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Consider Refraction at Spherical Surfaces: Starting point for the development of lens equations Vast majority of quality lenses that are used today have.

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Presentation on theme: "Consider Refraction at Spherical Surfaces: Starting point for the development of lens equations Vast majority of quality lenses that are used today have."— Presentation transcript:

1 Consider Refraction at Spherical Surfaces: Starting point for the development of lens equations Vast majority of quality lenses that are used today have segments containing spherical shapes. The aim is to use refraction at surfaces to simultaneously image a large number of object points which may emit at different wavelengths. Point V (Vertex) (object distance) (image distance)  i - Angle of incidence  t - Angle of refraction  r - Angle of reflection The ray SA emitted from point S will strike the surface at A, refract towards the normal, resulting in the ray AP in the second medium (n 2 ) and strike the point P.

2 All rays emerging from point S and striking the surface at the same angle  i will be refracted and converge at the same point P. Let’s return to Fermat’s Principal Using the Law of cosines:

3 Note: S i, S o, R are all positive variables here. Now, we can let d(OPL)/d  = 0 to determine the path of least time. Then the derivative becomes We can express this result in terms of the original variables l o and l i :

4 However, if the point A on the surface changes, then the new ray will not intercept the optical axis at point P. Assume new small vaules of the radial angle  so that cos   1, l o  s o, and l i  s i. This is known as the first-order theory, and involves a paraxial approximation. The field of Gaussian Optics utilizes this approach. Note that we could have also started with Snell’s law: n 1 sin  1 = n 2 sin  2 and used sin   . Again, subscripts o and i refer to object and image locations, respectively.

5 Using spherical (convex) surfaces for imaging and focusing i) Spherical waves from the object focus refracted into plane waves. Suppose that a point at f o is imaged at a point very far away (i.e., s i =  ). s o  f o = object focal length Object focus Suppose now that plane waves (parallel rays) are incident from a point emitting light from a point very far away (i.e., s o =  ). ii) Plane waves refracted into spherical waves.

6 Diverging rays revealing a virtual image point using concave spherical surfaces. Virtual image point Parallel rays impinging on a concave surface. The refracted rays diverge and appear to emanate from the virtual focal point F i. The image is therefore virtual since rays are diverging from it. R < 0 f i < 0 s i < 0 Signs of variables are important.


8 A virtual object point resulting from converging rays. Rays converging from the left strike the concave surface and are refracted such that they are parallel to the optical axis. An object is virtual when the rays converge toward it. s o < 0 here.

9 The combination of various surfaces of thin lenses will determine the signs of the corresponding spherical radii.

10 S (a) (b) (c) As the object distance s o is gradually reduced, the conjugate image point P gradually changes from real to virtual. The point P’ indicates the position of the virtual image point that would be observed if we were standing in the glass medium looking towards S.

11 We will use virtual image points to locate conjugate image points.

12 In the paraxial approximation: The 2 nd surface “sees” rays coming towards it from the P’ (virtual image point) which becomes the 2 nd object point for the 2 nd surface. Therefore (A) (B) Thus, at the 2 nd surface: Add Equations (A) & (B) 

13 Let d  0 (this is the thin lens approximation) and n m  1: and is known as the thin-lens equation, or the Lens maker’s formula, in which s o1 = s o and s i2 = s i, V 1  V 2, and d  0. Also note that For a thin lens (c)  f i = f o = f and Convex  f > 0 Concave  f < 0 Also, known as the Gaussian lens formula

14 Location of focal lengths for converging and diverging lenses

15 If a lens is immersed in a medium with f 2f2f 2f2f f Object Real image Convex thin lens Simplest example showing symmetry in which s o = 2f  s i = 2f Concave, f < 0, image is upright and virtual, |s i | < |f| f f Object Virtual image 2 3 1 sisi soso

16 Note that a ray passing through the center is drawn as a straight line. Ideal behavior of 2 sets of parallel rays; all sets of parallel rays are focused on one focal plane.


18 For case (b) below


20 Tracing a few key rays through a positive and negative lens yoyo S2S2 S1S1 Consider the Newtonian form of the lens equations. From the geometry of similar triangles:

21 Newtonian Form: x o > 0 if the object is to the left of F o. x i > 0 if the image is to the right of F i. The result is that the object and image must be on the opposite sides of their respective focal points.  Define Transverse (or Lateral) Magnification:


23 2f f f 2f Image forming behavior of a thin positive lens.

24 M T > 0  Erect image and M T < 0  Inverted image. All real images for a thin lens will be inverted. f 2f2f 2f2f f Simplest example 2f-2f conjugate imaging gives Define Longitudinal Magnification, M L This implies that a positive dx o corresponds to a negative dx i and vice versa. In other words, a finger pointing toward the lens is imaged pointing away from it as shown on the next slide.

25 The number-2 ray entering the lens parallel to the central axis limits the image height. The transverse magnification (M T ) is different from the longitudinal magnification (M L ). Image orientation for a thin lens:

26 (a) The effect of placing a second lens L 2 within the focal length of a positive lens L 1. (b) when L 2 is positive, its presence adds convergence to the bundle of rays. (c) When L 2 is negative, it adds divergence to the bundle of rays.

27 Two thin lenses separated by a distance smaller than either focal length. Note that d < s i1, so that the object for Lens 2 (L 2 ) is virtual. Note the additional convergence caused by L 2 so that the final image is closer to the object. The addition of ray 4 enables the final image to be located graphically.

28 Fig. 5.30 Two thin lenses separated by a distance greater than the sum of their focal lengths. Because the intermediate image is real, you could start with point P i ’ and treat it as if it were a real object point for L 2. Therefore, a ray from P i ’ through F o2 would arrive at P 1. Note that d > s i1, so that the object for Lens 2 (L 2 ) is real.

29 For the compound lens system, s o1 is the object distance and s i2 is the image distance. The total transverse magnification (M T ) is given by

30 For this two lens system, let’s determine the front focal length (ffl) f 1 and the back focal length (bfl) f 2. Let s i2   then this gives s o2  f 2. s o2 = d – s i1 = f 2  s i1 = d – f 2 but From the previous slide, we calculated s i2. Therefore, if s o1   we get, f ef = “effective focal length”

31 Suppose that we have in general a system of N lenses whose thicknesses are small and each lens is placed in contact with its neighbor. 1 2 3 ………N Then, in the thin lens approximation: Fig. 5.31 A positive and negative thin lens combination for a system having a large spacing between the lenses. Parallel rays impinging on the first lens enable the position of the bfl.

32 Example A Example B Example A: Two identical converging (convex) lenses have f 1 = f 2 = +15 cm and separated by d = 6 cm. s o1 = 10 cm. Find the position and magnification of the final image. s i1 = -30 cm at (O’) which is virtual and erect Then s o2 = |s i1 | + d = 30 cm + 6 cm = 36 cm s i2 = i’ = +26 cm at I’ Thus, the image is real and inverted.

33 The magnification is given by Thus, an object of height y o1 = 1 cm has an image height of y i2 = -2.17cm Example B: f 1 = +12 cm, f 2 = -32 cm, d = 22 cm An object is placed 18 cm to the left of the first lens (s o1 = 18 cm). Find the location and magnification of the final image. s i1 = +36 cm in back of the second lens, and thus creates a virtual object for the second lens. s o2 = -|36 cm – 22 cm| = -14 cm s i2 = i’ = +25 cm; The magnification is given by Thus, if y o1 = 1 cm this gives y i2 = -3.57 cm Image is real and Inverted

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