Thick Lens Formula

11 Key excerpts on "Thick Lens Formula"

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  Visual Optics and the Optical Space Sense

  • Hugh Davson(Author)
  • 2013(Publication Date)
  • Academic Press

    (Publisher)

  5. OPTICAL SYSTEMS IN GENERAL 69 IX. The Thick Lens A thick lens is one whose centre thickness has a significant effect upon its focal properties. In practice, most lenses have to be treated as such. Even the great majority of spectacle lenses would show inadmissible errors of power if their radii were computed on the basis of the thin lens theory. The most important formulae relating to thick lenses can readily be es-tablished from first principles. Let us denote the front- and back-surface powers of the lens by F 1 and F 2 respectively, the centre thickness by t (metres) and the refractive index of the material by n. Then, if we trace through the lens an initially parallel pencil we shall arrive at the following sequence of expressions: L x = 0 1ι λ = L ± + F ± = F ± L = L[ = Fl n n F 1 + F i -t -F 1 F i n L 2 = L 2 + F2 = n The expression for L 2 is a straightforward application of the effectivity formula (Eq. 1.11). By definition, L! 2 when L x = 0 is the back vertex power of the lens, F' v . Hence, Fl = -^ (5.16) 1-U n It might be mentioned at this stage that ophthalmic lenses are now uni-versally numbered in terms of their back vertex power. The corresponding expression for the front vertex power, F v , is F 1 + F 2 --F 1 F 2 F v = (5.17) I -' -* , n 70 A. G. BENNETT AND J. L. FRANCIS The equivalent power, F, can be obtained by substituting the above values for L' v L 2 > etc., in Eq. (5.15), which in this case reduces to F A . L 2 = F 1 + F 2 - -F X F 2 (5.18) It is sometimes useful to know the position of the principal points of a lens. As we have already noted, the distance from the first vertex A ± to the first FIG. 10. Principal points of a thick lens. principal point P is usually denoted by e, and the distance from the last vertex (in this case, A 2 ) to the second principal point, P', by e' (Fig. 10). For a single lens we have, and e = t F A n F - t F 1 V F (5.19) (5.20) FIG.

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  Optical Principles and Technology for Engineers

  • James Stewart(Author)
  • 2018(Publication Date)
  • CRC Press

    (Publisher)

  The focal length of a thin lens in air is given by the equation (4.1) The refractive index of the material (relative to air) is n and the radii of curvature of the two surfaces are Rl and R 2. I shall use the convention that radiation propagates from left to right and the sign of a radius of curvature is positive if the center of curvature is to the right of the surface. If a surface is flat its radius of curvature is infinity and 1 /R = 0. The optical power of a lens is the reciprocal of its focal length. If the focal length is in meters the unit of power is the diopter . As an example, consider a thin lens made of the common optical glass BK7 which has a refractive index of 1.51680 at a wavelength of 587.56 nm (the d-line of helium). If the second surface is flat with a radius of curvature of R2 = infinity and the first surface is convex as viewed from the left, with a radius of curvature of R j = 100 mm, then the focal length of the lens is calculated to be / = 193.5 mm and its power is 5.17 diopters. A bundle of parallel rays arriving at the lens from a very great distance are converged and focused at a distance of 193.5 mm from the lens (Fig. 4.1). The rays define a real image . On the other hand, if the first surface is made concave as viewed from the left, with a radius of curvature having a value Rx = -100 mm, then the focal length becomes / = -193.5 mm. The bundle of parallel rays from infinity diverges after passing through the lens and appear to come from a point 193.5 mm in front of the lens. The image which appears to be formed in front of the lens is a virtual image (Fig. 4.2). Lenses 47 Figure 4.2 Diverging lens. Parallel rays from a distant object point diverge as though coming from a virtual image point. Now suppose that the second surface of the lens in the first example is no longer a plane and that the two surfaces have equal and opposite curvatures R j = 200 mm and R2 = -200 mm. The focal length remains the same, and / = 193.5 mm.

