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Chapter 3:
Aberrations of a Plane-Parallel Plate
Abstract
3.1 Introduction In Chapter 2, we considered the imaging properties of a thin lens consisting of two spherical surfaces. Now, we consider “imaging” by a plane-parallel plate, i.e., a plate whose two surfaces are parallel to each other, and each with a radius of curvature of infinity. Unlike a lens, such a plate is not used for imaging per se, but it is often used in imaging systems, for example, as a beam splitter or a window. The imaging relations and aberrations of a plane-parallel plate cannot be obtained from those for a thin lens in Chapter 2 by letting the radii of curvature of its surfaces approach infinity, since we neglected its thickness. However, as discussed below, they can be obtained by applying the results of Section 1.8 to its two surfaces and combining the results obtained according to the discussion of Section 1.9. It is shown that the distance between an object and its image formed by the plate, called the image displacement, is independent of the object position, and the aberration produced by it approaches zero as the object distance approaches infinity. Thus, a plane-parallel plate placed in the path of a converging beam not only displaces its focus by a certain amount but also introduces aberrations into it. In the case of a collimated beam, it only shifts the beam without introducing any aberrations. 3.2 Gaussian Imaging Consider, as indicated in Figure 3-1, a plane-parallel plate of thickness t and refractive index n forming an image of a point object lying at a distance S from its front surface and at a height h from its axis. Let the aperture stop of the plate be of radius a located at its front surface. First, we determine the location of the image formed by the plate. Using Eqs. (1–17) and (1–18) we determine the location and height of the image. For the first surface, n=1 , n ′ =n , and R 1 =∞ . Accordingly, it forms the image of P at P ′ such that S ′ 1 =−nS 1 ≡−nS and M 1 =h ′ 1 ∕h 1 =1, where h 1 ≡h . For the second surface, n=n , n ′ =1 , R 2 =∞ , and S 2 =−S ′ 1 +t . Hence, it forms the image of P ′ at P ″ such that S ′ 2 =−S 2 ∕n=(S ′ 1 −t)∕n and M 2 =h ′ 2 ∕h ′ 1 =1.
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CHAPTER 3


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