \frac{-y_2}{y_1 - y_2}\frac{f}{d_i} =  \frac{y_1}{y_1 - y_2}

,兩式相加後即是『成像公式』

\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}

的了。作點簡單代數運算,即可將此式改寫為

(d_o - f ) (d_i - f) = f^2

。這就是牛頓成像公式的哩。

若問這麼清楚明白的事能有懸念乎?假設位於透鏡前焦距之前之物 ,形成倒立實像,人可以見得著嗎??相機能夠拍得到耶??!!何不秉持實證精神,自己檢驗一番呢!!??

Image formation with a lens

Demonstration

A set of experiments to introduce real and virtual images formed by a convex lens.

Apparatus and materials

Lens (+7D or 150 mm focal length)

Sheet of plain white paper

Greaseproof paper, small pieces

Floodlights (optional)

Health & Safety and Technical notes


If the Sun is visible from the laboratory windows, it is essential for the teacher to remind students that looking at the Sun through a lens will cause blindness.

In the following description, the source of light is referred to as a window. In some cases an electric lamp may be better, as it can be used in a room that is at least half dark. A carbon filament lamp and mounted lampholder are suitable.

Procedure


a Lenses forming real images

Student with lens back to window

Face away from the window, holding a sheet of paper at arm’s length. Hold the lens in front of the paper and move it to and from the paper (towards you and back towards the paper) until you see an image of the window on the paper.

b Real image with and without screen 

Student with lens facing window
Hold a lens at arm’s length towards the window. Hold a piece of greaseproof paper in the other hand and find the place between the lens and your head where there is a clear image of the window on the paper. You are looking at that image through the paper. Your eyes should be focused on the paper itself.

Remove the paper and look at the image in space. If you cannot find the image put the paper back and repeat the process. It may be helpful to catch half the image on the edge of the paper and the other half in space. Concentrate on the image on the paper and slide the paper away.
………

 

請讀者然後細思

  \left( \begin{array}{cc} A_2 &  0  \\ C_2 & D_2 \end{array} \right)  \left( \begin{array}{cc} A_1 &  0  \\ C_1 & D_1 \end{array} \right) = \left( \begin{array}{cc} A_2  A_1 &  0  \\ C_2 A_1 + D_2 C_1 & D_2  D_1 \end{array} \right)

 

,到底宣說何理呦???

【※ 隨文附註】

pi@raspberrypi:~ $ ipython3
Python 3.4.2 (default, Oct 19 2014, 13:31:11) 
Type "copyright", "credits" or "license" for more information.

IPython 2.3.0 -- An enhanced Interactive Python.
?         -> Introduction and overview of IPython's features.
%quickref -> Quick reference.
help      -> Python's own help system.
object?   -> Details about 'object', use 'object??' for extra details.

In [1]: from sympy import *

In [2]: from sympy.physics.optics import FreeSpace, FlatRefraction, ThinLens, GeometricRay, CurvedRefraction, RayTransferMatrix

In [3]: init_printing()

In [4]: do1, di1, f1, do2, di2, f2 = symbols('do1, di1, f1, do2, di2, f2')

In [5]: RealImage = RayTransferMatrix(-di1 / do1, 0, - 1 / f1, -do1 / di1)

In [6]: RealImage
Out[6]: 
⎡-di₁        ⎤
⎢─────    0  ⎥
⎢ do₁        ⎥
⎢            ⎥
⎢ -1    -do₁ ⎥
⎢ ───   ─────⎥
⎣  f₁    di₁ ⎦

In [7]: ViewImage = FreeSpace(di2) * ThinLens(f2) * FreeSpace(do2) * RealImage

In [8]: ViewImage
Out[8]: 
⎡      ⎛  di₂    ⎞             ⎛  di₂    ⎞       ⎛          ⎛  di₂    ⎞⎞ ⎤
⎢  di₁⋅⎜- ─── + 1⎟   di₂ + do₂⋅⎜- ─── + 1⎟  -do₁⋅⎜di₂ + do₂⋅⎜- ─── + 1⎟⎟ ⎥
⎢      ⎝   f₂    ⎠             ⎝   f₂    ⎠       ⎝          ⎝   f₂    ⎠⎠ ⎥
⎢- ─────────────── - ─────────────────────  ─────────────────────────────⎥
⎢        do₁                   f₁                        di₁             ⎥
⎢                                                                        ⎥
⎢                      do₂                             ⎛  do₂    ⎞       ⎥
⎢                    - ─── + 1                    -do₁⋅⎜- ─── + 1⎟       ⎥
⎢            di₁        f₂                             ⎝   f₂    ⎠       ⎥
⎢           ────── - ─────────                    ─────────────────      ⎥
⎣           do₁⋅f₂       f₁                              di₁             ⎦