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  Principles of Physical Optics

  • Charles A. Bennett(Author)
  • 2022(Publication Date)
  • Wiley

    (Publisher)

  Similarly, f b is located f e to the right of h b , or 6.00 cm behind lens 2, as before. The effective object distance s o must locate the object relative to h f ; thus, s o = 12.0 cm. The thin-lens equation with effective quantities gives s i = f e s o s o − f e = (8.00 cm) (12.0 cm) (12 − 8) cm = 24.0 cm or 22.0 cm to the right of lens 2, as illustrated in Figure 5.7. The magnification is m = − s i s o = −2.00 Thus the image size is 6.00 cm and it is inverted. f f f b h f h b s o s i Figure 5.7 An image formed by two thin, symmetric biconvex lenses. 148 5 Geometric Optics II 5.4 Thick Paraxial Lenses A thick lens consists of a combination of two refracting surfaces separated by a distance d. If the paraxial approximation remains valid (and this is not always the case, especially with thick lenses), we can use the ideas and methods just developed for thin-lens combinations. In particular, by using principal points and effective focal lengths, we may continue to use the thin-lens equation even with lenses that are thick. Equation 4.21 describes refraction of light rays at the first lens surface. Divide both sides by n m to give 1 s o1 + n s i1 = n − 1 R 1 (5.15) where n = n 𝓁 ∕n m is the relative index of refraction and R 1 is the radius of curvature of side 1 of the thick lens. We define a focal length for the first surface by letting s i1 → ∞ in Equation 5.15 (see Figure 4.10): f 1 = R 1 n − 1 (5.16) Combining Equations 5.15 and 5.16 gives 1 s o1 + n s i1 = 1 f 1 (5.17) Equation 5.17 looks like the thin-lens equation with s i1 replaced by s i1 ∕n.

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  Microoptics Technology

  Fabrication and Applications of Lens Arrays and Devices

  • Nicholas F. Borrelli(Author)
  • 2017(Publication Date)
  • CRC Press

    (Publisher)

  2

  Refractive Elements

  2.1 Optics Review

  2.1.1 Basics

  In order to follow the development and performance of refractive microlens elements, some rudimentary understanding of geometric optics is required. A number of references are given for the reader to consider. What we need here, at the very least, is an understanding of the basic terminology. It might be useful to bear in mind that there is really no conceptual difference in the optical principles relating to small refractive lenses as compared with large lenses; however, there is in the way one formulates the optical design. In the simplest case of paraxial ray tracing for large-diameter lenses, the so-called thin-lens approximation [1 ] is used, which simply means that the deviation of the rays through the lens thickness can be ignored. The paraxial specification means that Snell’s law can be written as

  ϕi

  = nϕ r . Here

  ϕi

  and

  ϕr

  are the angles of incidence and refraction measured from the surface normal as shown in Fig. 2.1 . When the lens thickness is comparable, or exceeds, the lens diameter, one has to resort to what is often called the thick-lens formulation. Referring to Fig. 2.2 , one has a set of definitions that derive from the desire to maintain the simple lens maker’s formula relating the focal length of the lens to the object and image distances

  Figure 2.1 Definition of the salient terms to describe the ray path through a spherical lens. Ray incident from left in medium of index n 1 making an angle with respect to the normal at that point of

  ϕ1

  , refracting to an angle

  ϕ2

  in the lens medium of refractive index n 2 . The exact form of Snell’s law is given relating the incident angle to the refracted angle, as well as the paraxial approximation.

  1 f

    =  

  1

  s 0

    +  

  1

  s i

  ⁢ ( 2.1 )

  Figure 2.2 Diagram defining the various terms and distances relating to a thick lens which are included in Table 2.1 .

  Here, f represents the focal length of the lens, s i is the image, and s 0 is the object distance. In the thick-lens formulation, one must use a new set of definitions of the various common lens terms. These definitions are summarized in Table 2.1 based on the description of Figs. 2.1 and 2.2. One can see that the principal planes, P 1 , P 2 , are defined in such a way as to make Eq. (2.1) valid if the object and image distances are measured from them and not the lens boundaries. One should realize from the expression in Table 2.1 that the position of the principal planes can lie outside the physical lens. We can see from the drawing of Fig. 2.2

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  College Physics

  • Michael Tammaro(Author)
  • 2019(Publication Date)
  • Wiley

    (Publisher)

