By Virendra N. Mahajan

This e-book presents a transparent, concise, and constant exposition of what aberrations are, how they come up in optical imaging structures, and the way they impact the standard of pictures shaped by means of them. The emphasis of the publication is on actual perception, challenge fixing, and numerical effects, and the textual content is meant for engineers and scientists who've a necessity and a hope for a deeper and higher figuring out of aberrations andRead more...

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CHAPTER 2. ABERRAITONS OF A THIN LENS W(Q) = {[PQP'] - [POP']] + {[P'QP" ] - [P'OP" ]} . 9, it can be shown that the primary aberration function of the thin lens is given by W (r, B; h') = al + a ch'r 3 cos 6 + a h' 2r 2cos2 B + adh' 2r 2 , 4 (2-3) where 3 as = __32n(nl-1) f 3 n+ i n1 nn 1 + (3n + 2Xn -1)p 2 + a, _f 2S [( 2n , + 1)p + (2-4a) q2 + 4(n + 1)pq + q (2-4b) ] a = - 1/2f ' 2 (2-4c) Q and a = 23 d (2-4d) = - (n + 1)/4nfSr 2 . Note that there is no distortion term in Eq. , a thin lens with an aperture stop at the lens does not produce any distortion.

The magnitude of a primary aberration is independent of whether the mirror is concave or convex, but its sign depends on its type. Spherical aberration is the dominant aberration in Table 4-2. Of course, the field curvature and distortion are zero once again. If the aperture stop of the mirror is moved to its center of curvature, the peak value A, of its spherical aberration does not change. However, its coma and astigmatism reduce to zero, but its field curvature becomes nonzero. The radius of the exit pupil a te , the field curvature coefficient ad, and the peak value of field curvature for the problems under consideration are given in Table 4-3.

Let the aperture stop and the corresponding exit pupil of the system be located as indicated in Figure 4-1. The line joining the center of curvature C of the mirror and the center of the aperture stop (and, therefore, the center 0 of the exit pupil) defines the optical axis of the system. Consider an object lying at a distance S from the vertex V0 of the mirror. Let the height of an object point P from the optical axis be h. The distance S' and height h' of its Gaussian image P' are given by —+—=--= 1 (4-1) and M = h S + R = -S'/S , respectively, where M is the magnification of the image.