Optics Course (Phys ) Geometrical Optics Refraction through Lenses Lecturer: Dr Zeina Hashim
Slide 1 Objectives covered in this lesson : 1. The refracting power of a thin lens. 2. Thin lens combinations. 3. Refraction through thick lenses.
Slide 2 The Refracting Power of a Thin Lens: The refracting power of a thin lens is given by: P = 1 f Vergence: is the convergence or divergence of rays: V = 1 p and V = 1 i V + V = P A diopter (D): is a unit used to express the power of a spectacle lens, equal to the reciprocal of the focal length in meters.
Slide 3 The Refracting Power of a Thin Lens: Individual Activity Q: What is the refracting power of a lens in diopters if the lens has a focal length = 20 cm?
Slide 4 Thin Lens Combinations: If the optical system is composed of more than one lens (or a combination of lenses and mirrors) which are located so that their optical axes coincide: the final image can be obtained by working in steps: 1. Consider the nearest lens only, find the image of the object through this lens. 2. The image in step 1 is the object for the second (adjacent) optical component: find the image of this object. This can be done both geometrically or numerically 3. Do the same for all optical components, nearest first. 4. The final image is the image of the original object through this optical system.
Slide 5 Thin Lens Combinations: For a system composed of N lenses (and/or mirrors), the effective focal length of the system is given by: 1 f = 1 + 1 + + 1 f 1 f 2 f N The refractive power of the optical system is therefore: P = P 1 + P 2 + + P N The lateral magnification of the optical system is: Study Sample Problem 34-5 in Halliday (8 th ed.) to get familiar with how to solve problems of thin lens combinations. M = m 1 m 2 m N
Slide 6 Refraction Through Thick Lens: Description of a thick lens: A thick lens is a refracting material which is bounded by two spherical refracting surfaces. The image of a given object, formed by refraction at the first surface, becomes the object for refraction at the second surface. The image formed by the second surface is then the final image.
Slide 7 Refraction Through Thick Lens: Radii of Curvature and refractive indices Lens thickness (d) F 1 V 1 H 1 N 1 N 2 H 2 V 2 F 2 n 1 n L n 2 r 1 r 2 n 1 n 2 n L is the radius of curvature of the first refracting surface that faces the object. is the radius of curvature of the second refracting surface. is the refractive index of the medium at the side of the lens of where the object is. is the refractive index of the medium at the other side. is the refractive index of the lens.
Slide 8 Refraction Through Thick Lens: The 6 Cardinal Points Cardinal Points: Lens thickness (d) are points, on the optical axis of a thick lens, from which its imaging properties can be deduced. F 1 n 1 V 1 H 1 N 1 n L N 2 H 2 V 2 n 2 F 2 The 6 Cardinal Points: F 1 and F 2 : 1 st and 2 nd focal points of the thick lens. H 1 and H 2 : 1 st and 2 nd principal points. N 1 and N 2 : 1 st and 2 nd nodal points. Cardinal Planes: are planes normal to the optical axis at the cardinal points.
Slide 9 Refraction Through Thick Lens: Principal Planes If a ray emerges the lens from F 1 : it is rendered parallel to the optical axis. the extensions of the incident and resultant rays intersect in the 1 st principal plane. F 1 H 1 F 2 If a ray emerges the lens parallel to the optical axis: it is refracted by the lens through F 2. F 1 F 2 the extensions of the incident and resultant rays intersect in the 2 nd principal plane. H 2
Slide 10 Refraction Through Thick Lens: Principal Planes If the thick lens were a single thin lens, the two principal planes would coincide at the vertical line drawn in the middle of the thin lens. Principal planes in thick lenses usually do not coincide, and they may be located outside the optical system. Once the locations of the principal planes are known, accurate raydiagrams can be drawn.
Slide 11 Refraction Through Thick Lens: Nodal Points A ray directed towards the first nodal point (N 1 ): emerges from the optical system parallel to the incident ray, but displaced, so that it appears to come from N 2. F 1 N 1 N 2 F 2
Slide 12 Refraction Through Thick Lens: Positions and signs Lens thickness (d) Distances shown in the figure (and their values) f 1 f 2 are: Positive: if arrow is to the right F 1 V 1 H 1 N 1 N 2 H 2 V 2 F 2 (i.e. from object side to the other side). n 1 n L n 2 Negative: if arrow is to the left (i.e. to object) s 1 s 2 v 1 v 2
Slide 13 Refraction Through Thick Lens: Positions and signs Lens thickness (d) f 1 is the distance from H 1 to F 1 f 1 f 2 f 2 is the distance from H 2 to F 2 s 1 is the distance from V 1 to H 1 F 1 V 1 H 1 N 1 N 2 H 2 V 2 F 2 s 2 is the distance from V 2 to H 2 n 1 n L n 2 v 1 is the distance from V 1 to N 1 s 1 s 2 v 2 is the distance from V 2 to N 2 v 1 v 2
Slide 14 Refraction Through Thick Lens: Positions and signs f (-) F is to the left of H Lens thickness (d) f (+) F is to the right of H f 1 f 2 s (-) H is to the left of V F 1 V 1 H 1 N 1 N 2 H 2 V 2 F 2 s (+) H is to the right of V n 1 n L n 2 v (-) N is to the left of V s 1 s 2 v (+) N is to the right of V v 1 v 2
Slide 15 Thick Lens Equations: Focal Lengths: 1 f 1 = n L n 2 n 1 r 2 n L n 1 n 1 r 1 (n L n 1 )(n L n 2 ) n 1 n L d r 1 r 2 These equations are NOT to memorize! f 2 = n 2 n 1 f 1 f 2 = f 1 if the lens is surrounded by a single medium (at which n 2 = n 1 )
Slide 16 Thick Lens Equations: Principal Planes Positions from the Vertices: s 1 = n L n 2 n L r 2 f 1 d Positions of Nodal Points from the Vertices: s 2 = n L n 1 n L r 1 f 2 d These equations are NOT to memorize! v 1 = 1 n 2 n 1 + n L n 2 n L r 2 d f 1 v 2 = 1 n 1 n 2 + n L n 1 n L r 1 d f 2
Slide 17 Thick Lens Equations: Image and Object Distance: p > 0 p < 0 f 1 p + f 2 i = 1 This equation is NOT to memorize! i > 0 i < 0 The distances p, i, and f are measured relative to their corresponding principal planes.
