工程光学练习题(英文题加中文题含答案)

English Homework for Chapter 1

1.In ancient times the rectilinear propagation of light was used to measure the height of objects by comparing the length of their shadows with the length of the shadow of an object of known length. A staff 2m long when held erect casts a shadow 3.4m long, while a building’s shadow is 170m long. How tall is the building?

Solution. According to the law of rectilinear propagation, we get, x=100 (m)

So the building is 100m tall.

2.Light from a water medium with n=1.33 is incident upon a water -glass interface at an angle of 45o. The glass index is 1.50. What angle does the light make with the normal in the glass? Solution. According to the law of refraction, We get,

So the light make 38.8o with the normal in the glass.

3. A goldfish swims 10cm from the side of a spherical bowl of water of radius 20cm. Where does the fish appear to be? Does it appear larger or smaller? Solution. According to the equation.

and n ’=1 , n=1.33, r=-20

we can get

So the fish appears larger.

4.32170=

x '

'sin sin I n I n =626968

.05.145

sin 33.1sin =?=

'ο

I ο

8.38='I r n

n l n l n -'=-''11416.110

133

.15836.8)(5836.81165.02033.01033.11>-=??-=''=

-='∴-=--+-=-'+='l n l n cm l r n n l n l βΘn′=1.50

n=1.33

water

45o

I′

A

4.An object is located 2cm to the left of convex end of a glass rod which has a radius of curvature of 1cm. The index of refraction of the glass is n=1.

5. Find the image distance. Solution. Refer to the figure. According to the equation

and n=1, n ’=1.5, l 1=-2cm, r 1=1cm , we get

English Homework for Chapter 2

1.An object 1cm high is 30cm in front of a thin lens with a focal length of 10cm. Where is the image? Verify your answer by graphical construction of the image. Solution. According to the Gauss’s equation,

r n n l n l n -'=-''cm l l d l l l 20

21115.15.12

121

1='∴-∞='-=∞='∴=-+-='R 2=-20cm R 1=20cm

A

-10cm

r 1=1cm

A

A′

-l 1=2cm

l 2′

′

and l=-30cm f ’=10 cm.

we get

Others are omitted.

2.A lens is known to have a focal length of 30cm in air. An object is placed 50cm to the left of the lens. Locate the image and characterize it. Solution. According to Gauss’s equation,

and f′=30cm l =-50cm

we get

The image is a real, larger one.

3.The object is transparent cube, 4mm across, placed 60cm in front of 20cm focal length. Calculate the transverse and axial magnification and describe what the image looks like? Solution. From Gauss’s equation, we find for the rear surface of the cube (the face closer to the lens)

that,

For the front surface (the face farther away from the lens),

The transverse magnification for the rear surface is

But the axial magnification is

Since

,the cube doesn’t look like a cube.

f l l '=-'11)(15)30(10)

30(10cm l f l f l =-+-?=+''=

'f l l '=-'11)(75)50(30)

50(30cm l f l f l =-+-?=+''=

'5

.15075

-=-='=l l β)(3020

)60()

20()60(111

cm f l f l l +=+-?-='+'=')

(9.29204.6020

)4.60(2

cm l +=+-?-='?

-=-+=

5.06030

t M ?+=----=?'?=

25.0)4.60(609

.2930l l M a a

t M M ≠-l =50cm

f ′=30cm

4.A biconvex lens is made out of glass of n=1.52. If one surface has twice the radius of curvature of the other, and if the focal length is 5cm, what are the two radii? Solution. Supposing r 1= -2r 2 (ρ2=-2ρ1),according to the lens equation

we get,

∴r 1=7.8(cm) r 2=- 3.9(cm)

返回

English Homework for Chapter 4

1. A stop 8mm in diameter is placed halfway between an extended object and a large -diameter lens of 9cm focal length. The lens projects an image of the object onto a screen 14cm away. What is the diameter of the exit pupil?

))(1(21ρρ?--=n ))(152.1(51

21ρρ+-=1282.01=∴ρ2564.02-=ρr 1

-r 2

l ’

-l

Image

Lens

Stop

Object

Solution. Refer to the figure. First, from the known focal length and the image distance,

we find the object distance.

and l ’=14 f ’=9 l =-25.2(cm)

The stop is one -half that distance is front of the lens, so l s =12.6(cm) ∴l s ’=31.5(cm)

∴

2. Two lenses, a lens of 12.5cm focal length and a minus lens of unknown power, are mounted coaxially and 8 cm apart. The system is a focal, that is light entering the system parallel at one side emerges parallel at the other. If a stop 15mm in diameter is placed halfway between the lenses: 1) Where is the entrance pupil? 2) Where is the exit pupil? 3) What are their diameters?

