Wave optics treats light as a wave and uses a similar analytical framework as sound waves and other mechanical waves.

Transcription

1 Otics 1. Intro: Models o Light 2. The Ray Model 1. Relection 2. Reraction 3. Total Internal Relection 3. Images 1. The lane mirror 2. Sherical Mirrors (concave) 4. Lenses: an introduction 1. The arameters o a lens 2. Real Images rom lenses 3. Virtual Images 4. Reraction at a Sherical Boundary 5. Two lens systems 5. Otical Instruments 1. Intro: Models o Light The Wave Model Wave otics treats light as a wave and uses a similar analytical ramework as sound waves and other mechanical waves. The Ray Model Ray otics uses the act that light seems to travel in a straight line. This makes understanding some henomenon easier. The Particle/Photon Model Modern Physics has shown that treating light at a article, a hoton, can be a very successul aroach as well. (Requires quantum theory) Some acts about light 1. Light can be considered a wave henomenon. It will be shown to be created by oscillations o electric and magnetic ields. 2. The seed o light in vacuum is constant: c = 299,792,458 m/s 3. In vacuum, our same relationshi between seed, requency, and wavelength holds: c = λ 4. The visible light we see is just a small section o the whole sectrum o electromagnetic oscillation requencie. udated on J. Hedberg 2018 Page 1

2 Quick Question 1 In 1667, Galileo attemted to measure the seed o light by having two eole hold covered lanterns on hills that were about 1.5 km aart. One erson would measure time. One o the eole with a lantern would uncover it. The other erson would then uncover his lantern when he saw the light rom the irst lantern. Reeated attemts ailed. To see why, determine the aroximate time it takes light to travel the 1.5 km distance μs μs 3. 5 ms ns 5. 5 ns 2. The Ray Model Euclid treated lights as a line, called a light ray. This is a retty amazing concetual accomlishment considering they didn t have lasers or lashlights or other line like sources o light. Like we did beore, we have to choose a lot o things to ignore when trying to describe the natural world. For the ray model, we will ignore the act that light is a wave henomena. That means we ll orget all about diraction and intererence and other wave related issues. This can be justiied i we restrict our discussions to the interaction between light and s that are large comared to the wavelength o the light, like most everyday things. 1. Light Rays travel in straight lines Light rays travel through a vacuum or a transarent material in straight lines called light rays. The seed o the light in the material is given by v = c/n, where n is the index o reraction. Light Rays can cross 2. Light rays do not interact with each other. Two rays can cross without being aected in any way. udated on J. Hedberg 2018 Page 2

3 3. A light ray travels orever, unless it interacts with matter vacuum 4. An is a source o light rays Rays originate rom every oint on an, and each oint send rays in all directions. relection vacuum material A light ray continues on orever unless it has an interaction with matter. This interaction may cause the ray to be redirected, or absorbed. We ll discuss these interactions later. Sources o rays Objects in the ray model can either be (a) sources o light (sel-luminous s): these are like the sun, or lightbulbs. 1. Ray Source: like a laser ointer. These emit a single ray in one direction. 2. Point Source: little light sources that roject in all directions (christmas tree lights) 3. Extended Source: These also roject udated on J. Hedberg 2018 Page 3

4 rays in all directions, but have a nonoint like extension. (Large light bulbs) 4. Parallel Ray source: These sources create a beam o arallel rays. (lashlights, ar away stars) Relection... or how do you see anything. The light rom sel luminous s will relect o other s creating essentially extended sources o light. In order to see something, a ray must travel rom an to our eye. In the case o the laser, the only way you can tell that it is on, is i the light rom the laser bounces o something and strikes your eye. Relection (Diusive) When light rays meet rough suraces, such as a iece o aer, they still obey the lay o relection microscoically, however, since the surace normals are changing, the relected light rays scatter at many angles. udated on J. Hedberg 2018 Page 4

5 Relection (Secular) incident ray angle o incidence normal angle o relection relected ray The Law o Relection: The incident ray and the relected ray are both in the same lane, which is erendicular to the surace The angle o incidence equals the angle o relection: relective surace θ i = θ r (For otics measurements, we will use θ as deined between the ray and the normal, not the surace.) Quick Question 2 Here is a light ray incident on a relective surace as shown. Which, i any, o the ollowing are true? light ray α = θ 2. α = θ 3. α + θ = α θ 5. Lies! These are all lies I say. Examle Problem #1: At what angle should the laser beam be aimed at the mirrored ceiling so that it hits the wall hal way u? udated on J. Hedberg 2018 Page 5

