傅立叶光学

1． 计算下列各式的一维卷积 解：（1）

Frectxcombxsincfcombffrectxcombx1（2）

x3Frect*x4*x122sinc2fexpj2f3expj2f4expj2f12sinc2fexpj2f2x3x2rect*x4*x1rect22

2． 已知一平面波的复振幅表达式为

112Ux,y,zAexpjxyz

666试计算此波的波长以及在x,y,z各方向的空间频率(波长在可见光范围内)。 解：平面波复振幅的一般表达式为U(x,y,z)aexp[j2(xfxyfyzfz)] 可知 2fx1,622fy1,62fz22 622112()coscoscos1666

26.2812222由于波长在可见光范围内，故 取0.628 ，此时圆频率的单位为10k因而系数的单位是10k(弧度）

mm(弧度） ，对应的空间频率为

mmmm,fy0.65klmm,fz1.30klmm

fx1260.65kl

3． 采用单位振幅的单色平面波垂直照明透过率函数为

1tx0,y0022ax0y01的孔径，求出相距孔径为z的观察平面上夫朗和

其他费衍射图样的强度分布。

y0

a 1 x0

解：孔径的透过率函数可表示为tx0,y0cirxy2020x2y20cir0a 用单位振幅的单色平面波垂直照明孔径，可得孔径的频谱为

Tfx,fyJ12fx2fy2fx2fy2aJ2a1fx2fy2fx2fy2

则夫琅禾费衍射图的复振幅分布为

Ux,ykexpjkzexpjx2y2Tfx,fyxyfx,fyjz2zzz122J2f2f2aJ2aff1xy1xy1kexpjkzexpjx2y22222jz2zffffxyxy

观察平面上夫朗和费衍射图样的强度分布为

22J2f2f2aJ2aff1xy1xy1Ix,y222222zffffxyxy2

222221xyxy2J12aJ12a2xyzzzz

4． 如图所示的正方形孔径放在透镜的前焦面上，以单位振幅的单色平面波垂直照明，

波长为，试求透镜后焦面上的夫朗和费衍射图样的复振幅分布。 Y0

a

-a a X0

-a

解：图示正方形四个边的方程分别为y0a,x0a，代入输入面位于透镜前焦面时衍射场的复振幅分布公式得

Ux,yc'tx0,y0expj2fxx0fyy0dx0dy0cexpj2fxx0dx0expj2fyy0dy0aaaaexpj2fxaexpj2fxaexpj2fyaexpj2fyacj2fxj2fycsin2fxasin2fyafxfy

csin2fxasin2fya2fxfy其中fx

xy,fy ff5． 有一种成象系统可以用来实现非相干低通滤波，其出瞳由大量随机分布的小圆孔组

成，小圆孔的直径都是D，出瞳到像面的距离为di，光波长为，问系统的截止频率近似为多大？

解：出瞳由大量随机分布的小圆孔构成，则其光瞳函数

x2y2 Px,ycircD2*xxn,yyn  当两圆重叠面积为0时，系统的OTF＝0，这时所对应的空间频率就是系统的截止频率 fcutDdi

7.用照相机拍摄某物体时，不慎因手动摄下重叠影像，运动的结果使像点在底片上的位移为a。为了改善此照片质量，对该相片进行消模糊处理，试设计一个逆滤波器，给出滤波函数。

解：由于重叠影像横向错开距离a，具有模糊缺陷的点扩散函数为 hIx1xrect aa 带有模糊缺陷的传递函数为

HfF1xrectsinc afaa11

sincaf 故所设计的逆滤波器的滤波函数为 H1fHf一、选择题（每题2分，共40分）

1．三角函数可以用来表示光瞳为________________的非相干成像系统的光学传递函数。

A、矩形

B、圆孔

C、其它形状

2．Sinc函数常用来描述________________的夫琅和费衍射图样

A、圆孔

B、矩形和狭缝

C、其它形状

3．高斯函数exp[(x2y2)]常用来描述激光器发出的________________

A、平行光束

B、高斯光束

C、其它光束

4．圆域函数Circ(r)常用来表示________________的透过率

A、圆孔

B、矩孔

C、方孔

5．卷积运算是描述线性空间不变系统________________的基本运算

A、输出-输入关系 B、输入-输出关系

C、其它关系

6．相关（包括自相关和互相关）常用来比较两个物理信号的________________

A、相似程度

B、不同程度

C、其它关系

7．卷积运算有两种效应，一种是展宽，还有一种就是被卷函数经过卷积运算，其细微结构在一定程度上被消除，函数本身的起伏振荡变得平缓圆滑，这种效应是________________

