mirrors and lenses PHY 232 Spring 2007 Jon

  • Slides: 37
Download presentation
mirrors and lenses PHY 232 – Spring 2007 Jon Pumplin http: //www. pa. msu.

mirrors and lenses PHY 232 – Spring 2007 Jon Pumplin http: //www. pa. msu. edu/~pumplin/PHY 232 (Ppt courtesy of. PHY 232 Remco Zegers) - Pumplin - Mirrors and lenses 1

we saw… Ø that light can be reflected or refracted at boundaries between material

we saw… Ø that light can be reflected or refracted at boundaries between material with a different index of refraction. Ø by shaping the surfaces of the boundaries we can make devices that can focus or otherwise alter an image. Ø Here we focus on mirrors and lenses for which the properties can be described well by a few equations. PHY 232 - Pumplin - Mirrors and lenses 2

the plane (=flat) mirror Ø in the previous chapter we already studied flat mirrors.

the plane (=flat) mirror Ø in the previous chapter we already studied flat mirrors. Ø Distance from the object to the mirror is the object distance p Ø Distance from the image to the mirror is the image distance -q Ø in case of a flat mirror, an observer sees a virtual image, meaning that the rays do not actually come from it. Ø the image size (h’ ) is the same as the object size (h), meaning that the magnification h’/h=1 Ø 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. PHY 232 - Pumplin - Mirrors and lenses 3

question Ø You are standing in front (say 1 m) of a mirror that

question Ø 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? Ø a) yes b) no object PHY 232 - Pumplin - Mirrors and lenses image 4

ray diagrams Ø to understand the properties of optical elements we use ray diagrams,

ray diagrams Ø 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 PHY 232 - Pumplin - Mirrors and lenses -q 5

concave mirrors M F C C: center of mirror curvature F: focal point a

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. PHY 232 - Pumplin - Mirrors and lenses 6

concave mirrors: an object outside F O I F step 1: draw the ray

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 PHY 232 - Pumplin - Mirrors and lenses 7

concave mirrors: an object outside F O I F The image is: a) inverted

concave mirrors: an object outside F O I F 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. PHY 232 - Pumplin - Mirrors and lenses 8

concave mirrors: object outside F O I F distance object-mirror: p distance image-mirror: q

concave mirrors: 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 PHY 232 - Pumplin - Mirrors and lenses 9

example Ø An object is placed 12 cm in front of a a concave

example Ø An object is placed 12 cm in front of a a concave mirror with focal length 5 cm. What are: Ø a) the location of the image Ø b) the magnification a) 1/p + 1/q = 1/f so 1/12 + 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 size of the object and that it is inverted. c) Image is real (whenever q>0, I. e. , whenever M > 0) PHY 232 - Pumplin - Mirrors and lenses 10

concave mirrors: an object inside F the image is: a) not inverted b) virtual

concave mirrors: an object inside F the image is: a) not inverted b) virtual c) magnified F O I 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. creates a magnified image (“shaving mirror”) PHY 232 - Pumplin - Mirrors and lenses 11

concave mirrors: an object inside F the image is: a) not inverted b) virtual

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, its sign should be negative PHY 232 - Pumplin - Mirrors and lenses 12

example Ø an object is placed 2 cm in front of a lens with

example Ø 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 - 1/2 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 (+). PHY 232 - Pumplin - Mirrors and lenses 13

demo: the virtual pig PHY 232 - Pumplin - Mirrors and lenses 14

demo: the virtual pig PHY 232 - Pumplin - Mirrors and lenses 14

convex mirrors: (p>|f|or p<|f| doesn’t matter) O I F F is now located on

convex mirrors: (p>|f|or p<|f| doesn’t matter) O I F F is now located on the other side of the mirror 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 (examples: garden ball, rearview mirrors that warn “objects are closer than they appear”) PHY 232 - Pumplin - Mirrors and lenses 15

convex mirrors O I F F is now located on the other side of

convex mirrors O I F 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 PHY 232 - Pumplin - Mirrors and lenses 16

example Ø an object with a height of 3 cm is placed 6 cm

example Ø 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. PHY 232 - Pumplin - Mirrors and lenses 17

Mirrors: an overview type p? image direction M q f concave p>f real inverted

Mirrors: an overview type p? image direction M q f concave p>f real inverted |M|>0 M - + + concave p<f virtual not inverted |M|>1 M + - + convex any p virtual not inverted |M|<1 M + - - Ø mirror equation 1/p + 1/q = 1/f Ø F = R/2 where R is the radius of the mirror Ø magnification: M = -q/p PHY 232 - Pumplin - Mirrors and lenses 18

