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1 mirrors and lenses PHY232 Remco Zegers Room W109 cyclotron building

2 quiz (extra credit) a ray of light moves from air to a material with n>1, at an initial angle of 30 0 with the normal. After reaching the material, the ray bends and has an outgoing angle of 20 0 with the normal. Now consider the reverse case: a ray of light travels from the same material with n>1 to air. The incident angle is After reaching air, the angle with the normal will be a) 20 0? b) between 20 0 and 30 0 c) 30 0 d) greater than PHY232 - Remco Zegers - Mirrors and lenses 2

3 an important point objects do not emit rays of light that get seen by your eye. Light (from a bulb or the sun) gets reflected off the object towards your eye. PHY232 - Remco Zegers - Mirrors and lenses 3

4 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. PHY232 - Remco Zegers - Mirrors and lenses 4

5 the flat mirror in the previous chapter we already saw flat mirrors. The distance from the object to the mirror the object distance p The distance from the image to the mirror is the image distance q p 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 NOTE: a virtual image cannot be projected on a screen but is visible by the eye or another optical instrument. PHY232 - Remco Zegers - Mirrors and lenses 5

6 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 PHY232 - Remco Zegers - Mirrors and lenses 6

7 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 h h p q PHY232 - Remco Zegers - Mirrors and lenses 7

8 concave mirrors M F C C: center of mirror curvature F: focal point a light ray passing through the center of curvature will be reflected back upon itself because it strikes the mirror normally to the surface. 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. PHY232 - Remco Zegers - Mirrors and lenses 8

9 concave mirrors: an object outside F O I 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) the image of the top of the object is located where the reflected rays meet step 3: note that a ray from the bottom of the object just reflects back. construct the image I PHY232 - Remco Zegers - Mirrors and lenses 9

10 concave mirrors: an object outside F O I F The image is: a) inverted (upside down) b) real (light rays pass through it) c) smaller than the object PHY232 - Remco Zegers - Mirrors and lenses 10

11 concave mirrors: an object outside F O I F 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 PHY232 - Remco Zegers - Mirrors and lenses 11

12 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. PHY232 - Remco Zegers - Mirrors and lenses 12

13 concave mirrors: an object inside F the image is: a) not inverted b) virtual c) magnified 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) 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 step 3: note that a ray from the bottom of the object just reflects back. create the image F PHY232 - Remco Zegers - Mirrors and lenses 13 O I

14 concave mirrors: an object inside F the image is: a) not inverted b) virtual c) magnified F O I The lens equation and equation for magnification are still valid. However, since the image is now on the other side of the mirror, the sign of the image distance should be negative PHY232 - Remco Zegers - Mirrors and lenses 14

15 example an object is placed 2 cm in front of a mirror 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 (+). PHY232 - Remco Zegers - Mirrors and lenses 15

16 demo: the virtual pig PHY232 - Remco Zegers - Mirrors and lenses 16

17 convex mirrors: an object outside F (p> f ) O F is now located on the other side of the mirror I 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) the image of the top of the object is located where the reflected rays meet step 3: note that a ray from the bottom of the object just reflects back. construct the image I PHY232 - Remco Zegers - Mirrors and lenses 17

18 convex mirrors: an object outside F (p> f ) O F is now located on the other side of the mirror the image is: a) not inverted b) virtual c) demagnified I F 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 PHY232 - Remco Zegers - Mirrors and lenses 18

19 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. PHY232 - Remco Zegers - Mirrors and lenses 19

20 convex mirrors with p < f the situation is exactly the same as for the situation with p > f. The demagnification will be different though F O I F PHY232 - Remco Zegers - Mirrors and lenses 20

21 Mirrors: an overview type p? image image M q f direction concave p>f real inverted M >0 M concave p<f virtual not inverted M >1 M convex p> f virtual not inverted M <1 M convex p< f virtual not inverted M <1 M mirror equation 1/p + 1/q = 1/f f=r/2 where R is the radius of the mirror magnification: M=-q/p PHY232 - Remco Zegers - Mirrors and lenses 21

22 lon-capa now do problems 7,8,11 of lon-capa 8 PHY232 - Remco Zegers - Mirrors and lenses 22

23 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 PHY232 - Remco Zegers - Mirrors and lenses 23

24 lensmakers equation object 1 2 R 2 R 1 f: focal length of lens n: refractive index of lens R 1 radius of front surface R 2 radius of back surface 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 if the lens is not in air then (n lens -n medium ) PHY232 - Remco Zegers - Mirrors and lenses 24