A、锐化 B、平滑化

C、其它

8互相关是两个信号之间存在多少相似性的量度。两个完全不同的，毫无关系的信号，对所有位置，它们互相关的结果应该为________________

A、0

B、无穷大

C、其它

9．周期函数随着其周期逐渐增大，频率（即谱线间隔）________________。当函数周期变为无穷大，实质上变为非周期函数，基频趋于零

A．愈来愈小 B、愈来愈大

C、不变

10．圆对称函数的傅立叶变换式本身也是圆对称的，它可通过一维计算求出，

我们称这种变换的特殊形式为________________。这种变换只不过是二维傅立叶变换用于圆对称函数的一个特殊情况

A、贝塞尔变换

11．函数(xa)(xb)的傅立叶变换为________________

A、1

B、(fx,fy) B、(fx,fy)

B、1

C、exp[i2(fxafyb)] C、sgn(x)sgn(y)

B、傅立叶-贝塞尔变换

C、贝塞尔-傅立叶变换

12．函数(x,y)的傅立叶变换为________________

A、1

(x)Comb(y)的傅立叶变换为________________ 13．函数CombA、(fx,fy) A、(fx,fy)

C、Comb(fx)Comb(fy) C、sinc(fx)sinc(fy)

14．函数rect(x)rect(y)的傅立叶变换为________________

B、1

15．对于电路网络 ，输入和输出都是一维 的实值函数 ，即随时间变化的 电压 或电流信号 。对光学系统 来说 ，输入和输出可能是二维的实值或复值函数 ，即分别对应 ________________

A、光强或位相

位相

16．一个 空间 脉冲 在输入平面位移，线性系统的响应函数形式不变 ，只产生 相应的位移 ，这样的 系统称为________________

A、时不变系统 统

17．线性空间不变系统的脉冲响应的傅立叶变换称为系统的________________。它表示系统在频率中的效应 ，它决定输入频谱中各种 频率成分 通过系统时将发生什么样的变化

A、传递函数

B、脉冲函数 C、矩形函数

B、空间不变系统或位移不变系统 C、其它系

B、光强或复振幅

C、振幅或

18．对于线性不变系统 ，输入某一函数 ，如果相应的输出函数仅等于 一个 复比例常数的乘积，这个输入函数 就称为 这种系统的________________

A、脉冲函数

B、本征函数

C、响应函数

19．透镜成像系统是把物面 上物点的集合变为象面 上重叠的衍射斑的集合 。因而 象不再是 物体的准确 复现，而是物体的平滑变形，孔径愈小，变形愈厉害 ，这种效应就是

________________

A、卷积效应

B、相关效应

C、干涉效应

20．非相干照明下，物体上所有点的振幅和位相随时间变化的方式是统计无关的或无关联的。因此象面上各个脉冲相应的变化也是统计无关的，必须按强度叠加。非相干成像系统对________________

A、振幅

二、判断题（画√或×，每题1分共10分）

1、空间相干照明条件下物体上每一点光的振幅和位相尽管都随时间做无规变化，但所有点随时间变化的方式都是相同的，各物点 在象面上的脉冲响应也

B、光强度 C、位相

以同一方式随时间作无规变化，总的光场按光强叠加 （ ）

2、同轴全息是在记录物体的全息图时，参考光和物光波来自同轴方向，光照射全息图的透射光波中包含四项，因为都在同一方向而无法分离。

（ ）

3、全息技术分为两个过程，第一个过程是利用干涉原理将物光波前以干涉条纹的形式记录下来，再用光波照射全息图，可以再现原始物光波。离轴全息消除了同轴全息图孪生像的相互干扰，离轴全息图在记录过程中，参考光和信号光不在同一方向。

（ ）

4、如果光学系统有像差，则入射的球面波经过系统后，由出瞳射出时已不再是球面波，是一个发生了畸变的 波面，与理想球面波的位相分布不相同。但像差的存在并不影响相干传递函数的通频带宽度，仅在通频带内引入了位相畸变。（ ）

5、光学信息处理是指采用光学方法实现对输入信息的各种交换或处理，来抑制噪声、检出信号或复原失真的图像。

（ ）

6、相干系统的截止频率为非相干系统的截止频率的两倍，我们可以得出结论：对同一个光学成像系统，使用相干照明一定要比使用非相干照明能得到更好的象。 （ ）

7、阿贝（ABBE）基于对显微镜成像的研究 ，他认为 成像过程包含了两次衍射过程。物体是一个复杂的衍射光栅，衍射光波在透镜后焦面形成物体的夫郎 和费衍射图样，把后焦面上的点看作 相干的次级波源，在象面上相干叠加产

