1. mirrors and lenses PHY232 – Spring 2007 Jon Pumplin http://www.pa.msu.edu/~pumplin/PHY232 (Ppt courtesy of Remco Zegers)

2. PHY232 - Pumplin - Mirrors and lenses 2 we saw…  that light can be reflected or refracted at boundaries between material with a different index of refraction.  by shaping the surfaces of the boundaries we can make devices that can focus or otherwise alter an image.  Here we focus on mirrors and lenses for which the properties can be described well by a few equations.

3. PHY232 - Pumplin - Mirrors and lenses 3 the plane (=flat) mirror  in the previous chapter we already studied flat mirrors.  Distance from the object to the mirror is the object distance p  Distance from the image to the mirror is the image distance -q  in case of a flat mirror, an observer sees a virtual image, meaning that the rays do not actually come from it.  the image size (h ’ ) is the same as the object size (h), meaning that the magnification h ’ /h=1  the image is not inverted p-q NOTE: a virtual image cannot be projected on a screen but is ‘visible’ by the eye or another optical instrument.

4. PHY232 - Pumplin - Mirrors and lenses 4 question  You are standing in front (say 1 m) of a mirror that is less high than your height. Is there a chance that you can still see your complete image?  a) yes b) no object image

5. PHY232 - Pumplin - Mirrors and lenses 5 ray diagrams  to understand the properties of optical elements we use ray diagrams, in which we draw the most important elements and parameters to understand the elements p-q h h’

6. PHY232 - Pumplin - Mirrors and lenses 6 concave mirrors C C: center of mirror curvature a light ray passing through the center of curvature will be reflected back upon itself because it strikes the mirror normally to the surface. F: focal point F a light ray traveling parallel to the central axis of the mirror will be reflected to the focal point F, with FM=CM/2 The distance FM is called the focal length f. M

7. PHY232 - Pumplin - Mirrors and lenses 7 concave mirrors: an object outside F O F step 1: draw the ray from the top of the object parallel to the central axis and its reflection (through F). step 2: draw the ray from the top of the object through F and its reflection (parallel to the central axis) step 3: note that a ray from the bottom of the object just reflects back. the image of the top of the object is located where the reflected rays meet construct the image I I

8. PHY232 - Pumplin - Mirrors and lenses 8 concave mirrors: an object outside F O FI The image is: a)inverted (upside down) b)real (light rays pass through it) c)For this location, image is smaller than object d)If interchange image and object, image would be bigger than object.

9. PHY232 - Pumplin - Mirrors and lenses 9 concave mirrors: object outside F O FI distance object-mirror: p distance image-mirror: q distance focal point-mirror: f mirror equation: 1/p + 1/q = 1/f given p,f this equation can be used to calculate q magnification: M = -q/p can be used to calculate magnification. if negative: the image is inverted if smaller than 1, object is demagnified

10. PHY232 - Pumplin - Mirrors and lenses 10 example  An object is placed 12 cm in front of a a concave mirror with focal length 5 cm. What are:  a) the location of the image  b) the magnification a)1/p + 1/q = 1/f so 1/12 + 1/q = 1/5 1/q = 1/5 - 1/12 so q = 8.57 cm b) M = -q/p = -8.57/12 = -0.71 this means that size of the image is only 71% of the the size of the object and that it is inverted. c) Image is real (whenever q>0, I.e., whenever M > 0)

11. PHY232 - Pumplin - Mirrors and lenses 11 concave mirrors: an object inside F OF step 1: draw the ray from the top of the object parallel to the central axis and its reflection (through F). step 3: note that a ray from the bottom of the object just reflects back. step 2: draw the ray from the top of the object through F and its reflection (parallel to the central axis) I the image of the top of the object is located where the reflected rays meet: in this you must draw virtual rays on the other side of the lens creates a magnified image (“shaving mirror”) the image is: a)not inverted b)virtual c)magnified

12. PHY232 - Pumplin - Mirrors and lenses 12 concave mirrors: an object inside F OFI the image is: a)not inverted b)virtual c)magnified The lens equation and equation for magnification are still valid. However, since the image is now on the other side of the mirror, its sign should be negative

13. PHY232 - Pumplin - Mirrors and lenses 13 example  an object is placed 2 cm in front of a lens with a focal length of 5 cm. What are the a) image distance and b) the magnification? a)1/p + 1/q = 1/f so 1/2 + 1/q = 1/5 1/q = 1/5 - 1/2 so q = -3.3 cm (note the ‘-’ sign!) b) M = -q/p = -(-3.3)/2 = +1.65 this means that size of the image is 65% larger than the size of the object and that it is not inverted (+).

