1. Lecture 25-1 Locating Images Real images form on the side of a mirror where the objects are, and virtual images form on the opposite side. only using the parallel, focal, and/or radial rays.

2. Lecture 25-2 Mirror Equation and Magnification s is positive if the object is in front of the mirror (real object) s is negative if it is in back of the mirror (virtual object) s’ is positive if the image is in front of the mirror (real image) s’ is negative if it is in back of the mirror (virtual image) m is positive if image and object have the same orientation (upright) m is negative if they have opposite orientation (inverted) f and r are positive if center of curvature in front of mirror (concave) f and r are negative if it is in back of the mirror (convex) (f = r/2)

3. Lecture 25-3 DOCCAM 2 7A-10 F and 2F Mirror

4. Lecture 25-4 READING QUIZ 1 Which of the following statements is incorrect ? The statements describe the optical characteristics of the human eye. A| A near sighted eye focuses in front of the retina when observing an object located at infinity. B| A far sighted eye focuses behind the retina when observing an object located at infinity. C| A converging lens is used to correct the vision of a near sighted eye. D| Astigmatism is a condition where the eye does not have cylindrical symmetry about the central optical axis. E| A cataract operation of the second lens inside the eye removes the clouded lens and replaces it with a plastic implant.

5. Lecture 25-5 s S’ Refracting Surface Formula A point object O is placed on the central axis of a convex refracting surface. The center of curvature of the surface is at C. It is easy to see (for small angles) Derivation: But Setting for parallel rays, the focal length is f (and r)>0 for convex surface, f (and r) n 1, where light goes 1-->2 Parallel ray refracts through the focal point. A ray through the focal point refracts parallel to the central axis. A ray through the center of curvature refracts straight.

6. Lecture 25-6 Thin Lens Formulas Relabeling,

7. Lecture 25-7 DOCCAM 2 7A-31 Under Water Lens Demonstration

8. Lecture 25-8 Lens Equation But ( < 0 )  True for thin lens and paraxial rays.  magnification m = h’/h = - q/p F s s’

9. Lecture 25-9 Thin Lenses nomenclature A lens is a piece of transparent material with two refracting surfaces whose central axes coincide. A lens is thin if its thickness is small compared to all other lengths (s, s’, radii of curvature). Net convex – thicker in the middle Parallel rays converge to real focus. f > 0 Net concave – thinner in the middle Parallel rays diverge from virtual focus. f < 0 f > 0 Convergent lens r 1 >0 r 2 <0 Divergent lens f < 0 r 1 <0 r 2 >0

10. Lecture 25-10 Thin Lenses nomenclature A lens is a piece of transparent material with two refracting surfaces whose central axes coincide. A lens is thin if its thickness is small compared to all other lengths (s, s’, radii of curvature). f > 0 Convergent lens r 1 >0 r 2 <0 The sign convention for radii is if it’s convex towards the light rays, it’s positive, and vice versa. So for example the lens below has both terms in the brackets POSITIVE (two minuses make a +.) Reciprocally passing light rays from right to left, the two radii both change sign, BUT since the right-most surface is hit FIRST, both terms are again POSITIVE. So this is a converging lens, and reciprocity says it’s converging in either direction!!

11. Lecture 25-11 Signs in the Lens Equation for Thin Lenses p is positive for real object q is positive for real image q is negative for virtual image m is positive if image is upright m is negative if image is inverted f is positive if converging lens f is negative if diverging lens p is negative for virtual object

12. Lecture 25-12 Properties of Images - Summary For converging lenses ( f > 0): If the object is inside the focal point, the image is virtual (q < 0), enlarged, has the same orientation, and is farther from the lens. If the object is outside the focal point, the image is real (q > 0), reduced or enlarged (depending on the object distance 2f), inverted, and closer to or farther from the lens on the other side. If the object is at the focal point, no image is formed. Not so for diverging lens For diverging lenses ( f < 0): The image is always virtual (q < 0), reduced, has the same orientation, and is closer to the lens. Object could be at focal point and still form a virtual image Power of a lens = 1/f (m -1 ) (diopters or D)