生物体的象。 （ ）

8、用光学信息处理系统可以实现图像的振幅和位相滤波，图像相关，图像卷积，图像相加和相减运算及微分，边缘检测，消模糊等光算及光学图像处理。

（ ）

9、图像识别是指检测和判断图像中是否包含有某一特定的信息，例如大量指纹档案中检查出罪犯的指纹；在病理照片中识别出癌变细胞；在军事侦查照片中检出特定目标，及文字识别等。

（ ）

10、匹配滤波器实际上是复数滤波器，可以用全息方法制作，也可以用计算全息方法制作，实际上为一张傅立叶变换全息图。它可以用来进行图像识别。但是用傅立叶变换匹配滤波手段进行图像的特征识别处理有其局限性，这是因为匹配滤波器对被识别图像的尺寸缩放和方位旋转都极为敏感 （ ）

三、名词解释（每题3分，共15分） 1、脉冲响应，点扩散函数 2、相干传递函数，光学传递函数

3、 光学信息处理中的4 f系统，说明相应各部分的功能 4、同轴全息，傅立叶变换全息 5、空间滤波

四、计算题（每题7分）

1、作出下列函数的图形 （1）（2）

xx1f1(x)comb()rect(a12a) a

xx7a1f2(x)comb()rect(aa3a) f2、已知线性不变系统的输入为g(x)Comb(x)，系统的传递函数为rect()，若bb取下列值，求系统的输出g'(x)。并作图 （1）b1

（2）b6

3、非相干照明的衍射成象系统，光阑逢宽l10mm，透镜焦距

f50mm，照明光波长103mm，成象倍率M1，物体是强度透过率

t0(x0)n(x0nd)的理想光栅，周期d0.02mm，求象的光强分布。

4、衍射受限的相干成像系统①其出瞳为边长L的正方形，求其光瞳函数，相干传递函数，及截止频率②若出瞳为长a，宽b的矩形，求其光瞳函数，相干传递函数，及截止频率。5．（1）计算余弦函数和脉冲函数的傅立叶变换

（2）设计一个孔径，写出其相应的复振幅透过率，并计算其夫琅和费衍射光

场分布

6． 全息图的光学复制一般采用什么方法？试简述其原理。 答：

H 图9.20反射全息的复制光路 HM 扩束激光 O 全息图的光学复制一般仍采用干涉的方法，用激光照明原始全息图，以再现的像光束作为物光，直射光作为参考光，记录全息图。这样在获得一张优质的母全息图后，就可以用一束光照明进行复制，反射全息和透射全息都可以用这一方法进行复制。图9.20显示的是反射全息的复制光路，其中HM是母全息图，H是复制全息干板。母全息图由图9.19的方法制作。再现时将母全息图翻转180，以母全息图的原背光面变成迎光面，全息图像被H的透过光再现，得到凹凸与原物相反的共轭像。入射激光直接入射至H的光作为参考光，HM的再现像与参考光干涉形成反射全息。

1.给出傅立叶变换

(1) rect(t)1,1/2t1/2[sinc(/2)] 5points

others0,(2) exp(i0t)[2(0)] 5points

2 (15分): 傅立叶变换性质 如果F{f(t)}F(),则

(1) F{f(at)}=[F(/a)/a] 5points (2) F{f(ta)}=[eiaF()] 5points (3) F{f(t)}=[F()] 5points 3(10分):

请给出通过下面孔径的夫琅和费衍射谱分布

A,A(z)00,A0(za/2b/2othersa/2b/2

eikZzdz) 5points

a/2b/2eikZzdza/2b/2a/2b/2A0bsinc(kZb/2)(eikZa/2eikZa/2)2A0bsinc(kZb/2)cos(kZa/2)4(10分): 光源的那个参数与光场的空间相干性有关？哪个与时间相干性有关？ 空间相干性是直接关系到有限的程度上的空间;5points 光源时间相干性是直接关系到有限的带宽的光源。

5(10 分): 证明f(t)Asin(t)的自相关函数为Cff()(A2/2)cos()

Cff()f(t)f(t)dt 3points

Tlim1T2T1limT2TTTf(t)f(t)dt

Asin(t)Asin(t)dt

TA2limT2T1T2[cos()cos(2t2)]dt 4points

T(A2/2)cos() 3points

6(10分):

有两个非相干光源照明一个双缝屏，在什么情况下在接收屏上P点光强等于4I0，这里I0是单独存在一个非相干光源时接收屏上P点的光强。

P O S S’ Q1 Q2 0 Solution:

When SS1PSQ2P/2,3/2,5/2, 3points The irradiance due to S is given by

I'4I0cos2('/2)2I0(1cos') 3points While the irradiance due to S” is

I\"4I0cos2(\"/2)4I0cos2[(')/2]2I0(1cos') 4points Hence I’+I”=4I0 7(10分):