In [9]: ViewImage.B
Out[9]: 
     ⎛          ⎛  di₂    ⎞⎞ 
-do₁⋅⎜di₂ + do₂⋅⎜- ─── + 1⎟⎟ 
     ⎝          ⎝   f₂    ⎠⎠ 
─────────────────────────────
             di₁             

In [10]: 

 

 

 

 

 

 

 

 

 

 

光的世界︰矩陣光學六丙

若問一個『厚透鏡』

  \left( \begin{array}{cc} 1 - \alpha &  \beta  \\ - \frac{1}{f} & 1 + \gamma \end{array} \right)

,此處

\alpha = \frac{(n - 1) d}{n R_1}\beta = \frac{d}{n}

\frac{1}{f} = (n-1) \left( \frac{1}{R_1} - \frac{1}{R_2} + \frac{(n - 1) d}{n R_1 R_2} \right)\gamma = \frac{(n - 1) d}{n R_2}

 

。是否也如『薄透鏡』 d \to 0 一般,具有『焦距』的呢?

pi@raspberrypi:~ ipython3 Python 3.4.2 (default, Oct 19 2014, 13:31:11)  Type "copyright", "credits" or "license" for more information.  IPython 2.3.0 -- An enhanced Interactive Python. ?         -> Introduction and overview of IPython's features. %quickref -> Quick reference. help      -> Python's own help system. object?   -> Details about 'object', use 'object??' for extra details.  In [1]: from sympy import *  In [2]: from sympy.physics.optics import FreeSpace, FlatRefraction, ThinLens, GeometricRay, CurvedRefraction, RayTransferMatrix  In [3]: init_printing()  In [4]: f, z, α, β, γ = symbols('f, z, α, β, γ')  In [5]: 厚透鏡 = RayTransferMatrix(1 - α, β, -1/f, 1 +γ)  In [6]: 厚透鏡 Out[6]:  ⎡-α + 1    β  ⎤ ⎢             ⎥ ⎢ -1          ⎥ ⎢ ───    γ + 1⎥ ⎣  f          ⎦  In [7]: 行經距離Z = FreeSpace(z)  In [8]: Z處匯聚現象 = 行經距離Z * 厚透鏡  In [9]: Z處匯聚現象 Out[9]:  ⎡         z               ⎤ ⎢-α + 1 - ─  z⋅(γ + 1) + β⎥ ⎢         f               ⎥ ⎢                         ⎥ ⎢   -1                    ⎥ ⎢   ───          γ + 1    ⎥ ⎣    f                    ⎦  In [10]: h, θ = symbols('h, θ')  In [11]: 平行光 = GeometricRay(h, θ)  In [12]: 平行光 Out[12]:  ⎡h⎤ ⎢ ⎥ ⎣θ⎦  In [13]: Z處光束 = Z處匯聚現象 * 平行光  In [14]: Z處光束 Out[14]:  ⎡  ⎛         z⎞                    ⎤ ⎢h⋅⎜-α + 1 - ─⎟ + θ⋅(z⋅(γ + 1) + β)⎥ ⎢  ⎝         f⎠                    ⎥ ⎢                                  ⎥ ⎢                      h           ⎥ ⎢          θ⋅(γ + 1) - ─           ⎥ ⎣                      f           ⎦  In [15]: 前焦點現象 = 厚透鏡 * 行經距離Z  In [16]: 前焦點現象 Out[16]:  ⎡-α + 1  z⋅(-α + 1) + β⎤ ⎢                      ⎥ ⎢ -1               z   ⎥ ⎢ ───      γ + 1 - ─   ⎥ ⎣  f               f   ⎦  In [17]: 方將通過前焦點之任意光線 = GeometricRay(0, θ)  In [18]: 前焦點現象 * 方將通過前焦點之任意光線 Out[18]:  ⎡θ⋅(z⋅(-α + 1) + β)⎤ ⎢                  ⎥ ⎢    ⎛        z⎞   ⎥ ⎢  θ⋅⎜γ + 1 - ─⎟   ⎥ ⎣    ⎝        f⎠   ⎦  In [19]:  </pre>    <span style="color: #003300;">簡單計算後,由下式</span>h \cdot \left( - \alpha + 1 -\frac{z}{f} \right) + \theta \cdot \left( z \cdot ( \gamma + 1 ) + \beta \right) <span style="color: #003300;">可知,當z = (1 - \alpha) \cdot f 時,平行光束將聚焦於『焦平面』上。再從</span> \theta \cdot \left( \gamma + 1 - \frac{z}{f} \right) <span style="color: #003300;">這個式子可曉,當z = (1 + \gamma) \cdot f 時,穿越『前焦點』之光束經過『厚透鏡』後,將成軸向平行光的了。</span>  <span style="color: #003300;">所以<a style="color: #003300;" href="https://en.wikipedia.org/wiki/Lens_%28optics%29">透鏡</a>的厚、薄實無礙其光學之性質也。雖然因為\alpha = \frac{(n - 1) d}{n R_1} 與\gamma = \frac{(n - 1) d}{n R_2} 項之存在,前、後焦距略有差異,果無妨論理乎!如是為著簡化複雜的數學表達式,除非真有必要,僅以『薄透鏡』說事的了!! </span>  <span style="color: #808080;">※ 註︰ 此處雖用『聚焦』說法,只是依循『慣例』,將未明說的『符號』f
*** QuickLaTeX cannot compile formula:
之值取『正號』而已。若是實際值『為負』,意味著『發散』的也。</span>

