Optics• Reflection • Prisms• Diffuse reflection • Rainbows• Refraction • Plane mirrors• Index of refraction • Spherical aberration• Speed of light • Concave and convex mirrors• Snell’s law • Focal length & radius of curvature• Geometry problems • Mirror / lens equation• Critical angle • Convex and concave lenses• Total internal reflection • Human eye• Brewster angle • Chromatic aberration• Fiber optics • Telescopes• Mirages • Huygens’ principle• Dispersion • Diffraction
ReflectionMost things we see are thanks to reflections, since most objectsdon’t produce their own visible light. Much of the light incidenton an object is absorbed but some is reflected. the wavelengths ofthe reflected light determine the colors we see. When white lighthits an apple, for instance, primarily red wavelengths arereflected, while much of the others are absorbed.A ray of light heading towards an object is called an incident ray.If it reflects off the object, it is called a reflected ray. Aperpendicular line drawn at any point on a surface is called anormal (just like with normal force). The angle between theincident ray and normal is called the angle of incidence, i, andthe angle between the reflected ray and the normal ray is calledthe angle of reflection, r. The law of reflection states that theangle of incidence is always equal to the angle of reflection.
Law of Reflection Normal line (perpendicular to surface)inc ys i r ra ide t ed nt c fle ra y re s i=r
Diffuse ReflectionDiffuse reflection is when light bounces off a non-smooth surface. Each ray of light still obeys the law of reflection, but because thesurface is not smooth, the normal can point in a different forevery ray. If many light rays strike a non-smooth surface, theycould be reflected in many different directions. This explainshow we can see objects even when it seems the light shining uponit should not reflect in the direction of our eyes. It also helps toexplain glare on wet roads: Water fills in and smoothes out therough road surface so that the road becomes more like a mirror.
Speed of Light & RefractionAs you have already learned, light is extremely fast, about3 × 108 m/s in a vacuum. Light, however, is slowed down by thepresence of matter. The extent to which this occurs depends onwhat the light is traveling through. Light travels at about 3/4 of itsvacuum speed (0.75 c ) in water and about 2/3 its vacuum speed(0.67 c ) in glass. The reason for this slowing is because whenlight strikes an atom it must interact with its electron cloud. Iflight travels from one medium to another, and if the speeds inthese media differ, then light is subject to refraction (a changingof direction at the interface). Refraction of Refraction of light waves light rays
Reflection & RefractionAt an interface between two media, both reflection and refraction canoccur. The angles of incidence, reflection, and refraction are all measuredwith respect to the normal. The angles of incidence and reflection arealways the same. If light speeds up upon entering a new medium, the angleof refraction, θr , will be greater than the angle of incidence, as depicted onthe left. If the light slows down in the new medium, θr will be less thanthe angle of incidence, as shown on the right.Inc Ray Inc ay ide te d ide dR nt lec nt c te Ra ef Ra fle y R y Re θr Re Refr normal normal ac fr ted R act ay θr ed R ay
Axle AnalogyImagine you’re on a skateboard heading from the sidewalk toward somegrass at an angle. Your front axle is depicted before and after entering thegrass. Your right contacts the grass first and slows, but your left wheel isstill moving quickly on the sidewalk. This causes a turn toward the normal.If you skated from grass to sidewalk, the same path would be followed. Inthis case your right wheel would reach the sidewalk first and speed up, butyour left wheel would still be moving more slowly. The result this timewould be turning away from the normal. Skating from sidewalk to grass islike light traveling from air to a more overhead view“optically dense” medium like glassor water. The slower light travels inthe new medium, the more it bendstoward the normal. Light traveling sidewalkfrom water to air speeds up and grassbends away from the normal. Aswith a skateboard, light travelingalong the normal will change speed θrbut not direction.