想象我们具有一个杨氏实验装置，其中一个孔被一块减光板遮挡，减光板可以将光强衰减10倍，而另一个孔覆盖一块透明板来补偿光程。计算在相干光照明时，这种情况得到的条纹对比度。 Solution:

V2(10I)I 7points

(10II)2100.57 3points 118(10分):

证明偏振方向与x轴夹角为 的偏振片琼斯矩阵为

cos2Mpol()sincossincos. 2sincosMrot()sinsin 5points cosMpol()Mrot()Mpol,x'Mrot()cossincossinsin10coscos00sinsincossincos00sincos2sinsincos 5points

cos2sincos9(15分):

(1) 考虑一个相干光处理器。空间滤波器为一维正弦光栅

H(p)31cos(ap)， 这里 a 等于入射面上两个函数f1(x, y)和f2(x, y)的间距22的一半。试计算在输出平面(,)上的复光场。

(2) 设计一套4f相干光处理器并说明其应用。 Solution: (1) The input function byf(x,y)f1(x,y)(xa)f2(x,y)(xa) Its by

Fourier

transform

iap

is given

is

given

F(p,q)F{f1(x,y)(xa)f2(x,y)(xa)}F1(p,q)ethrough

iapF2(p,q)efilter,

Pass bySo

the the output spectrum is given

G(p,q)F(p,q)H(p,q)[F1(p,q)eiapF2(p,q)eiap 31][cos(ap)]22G(p,q)F(p,q)H(p,q)31[F1(p,q)eiapF2(p,q)eiap][cos(ap)]2231eiapeiapiapiap[F1(p,q)eF2(p,q)e][] 5points

22232[F1(p,q)eiapF2(p,q)eiap]14[Fi21(p,q)(e0p1)F2(p,q)(1ei20p)]The irradiance

on the output plane g(,)F1{G(p,q)}F1{3[F1(p,q)eiapFiap22(p,q)e]1[F2ap1(p,q)(ei1)Fi2ap42(p,q)(1e)]}by

32[f1(,)(a)f2(,)(a)] 14[f1(,)(2a)f2(,)(2a)]14[f1(,)f2(,)](2) Any 4f coherent optical processor and explain its application.

is given

5points 5points 填空题(15分)

(2) 一个波带宽度为,的准单色光，中心波长为0，则其相干长度为 [20/]. (3) 复相干度表明了光的相干性，即：

~[1] 完全非相干光: ~[0] 完全相干光: 1212~1] 部分相干光: [ 0122 (10分) 已知f(x)h(x)g(x)，证明若其中一个函数发生x0的位移，证明 f(xx0)h(x)g(xx0).

证：f(x)h(x)f(t)h(xt)dt --3

So

f(xx0)h(x)t'tx0f(tx0)h(xt)dt

f(t')h(xt'x0)dt'f(t')h(xx0t')dt'g(xx0)3 (10分) 由一个宽为b的线光源发出的光，经过两个小孔后达到衍射屏（如图），证明在衍射屏上光强分别满足 I(Y)b y b S sinab(/l)cos2(aY/s)

a/lY O l s S2 a S1 Solution: The irradiance at S0 arising from a point source is

4I0cos2(/2)2I0(1cos) --2

For different source element of width dy at point S’, y from axis, the OPD to P at Y via the two slit is

(S'S1S1P)(S'S2S2P)(S'S1S'S2)(S1PS2P)ay/laY/sThe contribution of the irradiance from dy is then

--2

dI(1cosk)dy --2

So

Ib/2b/2(1cosk)dylkaYkabkaYkab[sin()sin()]kas2ls2llkaYkabkaYkabb[sin()cos()cos()sin()

kas2ls2lkaYkabkaYkabsin()cos()cos()sin()s2ls2l2lkaYkabbcos()sin()kas2lk2/bThen

I(Y)bsin(ab/l)cos(2aY/s)

a/l6 (15分) 如图电磁波在空中传播，若在点1和2处的复振幅为

u1(t)3eit 波前 1 2 和u2(t)2ei(t)，其中和 分别为时间角频率和位相因

子，求：(a) 互相干函数；(b) 复相干度。

Solution:

The mutual coherence function is given by

12()u1(t)u2(t)TT2ei(t)3ei(t)6ei()T

6ei()The degree of complex coherence is given by

~()1212()R11(0)R22(0)6ei()I1I26ei()2232ei() --7

wavefront 1 2

7: (15分) 考虑一个相干光学处理器。其中空间滤波器为一维正弦函数

H(p,q)1[1sin(0p)]，其中0 为两图案中心距离的一半（如图），两个图2案的透过率函数分别为f1(x, y) 和f2(x, y)。求在垂直入射平面波照射下，在输出平面P3上的复振幅光场。