<img class="alignnone size-full wp-image-57888" src="http://www.freesandal.org/wp-content/uploads/522px-Lens1b.svg.png" alt="522px-Lens1b.svg" width="522" height="345" />

 

<img class="alignnone size-full wp-image-57887" src="http://www.freesandal.org/wp-content/uploads/Concave_lens.jpg" alt="Concave_lens" width="800" height="378" />

 

<img class="alignnone size-full wp-image-57892" src="http://www.freesandal.org/wp-content/uploads/Lens_shapes.svg.png" alt="Lens_shapes.svg" width="553" height="406" />

<span style="color: #808080;">1-4為<a style="color: #808080;" title="凸透鏡" href="https://zh.wikipedia.org/wiki/%E5%87%B8%E9%80%8F%E9%95%9C">凸透鏡</a>,5-8為<a style="color: #808080;" title="凹透鏡" href="https://zh.wikipedia.org/wiki/%E5%87%B9%E9%80%8F%E9%95%9C">凹透鏡</a>,其中:</span>
<span style="color: #808080;"> <b>1</b> - 對稱雙凸透鏡,<b>2</b> - 非對稱雙凸透鏡,<b>3</b> - 平凸透鏡</span>
<span style="color: #808080;"> <b>4</b> - 凹凸透鏡(凸度大於凹度),<b>5</b> - 對稱雙凹透鏡</span>
<span style="color: #808080;"> <b>6</b> - 非對稱雙凹透鏡, <b>7</b> - 平凹透鏡,<b>8</b> - 凸凹透鏡(凹度大於凸度)
</span>

 

<span style="color: #003300;">所謂</span>
<h1 id="firstHeading" class="firstHeading" lang="zh-TW"><span style="color: #003300;"><a style="color: #003300;" href="https://zh.wikipedia.org/zh-tw/%E7%84%A6%E9%BB%9E">焦點</a></span></h1>
<span style="color: #808080;"><b>焦點</b>,在<a style="color: #808080;" title="幾何光學" href="https://zh.wikipedia.org/wiki/%E5%87%A0%E4%BD%95%E5%85%89%E5%AD%A6">幾何光學</a>中有時也稱為<b>像點</b>,是源頭的<a class="mw-redirect" style="color: #808080;" title="光線" href="https://zh.wikipedia.org/wiki/%E5%85%89%E7%B7%9A">光線</a>經過物鏡後匯聚的點。<sup id="cite_ref-1" class="reference"><a style="color: #808080;" href="https://zh.wikipedia.org/wiki/%E7%84%A6%E9%BB%9E#cite_note-1">[1]</a></sup>. 然而,焦點只是概念上的點,實際上在空間上有一個範圍,稱為<a class="mw-redirect" style="color: #808080;" title="朦朧圈" href="https://zh.wikipedia.org/wiki/%E6%9C%A6%E6%9C%A7%E5%9C%88">朦朧圈</a>。這種非理想的焦點也許會導致光學影像的<a style="color: #808080;" title="像差" href="https://zh.wikipedia.org/wiki/%E5%83%8F%E5%B7%AE">像差</a>,在沒有明顯的像差下,最小的朦朧圈是<a class="mw-redirect" style="color: #808080;" title="艾里盤" href="https://zh.wikipedia.org/wiki/%E8%89%BE%E9%87%8C%E7%9B%A4">艾里盤</a>,是因為光學系統的<a style="color: #808080;" title="口徑" href="https://zh.wikipedia.org/wiki/%E5%8F%A3%E5%BE%91">開口</a>產生<a class="mw-redirect" style="color: #808080;" title="繞射" href="https://zh.wikipedia.org/wiki/%E7%B9%9E%E5%B0%84">繞射</a>造成的。當口徑加大時,像差也會變得更為嚴重,而艾里圈是在大口徑下最小的。</span>