13. Lecture 25-13 SKETCH DOCCAM 2 F AND 2F FOR LENSES

14. Lecture 25-14 Warm up quiz 2 An object 1 cm tall is 20 cm in front (i.e., left) of a lens of focal length -20 cm and has an image 10cm also in front (left) of the lens. Which of the following is a correct description of the lens and its image? a)Diverging lens. The image size is magnified by a factor of 2. Image has the same orientation as the object. b). Diverging lens. The image size is magnified by a factor of 2 and inverted c). Diverging lens. The image size is reduced by a factor of 2. Image has the same orientation as the object. d). Diverging lens. The image size is reduced by a factor of 2 and inverted

15. Lecture 25-15 The Eye ≈ 2.5 cm f depends on p p ↓ => f↓ to keep q at ≈ 2.5 cm

16. Lecture 25-16 Lenses in Combination First lens: Second lens: Total transverse magnificationIn this example, p 1 > 0, q 1 > 0, p 2 > 0, q 2 < 0 ( < f 2 here) What if p 2 > f 2 ?

17. Lecture 25-17 Corrective Lenses Correct to 25 cm by conv. lens Hyperopia: Farsightedness: example- - near point = 75 cm Can’t focus near. Corrected by diverging lens Myopia(nearsighted) e.g., far point = 40 cm Can’t focus far. Use a lens of +2.67D Use a lens of -2.5D Presbyopia: stiff lens, can’t accommodate (be focused) What’s wrong with this picture?

18. Lecture 25-18 Magnifying Lens An object is placed near the focal point of a magnifying lens. The angle subtended by the image is tanθ = y/f. Without the lens, the largest angle subtended by the object is achieved when the object is placed at the near point, tanθ’ = y/x np. For small angles, the angular magnification f θ’  x np

19. Lecture 25-19 Compound Microscope A microscope consists of two converging lenses: an objective (the front lens) and an eyepiece. An object is placed near the first focal point of the objective. The separation of the lenses is adjusted so that the image produced by the objective is formed just inside the first focal point of the eyepiece. The lateral magnification of the objective is The overall magnifying power is defined as The eyepiece angular magnification (eye near point = x np )

20. Lecture 25-20 Astronomical Telescopes  The angular magnification M of the telescope is defined as θ e /θ o Refractor Telescope  Same combination (except image at ∞) as compound microscope: Objective creates a real image which allows the eyepiece to magnify.

21. Lecture 25-21 Aberrations Chromatic aberration correct Spherical aberration n blue > n red Parabolic mirrror Cameras, … Large telescopes, …

22. Lecture 25-22 Reflector Telescope  No chromatic aberration  Large mirrors can be made  (large amount of light gathered)  Easier to support  View center blocked off Whipple Telescopes (segmented lens)

23. Lecture 25-23 Physics 241 9:30 Quiz 3, April 19, 2011 An object 2 cm tall is 10 cm in front (i.e., left) of a convex (converging) lens of focal length 15 cm. Which of the following is a correct description of its image? a) The image is real. b) The image is inverted. c) The image is enlarged. d) The image is in back (right) of the lens. e) None of the above is correct.

24. Lecture 25-24 Physics 241 10:30 Quiz 3, April 19, 2011 An object 2 cm tall is 15 cm in front (i.e., left) of a concave (diverging) lens of focal length −10 cm. Which of the following is a correct description of its image? a) The image is enlarged. b) The image is real. c) The image is upright. d) The image is in back (right) of the lens. e) None of the above is correct.

25. Lecture 25-25 Physics 241 11:30 Quiz 3, April 19, 2011 An object 2 cm tall is 15 cm in front (i.e., left) of a convex (converging) lens of focal length 10 cm. Which of the following is a correct description of its image? a) The image is diminished. b) The image is inverted. c) The image is virtual. d) The image is in front (left) of the lens. e) None of the above is correct.