Fundamentals of geometrical optics by Virendra N. Mahajan

By Virendra N. Mahajan

Optical imaging starts off with geometrical optics, and ray tracing lies at its vanguard. This publication starts off with Fermat's precept and derives the 3 legislation of geometrical optics from it. After discussing imaging through refracting and reflecting platforms, paraxial ray tracing is used to figure out the scale of imaging parts and obscuration in replicate platforms. Stops, students, radiometry, and optical tools also are mentioned. The chromatic and monochromatic aberrations are addressed intimately, by way of spot sizes and see diagrams of aberrated photos of element gadgets. each one bankruptcy ends with a precis and a collection of difficulties. The ebook ends with an epilogue that summarizes the imaging method and descriptions the subsequent steps inside and past geometrical optics

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The z coordinate z1 of a point on the surface represents the sag of the surface at that point. 2 Rectilinear Propagation from the Object Plane to the First Refracting Surface r v It is evident from Figure 1-11 that the position vectors r1 and r0 are related to each other according to r v r1 = r0 + S01eˆ0 , (1-32) where S01 = [(x - x ) 1 0 2 2 + ( y1 - y 0 ) + (D01 + z1) 2 12 ] (1-33) is the distance between A0 and A1 . The sign of S01 is the same as that of D01 + z1, as may be seen by considering a ray incident at the vertex V1.

In practice, the wavefront exiting from an imaging system is rarely spherical. Its deviations from being spherical represent its wave aberrations. The possible aberrations of a system with an axis of rotational symmetry are discussed in Chapter 8. When the wavefront is not spherical, the rays intersect the image plane in the vicinity of the Gaussian image point. , the spot diagrams, are discussed in Chapter 9. A lens designer resorts to using nonspherical surfaces to reduce or eliminate the aberrations over a certain field of view.

1-80), Eq. (1-79) reduces to Ê 2D ˆ x 2 = Á 1 - 12 ˜ x 0 , R1 ¯ Ë (1-82) which gives the height x 2 of the image of an object of height x 0 . The ratio of x 2 to x 0 is called the transverse magnification of the image. Thus, Mx = x2 x0 = 1- (1-83a) 2D12 R1 (1-83b) or Mx = D12 D01 , (1-83c) where we have used Eq. (1-81) in the last step. Equations (1-81) and (1-83) describe the location and size of the image in terms of the corresponding quantities for the object. Substituting for x1 from Eq. (1-76) into Eq.

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