<span style="color: #808080;">一個影像,點像或區域如果能很好的被收歛就是<i><b>對焦</b></i>,如果未能良好的匯聚就是<i><b>失焦</b></i>。兩者之間的邊界有時被用來作為<a style="color: #808080;" title="模糊圈" href="https://zh.wikipedia.org/wiki/%E6%A8%A1%E7%B3%8A%E5%9C%88">模糊圈</a>的定義 。</span>

<span style="color: #808080;"><b>主焦點</b>或<b>焦點</b>是球面的焦點:</span>
<ul>
	<li><span style="color: #808080;">對一個<a class="mw-redirect" style="color: #808080;" title="透鏡" href="https://zh.wikipedia.org/wiki/%E9%80%8F%E9%8F%A1">透鏡</a>、<a style="color: #808080;" title="曲面鏡" href="https://zh.wikipedia.org/wiki/%E6%9B%B2%E9%9D%A2%E9%8F%A1">球面鏡</a>、或<a class="new" style="color: #808080;" title="拋物面鏡(頁面不存在)" href="https://zh.wikipedia.org/w/index.php?title=%E6%8B%8B%E7%89%A9%E9%9D%A2%E9%8F%A1&action=edit&redlink=1">拋物面鏡</a>,是被<a class="new" style="color: #808080;" title="瞄準(頁面不存在)" href="https://zh.wikipedia.org/w/index.php?title=%E7%9E%84%E6%BA%96&action=edit&redlink=1">瞄準</a>平行於光軸的光被聚集在光軸上的點。由於光可以從任何一個面穿越透鏡,因此透鏡在倆側各有一個焦點。在空氣中,透鏡或面鏡的<a class="new" style="color: #808080;" title="主平面(頁面不存在)" href="https://zh.wikipedia.org/w/index.php?title=%E4%B8%BB%E5%B9%B3%E9%9D%A2&action=edit&redlink=1">主平面</a>至焦點的距離稱為<i><a style="color: #808080;" title="焦距" href="https://zh.wikipedia.org/wiki/%E7%84%A6%E8%B7%9D">焦距</a></i>。</span></li>
	<li><span style="color: #808080;"><a class="mw-redirect" style="color: #808080;" title="橢圓" href="https://zh.wikipedia.org/wiki/%E6%A9%A2%E5%9C%93">橢圓的</a>面鏡有兩個焦點,穿過其中一個焦點後抵達鏡面上的光線 ,反射後會通過另一個焦點。</span></li>
	<li><span style="color: #808080;"><a class="mw-redirect" style="color: #808080;" title="雙曲線" href="https://zh.wikipedia.org/wiki/%E9%9B%99%E6%9B%B2%E7%B7%9A">雙曲線的</a>面鏡也有兩個焦點,他的特性是經過一個焦點後被反射的光,像是來自另一個焦點。</span></li>
</ul>
<span style="color: #808080;">一個發散(負)透鏡或凸面鏡不能將瞄準的光線匯聚在一個點上,換言之,焦點是被面鏡反射或穿透透鏡的光線看起來的發射點。凹的拋物面鏡可以將平行瞄準射入 的光線反射成看似由焦點發射出來的光;反之,經過焦點抵達拋物面鏡的光會被反射成平行的光,可以作為瞄準用的射線。凹的橢圓面鏡會將經過其中一個焦點的光 線反射至另一個焦點,好像是從那個焦點射出的,而兩個焦點都在鏡後。一個凹面的雙曲面鏡會將經過鏡前焦點的光反射,而看似是由鏡後的焦點直接發射出來的。 相反的,在<a style="color: #808080;" title="卡塞格林反射鏡" href="https://zh.wikipedia.org/wiki/%E5%8D%A1%E5%A1%9E%E6%A0%BC%E6%9E%97%E5%8F%8D%E5%B0%84%E9%8F%A1">卡塞格林反射望遠鏡</a>,會將光線匯聚在鏡後的焦點上,而這個焦點接近另一面鏡子的鏡前焦點。</span>