6 mirror 3.0 m wall Reraction 5.0 m A light ray (the incident ray) encounters a boundary between two media. At that boundary, some o the light is relected, but most is transmitted. Observation clearly shows that the light inside new medium (the glass) has changed direction. relected ray incident ray Index o Reraction reracted ray (air-glass) reracted ray (glass-air) The index o reraction (or reractive index) o a material is a dimensionless arameter, n, used to describe how light moves in a articular medium. c n = = v seed o light in vacuum seed o light in material The index o reraction o vacuum is n = and is exactly 1. Medium Index Vacuum 1 (exactly) Air (0ºC, 1 atm) Water 1.33 Glass 1.52 Sahire 1.77 Diamond 2.42 [*Note or labs: The index o reraction can change based on the wavelength o the light] udated on J. Hedberg 2018 Page 6

7 Quick Question 3 In which one o the ollowing substances does light have the largest seed? Medium Index Vacuum 1 (exactly) Air (0ºC, 1 atm) Water 1.33 Glass 1.52 Sahire 1.77 Diamond Diamond 2. Sahire 3. Glass 4. Water 5. None o the above. The seed o light has the same value everywhere in the Universe. air glass angle o incidence normal relected ray medium 1 medium 2 n 2 > n 1 Many arallel rays enter a new medium. They all change direction in the same manner. In case, we see rays traveling through air, with n = 1, and entering into glass with n = 1.5. angle o reraction We can look closely at one ray, and label the interesting arts o the reraction henomenon. We ll call θ 1 the angle between the incident ray and the normal o the surace. θ 2 is the angle between the reracted ray and the normal inside the 2nd medium. Snell's Law udated on J. Hedberg 2018 Page 7

8 angle o incidence normal relected ray Snell s law describes the geometry o this situation. n 2 sin θ 2 = n 1 sinθ 1 medium 1 medium 2 n 1 n 2 is the index o reraction o medium 1 is the index o reraction o medium 2 n 2 > n 1 angle o reraction I the direction o the ray is reversed (i.e. now it is traveling rom medium 2 into medium 1) we change the labels o reraction and incidence, but kee the angular deinitions and indices o reraction the same. angle o reraction normal medium 1 n 2 sin θ 2 = n 1 sinθ 1 medium 2 Snell's Law still holds. relected ray n 2 > n 1 Examle Problem #2: angle o incidence A beam o light o wavelength 550 nm in air strikes a slab o transarent material at an angle o 40º to the normal, and the reracted beam makes an angle o 26º to the normal. Find the index o reraction o the material.... a limit is reached Let's calculate θ 1 or this case =? n = 1 45º n = 2 Total Internal Relection When angle o incidence becomes too large, udated on J. Hedberg 2018 Page 8

9 none o the light will be able to be transmitted through the boundary. When this oint is reached, we have total internal relection. n 2 We can ind the critical angle when total internal relection begins by utting = 90 into Snell s Law: θ 1 45º 45º n 1 n 2 > n 1 n 1 n 2 Total internal relection occurs only when light attemts to move rom a medium o higher index o reraction ( n 2 ) to a medium o lower index o reraction ( n 1 ). The critical angle θ c is determined by Snell s Law, using 90 degrees or or the reracted angle: Total internal relection underwater image source θ c = sin 1 n ( 1 ) n 2 Examle Problem #3: A lightblub is set in the bottom o a 3.0 m dee swimming ool. What is the diameter o the circle o light seen on the water's surace rom above? 3. Images 1. Real Images: A real images exists at a osition in sace indeendent o i there is an observer. Ex: images on a movie screen, ocused images o sun using a magniying glass. 2. Virtual Images: These are images which do not really exist at a lace in sace -- we only erceive them to exist. Ex: images in lane mirrors, what you see when you look through binoculars. The lane mirror Light strikes a mirror. Each ray coming rom the obeys the law o relection. udated on J. Hedberg 2018 Page 9

10 mirror Quick Question 4 As I walk away rom a lane mirror what will haen to the image o my ace in the mirror. 1. It will take u more and more o the mirror surace, i.e. get bigger 2. Its size won't change 3. It will take u less and less o the mirror, i.e. get smaller The lane mirror i O I I we didn t know there was a mirror there, we might be temted to say the was actually located at oint P. Images in lane mirrors udated on J. Hedberg 2018 Page 10

11 O i I virtual image Plane Mirror - Virtual Image with extended I the is not a oint, but has a satial extent, then we have to consider each oint o the. Each oint on the will have a corresonding image oint on the oosite side o the mirror. i virtual image udated on J. Hedberg 2018 Page 11

12 Quick Question 5 A ball is held 1.5 m in ront o a lane mirror. How ar is the image o the ball rom the ball? 1. 0 m m m m m Conventions O i I virtual image We have to establish several sign conventions when dealing with ray otics. The irst will be to always call the distance ositive and the image distance or virtual images negative. = i Quick Question 6 An image in a lane mirror is lied let to right 2. lied to to bottom 3. lied ront to back 4. not lied at all. Sherical Mirrors (concave) udated on J. Hedberg 2018 Page 12

13 radius o curvature ocal oint ( ) mirror lane otical axi C Note: The otical axis (central axis) extends through the center o curvature and the center o the mirror. The ocal oint is where the relected rays rom two arallel incident rays meet and is also on the otical axis. It's ositive in the case o a concave mirror. Sherical Mirrors (convex) mirror lane otical axis ocal oint ( ) radius o curvature C Here the ocal oint is virtual and it gets a negative sign.( < 0) udated on J. Hedberg 2018 Page 13

14 The mirror lane sets the origin or all distance measurements. For both convex and concave mirrors, the ocal length is hal the radius o curvature. at a distance inside the ocal oint virtual image F C i We start with an inbetween the ocal oint and the lane o the mirror. This situation creates a virtual image which aears to be on the othe side o the mirror. at the ocal oint C F arallel rays Now we bring the exactly to the ocal oint. udated on J. Hedberg 2018 Page 14

15 When the rays originate rom that distance, then their relected ath are arallel. Thus, they do not converge until. No image is roduced. at a distance outside the ocal oint C real image F i Outside the ocal distance, and in ront o a concave mirror will create a real image. The real image is inverted. To calculate the ositions o the images, we can use the ollowing: 1 + = i Here: is the distance (always ositive) i is the image distance ( + or real images, or virtual images), and is the ocal length, ( + or concave mirrors). 1 1 udated on J. Hedberg 2018 Page 15

16 h h virtual image C i Deine a magniication value: m = h h Which can also be ound by using: m = i Convex mirror mirror lane virtual image otical axis ocal oint Examle Problem #4: i An is 30 cm in ront o a convex mirror with a ocal length o -20cm. Locate the image. Is it uright or inverted? 4. Lenses: an introduction udated on J. Hedberg 2018 Page 16

17 double diverging lens otical axis Lenses can be analyzed much the same as mirrors. 1 1 = + 1 i ocal oint double diverging lens otical axis ocal oint Chart o sign conventions Sherical Mirrors Thin Lenses + or concave mirror Focal Length or convex mirror + i is in ront mirror Object Distance i is behind the mirror + i image is in ront mirror (real image) Image Distance i image is behind the mirror (virtual) + or an uright image with resect to the Magniication or an inverted image with resect to the Imagine an incident ray hitting a iece o glass like this: A closer view shows the normals and angles o interest. tangent tangent incident ray normal normal n = 1 n > 1 n = 1 Snell's Law tells us how to deal with this situation. An aroximation is needed. udated on J. Hedberg 2018 Page 17

18 I our lens is shaed like so, and we trace all the incoming rays (assuming they are arallel) we end u noticing that they all converge at a oint. arallel rays otical axis double converging lens ocal oint The oint is located along the otical axis and is given the name ocal oint. This lens is called a converging lens. Another tye o lens we have looks like this: A closer view shows the normals and angles o interest. tangent tangent incident ray normal normal Again, Snell's Law tells us how to deal with this situation. n = 1 n > 1 n = 1 double diverging lens otical axis ocal oint Secial Rays udated on J. Hedberg 2018 Page 18

