实验五 菲涅耳双棱镜干涉[实验目的]1． 观察和研究菲涅耳双棱镜产生的干涉现象； 2． 测量干涉滤光片的透射波长(λ0)。

[仪器和装置]白炽灯，干涉滤光片，可调狭缝，柱面镜，菲涅耳双棱镜，双胶合成像物镜，测微目镜。

[实验原理]如图1a 所示，菲涅耳双棱镜装置由两个相同的棱镜组成。

两个棱镜的折射角α很小，一般约为5 ~ 30'。

从点（或缝）光源S 发出的一束光，经双棱镜折射后分为两束。

从图中可以看出，这两折射光波如同从棱镜形成的两个虚像S 1和S 2发出的一样。

S 1和S 2构成两相干光源，在两光波的迭加区产生干涉。

a、从图1b 看出，若棱镜的折射率为n ，则两虚像S 1、S 2之间的距离a n l d )1(2-= （5-1）干涉条纹的间距λan l l l e )1(2'-+=（5-2）式中，λ为光波的波长。

对于玻璃材料的双棱镜有n =1.50，则λal l l e '+=（5-3） 可得到e l l la'+=λ （5-4） 在迭加区内放置观察屏E ，就可接收到平行于脊棱的等距直线条纹。

若用白光照明，可接收到彩色条纹。

对于扩展光源，由图2可导出干涉孔径角：''l l al +=β （5-5） 和光源临界宽度：⎪⎭⎫⎝⎛+=='1l l a b λβλ （5-6） 从式（5-5）和（5-6）看出，当l'=0时，β=0，则光源的临界宽度b 变为无穷大。

此时，干涉条纹定域在双棱镜的脊棱附近。

b 为有限值时，条纹定域在以下区域内:λαλ-≤b ll ' （5-7）a) 图 1 双棱镜干涉原理图[内容和步骤]1．调整光路，观察和研究双棱镜干涉现象(1) 按图3所示，将光学元件置于光学平台上。

调整光学元件，使其满足同轴等高的要求。

(2) 取l ≈200mm ，l '≈1200mm ，按λ=550nm ，α=30'，n =1.50计算出b 的数值。

置狭缝宽度b t =b /4，调节棱镜的脊棱与狭缝方向平行，直到使得测微目镜视场里出现清晰的干涉条纹为止。

增大或减小狭缝宽度，观察干涉条纹对比度的变化，并给予解释。

2．测量干涉滤光片中心透射长λ0。

由式（5-3）看出，为了测量λ0，需要在一定的精度范围测定d 、l 、l'与e 值。

(1) 测定d 值如图4所示，通常S 1、S 2和S 并不在与图面垂直的同一平面内，而D 和A 又应从S 1S 2处测量才算准确。

故测量d 时，采用二次（共轭）成像法，即保持物像位置不变，移动成像物镜。

当成像物镜6在第一个位置时，若从测微目镜中测得S 1，S 2的两个实像'S 1，'S 2之间的距离d 1，据物像关系，则有BAd d =1 （5-8） 物镜6在第二个（共轭）位置成像时，则有ABd d =2 （5-9） 由上两式可解出21d d d = (5-10)实验时，对d 值的测量不应少于三次，然后取其平均值。

图2 双棱镜干涉的几何关系图图3 双棱镜干涉实验装置图 1—白炽灯 2—滤光片 3—柱面镜 4—狭缝 5—双棱镜 6—成象物镜 7—测微目镜(2) D 的计算设物镜6从第一个位置移置至第二个（共轭）位置的位移量是C ，则C =B －A ，而D =l +l '=A +B ，再与式（5-9）和（5-10）联立，消去A 、B ，可得到：2121d d d d CD -+= （5-11）由各次测量C 、d 1、d 2的值，计算相应的D ，然后取其平均值D 。