14. PHY232 - Pumplin - Mirrors and lenses 14 demo: the virtual pig

15. PHY232 - Pumplin - Mirrors and lenses 15 step 2: draw the ray from the top of the object through F and its reflection (parallel to the central axis) convex mirrors: (p>|f|or p<|f| doesn’t matter) O F step 3: note that a ray from the bottom of the object just reflects back. the image of the top of the object is located where the reflected rays meet construct the image I (examples: garden ball, rearview mirrors that warn “objects are closer than they appear”) I step 1: draw the ray from the top of the object parallel to the central axis and its reflection (through F). F is now located on the other side of the mirror

16. PHY232 - Pumplin - Mirrors and lenses 16 convex mirrors O FI F is now located on the other side of the mirror the image is: a)not inverted b)virtual c)demagnified The lens/mirror equation and equation for magnification are still valid. However, since the image and focal point are now on the other side of the mirror, their signs should be negative

17. PHY232 - Pumplin - Mirrors and lenses 17 example  an object with a height of 3 cm is placed 6 cm in front of a convex mirror, with f=-3 cm. What are a) the image distance and b) the magnification? answer a): 1/p+1/q=1/f with p=6 cm f=-3 cm 1/6 + 1/q = -1/3 so q=-2 cm b) M=-q/p=-(-2)/6=1/3 the image is only 33% of the height of the object.

18. PHY232 - Pumplin - Mirrors and lenses 18 Mirrors: an overview  mirror equation 1/p + 1/q = 1/f  F = R/2 where R is the radius of the mirror  magnification: M = -q/p typep?imageimage direction Mqf concavep>frealinverted|M|>0 M -++ concavep<fvirtualnot inverted |M|>1 M +-+ convexany pvirtualnot inverted |M|<1 M +--

19. PHY232 - Pumplin - Mirrors and lenses 19 Lenses  Lenses function by refracting light at their surfaces  Their action depends on  radii of the curvatures of both surfaces  the refractive index of the lens  converging (positive lenses) have positive focal length and are always thickest in the center  diverging (negative lenses) have negative focal length and are thickest at the edges + - used in drawings

20. PHY232 - Pumplin - Mirrors and lenses 20 lensmakers equation R1R1 R2R2 f: focal length of lens n: refractive index of lens R 1 radius of front surface R 2 radius of back surface 1 2 R 2 is negative if the center of the circle is on the left of curvature 2 of the lens R 1 is positive if the center of the circle is on the right of curvature 1 of the lens object if the lens is not in air then (n lens -n medium )

21. PHY232 - Pumplin - Mirrors and lenses 21 example  Given R 1 =10 cm and R 2 =5 cm, what is the focal length? The lens is made of glass (n=1.5) R1R1 R2R2 1 2 R 1 is on the right of the curvature, so positive +10 cm R 2 is on the left of the curvature, so negative –5 cm n=1.5 1/f=0.5(0.1-(-0.2))=0.15 f=+6.67 cm object

22. PHY232 - Pumplin - Mirrors and lenses 22 example 2  Given R 1 =5 cm and R 2 =10 cm, what is the focal length? The lens is made of glass (n=1.5) R2R2 R1R1 1 2 object R 1 is on the left of curvature 1 so R 1 =-5 cm R 2 is on the right of curvature 2 so R 2 =+10 cm n=1.5 1/f=0.5(-0.2-0.1)=-0.15 f=-6.67 cm

23. PHY232 - Pumplin - Mirrors and lenses 23 converging lens p>f OF + F 1) A ray parallel to the central axis will be bend through the focal point 2) A ray through the center of the lens will continue unperturbed 3) A ray through the focal point of the lens will be bend parallel to the central axis I 4) the image is located at the crossing of the above 3 rays (you need just 2 of them). A real inverted image is created. The magnification depends on p: |M| can be 1

24. PHY232 - Pumplin - Mirrors and lenses 24 lens equation OF + F I The equation that connects object distance p, image distance q and focal length f is (just like for mirrors): 1/p + 1/q = 1/f Similarly for the magnification: M=-q/p q is positive if the image is on the opposite side of the lens as the object NOTE THAT THIS IS DIFFERENT THAN THE CASE FOR MIRRORS. In either case, q >0 means the image is on the side where the light from the lens or mirror is.

25. PHY232 - Pumplin - Mirrors and lenses 25 example  an object is put 20 cm in front of a positive lens, with focal length of 12 cm. a) What is the image distance q? b) What is the magnification? a) P = 20 cm, f = 12 cm use 1/p + 1/q = 1/f 1/20 + 1/q = 1/12 solve for q gives q=30 cm b) M = -q/p = -30/20 = -1.5 The image is inverted (M negative). The image is magnified (|M|>1)

26. PHY232 - Pumplin - Mirrors and lenses 26 converging lens p<f OF + F A virtual non-inverted image is created. Magnification >1 1) A ray parallel to the central axis will be bend through the focal point 3) A ray through the focal point of the lens will be bend parallel to the central axis 2) A ray through the center of the lens will continue unperturbed 4) the image is located at the crossing of the above 3 rays (you need just 2 of them). I