Slide 18 Thick Lens Equations: Image and Object Distance: Is the image (and object) real or virtual?? In thin lenses: Objects are always real. Image is real : if it is on the opposite side of the lens of where the object is. Image is virtual: if it is on the same side of the lens where the object is. In thick lenses: It is more complicated we will not study it in this course. The object can be virtual with thick lenses!
Slide 19 Thick Lens Equations: Lateral Magnification: m = n 1 i n 2 p This equation is NOT to memorize! When m is positive object and image have the same orientation. When m > 1 image > object. When m < 1 image < object.
Slide 20 Refraction through thin lenses: Group Work Work in pairs: (a) What happens to the positions of the nodal points and positions of the principal points if the lens was in air?
Slide 21 Thick Lens Equations: Final notes: When the lens is in air: the focal lengths will have the same magnitudes, and the focal length equation will become: 1 f = 1 = (n f L 1)( 1 1 n L 1 2 r 2 r 1 n L d r 1 r 2 ) the usual thin-lens equations will be valid; 1 p + 1 i = 1 f, m = i p where f = f 2 = f 1
Exercise: 1 f 1 = n L n 2 n 1 r 2 n L n 1 n 1 r 1 (n L n 1 )(n L n 2 ) n 1 n L d r 1 r 2 Slide 22 f 2 = n 2 n 1 f 1 Determine the focal lengths and the principal points for a 4-cm thick, biconvex, lens with refractive index of 1.52 and radii of curvature of 25 cm, when the lens caps the end of a long cylinder filled with water ( n= 1.33)? Using the thick lens equations: 1 f 1 = 1.52 1.33 1( 25) f 1 = 35.74 cm 1.52 1 1 +25 to the left of the first principal plane (1.52 1)(1.52 1.33) 1(1.52) 4 (+25)( 25) f 2 = 1.33 35.74 = 47.53 cm 1 to the right of the second principal plane
Slide 23 Exercise: s 1 = n L n 2 f n L r 1 d s 2 = n L n 1 f 2 n L r 2 d 1 Determine the focal lengths and the principal points for a 4-cm thick, biconvex, lens with refractive index of 1.52 and radii of curvature of 25 cm, when the lens caps the end of a long cylinder filled with water ( n= 1.33)? Using the thick lens equations: s 1 = 1.52 1.33 1.52 25 35.74 4 = 0.715 cm, s 2 = 1.52 1 1.52 +25 47.53 4 = 2.60 cm Thus, the principal point H 1 is situated 0.715 cm to the right of the left vertex of the lens, and H 2 is situated 2.60 cm to the left of the right vertex.
Slide 24 How to draw Rays with Thick Lenses:
Slide 25 Homework : Q1: Two-lens systems: In Figure, stick figure O (the object) stands on the common central axis of two thin, symmetric lenses, which are mounted in the boxed regions. Lens 1 is mounted within the boxed region closer to O, which is at object distance p 1. Lens 2 is mounted within the farther boxed region, at distance d. Each problem in the table (labeled 80 87) refers to a different combination of lenses and different values of distances, which are given in centimeters. The type of lens is indicated by C for converging and D for diverging; the number after C or D is the distance between a lens and either of its focal points (the proper sign of the focal distance is not indicated). Find (a) the image distance i 2 for the image produced by lens 2 (the final image produced by the system) and (b) the overall lateral magnification M for the system, including signs. Also, determine whether the final image is (c) real (R) or virtual (V), (d) inverted (I) from object O or noninverted (NI), and (e) on the same side of lens 2 as object O or on the opposite side.
Slide 26 Homework : Q1:
Slide 27 Homework : Q2:
Slide 28 Homework : Q3: A small object is placed 20 cm from the first of a train of three lenses with focal lengths, in order, of 10, 15, and 20 cm. The first two lenses are separated by 30 cm and the last two lenses are separated by 20 cm. Calculate the final image position relative to the last lens and its linear magnification relative to the original object when (a) all three lenses are positive, (b) the middle lens is negative, (c) the first and last lenses are negative. Provide ray diagrams for each.
Slide 29 Homework : Q4: Go back to the thick-lens example given in the previous slides and study it well.
Slide 30 (last) Summary: Next lesson will cover: 1. The refracting power of a thin lens. 2. Thin lens combinations. 3. Refraction through thick lenses. Field of View + Aberrations Any Questions?