  26.5 The Thin Lens Equation and Magnification Suppose that an object is placed beyond the focal point of a converging lens (Figure 26.5.1). The focal length f is the distance from the lens to either focal point. The symbols d o and d i denote the object and image distances, respectively. These quantities are related by the thin lens equation: f d d 1 1 1 o i = + (26.5.1) The lateral magnification m is m d d i o = − (26.5.2) Equation 26.5.2 is called the magnification equation and has the same meaning for lenses as it does for mirrors—that is, the object and image heights, h o and h i , are related by h m h i o = . Moreover, a positive magnification means that the image is upright, while a negative mag- nification means that the image is inverted. Virtual images and diverging lenses can be described by the thin lens and magnification equations if we adhere to the sign conventions for lenses. 26.6 Lenses in Combination and the Human Eye When two lenses are used in combination, the image formed by the first lens serves as the object for the second lens. This principle can be applied to determine the focal length of corrective eyeglass lenses, where the first lens is the eyeglass lens and the second lens is the eye, with the final image formed on the retina. The near point of an eye is the nearest point from the eye that an object can be placed so that it remains in focus. A near point of a healthy unaided eye is about 25 cm from the eye. The near point of a farsighted eye is too far from the eye. A converging corrective lens may be used, which places an image of a close object at the near point of the farsighted eye (Figure 26.6.4). The far point of an eye is the furthest point from the eye that an object can be placed so that it remains in focus. A healthy eye should be able to focus on objects that are effec- tively infinitely far away, so the ideal far point is infinitely far from the eye.

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  Applied Optics and Optical Engineering V9

  • Robert Shannon(Author)
  • 2012(Publication Date)
  • Academic Press

    (Publisher)

  All rights of reproduction in any form reserved. ISBN 0-12-408609-8 2 GEORGE W. HOPKINS These practices result in more code than required for computation only, but the benefits when using or changing a stale program are suffi-ciently great that the author recommends them even for engineers who expect to be the sole users of their programs. II. THICK LENS PROGRAM This program uses algebraic formulas (Kingslake, 1965) to calculate thick lens parameters and to determine the thick lens equivalent of a thin lens, preserving power and shape factor. These formulas are available in the program listing. Lens parameters are entered using the following prompts: Cl, First (left) optical surface curvature. Default, Cl = 0. C2, Second (right) optical surface curvature. Default, C2 = 0. D, Outer diameter or clear aperture of lens. Default, D = 0. N, Index of refraction referred to medium surrounding the lens. De-fault, N = 1.5168. R, Outer radius of lens (D/2). Default, R = 0. Rl, First (left) optical surface radius of curvature. Default, Rl infinite (flat surface). R2, Second (right) optical surface radius of curvature. Default, R2 infinite (flat surface). T, Center or axial thickness of lens. Default, T = 0. These parameters are illustrated in Fig. 1 and follow Cartesian sign con-ventions. The parameters remain at default values, previously entered values, or values calculated by the program unless changed using the prompts above. Calculations are initiated with the following commands: LENS Lens parameters are displayed together with cardinal point data if the lens is not afocal. SAGS Sagittas for both surfaces, center and edge thicknesses, and thickness-to-diameter ratio are displayed. THICKEN A thick lens with center thickness T is calculated assum-ing that values of Cl and C2 in memory represent thin lens (T = 0) values. Power and shape factor, (C1-C2)/(C1+C2), are preserved. The command HELP displays a menu of the prompts and commands.

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  Principles of Physical Optics

  • Charles A. Bennett(Author)
  • 2015(Publication Date)
  • Wiley

    (Publisher)

  Equation 4.21 describes refraction of light rays at the first lens surface. Divide both sides by n m to give 1 s o1 + n s i 1 = n − 1 R 1 (4.101) where n = n  / n m is the relative index of refraction, and R 1 is the radius of curvature of side 1 of the thick lens. We define a focal length for the first surface by letting s i 1 → ∞ in Equation 4.101 (see Figure 4.10): f 1 = R 1 n − 1 (4.102) ∗ THICK LENSES 175 Combining Equations 4.101 and 4.102 gives 1 s o1 + n s i 1 = 1 f 1 (4.103) Equation 4.103 looks like the thin-lens equation with s i 1 replaced by s i 1 / n. Thus, surface 1 produces a magnification given by m 1 = − s i 1 n s o1 (4.104) For surface 2, the incident medium has index n  : n s o2 + 1 s i 2 = 1 − n R 2 = − n − 1 R 2 (4.105) In Equation 4.105, an infinite object distance produces an image at the focal point of the second surface f 2 = − R 2 n − 1 (4.106) Combining Equations 4.105 and 4.106 gives n s o2 + 1 s i 2 = 1 f 2 (4.107) With this definition, Equation 4.107 looks like the thin-lens equation with s o1 replaced by s o1 / n. Thus, surface 2 produces a magnification given by m 2 = − n s i 2 s o2 (4.108) As a check, note that for a thin lens, we can combine Equations 4.102 and 4.106 using the formula for close contact: 1 f = 1 f 1 + 1 f 2 = (n − 1)  1 R 1 − 1 R 2  as in Equation 4.25. The front and back focal points of a thick lens may be obtained as was done for lens combinations. According to Equation 4.103, surface 1 produces an image located by s i 1 = n f 1 s o1 s o1 − f (4.109) This image becomes the object for surface 2, with object distance given by s o2 = d − s i 1 . According to Equations 4.107 and 4.103, the image distance for surface 2 is given by s i 2 = f 2 s o2 s o2 − nf 2 = f 2 (d − s i 1 ) d − s i 1 − nf 2 = f 2 d − f 2  n f 1 s o1 s o1 − f 1  d − n f 2 − n f 1 s o1 s o1 − f 1 (4.110) Taking the limit as s o1 → ∞ gives the back focal point: f b = f 2 (d − n f 1 ) d − n ( f 1 + f 2 ) (4.111)

