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Physics 1202: Lecture 22 Today’s Agenda Announcements: –Lectures posted on: www.phys.uconn.edu/~rcote/ www.phys.uconn.edu/~rcote/ –HW assignments, etc.

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Presentation on theme: "Physics 1202: Lecture 22 Today’s Agenda Announcements: –Lectures posted on: www.phys.uconn.edu/~rcote/ www.phys.uconn.edu/~rcote/ –HW assignments, etc."— Presentation transcript:

1 Physics 1202: Lecture 22 Today’s Agenda Announcements: –Lectures posted on: www.phys.uconn.edu/~rcote/ www.phys.uconn.edu/~rcote/ –HW assignments, etc. Homework #7:Homework #7: –Due next Friday

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3 Prisms 11 22 33 44 Overall Deflection At both deflections the amount of downward deflection depends on n (and the prism apex angle,  ). Each color has a different speed or n. Different colors will bend different amounts ! 

4 o i f h’h’ h R   h h’h’ o-R R-i o i &

5 Concave Spherical Mirrors

6 The Mirror Equation We will now transform the geometric drawings into algebraic equations: R     object  h image o i from triangles, eliminating , Now we employ the small angle approximations: Plugging these back into the above equation relating the angles, we get: Defining the focal length f = R/2, This eqn is known as the mirror eqn. Note that there is no mention of  in this equation. Therefore, this eqn works for all , ie we have an image!

7 Magnification We have derived the mirror eqn which determines the image distance in terms of the object distance and the focal length: What about the size of the image? How is h’ related to h?? From similar triangles: Now, we can introduce a sign convention. We can indicate that this image is inverted if we define its magnification M as the negative number given by:   R h o h’h’ i

8 More Sign Conventions Consider an object distance s which is less than the focal length:   h’h’ i Ray Trace: Ray through the center of the sphere (light blue) is reflected straight back. R h o f We call this a virtual image, meaning that no light from the object passes through the image point. Proof left to student: This situation is described by the same mirror equations as long as we take the convention that images behind the mirror have negative image distances s’. ie: In this case, i 0, indicating that the image is virtual (i 0). Ray parallel to axis (red) passes through focal point f. These rays diverge! ie these rays look they are coming from a point behind the mirror.

9 Concave-Planar-Convex What happens as we change the curvature of the mirror? –Plane mirror: »R =  IMAGE: virtual upright (non-inverted) h’h’  h  o i f IMAGE: virtual upright (non-inverted) –Convex mirror: »R < 0

10 Lecture 22, ACT 1 In order for a real object to create a real, inverted enlarged image, a) we must use a concave mirror. b) we must use a convex mirror. c) neither a concave nor a convex mirror can produce this image.

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14 Mirror – Lens Definitions Some important terminology we introduced last class, –o = distance from object to mirror (or lens) – i = distance from mirror to image o positive, i positive if on same side of mirror as o. –R = radius of curvature of spherical mirror –f = focal length, = R/2 for spherical mirrors. –Concave, Convex, and Spherical mirrors. –M = magnification, (size of image) / (size of object) negative means inverted image R     object  h image o i

15 Lenses A lens is a piece of transparent material shaped such that parallel light rays are refracted towards a point, a focus: –Convergent Lens »light moving from air into glass will move toward the normal »light moving from glass back into air will move away from the normal »real focus –Divergent Lens »light moving from air into glass will move toward the normal »light moving from glass back into air will move away from the normal »virtual focus

16 1) Rays parallel to principal axis pass through focal point. 2) Rays through center of lens are not refracted. 3) Rays through F emerge parallel to principal axis. Assumptions: monochromatic light incident on a thin lens. rays are all “ near ” the principal axis. F F Object P.A. Image is: real, inverted and enlarged (in this case). Image Converging Lens Principal Rays

17 The Lens Equation We now derive the lens equation which determines the image distance in terms of the object distance and the focal length. –Convergent Lens: i f h’h’ o h Ray Trace: Ray through the center of the lens (light blue) passes through undeflected. two sets of similar triangles: eliminating h’/h: same as mirror eqn if we define i > 0 f > 0 magnification: also same as mirror eqn!! M < 0 for inverted image. Ray parallel to axis (white) passes through focal point f.