25 object R R 1 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) 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 PHY232 - Remco Zegers - Mirrors and lenses 25

26 object R R 2 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) 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( )=-0.15 f=-6.67 cm PHY232 - Remco Zegers - Mirrors and lenses 26

27 example 3 object R R 2 Given R 1 =5 cm and R 2 =, what is the focal length? The lens is made of glass (n=1.5) R 1 is on the left of curvature 1, so R 1 =-5 cm R 2 is infinity (either positive or negative) n=1.5 1/f=0.5( )=-0.1 f=-10 cm PHY232 - Remco Zegers - Mirrors and lenses 27

28 question A person is trying to make a lens but decides to make both surfaces flat, resulting in essentially a flat piece of glass on both sides. What is the focal length of this lens? a) infinity b) 0 c) cannot say, depends on the index of refraction n R 1 =infinity R 2 =infinity so 1/f=0 and f is infinity a focal length of infinity means that there is no focus PHY232 - Remco Zegers - Mirrors and lenses 28

29 O converging lens p>f F F I 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 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, 1 or >1 + PHY232 - Remco Zegers - Mirrors and lenses 29

30 O lens equation F 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 PHY232 - Remco Zegers - Mirrors and lenses 30

31 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) PHY232 - Remco Zegers - Mirrors and lenses 31

32 converging lens p<f I F O 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 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. Magnification >1 PHY232 - Remco Zegers - Mirrors and lenses 32

33 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 PHY232 - Remco Zegers - Mirrors and lenses 33

34 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. O F F I rays of light can still make it to any point on the image, but there are less of them, so less bright (only) + screen PHY232 - Remco Zegers - Mirrors and lenses 34

35 NOT CORRECT PHY232 - Remco Zegers - Mirrors and lenses 35

36 diverging lens p> f O F I 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 2) A ray through the center of the lens will continue unperturbed 3) A ray aimed at the focal point on the other side of the lens will be bent parallel to the central axis 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 - PHY232 - Remco Zegers - Mirrors and lenses 36

37 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= The image is non-inverted (M positive). The image is demagnified ( M <1) PHY232 - Remco Zegers - Mirrors and lenses 37

38 diverging lens p< f F O I 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 2) A ray through the center of the lens will continue unperturbed 3) A ray aimed at the focal point on the other side of the lens will be bent parallel to the central axis 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 similar to case with p> f - PHY232 - Remco Zegers - Mirrors and lenses 38

39 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) PHY232 - Remco Zegers - Mirrors and lenses 39

40 lenses, an overview type p? image image M q f direction converging p>f real inverted M >0 M converging p<f virtual not inverted diverging p> f virtual not inverted diverging p< f virtual not inverted M >1 M + M <1 M + M <1 M + mirror equation 1/p + 1/q = 1/f magnification: M=-q/p lens makers equation: 1/f=(n-1)(1/R 1-1/R 2 ) 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 PHY232 - Remco Zegers - Mirrors and lenses 40

41 type of lens p? image image direction real converging p<f virtual not inverted diverging p> f virtual not inverted diverging p< f virtual not inverted M q f converging p>f inverted M >0 M - type of mirror p? image image direction real M >1 M + M <1 M + M <1 M M q f concave p>f inverted M >0 M - concave p<f virtual not inverted convex p> f virtual not inverted convex p< f virtual not inverted M >1 M + M <1 M + M <1 M + PHY232 - Remco Zegers - Mirrors and lenses

42 spherical aberrations: Hubble space telescope spherical aberrations are due to the rays hitting the lens at different locations have a different focal point perfect distorted example: Hubble before after correction PHY232 - Remco Zegers - Mirrors and lenses 42

43 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 PHY232 - Remco Zegers - Mirrors and lenses 43

44 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 lens 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? + - 3cm 5cm 5 cm 15 cm PHY232 - Remco Zegers - Mirrors and lenses 44

45 O answer + - 5cm 5 cm 3cm 5+7.5= 12.5 cm 15 cm 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 I 1 I =13.33 cm 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 PHY232 - Remco Zegers - Mirrors and lenses 45

46 lon-capa now do problems 9,10,12 of lon-capa 8 PHY232 - Remco Zegers - Mirrors and lenses 46

47 question (extra credit) a object is placed in front of a concave mirror, exactly at the location of the focal point (p=f). What is the image distance q? a) q=p b) q=f c) q=0 d) q= infinity 1/p+1/q=1/f if p=f then 1/q=0 and q is infinity. Meaning there is no image! PHY232 - Remco Zegers - Mirrors and lenses 47

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