Index of Refraction, n The index of refraction of a substance is the ratio of the speed in light in a vacuum to the speed of light in that substance: c n= v Medium n n = Index of Refraction Vacuum 1 c = Speed of light in vacuum Air (STP) 1.00029 v = Speed of light in medium Water (20º C) 1.33Note that a large index of refractioncorresponds to a relatively slow Ethanol 1.36light speed in that medium. Glass ~1.5 Diamond 2.42
θi Snell’s Law ni nr θrSnell’s law states that a ray of light bends insuch a way that the ratio of the sine of theangle of incidence to the sine of the angle ofrefraction is constant. Mathematically, ni sinθ i = nr sinθrHere ni is the index of refraction in the originalmedium and nr is the index in the medium thelight enters. θ i and θr are the angles ofincidence and refraction, respectively. Willebrord Snell
Snell’s Law Derivation Two parallel rays are shown. Points A and B are directly opposite one another. The top pair is at one point in time, and the bottom pair after time t. A θ1 The dashed lines connectingn1 x • the pairs are perpendicular to A the rays. In time t, point A d • •B travels a distance x, while yn2 point B travels a distance y. •B sinθ1 = x / d, so x = d sinθ1 θ2 sinθ2 = y / d, so y = d sinθ2 Speed of A: v1 = x / t Speed of B: v2 = y / t Continued…
Snell’s Law Derivation (cont.) A θ1n1 x • A d • •B v1 x/ t x sinθ1 y = = = So,n2 •B v2 y/ t y sinθ2 θ2 v1 / c sinθ1 1 / n1 sinθ1 n2 = ⇒ = = v2 / c sinθ2 1 / n2 sinθ2 n1 ⇒ n1 sinθ1 = n2 sinθ2
Refraction Problem #1 Goal: Find the angular displacement of the ray after having passed through the prism. Hints: 1. Find the first angle of refraction using Snell’s law. 19.4712º 2. Find angle ø. (Hint: Use Geometry skills.) 79.4712º Air, n1 = 1 30° 3. Find the second angle of incidence. 10.5288º 4. Find the second angle ofHoriz. ray, refraction, θ, using Snell’s Lawparallel to ø θ 15.9ºbase Glass, n2 = 1.5
Refraction Problem #2Goal: Find the distance the light ray displaced due to the thickwindow and how much time it spends in the glass. Some hints aregiven. 20º θ1 1. Find θ1 (just for fun). 20º H20 n1 = 1.3 2. To show incoming & outgoing rays are parallel, find θ. 20º 3. Find d. 0.504 m glass 10m n2 = 1.5 4. Find the time the light spends in 5.2 · 10-8 s d H20 the glass. θ Extra practice: Find θ if bottom medium is replaced with air. 26.4º
Refraction Problem #3Goal: Find the exit angle relative to the horizontal. θ = 19.8° 36°air glass θ=? The triangle is isosceles. Incident ray is horizontal, parallel to the base.
Reflection ProblemGoal: Find incident angle relative to horizontal so that reflected raywill be vertical. θ = 10º θ 50º center of semicircular mirror with horizontal base
Brewster AngleThe Brewster angle is the angle of incidence the produces reflectedand refracted rays that are perpendicular. From Snell, n1 sinθb = n2 sinθ. n2 θ α = θb since α + β = 90º, α and θb + β = 90º. β n1 θb θb β = θ since α + β = 90º, and θ + α = 90º. Thus, n1 sinθb = n2 sinθ = n2 sinβ = n2 cosθb tanθb = n2 / n1 Sir David Brewster
Critical AngleThe incident angle that causes nrthe refracted ray to skim right nialong the boundary of a θcsubstance is known as the criticalangle, θc. The critical angle is theangle of incidence that produces From Snell,an angle of refraction of 90º. If n1 sinθc = n2 sin 90°the angle of incidence exceedsthe critical angle, the ray is Since sin 90° = 1, wecompletely reflected and does have n1 sinθc = n2 andnot enter the new medium. A the critical angle iscritical angle only exists whenlight is attempting to penetrate a nrmedium of higher optical density θc = sin-1than it is currently traveling in. ni
Critical Angle Sample Problem Calculate the critical angle for the diamond-air boundary. Refer to the Index of Refraction chart for the information. air θc = sin-1 (nr / ni)diamond = sin-1 (1 / 2.42) θc = 24.4° Any light shone on this boundary beyond this angle will be reflected back into the diamond.
Total Internal ReflectionTotal internal reflection occurs when light attempts to passfrom a more optically dense medium to a less optically densemedium at an angle greater than the critical angle. When thisoccurs there is no refraction, only reflection. n1 n2 > n1 n2 θ θ > θcTotal internal reflection can be used for practical applicationslike fiber optics.
Fiber Optics Fiber optic lines are strands of glass or transparent fibers that allows the transmission of light and digital information over long distances. They are used for the telephone system, the cable TV system, the internet, medical imaging, and mechanical engineeringspool of optical fiber inspection. Optical fibers have many advantages over copper wires. They are less expensive, thinner, lightweight, and more flexible. They aren’t flammable since they use light signals instead of electric signals. Light signals from one fiber do not interfere with signals in nearby fibers, which means clearer TV A fiber optic wire reception or phone conversations. Continued…
Fiber Optics Cont.Fiber optics are often long strandsof very pure glass. They are verythin, about the size of a humanhair. Hundreds to thousands ofthem are arranged in bundles(optical cables) that can transmitlight great distances. There arethree main parts to an opticalfiber: • Core- the thin glass center where light travels. • Cladding- optical material (with a lower index of refraction than the core) that surrounds the core that reflects light back into the core. • Buffer Coating- plastic coating on the outside of an optical fiber to protect it from damage. Continued…
Light travels through the core of afiber optic by continually Fiber Optics (cont.)reflecting off of the cladding. Dueto total internal reflection, thecladding does not absorb any ofthe light, allowing the light to There are two types of opticaltravel over great distances. Some fibers:of the light signal will degrade • