Principle and application of optical information technology (Nine) Example and Exercise of Fraunhofer diffraction

Example of Fraunhofer diffraction E.g. 1 Intensity distribution of Fraunhofer diffraction of cosine type amplitude grating. 余弦型振幅光栅夫琅和费衍射的光强分布 The cosine type amplitude grating is in a square hole with width of . Spatial frequency of grating is . Transmission rate modulation(透过率调制度)is . (余弦型光栅振幅透过率函数)

Example of Fraunhofer diffraction The transmittance function of cosine type amplitude grating can be expressed as According to the Fourier transform of cosine function rectangular function, the spectrum of the grating can be

Complex amplitude distribution of Fraunhofer diffraction pattern can be expressed as

Example of Fraunhofer diffraction According to the distribution of function, the width of the main lobe of each function is proportional to . From the above equation, the distance between the main lobe of three functions(三个函数主瓣之间的距离) is . If is much larger than , namely, is much smaller than , there is no overlap between the three functions (the main lobe), and there is no cross term in the square. Therefore,

Thus, the optical energy of the grating is redistributed by using the plane wave illumination, and the energy is focused on the three diffraction orders. Obviously the Fourier analysis method is much simpler than the traditional optical path difference analysis method. 显然傅里叶分析方法比传统的光程差分析方法要简捷得多。

Light intensity distribution of Fraunhofer diffraction Cosine Amplitude Grating

Example of Fresnel diffraction As shown in the figure, the monochromatic spherical wave is converged to point P with illumination aperture . The point is located on the observation plane with the aperture back distance of Z, coordinate . If the observation plane is located in the Fresnel diffraction region , please testify that the intensity distribution of the observation plane is the Fraunhofer diffraction pattern of the aperture with the center of P. （观察平面上的强度分布是以P点为中心的孔径的夫琅和费衍射图样）

Establishing rectangular coordinates in the aperture plane, which is parallel to coordinate . Therefore, incident light field of illumination spherical wave converging to point P in the aperture plane can be expressed as 二项式近似，取一阶近似

The incident light field of illumination spherical wave converging to point P in the aperture plane can be simplified as It is assumed that the amplitude transmittance function of the aperture（孔径的复振幅透过率函数）is . Thus, the optical field distribution of the aperture under the spotlight illumination is

The field distribution of Fresnel diffraction in the observation plane can be calculated by the Fresnel diffraction formula. Further simplification

The intensity distribution is Therefore, the intensity distribution of the observation plane is the Fraunhofer diffraction pattern of the aperture with the center of P. 可见强度分布是以P点为中心的孔径的夫琅和费衍射图样。

The significance of this exercise is that, in any imaging system, all of the images are gathered through the pupil to the image. Imaging diffraction spots on the surface are Fraunhofer diffraction, which is the basis of the following study of coherent optical imaging and its transfer function. 本题的重要意义在于在任何成像系统中，像都是通过出瞳会聚到像面上才成像的，因此成像面上的衍射斑都是夫琅和费衍射，这是以下研究相干光成像过程及其传递函数的基础。

Exercise As shown in the figure, a single slit with width of . Phase difference is introduced between left and right parts. Single amplitude monochromatic plane wave illuminates vertically（采用单位振幅单色平面波垂直照明）. Please solve the intensity distribution of Fraunhofer diffraction on the observation screen with the distance of . Try to draw the intensity distribution along the direction of the cross section.

Solution Phase difference introduced between left and right parts can be expressed as the sum of the two slits with half width. For single amplitude monochromatic plane wave illumination, Fraunhofer diffraction on the observation screen with the distance of is the Fourier transform of this slit. 采用单位振幅单色平面波垂直照明孔径，平面波的振幅为1

在作傅里叶变换时，必须注意建立观察面坐标与频率坐标之间的关系

Then the Fraunhofer diffraction can be expressed as 利用傅氏变换的相似性定理和位移定理就可以求出衍射的复振幅分布，进而用复振幅的模平方可以算出夫琅和费衍射的强度分布。

Ignoring constant coefficients, the intensity distribution of the Fraunhofer diffraction can be expressed as

Exercise 2.4，2.5，2.6 on page 47