工程本科的物理实验室手册外文翻译资料

Physics Laboratory Manual for Engineering Undergraduates

Dr. P. K. Giri

Department of Physics

Indian Institute of Technology Guwahati

Experiment 21

Newtonrsquo;s Rings

Apparatus:

Traveling microscope, sodium vapour lamp, plano-convex lens, plane glass plate,

magnifying lens.

Purpose of the experiment:

To observe Newton rings formed by the interface of produced by a thin air film and to

determine the radius of curvature of a plano-convex lens.

Basic Methodology:

A thin wedge shaped air film is created by placing a plano-convex lens on a flat glass

plate. A monochromatic beam of light is made to fall at almost normal incidence on the

arrangement. Ring like interference fringes are observed in the reflected light. The

diameters of the rings are measured.

I. Introduction:

I.1 The phenomenon of Newtonrsquo; s rings is an illustration of the interference of light

waves reflected from the opposite surfaces of a thin film of variable thickness. The

two interfering beams, derived from a monochromatic source satisfy the coherence

condition for interference. Ring shaped fringes are produced by the air film existing

between a convex surface of a long focus plano-convex lens and a plane of glass

plate.

I.2. Basic Theory:

When a plano-convex lens (L) of long focal length is placed on a plane glass plate

(G) , a thin film of air I enclosed between the lower surface of the lens and upper

surface of the glass plate.(see fig 1). The thickness of the air film is very small at the

point of contact and gradually increases from the center outwards. The fringes

produced are concentric circles. With monochromatic light, bright and dark circular

fringes are produced in the air film. When viewed with the white light, the fringes

are coloured.

A horizontal beam of light falls on the glass plate B at an angle of 450. The plate B

reflects a part of incident light towards the air film enclosed by the lens L and plate

G. The reflected beam (see fig 1) from the air film is viewed with a microscope.

Interference takes place and dark and bright circular fringes are produced. This is

due to the interference between the light reflected at the lower surface of the lens and

the upper surface of the plate G.

For the normal incidence the optical path difference between the two waves is nearly

2, where is the refractive index of the film and t is the thickness of the air film.

Here an extra phase difference pi; occurs for the ray which got reflected from upper

surface of the plate G because the incident beam in this reflection goes from a rarer

medium to a denser medium. Thus the conditions for constructive and destructive

interference are (using = 1 for air)

2 t = mlambda; for minima; m =0,1,2,3hellip; hellip; hellip; hellip; hellip; .. -------------(1)

and 2 t =(m for maxima; m = 0,1,2,3hellip; hellip; hellip; hellip; hellip; . -------------(2)

Then the air film enclosed between the spherical surface of R and a plane surface

glass plate, gives circular rings such that (see fig 2)

= (2R-t)t

where is the radius of the order dark ring .(Note: The dark ring is the dark

ring excluding the central dark spot).

Now R is the order of 100 cm and t is at most 1 cm. Therefore Rgt;gt;t. Hence

=-gt;=(2R-t)t

(neglecting the term ), giving

2t

Putting the value of “ 2t” in eq(1) gives

m mR, m =0,1,2,3hellip; hellip; ----------------(3)

and eq (2) gives (for the radius of order bright ring )

=(m )(m ) R ----------------------(4)

Hence for dark rings

= --------------------------(5)

while for bright rings

= m =0,1,2,3hellip; hellip; ------------------(6)

With the help of a traveling microscope we can measure the diameter of the mth ring

order dark ring = . Then =and hence，

=4mR ----------------------(7)

So if we know the wavelength  , we can calculate R(radius of curvature of the lens).

II. Setup and Procedure:

1. Clean the plate G and lens L thoroughly and put the lens over the plate with the

curved surface below B making angle with G(see fig 1).

2. Switch in the monochromatic light source. This sends a parallel beam of light.

This beam of light gets reflected by plate B falls on lens L.

3. Look down vertically from above the lens and see whether the center is well

illuminated. On looking through the microscope, a spot with rings around it can

be seen on properly focusing the microscope.

4. Once good rings are in focus, rotate the eyepiece

such that out of the two perpendicular cross wires,

one has its length parallel to the direction of travel

of the microscope. Let this cross wire also passes

through the center of the ring system.

5. Now move the microscope to focus on a ring (say, the 20th order dark ring). On

one side of the center. Set the crosswire tangential to one ring as shown in fig 3.