19 Quick Question 7 We would like to create a beam o light that consists o arallel rays. Which one o the ollowing arrangements would make that haen? 1. A light bulb is laced at the ocal oint o a convex mirror. 2. A light bulb is laced at the ocal oint o a diverging lens. 3. A light bulb is laced at the ocal oint o a converging lens. 4. A light bulb is located at twice the ocal length rom a concave mirror. 5. A light bulb is located at twice the ocal length rom a converging lens. The arameters o a lens The lens has a shae and an index o reraction. same n, dierent r same r, dierent n ocal oint ocal oint ocal oint ocal oint udated on J. Hedberg 2018 Page 19

20 Quick Question 8 Parallel rays o violet light that are directed at a converging lens are ocused at a oint P on the central axis to the right o the lens when the lens is surrounded by air as shown. I the lens is surrounded by water instead o air, where will the red arallel rays be ocused relative to oint P? P 1. above oint P 2. below oint P 3. to the let o oint P 4. to the right o oint P 5. at oint P These two lenses are both converging, yet one has a shorter ocal length. This could be due to its geometry or its material makeu. arallel rays ocal oint The ocal length is a measure o the strength o the lens: A shorter ocal length imlies a stronger otical system, since the light is bent more. arallel rays ocal oint Real Images rom lenses udated on J. Hedberg 2018 Page 20

21 Regarding the images We can see that the real image o the is smaller than the. Additionally, it is inverted. The magniication is the same or lenses as it was or mirrors: i h m = = i and m = h lane image lane height = h otical axis image height = h i udated on J. Hedberg 2018 Page 21

22 Quick Question 9 What will haen to the image on the screen when hal the lens is covered? mask converging lens screen 1. It will invert 2. The image will get dimmer overall but remain the same otherwise 3. The uer hal will be darkened 4. The lower hal will be darkened 5. It will vanish entirely. converging lens wall Virtual Images In order to orm a real image, we must have converging rays. I the is ositioned between the near ocal oint and the lens, the rays through the lens will not converge. Thus, no real image is ormed. This situation creates a virtual image. udated on J. Hedberg 2018 Page 22

23 virtual image otical axis diverging rays Converging Lens summary real image virtual image Diverging Lens otical axis i No matter where the is, the image is virtual. The thin lens equation For a thin lens with ocal length, image distance i, and distance, we can use the thin lens equation: udated on J. Hedberg 2018 Page 23

24 1 1 = + 1 i (This was the same or mirrors) Examle Problem #5: lens is converging, = 30, = +50 Examle Problem #6: lens is converging, = 30, = +20 Examle Problem #7: lens is diverging, = -30, = +50 The Lensmaker's equation I we have a lens which has two radii o curvatures, then in order to ind the ocal length, we must use the lensmaker s equation. This assumes the lens is thin, and in air = (n 1) ( ) r 1 r 2 r 1 r 2 inal image image rom 1st boundary n t Sign Convention I the aces a concave surace, then that surace has a negative radius o curvature. udated on J. Hedberg 2018 Page 24

25 - concave + I the aces a convex surace, then that surace has a ositive radius o curvature. convex Examle Problem #8: Find the ocal Length o this lens n = cm 30 cm Find the ocal length o theses lenses: udated on J. Hedberg 2018 Page 25

26 50cm 20cm 30cm 20cm 20cm 60cm Chart o sign conventions Sherical Mirrors Thin Lenses + or concave mirror + or a converging lens Focal Length or convex mirror or a diverging lens Object + i is in ront mirror + or real s Distance i is behind the mirror or virutal s + i image is in ront mirror (real Image image) + or real image Distance i image is behind the mirror (virtual) or virtual image + or an uright image with resect + or an uright image with resect to the to the Magniication or an inverted image with resect or an inverted image with resect to the to the udated on J. Hedberg 2018 Page 26

27 h ar ocal oint near ocal oint h Reraction at a Sherical Boundary i Here we have a sherical boundary between two dierent materials with dierent index o reraction. Rays rom oint O will be reracted at the boundary and converge at oint I. n 1 n 2 R oint O C image oint I i sherical surace Snell s law can be used to determine the angles o reraction. There are however, many ossible ermutations or the s and images. We have to account or both indices o reraction or the two media, the location o the, and the location o the center o curvature. Here are two situations which roduce real images. udated on J. Hedberg 2018 Page 27

