Lenses

1. Lenses Types of Lenses Locating images using ray diagrams Calculations involving Lenses

2. Lenses Recall mirrors have one reflective & one opaque surface. Lenses on the other hand have no opaque surface. So instead of light being reflected as in mirrors, lights in lens will be refracted (as it goes through from air to the lens and back to air)

3. Basic Lens Shapes • Converging lens: • Thickest in the middle • Parallel incident rays will be refracted and converge at a single point

4. Basic Lens Shapes • Diverging lens: • Thinnest in the middle • Parallel incident rays will be refracted and diverge (spread apart)

5. Simplifying Path of Light Rays In reality, as light goes from AIR to LENS then out to AIR again, you see 2 refractions (bending of light). To simplify it, we only show 1 refraction through central line. Incident ray Emergent ray

6. Terminology of Converging Lens Different terminology than the ones used for mirrors. Optical Centre Principal axis F’ Secondary Principal focus O F Principal focus

7. Terminology of Diverging Lens Different terminology than the ones used for converging lens.

8. Converging Lens - Rules A ray through the secondary principal focus (F’) is refracted parallel to principal axis A ray parallel to principal axis is refracted through the principal focus (F) • F = principal focus • F’ = secondary principal focus • O = optical centre Principal axis O 2F’ F’ F 2F A ray through Optical centre (O) continues straight through without being refracted.

9. Converging Lens – Real vs. Virtual Image • If Image is on the OPPOSITE SIDE of the lens from the object = REAL Image • If Image is on the SAME SIDE as object = Virtual Image This is the opposite of mirrors! Don’t get them mixed up. VIRTUAL Image Object REAL Image

10. Converging Lens – Finding the image • Use 2 of the rules to find the image O 2F’ F’ F 2F

11. Converging Lens Investigation • Use the templates given to investigate and complete the following table:

12. Converging Lens -Object beyond 2F’ When object is beyond 2F’, the image will be: S – smaller A – inverted L – between F and 2F T – Real 2F’ F’ O 2F F

13. Converging Lens -Object at2F’ When object is at 2F’, the image will be: S – same size A – inverted L – at 2F T – Real 2F’ F’ O 2F F

14. Converging Lens -Object between 2F’ and F’ When object is between 2F’ and F’, the image will be: S – larger A – inverted L – beyond2F T – Real 2F’ F’ O 2F F

15. Converging Lens -Object at F’ When object is at F’ NO IMAGE – lines are parallel 2F’ F’ O 2F F

16. Converging Lens -Object at F’ When object is between F’ and the lens, the image will be: S – larger A – UPRIGHT L – same side as object T – VIRTUAL 2F’ F’ O 2F F Extend your refracted ray BACKWARDS to locate theimage

17. Remember: The only time you’ll get VIRTUAL image with converging lens is when the object is between F’ and O Extend your refracted ray BACKWARDS to locate virtual image O 2F’ F’ F 2F

18. Converging Lens Investigation:Results

19. Diverging Lens

20. Diverging Lens- Rules A ray that APPEARS TO PASS through the secondary principal focus (F’) is refracted parallel to p.a Then EXTEND refracted ray backwards! A ray parallel to principal axis is refracted AS IF IT HAD COME through the principal focus (F) Note the DIFFERENCE IN PLACEMENT of F and F’ in diverging lens Principal axis O 2F F F’ 2F’ A ray through Optical centre (O) continues straight through

21. Diverging Lens- Finding the image Use 2 of the rules to locate the image: Principal axis O 2F F F’ 2F’

22. Diverging Lens Investigation • Use the templates given to investigate and complete the following table:

23. Diverging Lens Investigation:Results Always the SAME image characteristics no matter where the object is located:

24. Calculations involving Lenses

25. Thin Lens Equation 1 + 1 = 1 dodif do di Image Object O F 2F’ F’ 2F f f

26. Thin Lens Equation 1 + 1 = 1 dodi f do is always + di is + for real image – for virtual image f is + for converging lens – for diverging lens

27. Thin Lens Equation Sample Problem: A converging lens has a focal length of 17 cm. A candle is located 48 cm from the lens. What type of image will be formed and where will it be located? 1 + 1 = 1 dodi f do is always + di is + for real image – for virtual image f is + for converging lens – for diverging lens

28. Thin Lens Equation Given: f = 17cm (converging lens so +) do = 48 cm (do always +) Required: type of image? d i? Analysis: Rearrange thin lens equation 1 = 1 - 1 di f do 1 + 1 = 1 dodi f do is always + di is + for real image – for virtual image f is + for converging lens – for diverging lens

29. Thin Lens Equation Solution: 1 = 1 – 1 . di17cm 48cm 1 = 0.038 cm-1 di Use inverse function: di = 26 cm Since di is positive, it’s real image. Statement: The image is real image and is located 26 cm from the lens, opposite to the object 1 + 1 = 1 dodi f do is always + di is + for real image – for virtual image f is + for converging lens – for diverging lens

30. Magnification Equation M = hi = – di ho do do di ho Image Object O F 2F’ F’ 2F hi f f

31. Magnification Equation M = hi = – di ho do h0 and h1 are: + when measured upward • – when measured downward • M is: • + for upright image • – for inverted image

32. Magnification Equation Sample Problem : A toy of height 8.4 cm is balanced in front of converging lens. An inverted, real image of height 23 cm is noticed on the other side of the lens. What is the magnification of the lens? M = hi = - di ho do h0 and h1 are: + when upward • – when downward • M is: • + for upright image • – for inverted image

33. Magnification Equation • Given: • ho = 8.4 cm (upward, so +) • hi = – 23 cm (inverted, so – ) • Required: M? • Analysis: • Use M = hi • ho M = hi = - di ho do h0 and h1 are: + when upward • – when downward • M is: • + for upright image • – for inverted image

34. Magnification Equation • Solution: • M = hi = – 23 cm • ho8.4 cm • M = – 2.7 • Statement: • The lens has a magnification of – 2.7 M = hi = - di ho do h0 and h1 are: + when upward • – when downward • M is: • + for upright image • – for inverted image