Note down the microscope reading . fig 3 _

(Make sure that you correctly read the least count of the vernier in mm units)

6. Move the microscope to make the crosswire tangential to the next ring nearer to

the center and note the reading. Continue with this purpose till you pass through

the center. Take readings for an eq

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工程本科的物理实验室手册

Dr. P. K. Giri

物理系

印度技术学院

实验21

牛顿环

装置：

读数显微镜，钠蒸气灯、平凸透镜、平面玻璃板，放大镜。

实验目的：

观察由一个薄膜的界面形成的牛顿环和确定一个平凸透镜的曲率半径。

基本方法：

薄的楔形空气薄膜是用一个平凸透镜放在平板玻璃上制造的。一个单色光束正常垂直射入，

环形的干涉条纹将在反射光中被观察到。环的直径可以被测量。

一、引言：

1.1牛顿环现象是由一个相对厚度改变的薄膜产生光的干涉现象后形成的。这个来自单色光源的两束干扰光束，满足干涉的相干性条件。环形条纹是由一长焦距平凸透镜与玻璃平面之间形成的空气薄膜产生的。

1.2基础理论：

当一个长聚焦的平凸透镜（L）放置在一个平面玻璃板（G）上，在透镜的下表面与玻璃板上表面之间形成了一层薄薄的空气膜（见图1）。中心点空气膜的厚度很薄并从中心切点向外逐渐增加。条纹产生的是同心圆。在单色光照射时，在空气膜中产生明暗相间的圆形条纹。当用白光照射时，条纹是彩色的。

一束水平的光束以45度角入射平面玻璃G。G板朝向由透镜和平面玻璃围成的空气膜反射入射光的一部分。用显微镜观察了空气膜的反射光束（见图1）。干扰发生的地方产生了明暗相间的圆形条纹。这是由于在透镜下表面的反射光与平面玻璃G上表面的反射光产生了干射。

图1

正入射产生的两束光之间的光程差为2t，是膜的折射率，t是膜的厚度。从平面玻璃G上表面反射的光有一个额外的光程差pi;，因为这束光是从光疏介质进入光密介质。因此有（空气时=1）：

2 t = mlambda; 暗环; m =0,1,2,3hellip; hellip; hellip; hellip; hellip; .. -------------(1)

2 t =(m 亮环; m = 0,1,2,3hellip; hellip; hellip; hellip; hellip; . -------------(2)

然后用半径为R的圆和平面玻璃围城了空气膜，如图2所示

图2

= (2R-t)t

是第m级暗环的半径（注意：第m级暗环要去除中间的暗斑）R是100cm而t至多1cm,则Rgt;gt;t,因此

=-gt;=(2R-t)t

（忽略）得到：

2t

将公式1中的2t代入得：

m mR, m =0,1,2,3hellip; hellip; ----------------(3)

并将公式2代入（其中为第m级亮纹的半径）

=(m )(m ) R ------------------(4)

因此暗环为：

= -----------------------(5)

亮环为：

= m =0,1,2,3hellip; hellip; ------------------(6)

在读书显微镜的帮助下我们可以测量各级圆环的直径，然后求出半径 =，因此

=4mR ----------------------(7)

因此当我们再知道波长后，就可以求出R(晶体曲率半径)。

二、装置和过程

1.将凸透镜L和平面镜G彻底清理干净，将凸透镜凸面向下放在平面玻璃上，与平面玻璃形成一定角度（如图1）。

2.打开单色光光源。使其发射出一平行光束，这束光由平板G反射到凸透镜L上。

3.从透镜上方垂直看下去，看看中心是否被照亮。调节显微镜，直至中心点及其周围的圆环可以被清楚的看到。

4.清晰地圆环出现在视野时，旋转目镜直至出现垂直的十字叉丝，它的一个边平行于显微镜的移动方向。让十字叉线也通过圆环中心。

5. 现在把显微镜的叉丝移到一个环上（比如说，第二十阶暗环），将十字叉丝移动到中心一侧的环上如图3所示，并记录显微镜读数。（确定你测得了最小读数，精确到mm）

图3

6.移动十字叉丝至靠近中心的下一圈，测量并记录数据。继续这一过程直到十字叉丝通过圆心，在中心两侧取相同数量的圆环的读数。

注意事项：

注意当你远离中心的暗斑时，条纹宽度会减小。在为了最大限度地减少测量环的直径时产生的错误，应采取以下措施：

1）显微镜应该平行于平面玻璃的边缘。

2）如果你把十字叉丝垂直于圆环放在这一圆环的外侧，那么在中心点另一端你要把叉丝放在同一环的内侧。

3）显微镜只能朝一个方向移动。

三、练习及问题

1.在这个实验中，造成干射的媒介是什么？为什么干涉产生于平面玻璃和透镜之间？

2.解释为什么干涉环是圆形的。

3.为什么环随着环的增加环间变得越来越紧密？

4.给出了相邻亮环间的半径差，通过下式：

Delta;r=-for m gt;gt;1.

5.给出A=R, for m gt;gt;1，以米为单位表明相邻环之间的独立区域。

6.为什么中心点是暗的？在实验中，中心点不是暗的的理由是？

7.如果是一个楔形的透镜倒置在平面玻璃上形成的干涉图形是什么形状的？

8.在实验中使用平凸透镜的效果是什么？给出明暗圆环半径的表达式。

9.用白光代替单色光会产生什么效果？

10在这个实验中为什么使用一个长聚焦的透镜？

参考：

1. “ Physics” ,M.Alonso and E.J.Finn, Addison-Wiley, 1992

2. “ Fundamentals of Physics” , D.Halliday , R.Resnick and J.Walker,

6th edition ,John-Wiley amp; sons , New York 2001.

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