相干 平面波 Solution:

The input function is given by

f(x,y)f1(x,y)(x0)f2(x,y)(x0) --3 Its Fourier transform is given by

F(p,q)F{f1(x,y)(x0)f2(x,y)(x0)}F1(p,q)ei0pF2(p,q)ei0p

Pass through the filter, the output spectrum is given by

G(p,q)F(p,q)H(p,q)ipip --3 1[F1(p,q)e0F2(p,q)e0]2[1sin(0p)]So

G(p,q)F(p,q)H(p,q)[F(p,q)ei0pFi0p]112(p,q)e2[1sin(0p)][Fipp,q)ei1ei0pei0p1(p,q)e0F2(0p]2[12i] 1[Fi1(p,q)ei0pF2(p,q)e0p2]1[F1(p,q)(ei20p1)F2(p,q)(1ei20p4i)]

The irradiance on the output plane is given by

g(,)F1{G(p,q)}F1{1[F1(p,q)ei0pFi2(p,q)e0p2]1[F1(p,q)(ei20p1)F2(p,q)(1ei20p4i)]}12[f1(,)(0)f2(,)(0)] 14i[f1(,)(20)f2(,)(20)]14i[f1(,)f2(,)]--3

1. (xx0)f(x)(f(xx0))(ax,by)(G(f))

1(x,y)ab)

If F{g(t)}G(f), then F{g(t)}(2. The Fresnel diffraction integral is

kkj(22)-j(xy)ejkzj2kz(x2y2)U(x,y)e[U(,)e2z]ezddjz- (

eejzjkzkj(x2y2)2z-[U(,)ekj(22)2z]exy-j2()zz)

ddThe Fraunhofer diffraction integral is

-j2()ejkzj2kz(x2y2)zz (U(x,y)eU(,)edd) jz-xy3. In the figure below, neglect the influence of the size of the lens and assume that

the amplitude transmittance of the input transparency is tA(x,y). The incident wave is a normal incident monochromatic unit-amplitude plane wave. Please give the amplitude distribution at the rare focal plane.

Uf(u,v)(A22exp[j2kf(1df)(uv)]jf-tA(,)ej2(uv)fdd)

Uf(u,v)(Aj2d(u2v2)fejddk-tA(,)e-j2(uv)dddd)

4. A coherent imaging system is linear in ( amplitude), while an incoherent imaging system is linear in ( intensity ). The impulse response of an incoherent system is the ( squared magnitude ) of the amplitude impulse response.

5. Three general properties of the OTF are

((( H(0,0)1)) ) H(fX,fY)H(fX,fY) H(fX,fY)H(0,0)6. If the impulse response function of a system is h(u,v), the amplitude transfer

function (ATF) of this system is (H(fX,fY)-h(u,v)e-j2(fXufYv)dudv ), and the

optical transfer function (OTF) is (H(fX,fY)j2(fXufYv)h(u,v)edudv2).

-h(u,v)dudv2

Question 2: (10)

The “equivalent area” XY of a function g(x,y) can be defined by

XY-g(x,y)dxdyg(0,0), while the “equivalent bandwidth” fXfY of g is defined in

terms of its transform G by fXfYSolution:

-G(fX,fY)dfXdfY, Show that XYfXfY1.

G(0,0)Note that since G(fX,fY)-g(x,y)ej2(fXxfYy)dxdy

We see that G(0,0)-g(x,y)dxdy



………3

Similarly, since g(x,y)-G(fX,fY)ej2(fXxfYy)dfXdfY

We have g(0,0)-G(fX,fY)dfXdfY

……….3

Thus XY-g(x,y)dxdyg(0,0)G(0,0)g(0,0)G(0,0)-1fXfY………..3

G(fX,fY)dfXdfYHence XYfXfY1

………1

Question 3: (10)

A monochromatic point source is placed a fixed distance z1 to the left of a positive lens (focal length f), and a transparent object is placed a variable distance d to the left of the lens. The distance z1 is greater than f, the Fourier transform and the image of the object appear to the right of the lens. (a) How large should the distance d be (in term of z1 and f) to assure that the Fourier plane and the object are equidistant from the lens? (b) When the object has the distance found in part (a) above, how far to the right of the lens is its image and what is the magnification of the image. Solution:

The Fourier plane is found in the plane where the source is imaged. Therefore the distance zf of the Fourier plane to the right of the lens must satisfy