27. PHY232 - Pumplin - Mirrors and lenses 27 example  an object is put 2 cm in front of a positive lens, with focal length of 3 cm. a) What is the image distance q? b) What is the magnification? a) p=2 cm, f=3 cm use 1/p + 1/q = 1/f 1/2 + 1/q = 1/3 solve for q gives q=-6 cm NOTE q is negative which means it is on the same side of the lens as the object b) M=-q/p=-(-6)/2=+3 The image is not inverted (M positive). The image is magnified (|M|>1) The image is virtual

28. PHY232 - Pumplin - Mirrors and lenses 28 question  An object is placed in front of a converging (positive) lens with the object distance larger than the focal distance. An image is created on a screen on the other side of the lens. Then, the lower half of the lens is covered with a piece of wood. Which of the following is true:  a) the image on the screen will become less bright only  b) half of the image on the screen will disappear only  c) half of the image will disappear and the remainder of the image will become less bright. OF + F I screen rays of light can still make it to any point on the image, but there are less of them, so less bright (only)

29. PHY232 - Pumplin - Mirrors and lenses 29 NOT CORRECT

30. PHY232 - Pumplin - Mirrors and lenses 30 diverging lens p>|f| OF - F 2) A ray through the center of the lens will continue unperturbed I 4) the image is located at the crossing of the above 3 rays (you need just 2 of them). A virtual non-inverted image is created. The magnification |M|<1 1)A ray parallel to the central axis will be bend so that the ray passes through the focal point IN FRONT of the lens 3) A ray aimed at the focal point on the other side of the lens will be bent parallel to the central axis

31. PHY232 - Pumplin - Mirrors and lenses 31 example  an object is put 5 cm in front of a negative lens, with focal length of -3 cm. a) What is the image distance q? b) What is the magnification? a) p=5 cm, f=-3 cm use 1/p + 1/q = 1/f 1/5 + 1/q = -1/3 solve for q gives q=-1.88 cm b) M=-q/p=-(-1.88)/5=+0.375 The image is non-inverted (M positive). The image is demagnified (|M|<1)

32. PHY232 - Pumplin - Mirrors and lenses 32 diverging lens p<|f| O F - F 2) A ray through the center of the lens will continue unperturbed A virtual non-inverted image is created. The magnification |M| |f| 1)A ray parallel to the central axis will be bend so that the ray passes through the focal point IN FRONT of the lens 3) A ray aimed at the focal point on the other side of the lens will be bent parallel to the central axis I 4) the image is located at the crossing of the above 3 rays (you need just 2 of them).

33. PHY232 - Pumplin - Mirrors and lenses 33 example  an object is put 2 cm in front of a negative lens, with focal length of -3 cm. a) What is the image distance q? b) What is the magnification? a) p = 2 cm, f = -3 cm use 1/p + 1/q = 1/f 1/2 + 1/q = -1/3 solve for q gives q = -1.2 cm b) M = -q/p = -(-1.2)/2 = +0.6 The image is non-inverted (M positive). The image is demagnified (|M|<1)

34. PHY232 - Pumplin - Mirrors and lenses 34 lenses, an overview typep?imageimage direction Mqf converging p>frealinverted|M|>0 M -++ converging p<fvirtualnot inverted |M|>1 M +-+ diverging any pvirtualnot inverted |M|<1 M +--  mirror or lens equation 1/p + 1/q = 1/f  mirror or lens magnification: M = -q/p  lens makers equation: 1/f=(n-1)(1/R 1 -1/R 2 )

35. PHY232 - Pumplin - Mirrors and lenses 35 chromatic aberrations Chromatic aberrations are due to light of different wavelengths having a different index of refraction Can be corrected by combining lenses/mirrors If n varies with wavelength, the focal length f changes with wavelength

36. PHY232 - Pumplin - Mirrors and lenses 36 two lenses  an object, 1 cm high, is placed 5 cm in front of a converging mirror with a focal length of 3 cm. This setup is placed in front of a diverging mirror with a focal length of –5 cm. The distance between the two lenses is 10 cm. Where is the image located, and what are its properties? +- 5 cm 3cm 15 cm 5cm

37. PHY232 - Pumplin - Mirrors and lenses 37 answer +- 5 cm 3cm 15 cm 5cm consider the converging lens first: 1/p+1/q=1/f so 1/5+1/q=1/3 so q=7.5 cm M=-q/p=-7.5/5=-1.5, so 1.5x1cm=1.5 cm high A real inverted magnified image I 1 5+7.5= 12.5 cm I1I1 O consider the action of diverging lens on the image constructed above 1/p+1/q=1/f so 1/(2.5) + 1/q = 1/(-5) q=-1.67 cm M=-q/p=-(-1.67)/2.5=0.67, so 0.67x 1.5=1 cm high A virtual non-inverted demagnified image I 2 relative to I 1 I2I2 15-1.67=13.33 cm