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 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)<0 for concave surface 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.

4. Lecture 25-4 Thin Lens Formulas Relabeling,

5. Lecture 25-5 Lens Equation ( < 0 )  True for thin lens and paraxial rays.  magnification m = h’/h = - q/p

6. Lecture 25-6 Thin Lenses 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

7. Lecture 25-7 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

8. Lecture 25-8 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 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), inverted, and farther from or closer to the lens on the other side. If the object is at the focal point, no image is formed. For diverging lenses ( f < 0): The image is always virtual (q < 0), reduced, has the same orientation, and closer to the lens. Power of a lens = 1/f (m -1 ) (diopters or D)

9. Lecture 25-9 Warm up quiz 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

10. Lecture 25-10 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 ?

11. Lecture 25-11 Compound Lenses 1 2 p > 0 q 1 > 0 q > 0 p 2 < 0 Power adds up in compound lenses Virtual object For a virtual object, the object distance is negative.

12. Lecture 25-12 The Eye ≈ 2.5 cm f depends on p p ↓ => f↓ to keep q at ≈ 2.5 cm

13. Lecture 25-13 Corrective Lenses Correct to 25 cm by conv. lens Hyperopia: A person has 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 Forgot to bring the glass? Not a problem.

14. Lecture 25-14 Magnifying Lens An object is placed near the focal point of a magnifying lens. The angle subtended by the image is tan  = y/f. Without the lens, the largest angle subtended by the object is achieved when the object is placed at the near point, tan  ’ = y/x np. For small angles, the angular magnification f ’’  x np

15. Lecture 25-15 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 is:

16. Lecture 25-16 Astronomical Telescopes  The angular magnification M of the telescope is defined as  e /  o Refractor Telescope  Same idea as compound microscope: Objective creates a real image which allows the eyepiece to magnify.

17. Lecture 25-17 Aberrations Chromatic aberration correct Spherical aberration n blue > n red Parabolic mirrror Cameras, … Large telescopes, …

18. Lecture 25-18 Reflector Telescope  No chromatic aberration  Large mirrors can be made (large amount of light gathered)  Easier to support  View center blocked off Whipple Telescope

19. Lecture 25-19 Physics 241 –Quiz A 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.

20. Lecture 25-20 Physics 241 –Quiz B 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.

21. Lecture 25-21 Physics 241 –Quiz C 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.