111 z1zff

……………2

in which case zf is given by

zffz1

z1f

…………..1

For the distance of the object to the left of the lens to equal the distance of the Fourier plane to the right of the lens, we require

dzffz1 z1f

…………..2

d Source zf (b) Let zi represent the distance of the image from the lens. Then from the lens law,

111 dzif

……………..2

z1 zi Substitute the expression for d obtained in part (a) into this equation and solve for zi. The result is

1111111 dzifz1zfz1dziz1

…………….1

The magnification is given by

Mzidz1zf1 fz1fz1f

……………..2

Question 4: (10)

Find an expression for the intensity distribution in the Fraunhofer diffraction pattern of the aperture shown in figure below. Assume unit-amplitude, normally incident plane-wave illumination. The aperture is square and has a square central obscuration. Solution:

The amplitude transmittance of this aperture is given by

tA(x,y)rect(xyxy)rect()rect()rect() 2w02w02wi2wi

………….3

The Fourier transform of this transmittance function is

2F{tA(x,y)}4w0sinc(2w0fX)sinc(2w0fY)wi2sinc(2wifX)sinc(2wifY)…….

3

It follows that the Fraunhofer diffraction pattern of this aperture is

24w02wx2wyI(x,y)()2sinc2(0)sinc2(0)zzz4ww2wx2wy2wx2wy2(0i)2sinc(0)sinc(0)sinc(i)sinc(i)……………..4

zzzzz4wi222wx2wy()sinc2(i)sinc2(i)zzz

Question 6: (10)

Find an expression for the image intensity observed when the phase-shifting dot of the Zernike phase-contrast microscope is also partially absorbing, with intensity transmittance equal to  (0< <1).

Solution:

Assume that the phase shifting dot retards the phase by /2 radians. Represent the absorption of the dot by an amplitude transmittance constant component. The intensity of the image becomes

2 applied only to the

Iiej2jj()2

2

…….….5

Note that the contrast of the image variations,

C22

………..3

is increased when <1.

A similar argument applies when the phase shift is 3/2, yielding

Ii2 Question 8: (10)

………..2

1A grating with amplitude transmittance tA(x,y)[1cos(2fox)] is placed at the

2input to a standard“4f”coherent optical processing system of the kind illustrated as shown above. Specify the transfer function (as a function of fX) of a pure phase spatial filter that will completely suppress the spatial frequency component of output intensity having spatial frequency fo. Assume normally incident plane wave illumination, monochromatic light, and neglect the effects of finite lens apertures.

Solution:

The object amplitude transmittance is given by

11ej2foxej2foxtA(x,y)[1cos(2fox)](1)

222

………….2

Since we are restricted to using a pure phase filter, we represent the amplitude

transmittance of that filter in the frequency plane by exp[j(fX)]. Thus the image amplitude can be written

UiFF1{F{tA}ej(fX)}111{ej(fX)[(fX)(fXfo)(fXfo)]}244………….3 1j(fo)1j(fo)1j(0)1F{e(fX)e(fXfo)e(fXfo)}244111ej(0)ej(fo)ej2foxej(fo)ej2fox2441The image intensity is given by

312IiUicos[(fo)(0)2fox]8411cos[(fo)(0)2fox]cos[(fo)(fo)4fox]48…………….3

31cos[4fox(fo)(fo)]88(f0)(f0)(f0)(f0)1cos[(0)]cos[2fox]222We wish to cancel out the last term, since they are the only terms that have spatial frequency components corresponding to fo. One possibility is to let

(fo)/2(fo)/2 (0)0(0)

…………..2

22(f0)(f0)cos[(0)]0

2(Many other answers are possible.)

Note that we only need to know the phase of the filter at three points, 0,-fo,and fo, since the original object contains only these frequency components.

(f0)(f0)

Question 1：(40)

1. F{rect(x)rect(y)} ( sinc(fX)sinc(fY) )

x)rect(-y)rect(x)rect(y) ) F{F{rect(x)rect(y)}} ( rect(-2. An transparent object tA(,)ejoej(,)ejo[1j(,)] was observed

using the Zernike phase-contrast microscope. If a dot is centered on the optical axis

in the focal plane and has a thickness and index of refraction such that it retards the phase of the focused light by /2 radians, so we can get the image intensity distribution as ( j(1(,)12(,) ).

3 In an imaging system (as fig A below), we put a slit filter in the rare focal plan as figure B and C. Please draw the image pattern in the blank plane.