Solution. Refer to the figure. For the system to be a focal, the focal points of the two lenses must

coincide. Since f 1’=12.5cm, and the two lenses are 8cm apart, so f 2’=-4.5cm. The entrance pupil is the image of stop formed by the first lens.

f l l '=

-'111Θ

22.255

.31-

='==

s

s stop ex l l D D βΘ)

(28.05.2cm D ex =?=F 1’(F 2)

-l 2’

L 1’

Stop

f =12.5cm

8cm

According to Gauss’s equation,

and l 1’=4cm, f 1’=12.5cm. We get

The exit pupil’s location is

返回

English Homework for Chapter 7

1. A person wants to look at the image of his or her own eyes, without accommodation, using a concave mirror of 60cm radius of curvature. How far must the mirror be from the eye if the person has 1) Normal vision?

2) 4diopter myopia, without correction? 3) 4diopter hyperopia, without correction? Solution.

1) When the person has normal vision, according to the following scheme 1, we get

so,

2) According to the following scheme 2,

111111f l l '

=-'())(88.55

.84

5.1211111cm l f l f l =?='-'''=

)(05.224

88

.5151

mm D D stop

entrance =?=

=

β)(95.7154

12

.2)

(12.25

.818

)4()5.4()4()5.4(222222

mm D D cm f l l f l stop exit =?=

?=-=-=-+--?-='+'='β∞='l cm r

l 302==

l'=∞

l

Scheme 1

and

, or

So the mirror must be 75cm or 10cm from the eye. 3) According to the following scheme 3,

and

, or (Since the object is real, so we can give up this answer)

So the mirror must be 50cm from the eye.

141

-=m l r

cm

l l r 25-=='r l l 211=

+'Θ

)(25cm l l +'=cm r 60=265852

253048585025308522±=

??-±=

=?+-l l l Θ???==∴)

(50')(7511

cm l cm l ??

?-==)

(15')(1022cm l cm l r l l 211=

+'Θ

)(25'cm l l +=cm r 60=265352

253043535025303522±=

??+±=

=?--l l l Θ???==∴)(75')(5011cm l cm l ??

?=-=)(10')(1522cm l cm l l

l'

Scheme 2

25

l'

l

Scheme 3

25

2. Discussion: What differences between the following situations:

1) a microscope is used for projection;

2) the microscope is used for visual observation.

返回

工程光学（上）期末考试试卷

一．问答题：（共12分，每题3分）

1．摄影物镜的三个重要参数是什么？它们分别决定系统的什么性质？

2．为了保证测量精度，测量仪器一般采用什么光路？为什么？

3．显微物镜、望远物镜、照相物镜各应校正什么像差？为什么？

4．评价像质的方法主要有哪几种？各有什么优缺点？

二．图解法求像或判断成像方向：（共18分，每题3分）1．求像A'B'

2．求像A'B'

3．求物AB经理想光学系统后所成的像，并注明系统像方的基点位置和焦距4．判断光学系统的成像方向

5．求入瞳及对无穷远成像时50%渐晕的视场

6．判断棱镜的成像方向

三．填空：（共10分，每题2分）

1．照明系统与成像系统之间的衔接关系为：

①________________________________________________

②________________________________________________

2．转像系统分____________________和___________________两大类，其作用是：_________________________________________ 3．一学生带500度近视镜，则该近视镜的焦距为_________________,该学生裸眼所能看清的最远距离为_________________。

4．光通过光学系统时能量的损失主要有：________________________, ________________________和_______________________。

5．激光束聚焦要求用焦距较________的透镜，准直要用焦距较________的透镜。

四．计算题：（共60分）

1．一透镜焦距，如在其前边放置一个的开普勒望远镜，求组合后系

统的像方基点位置和焦距，并画出光路图。（10分）

2．已知，的双凸透镜，置于空气中。物A 位于第一球面前处，第二面镀反射膜。该物镜所成实像B 位于第一球面前，如图所示。若按薄透镜处理，求该透镜的折射率n 。（20分）