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  Cutnell & Johnson Physics, P-eBK

  • John D. Cutnell, Kenneth W. Johnson, David Young, Shane Stadler, Heath Jones, Matthew Collins, John Daicopoulos, Boris Blankleider(Authors)
  • 2020(Publication Date)
  • Wiley

    (Publisher)

  For an object placed in front of a lens, Snell’s law of refraction leads to the technique of ray tracing and to equations that are identical to the mirror and magnification equations. Thus, mirrors work because of the reflection of light, whereas lenses work because of the refraction of light, a distinction between the two devices that is important to keep in mind. The equations that result from applying Snell’s law to lenses are referred to as the thin‐lens equation and the magnification equation: Thin‐lens equation 1 d o + 1 d i = 1 f (26.6) Magnification equation m = Image height Object height = h i h o = - d i d o (26.7) 744 Physics Maths skills The thin‐lens equation ( 1 d o + 1 d i = 1 f ) is sometimes thought to imply that d o + d i = f. To emphasise that the focal length f does not equal the object distance d o plus the image distance d i , we can solve the thin lens equation for f. First, we multiply the left side of the thin‐lens equation by 1 in the form of d o d i d o d i : ( d o d i d o d i ) ⏟⏞⏞⏞⏞ ⏟⏞⏞⏞⏞ ⏟ 1 ( 1 d o + 1 d i ) = 1 f or ( 1 d o d i ) ( d o d i d o + d o d i d i ) = 1 f Simplifying this result gives ( 1 d o d i ) ( ✁ d o d i ✁ d o + d o✁ d i ✁ d i ) = 1 f or d i + d o d o d i = 1 f Taking the reciprocal of both sides of the simplified result shows that ( d i + d o d o d i ) -1 = ( 1 f ) -1 or d o d i d i + d o = f Clearly, it is not true that d o + d i = f. Do not make this mistake when solving problems. Figure 26.30 defines the symbols in these expressions with the aid of a thin converging lens, but the expressions also apply to a diverging lens, if it is thin. The derivations of these equations are presented at the end of this section. FIGURE 26.30 The drawing shows the focal length f, the object distance d o , and the image distance d i for a converging lens. The object and image heights are, respectively, h o and h i .

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  Light-Matter Interaction

  Fundamentals and Applications

  • John Weiner, P.-T. Ho(Authors)
  • 2008(Publication Date)
  • Wiley-VCH

    (Publisher)

  Figure 8.7b shows a nty passing through a plauoconvex thin leiis one of whose surfaces is plane and tlie other a sphere of radius R. The term “thin“ means the change in the distalice r inside the lens can be ignored. Applying tlie formula above to trace the ray through tlie lens, and comparing with thc ABCD matrix for a thin lciis, the 186 8.2. GEOMETRIC OPTICS . R , = -1.37 m m 4.9 m Figure 8.8: Optics of the Hubble Space Telescope focal length f of the lens can be found to be 1 n -1 Problem 8.2 The Hubble Space Telescope: The optics of the Hubble Space Telescope (HST) (see Fig. 8.8) consists of a concaiie mirror of radius of curvature 11.04 m and a convex mirror with radius of curvature 1.37 m. The mirrors are separated by 4.9 m. The diameter of the concave mirror is 2.4 m. (a) What is the equivalent lens system of the HST? (b) What is the ABCD matrix for the lens equivalent of the HST? The input is right i,n front of the first lens, an.d the output is right after the second lens. (c) An object of height h, is at a large distance d, from the HST. Use the ABCD matrix to find (i) the image distance di from the second lens; (ii) the image height k; (iii) the ang,ular magnification or power (h.i/d.i)/(h,/do). Yoii will find that the power is not verv diflerent from that of an inexpensive telescope. What do you think rnukes the HST such a valuable instrument? (d) The closest distance of the planet Jupiter to the earth is 45 million miles (72 million km). What is the diameter of the smallest area on Ji~piterthat HST cun resolve, and whnt is the size of the image o,f that area? The diameter of Jupiter is 86,000 miles (138,000 km). What is the size of the image of Jupiter? (e) Wouldn’t the front mirror and the hole in the back mirror make a dark spot in the center of the ,image? 187 CHAPTER. 8. ELEMENTS OF OPTICS 8.3 Wave Optics 8.3.1 General concepts and definitions in wave propaga- t ion We consider moiiochroinatic waves of frequency w.

