3. TopicSlidesMinutes 1 Displacement 927 2 Vectors 1339 3 Kinematics 1339 4 Graphs 1030 5 Energy 1030 6 Power 515 7 Springs 412 8 Shadows 39 9 Field of Vision 721 10 Colors 39 11 Concave mirrors 721 12 Convex mirrors 412 13 Refraction 515 14 1030 15 Optical Power 618 Lenses

4. Click Convex lenses are thicker in the middle and thus they converge light rays. Concave lenses are thinner in the middle and thus they diverge light rays.

5. Click Just as with concave mirrors, the characteristics of the image formed by a converging lens depend upon the location of the object. There are six "strategic" locations where an object may be placed. For each location, the image will be formed at a different place and with different characteristics. We will illustrate the six different locations and label them as CASE-1 to CASE-6. Case-1: Object at infinity Case-2: Object just beyond 2 F’ Case-3: Object at 2F’ Case-4: Object between 2F’ and F’ Case-5: Object at F’ Case-6: Object within focal length (f)

6. Click CASE-1 : Object at “infinity” No image formed (All rays pass through F) No image NOTE Since the object is at “infinity”, all rays arrive parallel. Infinity simply means “far away”. Object

7. Click CASE-2 : Object just beyond 2F’ Image is real (formed by refracted rays) Inverted (upside down) Reduced (smaller than object) Located between F and 2F ObjectImage Note-1 A ray that comes parallel is refracted through F. Note-2 A ray that goes through the vertex goes right through. Note-3 A ray that goes through F’ is refracted parallel. NOTE In order to establish an image point, all we need are two intersecting rays. This ray is extra in locating the image.

8. Click CASE-3 : Object at 2F’ Image is real (formed by refracted rays) Inverted (upside down) Same size as object Located at 2F Object Image Again: In order to establish an image point, all we need are two intersecting rays. This ray is extra.

9. Click CASE-4 : Object between 2F’ and F’ Image is real (formed by refracted rays) Inverted (upside down) Magnified (larger than object) Located beyond 2F ObjectImage

10. Click CASE-5 : Object at F’ No image is formed (rays refract parallel) Object No image

11. Click CASE-6 : Object is within focal length Image is virtual (formed by extended rays) Upright Magnified Located on same side as object Object Image

13. Click An object whose height is 0.10 m is placed 1.0 m from the focal point of a converging lens whose focal length is 0.50 m. Determine the height of the image. Calculation of image distance Calculation of object distance The negative sign indicates inversion. Given Lenses Slide: 14. 1

14. Click Which of the following lenses corrects the eye defect known as myopia? ABCD Lenses Slide: 14. 2

15. A) Plane B) Plano-concave C) Concavo-convex D) Convex-concave Which type of lens is used to correct hyperopia? Click Thicker in the middle than at the ends. Myopia (farsightedness) is corrected with a Concavo-concave lens. Nearsightedness Thicker at the ends than in the middle. Lenses Slide: 14. 3

16. Click Lenses Slide: 14. 4

17. This ray should go thru F since it emerges parallel to the principal axis. This ray should go thru F since it arrives parallel to the principal axis. parallel incident ray parallel refracted ray This ray should go thru V since it is refracted without being bent. Click Lenses Slide: 14. 5

18. An object that is 4.0 cm tall is placed 7.0 cm from a converging lens that has a focal length of 2.0 cm. Click Calculate the height of the image formed by this lens. The negative sign indicates inversion and may be ignored. The image is 1.6 cm tall. Lenses Slide: 14. 6

19. As illustrated below, an object is placed infront of a diverging lens. Click Which of the following statements correctly describes the characteristics of the image formed. A) Virtual, upright, smaller and on the same side as the object B) Real, upright, smaller and on the same side as the object C) Virtual, inverted, larger and on the opposide side of the object D) Real, inverted, larger and on the opposide side of the object Image Lenses Slide: 14. 7

20. Draw a ray connecting the object to the image in order to establish the center of the lens, V. Center of lens, V Converging lens F F’ V Click Lenses Slide: 14. 8

21. Karen wants to draw a mural of her dog on a wall. Using a projector she displays a picture of her dog on the wall. The focal length of the projector lens is 35 mm. She needs to position the projector so that the real image of the picture of the dog has a magnification of 40. How far from the wall must the projector lens be? Note that since the image is real, it is inverted and the magnification factor is negative. Lenses Slide: 14. 9

22. Lenses Slide: 14. 10 Construct a ray diagram to locate the image formed by the lens illustrated below. Click Top (object) point Top (image) pointBottom (object) point Bottom (image) point Image This is the symbol for a converging lens

23. … and good luck!

24. Lenses Slide: 14. 10 Construct a ray diagram to locate the image formed by the lens illustrated below. Click Top (object) point Top (image) pointBottom (object) point Bottom (image) point Image This is the symbol for a converging lens