28 n 1 n 2 C image In both o theses cases, the real images orms on the oosite side o the boundary rom the image. n 1 n 2 C image And these situations will roduce virtual images. n 1 n 2 n 1 n 2 image C O I C n 1 n 2 n 1 n 2 C O I O I C Note that or theses virtual images, the image is on the same side o the boundary as the. Fortunately, one equation rules them all! Conventions to use this equation: is ositive. It is the distance + = i i will be ositive or a real image, and negative or a virtual image n 1 n 2 n 2 n 1 r r will be ositive i the is acing a convex boundary and negative i the is acing a concave boundary. Examle Problem #9: This ly is in a ball o amber with a radius o curvature o 12 cm. The amber has an index o reraction o n = 1.7. To an observer outside the amber, the ly aears to be located 2 udated on J. Hedberg 2018 Page 28

29 cm rom the surace. Where is the ly actually located? n = 1 n = 1.7 r = 12cm Two lens systems The image rom the irst lens acts as the or the second lens. Examle: The ol' ashioned telescoe lens 1 real image (ive) Examle: The ol' ashioned telescoe, cont'd lens 1 lens 2 (ive) (eyeiece) virtual image 5. Otical Instruments What do we need to make an good image o something? We need all the rays coming rom a articular oint on an to meet again at another lace. Certain lenses can accomlish this. udated on J. Hedberg 2018 Page 29

30 Beore lenses were around... Or, we could just block all the other, diverging rays with a iece o material. Now, we can imagine only one ray leaving each oint, and eventually reaching the image lane. Historical udated on J. Hedberg 2018 Page 30

31 The camera obscura - a very old imaging tool To imrove the ocus, the in-hole needs to be smaller. But, then less light gets through. (Also, i the in-hole gets too small, then we get diraction). So, we can do better. The Modern Camera The essential eatures o a modern camera are the lens, which ocuses the light, and the image lane, where either the ilm, or a digital sensor is located. udated on J. Hedberg 2018 Page 31

32 The CCD Detector Charge-Couled Device The eye Nature has made many dierent eyeballs. Most oerate on the rinciles o lenses we ve just looked at. udated on J. Hedberg 2018 Page 32

33 iris cornea aqueous humor lens ocusing the eye retina otic nerve Focusing on s: We cannot adjust the osition o the lens with resect to the retina. So, the muscles around the eye change the shae o the lens, which then changes its ocal length. Why some o use ut lenses on our ace all day. 1. Normal vision. Here everything is set u just right 2. Hyeroia (ar-sightedness) My eyeball is too short 3. Myoia (near-sightedness) My eyeball is udated on J. Hedberg 2018 Page 33

34 too long These two are the same size, but since they are located in dierent ositions, the occuy dierent amounts o our ield o vision. Their corresonding images on our retinal surace are thereore dierent sizes. This give the aearance o dierent sizes. We can say the angle subtended by each is its angular size. We don't want to get too close though, because the eye can't ocus ast its near oint. On average, that's 25 cm. Simle Magniying Lens Placing on within the ocal length o a converging lens results in a virtual image. The aarent size o this virtual image is larger, and thus aears to be magniied. udated on J. Hedberg 2018 Page 34

35 virtual image i otical axis h near oint diverging rays Angular magniication θ m θ = θ Since θ h/25cm and θ h/, we can write: m θ = 25cm Comound Microscoe ive lens eyeiece diverging rays Virtual Image Magniication= M = m = The telescoe m θ s 25cm ob ey udated on J. Hedberg 2018 Page 35

36 The reracting telescoe: distance light source ive eyeiece arallel rays virtual image eyeiece The relecting telescoe eyeiece lat mirror arallel rays Aberrations arabolic mirror Lenses aren't erect. Since dierent colors will reract at dierent angles, the ocal oint will be slightly dierent or dierent wavelengths. This leads to Chromatic Aberration. udated on J. Hedberg 2018 Page 36

37 Sherical Aberration Earlier, our thin lens aroximation ignored the act that the thickness o the lens changed as a unction o distance away rom the central axis. blurry ocal oint This leads to rays having slightly dierent ocal oints deending on where they are incident on the lens. The urther the rays are away rom the central axis, the worse the Sherical Aberration eect is. Fixing aberrations Fortunately, by using a multi lens setu, we can correct these aberrations. For examle, to correct the chromatic aberration caused by a converging lens we can insert a diverging lens ater the converging lens to reocus the dierent colors back to the same oint. udated on J. Hedberg 2018 Page 37