3．已知物镜焦距为

，相对孔径，对无穷远物体成像时，由物镜第一面到像平面的距离为，物镜最后一面到像平面的距离为。 （1）按薄透镜处理，求物镜的结构参数；（8分）

（2）若用该物镜构成开普勒望远镜，出瞳大小为，求望远镜的视觉放大率；

（4分）

（3）求目镜的焦距、放大率；（4分）

（4）如果物镜的第一面为孔径光阑，求出瞳距；（6分） （5）望远镜的分辨率；（2分）

（6）如果视度调节为折光度，目镜应能移动的距离。（2分） （7）画出光路图。（4分）

工程光学（上）期末考试参考答案

一．简答题：（共12分，每题3分）

1．摄影物镜的三个重要参数是什么？它们分别决定系统的什么性质？

mm f 30'=x

6-=Γmm r 201=mm r 202-=mm 50mm 5mm 500101

mm 400mm 300mm 2

答：摄影物镜的三个重要参数是：焦距、相对孔径和视场角。焦距影响成像的大小，相对孔径影响像面的照度和分辨率，视场角影响成像的范围。

2．为了保证测量精度，测量仪器一般采用什么光路？为什么？

答：为了保证测量精度，测量仪器一般采用物方远心光路。由于采用物方远心光路时，孔径光阑与物镜的像方焦平面重合，无论物体处于物方什么位置，它们的主光线是重合的，即轴外点成像光束的中心是相同的。这样，虽然调焦不准，也不会产生测量误差。

3．显微物镜、望远物镜、照相物镜各应校正什么像差？为什么？

答：显微物镜和望远物镜应校正与孔径有关的像差，如：球差、正弦差等。照相物镜则应校正与孔径和视场有关的所有像差。因为显微和望远系统是大孔径、小视场系统，而照相系统则是一个大孔径、大视场系统。

4．评价像质的方法主要有哪几种？各有什么优缺点？

答：评价像质的方法主要有瑞利（Reyleigh）判断法、中心点亮度法、分辨率法、点列图法和光学传递函数（OTF）法等5种。瑞利判断便于实际应用，但它有不够严密之处，只适用于小像差光学系统；中心点亮度法概念明确，但计算复杂，它也只适用于小像差光学系统；分辨率法十分便于使用，但由于受到照明条件、观察者等各种因素的影响，结果不够客观，而且它只适用于大像差系统；点列图法需要进行大量的光线光路计算；光学传递函数法是最客观、最全面的像质评价方法，既反映了衍射对系统的影响也反映了像差对系统的影响，既适用于大像差光学系统的评价也适用于小像差光学系统的评价。

二．图解法求像或判断成像方向：（共18分，每题3分）

1．求像A'B'（图中C为球面反射镜的曲率中心）

2．求像A'B'