(3) 测量条纹间距e用测微目镜测出10条以上明（或暗）条纹的宽度，计算出干涉条纹间距e 。

多次重复测量，取其平均值e 。

(4) 将e D d 、、各值代入式（5-4）计算干涉滤光片中心透射波长λ0。

[思考题]如果狭缝方向与脊棱稍不平行，就观察不到干涉条纹，为什么？图4 二次（共轭）成象法测量d 值Experiment 5 Use of Fresnel's biprism to determine transmission wavelengthof an interferometric filter[Experimental Objectives]1. To observe and study the interference phenomenon by using a Fresnel's biprism2. To determine the transmission wavelength (λ0) of an interferometric filter[Apparatus and Setup]Incandescent lamp, interferometric filter, adjustable slit, cylindrical lens, Fresnel's biprism, doublet imaging objective, and micrometer eyepiece[Experimental Principle]As shown in Fig. 1a, the Fresnel's biprism system consists of two identical prisms (biprism), whose refractive angles are very small (normally ca 5 ~ 30'). Refracted by the biprism, a beam from a point (or slit) light source is split into two. It can be seen from Fig. 1a that two refracted beams seem to come from two virtual images, S 1 and S 2. Here S 1 and S 2 are coherent light sources, which may produce interference patterns in the overlapped region of two beams.Fig. 1 Principle of Fresnel's biprism interferenceFrom Fig. 1b, if the refractive index of the prism is n , the separation of two virtual images S 1 and S 2 can be expressed as ： a n l d )1(2-= (5-1) and the separation of two fringes is given byλan l l l e )1(2'-+=(5-2)where λ is the light wavelength. Given n =1.50, we haveλal l l e '+=(5-3)ore l l la'+=λ.(5-4)A screen E is placed in the overlapped region and some equi-spaced straight fringes, which are parallel to the prism spine, are produced. If an incandescent lamp is used as the light source, chromatic fringes can be observed.Based on Fig. 2, the interference aperture angle can be derived as''l l al +=β (5-5)and the critical width of the light source isa)⎪⎭⎫ ⎝⎛+=='1l l a b λβλ. (5-6)Fig. 2 Geometrical relationship of Fresnel's biprism interferenceFrom Eqs. (5-5) and (5-6), if l'=0, β=0, resulting in the critical width of the light source (b ) being infinite. In this case, interference fringes are produced near the biprism spine. If b is of a finite value, fringe localization is in the following region:λαλ-≤b ll '. (5-7) [Experimental Procedure]1. Adjust light path to observe biprism interference phenomenon(1) The optical apparatus are arranged on an optical bench as shown schematically in Fig. 3. Adjust the optical elements in order to meet the requirements of proper alignment and equal height.Fig. 3 Setup of Fresnel's biprism interference1-Incandescent lamp 2-Interferometric filter 3-Cylindrical lens 4-Adjustable slit 5-Fresnel'sbiprism 6-Objective 7-Micrometer eyepiece(2) Assuming l ≈200mm and l '≈1200mm, for λ=550nm, α=30', and n =1.50, the value of b can be calculated. Set the slit width b t =b /4, and adjust the prism spine parallel to the slit until clear interference fringes appear in the microscope field. Increase or decrease the slit width, see what will happen to the contrast of interference fringes and give your explanations.2. Determine the transmission wavelength (λ0) of an interferometric filterFrom Eq. (5-3), in order to determine λ0, it is necessary to measure the values of d , l , l', and e with relatively high accuracies.(1) Determination of dAs shown in Fig. 4, S 1, S 2 and S are usually not in the same plane perpendicular to the paper. However, D and A only can be measured accurately from the S 1S 2 position. Therefore, secondary (conjugate) imaging will be used for the determination of d . When the objective 6 is at the first position, if the distance d 1 of tworeal images 'S 1 and 'S 2 of S 1 and S 2 is measured by the micrometer eyepiece, based on the relationship of objective and image, the following expression can be obtained:BA d d =1. (5-8)While the objective is at the second (conjugate) position, we haveAB d d =2.(5-9) From the above two equations,21d d d =.(5-10)In this experiment, d is measured no less than three times and the average value d is calculated.Fig. 4 Determination of d by secondary (conjugate) imaging(2) Calculation of DIf the displacement of the objective 6 between the first position and the second (conjugate) position is C , C =B －A and D =l +l '=A +B . Combined with Eqs. (5-9) and (5-10), we have:2121d d d d CD -+=. (5-11) From the values of C , d 1 and d 2, the corresponding D value can be calculated and the average value D is obtained.(3) Measurement of fringe separation eMeasure the widths of more than 10 bright (or dark) fringes using the micrometer eyepiece and calculate the fringe separation e . Repeat the measurement for several times and calculate their average value e .(4) Substitute the values of d , D and e into Eq. (1-4) to calculate the central transmission wavelength (λ0) of the interferometric filter.[Questions]If the slit is oriented slightly unparallel to the prism spine, interference fringes cannot be observed. Why?。