2

4. Consider the optical system shown in figure below. A transparency with a real and nonnegative amplitude transmittance s1(,) is placed in plane P1 and coherently illuminated by a monochromatic, unit-intensity, normally incident plane wave. Lenses L1 and L2 are spherical with common focal length f. The intensity

1xyS(,) ). distribution incident on plane P2 is (Io(x,y)1(f)2ff2In plane P2, which is the rear focal plane of L1, a moving diffuser is placed. The

effect of the moving diffuser can be considered to be the conversion of spatially coherent incident light into spatially incoherent transmitted light, without changing the intensity distribution of the light in plane P2 and without appreciably broadening the spectrum of the light. In plane P3, in contact with L2, is placed a second transparency, this one with amplitude transmittance s2(x,y). The intensity distribution incident on plane P4 is

1uvuvS(,)S(,) ). ( Ii(u,v)h(u,v)Io(u,v)2144(f)2f2fff222

5. Given that f(x)h(x)g(x)，after shifting one of the functions an amount x0, we get f(xx0)h(x)( g(xx0) ).

6. If the impulse response function of a system is h(u,v), the amplitude transfer

function (ATF) of this system is (H(fX,fY)-j2(fXufYv)h(u,v)edudv ), and the optical transfer function (OTF) is (H(fX,fY)-j2(fufv)h(u,v)eXYdudv2 ).

-h(u,v)dudv27. If the input at the front focal plane of the optical system as below is U1(x1,y1)h(x1,y1Y/2)g(x1,y1Y/2) The complex amplitude at the rare focal plane is

( U2(x2,y2)1xy1xyH(2,2)ejy2Y/fG(2,2)ejy2Y/f ). ffffffThe intensity distribution at the rare focal plane is

I(x2,y2)1x2y2x2y2[H(,)G(,)22fffff22(

H(x2y2x2y2j2y2Y/fxyxy,)G(,)eH(2,2)G(2,2)ej2y2Y/f]ffffffff).

8. The types of holograms include ( Fresnel hologram, Fraunhofer hologram, Image hologram, Fourier hologram, Transmission hologram, Reflection hologram, Rainbow hologram),...... Question 2: (10)

An object has a periodic amplitude transmittance described by tA(,)tA()1

where tA() is shown above. The object is placed in the object plane of the optical

system shown below, and a tiny completely opaque stop is introduced on the optical axis in the focal plane, blocking only the spot on the optical axis. Sketch the intensity distribution observed in the image plane.

Solution:

144g(,)tA()[rect()comb()]45555144G(fX,fY)[4sinc(4fX)5comb(5fX)(fX)](fY)(fX)(fY)55G'(fX,fY)[4sinc(4fX)comb(5fX)(fX)](fY)5x1x4gi(x,y)rect()comb()4555After removing the

0th item

Ii(x,y)gi(x,y)2

1/5 gi(x, y) 0 x

-4/5

Question 3: (10)

For a wave that travel only in directions that have small angles with respect to the optical axis, the general form of the complex field may be approximated by

U(x,y,z)A(x,y,z)exp(jkz), where A(x,y,z) is a slowly varying function of z.

Show that for such a wave the Helmholtz equation can be reduced to

At2Aj2k0, where t22/x22/y2is the transverse portion of the

zLaplacian. This equation is known as the paraxial Helmholtz equation. Certification：

Substituting U(x,y,z) A(x,y,z)ejkz into the Helmholtz equation (2+k2)U=0,

222[222k2]A(x,y,z)ejkz0 xyzThen

22A[22]Aejkz[ejkzjkAejkz]k2Aejkz0 xyzzAe2tjkz2AjkzAjkz2e2jke(jk)2Aejkzk2Aejkz0 zzDividing by ejkz and simplifying，

2AAA22jk0

zz2tThe “slowly varying”

2AA2jk 2zzLeaving,

t2A2jkA0 z

Question 5: (10)

A normally incident, unit-amplitude, monochromatic plane wave illuminates a converging lens of 5cm diameter and 2 meters focal length. One meter behind the lens and centered on the lens axis is placed an object with amplitude transmittance

1tA(,)[1cos(2f0)]rect()rect()

2LLAssuming L=1m, =0.488m, and f0=10 circles/mm, sketch the intensity distribution across the u axis of the focal plane, labeling the numerical values of the distance between diffracted components and the width (between first zeros) of the individual components.

Solutions：

Since the projected pupil function of the lens is considerably larger than the finite size of the object, we can neglect it.