3．求物AB经理想光学系统后所成的像，并注明系统像方的基点位置和焦距4．判断光学系统的成像方向

5．求入瞳及对无穷远成像时50%渐晕的视场

6．判断棱镜的成像方向

>

三．填空：（共10分，每题2分）

1．照明系统与成像系统之间的衔接关系为：

①__照明系统的拉赫不变量要大于成像系统的拉赫不变量___

②__保证两个系统的光瞳衔接和成像关系_________________

2．转像系统分___棱镜式___________和___透镜式__________两大类，

其作用是：_使物镜所成的倒像转变为正像。

3．一学生带500度近视镜，则该近视镜的焦距为_____-0.2 米________,

该学生裸眼所能看清的最远距离为___眼前0.2米__________。

4．光通过光学系统时能量的损失主要有：_两透明介质面上的反射损失______, __介质吸收的损失_____________和_____反射面的光能损失__。

5．激光束聚焦要求用焦距较__短____的透镜，准直要用焦距较__长______的透镜。

四．计算题：（共60分）

1．一透镜焦距，如在其前边放置一个的开普勒望远镜，求组合后系统的像方基点位置和焦距，并画出光路图。（10分）

解：, 求得：

答：组合后的焦距是-180mm。基点位置如图所示。

其光路图如下所示：

2．已知，的双凸透镜，置于空气中。物A位于第一球面前处，第二面镀反射膜。该物镜所成实像B位于第一球面前，如图所示。若按薄透镜处理，求该透镜的折射率n。（20分）

解：

设：透镜的折射率为n

物点A经折射成像在A'处，将已知条件代入公式得

----①

A'经反射后，成像于B'点。故将，代入反射面公式，

得：

----②

B'点再经折射成像在B点。根据光路的可逆性，将B视为物，B'点视为像，有，代入折射公式，得：

----③

由①②③式解

得：

答：透镜的折射率为1.6。

3．已知物镜焦距为，相对孔径，对无穷远物体成像时，由物镜第一面到像平面的距离为，物镜最后一面到像平面的距离为。

（1）按薄透镜处理，求物镜的结构参数；（8分）

（2）若用该物镜构成开普勒望远镜，出瞳大小为，求望远镜的视觉放大率；（4分）

（3）求目镜的焦距、放大率；（4分）

（4）如果物镜的第一面为孔径光阑，求出瞳距；（6分）

（5）望远镜的分辨率；（2分）

（6）如果视度调节为折光度，目镜应能移动的距离。（2分）（7）画出光路图。（4分）

解：根据题意，画出物镜的结构图如下：

(1)将和代入公式，得：

将代入牛顿公式，得：

(2)因，则：

(3)，

(4)望远镜系统的结构如下图所示：

将和代入公式，得：

将和代入公式，得出瞳距：

(5)

(6)

(7)望远系统光路图如下：

? 物理光学试卷 ? 物理光学答案

工程光学（下）期末考试试卷

一、填空题（每题2分，共20分）

1．在夫琅和费单缝衍射实验中，以钠黄光（波长为589nm ）垂直入射，若缝宽为0.1mm ，则第1极小出现在（ ）弧度的方向上。 2．一束准直的单色光正入射到一个直径为1cm 的汇聚透镜，透镜焦距为50cm ，测得透镜焦平面上衍射图样中央亮斑的直径是cm ，则光波波长为（ ）nm 。

3．已知闪耀光栅的闪耀角为15o ，光栅常数d=1μm ，平行光垂直于光栅平面入射时在一级光谱处得到最大光强，则入射光的波长为（ ）nm 。 4．晶体的旋光现象是（ ），其规律是（ ）。 5．渥拉斯棱镜的作用（ ），要使它获得较好的作用效果应（ ）。

6．

利用此关系可（ ）。

7．波片快轴的定义：（ ）。

8．光源的相干长度与相干时间的关系为（ ）。 相干长度愈长，说明光源的时间相干性（ ）。

9．获得相干光的方法有（ ）和（ ）。

10． 在两块平板玻璃A 和B 之间夹一薄纸片G ，形成空气劈尖。用单色光垂直照射劈尖，如图1所示。当稍稍用力下压玻璃板A 时，干涉条纹间距（ ），条纹向（ ）移动。若使平行单色光倾斜照射玻璃板（入射角），形成的干涉条纹与垂直照射时相比，条纹间距（ ）。

3

1066.6-?()

=??

?

?????????-??????-?????

?110

01

011111i i 01>i α A

B

G 图1

二、问答题（请选作5题并写明题号，每题6分，共30分）

1． 简要分析如图2所示夫琅和费衍射装置如有以下变动时，衍射图样会发生怎样的变化？

1）增大透镜L 2的焦距； 2）减小透镜L 2的口径；

3

）衍射屏作垂直于光轴的移动（不超出入射光束照明范围）。

2． 以迈克尔逊（M ）干涉仪的等倾圆环和牛顿（N ）环为例，对“条纹形状”作一简要讨论，（从中央级次、条纹移动分析它们的相同点与不同点）。

3． 利用惠更斯作图法求下列方解石晶体中的双折射光（标出光线方向和光矢量方向）。

4.拟定部分偏振光和方位角为的椭圆偏振光的鉴别实验。（包括光路、器件方位、实验步骤。）

5． 试述如图3所示格兰－付科棱镜的结构原理（要求画出并标出o 光、e

光的传播方向，光矢量方向）特点，用途和使用方法，并说明此棱镜的透光轴方向。

6． A grating, used in the second order, diffracts light of 632.8nm wavelength

through an angle of 30o . How many lines per millimeter does the grating

have?

αθ

L 1

L 2

P

接收屏

衍射屏

光源

图2 夫琅和费衍射系统俯视图 θ

图3

最新中文地址如何翻译成英文(精)

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