Lenses Ø Lenses function by refracting light at their surfaces Ø Their action depends

Lenses Ø Lenses function by refracting light at their surfaces Ø Their action depends on Ø radii of the curvatures of both surfaces Ø the refractive index of the lens Ø converging (positive lenses) have positive focal length and are always thickest in the center + Ø diverging (negative lenses) have negative focal length and used in are thickest at the edges drawings PHY 232 - Pumplin - Mirrors and lenses 19

lensmakers equation object 1 R 2 2 R 1 f: focal length of lens

lensmakers equation object 1 R 2 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 is not in air then (nlens-nmedium) PHY 232 - Pumplin - Mirrors and lenses 20

example object 1 R 2 Ø Given R 1=10 cm and R 2=5 cm,

example object 1 R 2 Ø Given R 1=10 cm and R 2=5 cm, what is the focal length? The lens is made of glass (n=1. 5) 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 PHY 232 - Pumplin - Mirrors and lenses 21

example 2 object 1 R 1 Ø Given R 1=5 cm and R 2=10

example 2 object 1 R 1 Ø Given R 1=5 cm and R 2=10 cm, what is the focal length? The lens is made of glass (n=1. 5) 2 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 PHY 232 - Pumplin - Mirrors and lenses 22

converging lens p>f I O F F + 1) A ray parallel to the

converging lens p>f I O F 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 real inverted image is created. The magnification depends on p: |M| can be <1, 1 or >1 PHY 232 - Pumplin - Mirrors and lenses 23

lens equation I O F F + The equation that connects object distance p,

lens equation I O F F + 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. PHY 232 - Pumplin - Mirrors and lenses 24

example Ø an object is put 20 cm in front of a positive lens,

example Ø 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) PHY 232 - Pumplin - Mirrors and lenses 25

converging lens p<f I F O F + 1) A ray parallel to the

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 PHY 232 - Pumplin - Mirrors and lenses 26

example Ø an object is put 2 cm in front of a positive lens,

example Ø 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 PHY 232 - Pumplin - Mirrors and lenses 27

question Ø An object is placed in front of a converging (positive) lens with

question Ø 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: Ø a) the image on the screen will become less bright only Ø b) half of the image on the screen will disappear only Ø c) half of the image will disappear and the remainder of the image will become less bright. I O F F + rays of light can still make it to any point on the image, but there are less of them, so less bright (only) PHY 232 - Pumplin - Mirrors and lenses screen 28

NOT CORRECT PHY 232 - Pumplin - Mirrors and lenses 29

NOT CORRECT PHY 232 - Pumplin - Mirrors and lenses 29

diverging lens p>|f| O F F I - 1) A ray parallel to the

diverging lens p>|f| O F F I - 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 PHY 232 - Pumplin - Mirrors and lenses 30

example Ø an object is put 5 cm in front of a negative lens,

example Ø 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) PHY 232 - Pumplin - Mirrors and lenses 31

diverging lens p<|f| F O F I - 1) A ray parallel to the

diverging lens p<|f| F O F I - 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| PHY 232 - Pumplin - Mirrors and lenses 32

example Ø an object is put 2 cm in front of a negative lens,

example Ø 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) PHY 232 - Pumplin - Mirrors and lenses 33

lenses, an overview type p? image direction M q f converging p>f real inverted

lenses, an overview type p? image direction M q f converging p>f real inverted |M|>0 M - + + converging p<f virtual not inverted |M|>1 M + - + diverging any p virtual not inverted |M|<1 M + - - Ø mirror or lens equation 1/p + 1/q = 1/f Ø mirror or lens magnification: M = -q/p Ø lens makers equation: 1/f=(n-1)(1/R 1 -1/R 2) PHY 232 - Pumplin - Mirrors and lenses 34

chromatic aberrations Chromatic aberrations are due to light of different wavelengths having a different

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 PHY 232 - Pumplin - Mirrors and lenses 35

two lenses Ø an object, 1 cm high, is placed 5 cm in front

two lenses Ø 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? + 3 cm 5 cm 15 cm PHY 232 - Pumplin - Mirrors and lenses 36

answer + - O 5 cm 3 cm 5 cm I 1 I 2

answer + - O 5 cm 3 cm 5 cm I 1 I 2 5+7. 5= 12. 5 cm 15 -1. 67=13. 33 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. 5 x 1 cm=1. 5 cm high A real inverted magnified image I 1 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. 67 x 1. 5=1 cm high A virtual non-inverted demagnified image I 2 relative to I 1 PHY 232 - Pumplin - Mirrors and lenses 37