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  Foundations of Optical System Analysis and Design

  • Lakshminarayan Hazra(Author)
  • 2022(Publication Date)
  • CRC Press

    (Publisher)

  94 ].

  Note that although the equivalent focal length, and therefore some of the paraxial characteristics, can be kept the same, the thick system, as obtained above, does not automatically ensure that the path of paraxial rays through the thick lens remain unchanged from its thin counterpart. As a result, the primary aberrations of the thick lens often become significantly different from the thin lens primary aberrations. This is often cited as a major shortcoming of the practical use of the concept of thin lenses in practical use. However, it should be noted that use of generalized bending and suitable axial repositioning of the thick lens can circumvent this problem. Hopkins and Rao [83 ] suggested an iterative procedure for this purpose. In what follows, we present an analytical procedure that can significantly reduce the change in primary aberrations that occur after thickening. The method is adapted from a brief discussion of the topic by Blandford [89 ].

  For the sake of generality, Figure 14.7(a) shows a part of an optical system where a thin lens, of refractive index µ, is located at an intermediate position. Its axial distance from the last vertex on the left is l, and its axial distance from the next vertex on the right is

  l ′

  , so that

  A ^

  j

  A

  j − 1

  = l ,

  and

  A ^

  j + 1

  A

  j + 2

  =

  l ′

  The left and right vertices of the thin lens,

  A ^

  j

  and

  A ^

  j + 1

  , are superimposed on each other on the axis. Let the curvatures of the two refracting interfaces of the thin lens be

  c 1

  and

  c 2

  , and let the convergence angle of the paraxial ray incident on the lens be

  u 1

  , and the convergence angle of the corresponding emergent paraxial ray be

  u

  ′ 2

  . Let the convergence angle of the paraxial ray inside the thin lens (not shown in figure) be

  u

  ′ 1

    =

  u 2

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  Physics, Student Solutions Manual

  • John D. Cutnell, Kenneth W. Johnson, David Young, Shane Stadler(Authors)
  • 2018(Publication Date)
  • Wiley

    (Publisher)

  However, we must consider the phenomenon of total internal reflection. Object Scale: 20 cm F F Image 20 cm THE REFRACTION OF LIGHT: LENSES AND OPTICAL INSTRUMENTS 330 Some of the light will enter the liquid as long as the angle of incidence is less than or equal to the critical angle. At incident angles greater than the critical angle, total internal reflection occurs, and no light enters the liquid. Since the angle of incidence is 75.0º, the critical angle cannot be allowed to fall below 75.0º. The critical angle θ c is determined according to Equation 26.4: Liquid c Glass sin n n θ = As n Liquid decreases, the critical angle decreases. Therefore, n Liquid cannot be less than the value calculated from this equation, in which θ c = 75.0º and n Glass = 1.56. SOLUTION Using Equation 26.4, we find that ( ) Liquid c Liquid Glass c Glass sin or sin 1.56 sin 75.0 1.51 n n n n θ θ = = = ° = 109. REASONING The glasses form an image of a distant object at her far point, a distance L = 0.690 m from her eyes. This distance is equal to the magnitude |d i | of the image distance, which is measured from the glasses to the image, plus the distance s between her eyes and the glasses: i L d s = + (1) Because distant objects may be considered infinitely distant, the object distance in the thin-lens equation o i 1 1 1 d d f + = (Equation 26.6) is o d = ∞ , and we see that the image distance is equal to the focal length f of the eyeglasses: i i i i 1 1 1 1 1 0 or d f f d d d = + = + = = ∞ (2) The focal length f (in meters) is equal to the reciprocal of the refractive power (in diopters) of the glasses, according to 1 Refractive power f = (Equation 26.8). SOLUTION Solving Equation (1) for s yields i s L d = − . Substituting Equation (2) into this result, we obtain i s L d L f = − = − (3)

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