<img class="alignnone size-full wp-image-58071" src="http://www.freesandal.org/wp-content/uploads/250px-DOF-ShallowDepthofField.jpg" alt="250px-DOF-ShallowDepthofField" width="250" height="180" />

<span style="color: #808080;">部份對焦的影像,除了中間區域以外,其它大部份區域都落在焦點外(失焦)。</span>

<img class="alignnone size-full wp-image-58073" src="http://www.freesandal.org/wp-content/uploads/300px-Glasses_800_edit.png" alt="300px-Glasses_800_edit" width="300" height="225" />

<span style="color: #808080;">在玻璃上的朦朧圈是由<a class="new" style="color: #808080;" title="計算機產生影像(頁面不存在)" href="https://zh.wikipedia.org/w/index.php?title=%E8%A8%88%E7%AE%97%E6%A9%9F%E7%94%A2%E7%94%9F%E5%BD%B1%E5%83%8F&action=edit&redlink=1">計算機產生影像</a>模擬的,使用的軟體是<a style="color: #808080;" title="POV-Ray" href="https://zh.wikipedia.org/wiki/POV-Ray">POV-Ray</a>。</span>

 

<span style="color: #003300;">不單是『幾何光學』系統之重要概念,關乎『成像法則』矣!尚且可以通達『傅立葉光學』的哩??</span>
<h3><span id="Fourier_transforming_property_of_lenses" class="mw-headline" style="color: #003300;"><a style="color: #003300;" href="https://en.wikipedia.org/wiki/Fourier_optics">Fourier transforming property of lenses</a></span></h3>
<span style="color: #808080;">If a transmissive object is placed one focal length in front of a <a style="color: #808080;" title="Lens (optics)" href="https://en.wikipedia.org/wiki/Lens_%28optics%29">lens</a>, then its <a style="color: #808080;" title="Fourier transform" href="https://en.wikipedia.org/wiki/Fourier_transform">Fourier transform</a> will be formed one focal length behind the lens. Consider the figure to the right (click to enlarge)</span>

<span style="color: #808080;"><img class="alignnone size-full wp-image-58079" src="http://www.freesandal.org/wp-content/uploads/Lens_FT.jpg" alt="Lens_FT" width="934" height="384" /></span>

<span style="color: #999999;">On the Fourier transforming property of lenses</span>

<span style="color: #808080;">In this figure, a plane wave incident from the left is assumed. The transmittance function in the front focal plane (i.e., Plane 1) <i>spatially modulates the incident plane wave</i> in magnitude and phase, <i>like on the left-hand side of eqn. (2.1)</i> (specified to <i>z</i>=0), and <i>in so doing, produces a spectrum of plane waves</i> corresponding to the FT of the transmittance function, <i>like on the right-hand side of eqn. (2.1)</i> (for <i>z</i>>0). The various plane wave components propagate at different tilt angles with respect to the optic axis of the lens (i.e., the horizontal axis). The finer the features in the transparency, the broader the angular bandwidth of the plane wave spectrum. We'll consider one such plane wave component, propagating at angle θ with respect to the optic axis. It is assumed that θ is small (<a style="color: #808080;" title="Paraxial approximation" href="https://en.wikipedia.org/wiki/Paraxial_approximation">paraxial approximation</a>), so that</span>

<dl><dd><span style="color: #808080;"><img class="mwe-math-fallback-image-inline" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/660113056ed3c434620c1f1b279c75288373fe8b" alt="{\frac {k_{x}}{k}}=\sin \theta \simeq \theta " /></span></dd></dl><span style="color: #808080;">and</span>