We then have the following expression for the field in the focal plane,

Aexp[j2kd(u2v2)]fUf(u,v)jddtA(,)exp[j2(uv)dd d1For tA(,)[1cos(2f0)]rect()rect()

2LLAfexp2dUf(u,v)j2d2Afexpj2d2jk(u2v2)2[1cos(2f)]rect()rect()exp[j(uv)dd0LLd(fXf0)(fXf0)2]sinc(LfX)sinc(LfY)}j2kd(u2v2){[(fX)

Uf(u,v)Afexpj2d2j2kd(u2v2)[sinc(LfX)sinc(L(fXf0))sinc(L(fXf0))]sinc(LfY)2

HerefXu,dfYjkv dsinc(L(uf0d)1L(uf0d)LvLu1){sinc()sinc[]sinc[]}dd2d2dAfexp2dUf(u,v)j2d2(u2v2)Because there are many grating periods within the aperture, then f0>>1/L, and there will be negligible overlap of the three sinc functions, the intensity is given by

I(u,v)[Af22Lv2Lu]sinc(){sinc()22ddd

1L1Lsinc2[(uf0d)]sinc2[(uf0d)]}4d4dFor the particular parameter values given

d6.3310716.33105m63.3μm2 L10f0d1046.3310716.33103m6.33mmA plot of the (normalized) intensity pattern is shown below, with all distances expressed in meters.

Question 6: (10)

An incoherent imaging system has a square pupil function of width 2w. A square stop of width w is placed at the center of the pupil as shown above. Sketch cross section of the optical transfer function with and without the stop present.

w 2w

Solution:

(a) The fX-axis and fY-axis sections of the OTF of a clear square pupil are already known to be identical triangle functions, dropping linearly to zero at frequency 2fo = 2w/zi from value unity at the origin.

Such a curve is included in part (a) of the figure (red line).

More interesting is the case with a central obscuration.

We can calculate either the fX section or the fY section, since they are identical. Note that the total area of the obscured pupil is 4w2-w2 = 3w2, which must be used as a normalizing factor for the autocorrelation function.

In calculating the autocorrelation function of the pupil, we shift one version of the pupil in the x direction with respect to the other version.

As the shift takes place, the area of overlap drops from 3w2 with no shift, linearly to 3w2/2 at a shift of fo/2. With further shift, the curve changes slope, dropping linearly to value w2 at shift fo.

Continuing shift results in no change of overlap until the shift is 3fo/2 following which the curve falls linearly to zero at 2fo.

The figure above shows the properly normalized OTF that results.

(b) Suppose that the width of the stop is 2w-2.

The total clear area of the pupil become 4w2-(2w-2)2=8w-42 8w.

As the two pupils are shifted, the overlap area quickly drops to 2(2w-) 4w after a shift of .

The overlap then continues to drop linearly, but with a shallower slope, reaching value 42 for a shift of 2w-2.

Continued shifting results in a rapid linear rise in the overlap to a value of 2w when the displacement is 2w-, following which it falls linearly to zero at displacement 2w.

After proper normalization, the resulting OTF is as shown as figure above.

1．已知函数u(x)Aexp⑴u(x) ⑵u(x)u(x) ⑶u(x)u(x)*2(i2f0x) ，求下列函数：

2*

f2．已知线性不变系统的输入为g(x)comb(x)，系统的传递函数为rect(b)，

若b取下列值，求系统的输出g'(x)，并画出函数及其频谱的图形， ⑴b=2.5 ⑵b=1.5

3.衍射受限的相干成像系统，其出瞳分别为下列孔径时，分别写出出瞳函数

p(x,y)，相干传递函数Hc(fx,fy)的表达式，并求沿fx,fy方向的两个截止频率。

若改为非相干光学系统，再求沿fx,fy方向的两个截止频率。 ⑴边长为l的正方形 ⑵半径为3的圆形 ⑶长为5，宽为8的长方形

4.推导1和高斯函数的傅立叶变换式，并尽可能的写出傅立叶变换对表 二、作图：（20分）

1．画出阿贝成像的两次衍射原理图，并加以简单解释。

2．画出4f光学频谱分析系统光路图，指出各元件的名称；说明该光路的重要性 3．画图说明傅立叶变换光学全息的物光波的波前记录和波前再现的两个过程，并用公式表示上述两个过程。 4．画出下列函数的图形： ⑴

xx51xxfcomb()rect(f1(x)1comb()rect()⑵2xaa5aaa5a)

1．什么是空间线性不变系统?写出空间域中的一维和二维输入输出关系的卷积表达式。2．空间域中的输入输出关系的卷积表达式的物理意义是什么。 3．说明透镜在光学系统中的位相变换功能及公式．

4． 一维和二维空间坐标系下的傅立叶变换和反变换的表示式

5．说明观察面放在透镜的后焦面上，孔径面在透镜前焦面上的傅立叶变换性质及公式表达，与孔径面不在透镜前焦面上的变换有何不同？

6.分别写出相干系统和非相干系统的空间域和空间频谱域的输出输入关系表达式．说明光瞳函数，脉冲响应，相干传递函数的关系

7．写出线性不变系统相干传递函数的计算式和其物理意义，解释其厄米性．