18 Summary We have derived, in the paraxial (and thin lens) approximation, the same equations for mirrors and lenses: when the following sign conventions are used: Variable f > 0 f < 0 o > 0 o < 0 i > 0 i < 0 Mirror concave convex real (front) virtual (back) real (front) virtual (back) Lens converging diverging real (front) virtual (back) real (back) virtual (front)

19 This could be used as a projector. Small slide on big screen This is a magnifying glass This could be used in a camera. Big object on small film Upright Enlarged Virtual Inverted Enlarged Real Inverted Reduced Real 3 Cases for Converging Lenses ImageObject Inside F Object Image Past 2F Image Object Between F & 2F

20 1) Rays parallel to principal axis pass through focal point. 2) Rays through center of lens are not refracted. 3) Rays toward F emerge parallel to principal axis. F F Object P.A. Image is virtual, upright and reduced. Image Diverging Lens Principal Rays

21 Lecture 22, ACT 2 A lens is used to image an object on a screen. The right half of the lens is covered. –What is the nature of the image on the screen? (a) left half of image disappears (b) right half of image disappears (c) entire image reduced in intensity object lens screen

22 Multiple Lenses We determine the effect of a system of lenses by considering the image of one lens to be the object for the next lens. For the first lens: o 1 = +1.5, f 1 = +1 For the second lens: o 2 = +1, f 2 = -4   f = +1 f = -4 +3 +1 0 +2+6 +5+4

23 Multiple Lenses Objects of the second lens can be virtual. Let’s move the second lens closer to the first lens (in fact, to its focus): For the first lens: o 1 = +1.5, f 1 = +1 For the second lens: o 2 = -2, f 2 = -4   Note the negative object distance for the 2nd lens. f = +1 f = -4 +3 +1 0 +2+6 +5+4

24 Multiple Lenses If the two lenses are thin, they can be touching – i.e. in the same position. We can treat as one lens. f total = ?? ? Adding, For the first lens: o=o 1, i 1 and f 1 For the second lens: o 2 = -i 1, i 2 =i, f 2 As long as,

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27 The Lens Equation –Convergent Lens: i f h’h’ o h

28 The Lensmaker’s Formula So far, we have treated lenses in terms of their focal lengths. How do you make a lens with focal length f ? Start with Snell’s Law. Consider a plano-convex lens: Snell’s Law at the curved surface: The bend-angle  is just given by: The bend-angle  also defines the focal length f: The angle  can be written in terms of R, the radius of curvature of the lens : Putting these last equations together, R N air h     light ray Assuming small angles,

29 More generally…Lensmaker’s Formula Two curved surfaces… Two arbitrary indices of refraction R > 0 if convex when light hits it R < 0 if concave when light hits it The complete generalized case… Note: for one surface Planar,

30 Compound Microscope o1o1 h O I2I2 h2h2 f eye h1h1 I1I1 i1i1 Objective (f ob < 1cm) f ob L Eyepiece (f eye ~5cm) Magnification:

31 Refracting Telescope Star f eye I2I2 h2h2 f ob Objective (f ob ~ 250cm) Eyepiece (f eye ~5cm) i1i1 I1I1 h1h1 Angular Magnification:    

32 ~f e I1I1 eyepiece I2I2 ~f o objective L The EYE

33 Retina To brain The Eye What does the eye consist of? –Sphere (balloon) of water. - An aperture that controls how much light gets through – the Iris/pupil - Bulge at the front – the cornea - A variable focus lens behind the retina – the lens - A screen that is hooked up to your brain – the retina Cornea Iris Lens


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