<dl><dd><span style="color: #808080;"><img class="mwe-math-fallback-image-inline" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/f6ee18b2e3c3fb795c57a0fc55ad82b491a10ff8" alt="{\frac {k_{z}}{k}}=\cos \theta \simeq 1-{\frac {\theta ^{2}}{2}}" /></span></dd></dl><span style="color: #808080;">and</span>

<dl><dd><span style="color: #808080;"><span class="mwe-math-mathml-inline mwe-math-mathml-a11y">  </span><img class="mwe-math-fallback-image-inline" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/26ddbbaa9787d0eb59443e5a56f26ce9bbf5b7f9" alt="{\frac {1}{\cos \theta }}\simeq {\frac {1}{1-{\frac {\theta ^{2}}{2}}}}\simeq 1+{\frac {\theta ^{2}}{2}}" /></span></dd></dl><span style="color: #808080;">In the figure, the <i>plane wave</i> phase, moving horizontally from the front focal plane to the lens plane, is</span>

<dl><dd><span style="color: #808080;"><span class="mwe-math-mathml-inline mwe-math-mathml-a11y">  </span><img class="mwe-math-fallback-image-inline" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/ac612917b2bf37d17c372cfa288877c8ec0ea3d9" alt="e^{{jkf\cos \theta }}\," /></span></dd></dl><span style="color: #808080;">and the <i>spherical wave</i> phase from the lens to the spot in the back focal plane is:</span>

<dl><dd><span style="color: #808080;"><span class="mwe-math-mathml-inline mwe-math-mathml-a11y">  </span><img class="mwe-math-fallback-image-inline" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/1c4f4929e2694c32a82c3f5cb0e68ec62a902f32" alt="e^{{jkf/\cos \theta }}\," /></span></dd></dl><span style="color: #808080;">and the sum of the two path lengths is <i>f</i> (1 + θ<sup>2</sup>/2 + 1 - θ<sup>2</sup>/2) = 2<i>f</i> i.e., it is a constant value, independent of tilt angle, θ, for paraxial plane waves. Each paraxial plane wave component of the field in the front focal plane appears as a <a style="color: #808080;" title="Point spread function" href="https://en.wikipedia.org/wiki/Point_spread_function">point spread function</a> spot in the back focal plane, with an intensity and phase equal to the intensity and phase of the original plane wave component in the front focal plane. In other words, the field in the back focal plane is the <a style="color: #808080;" title="Fourier transform" href="https://en.wikipedia.org/wiki/Fourier_transform">Fourier transform</a> of the field in the front focal plane.</span>

<span style="color: #808080;">All FT components are computed simultaneously - in parallel - at the speed of light. As an example, light travels at a speed of roughly 1 ft (0.30 m). / ns, so if a lens has a 1 ft (0.30 m). focal length, an entire 2D FT can be computed in about 2 ns (2 x 10<sup>−9</sup> seconds). If the focal length is 1 in., then the time is under 200 ps. No electronic computer can compete with these kinds of numbers or perhaps ever hope to, although new supercomputers such as the petaflop <a style="color: #808080;" title="IBM Roadrunner" href="https://en.wikipedia.org/wiki/IBM_Roadrunner">IBM Roadrunner</a> may actually prove faster than optics, as improbable as that may seem. However, their speed is obtained by combining numerous computers which, individually, are still slower than optics. The disadvantage of the optical FT is that, as the derivation shows, the FT relationship only holds for paraxial plane waves, so this FT "computer" is inherently bandlimited. On the other hand, since the wavelength of visible light is so minute in relation to even the smallest visible feature dimensions in the image i.e.,</span>

<dl><dd><span style="color: #808080;"><img class="mwe-math-fallback-image-inline" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/7e16467ca1209514d7502ac3f415baf4dc6fc8cb" alt="k^{2}\gg k_{x}^{2}+k_{y}^{2}" /></span></dd></dl><span style="color: #808080;">(for all <i>k<sub>x</sub></i>, <i>k<sub>y</sub></i> within the spatial bandwidth of the image, so that <i>k<sub>z</sub></i> is nearly equal to <i>k</i>), the paraxial approximation is not terribly limiting in practice. And, of course, this is an analog - not a digital - computer, so precision is limited. Also, phase can be challenging to extract; often it is inferred interferometrically.</span>

<span style="color: #808080;">Optical processing is especially useful in real time applications where rapid processing of massive amounts of 2D data is required, particularly in relation to pattern recognition.</span>
<h3><span id="The_complete_solution:_the_superposition_integral" class="mw-headline" style="color: #808080;">The complete solution: the superposition integral</span></h3>
<span style="color: #808080;">A general solution to the homogeneous electromagnetic wave equation in rectangular coordinates may be formed as a weighted superposition of all possible elementary plane wave solutions as:</span>

<dl><dd><span style="color: #808080;"><img class="mwe-math-fallback-image-inline" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/d4c54785f5ea0b4c5681f8f8477ae3bfbedabe97" alt="\psi (x,y,z)=\int _{{-\infty }}^{{+\infty }}\int _{{-\infty }}^{{+\infty }}\Psi _{0}(k_{x},k_{y})~e^{{j(k_{x}x+k_{y}y)}}~e^{{\pm jz{\sqrt {k^{2}-k_{x}^{2}-k_{y}^{2}}}}}~dk_{x}dk_{y}~~~~~~~~~~~~~~~~~~(2.1)" /></span></dd></dl><span style="color: #808080;">Next, let</span>

<dl><dd><span style="color: #808080;"><span class="mwe-math-mathml-inline mwe-math-mathml-a11y">  </span><img class="mwe-math-fallback-image-inline" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/a2367e6d7c5e7a2897f7804f67342987943a4804" alt="\psi _{0}(x,y)=\psi (x,y,z)|_{{z=0}}" />.</span></dd></dl><span style="color: #808080;">Then:</span>

<dl><dd><span style="color: #808080;"><img class="mwe-math-fallback-image-inline" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/98c78685f09c423303a3d47d5a30d718f895e14c" alt="\psi _{0}(x,y)=\int _{{-\infty }}^{{+\infty }}\int _{{-\infty }}^{{+\infty }}\Psi _{0}(k_{x},k_{y})~e^{{j(k_{x}x+k_{y}y)}}~dk_{x}dk_{y}" /></span></dd></dl><span style="color: #808080;"><b>This plane wave spectrum representation of the electromagnetic field is the basic foundation of Fourier optics</b> (this point cannot be emphasized strongly enough), because when <i>z</i>=0, the equation above simply becomes a <b>Fourier transform (FT) relationship between the field and its plane wave content</b> (hence the name, "Fourier optics").</span>

<span style="color: #808080;">Thus:</span>

<dl><dd><span style="color: #808080;"><img class="mwe-math-fallback-image-inline" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/e2ec51b08a45661d3119a2fd55be1e36838b431b" alt="\Psi _{0}(k_{x},k_{y})={\mathcal {F}}\{\psi _{0}(x,y)\}" /></span></dd></dl><span style="color: #808080;">and</span>

<dl><dd><span style="color: #808080;"><img class="mwe-math-fallback-image-inline" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/587fec06dfef480e3c09639c1d1236bd016eea44" alt="\psi _{0}(x,y)={\mathcal {F}}^{{-1}}\{\Psi _{0}(k_{x},k_{y})\}" /></span></dd></dl><span style="color: #808080;">All spatial dependence of the individual plane wave components is described explicitly via the exponential functions. The coefficients of the exponentials are only functions of spatial wavenumber <i>k<sub>x</sub></i>, <i>k<sub>y</sub></i>, just as in ordinary <a style="color: #808080;" title="Fourier analysis" href="https://en.wikipedia.org/wiki/Fourier_analysis">Fourier analysis</a> and <a style="color: #808080;" title="Fourier transform" href="https://en.wikipedia.org/wiki/Fourier_transform">Fourier transforms</a>.</span>

 

<span style="color: #003300;">這事提倡『<a style="color: #003300;" href="http://www.freesandal.org/?p=11232">光粒子說</a>』之牛頓固不知曉,他可是早就用『焦平面』</span>

<img class="alignnone size-full wp-image-58019" src="http://www.freesandal.org/wp-content/uploads/Opticks.jpg" alt="Opticks" width="738" height="1000" />

<span style="color: #808080;">Cover of the first edition of Newton's <i>Opticks</i></span>

 

<span style="color: #003300;">論述『成像法則』的人!!</span>

這裡且與一圖︰

 

<img class="alignnone size-full wp-image-58035" src="http://www.freesandal.org/wp-content/uploads/牛頓成像公式.png" alt="牛頓成像公式" width="1310" height="581" />

 

讀者自可證明

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的吧!!!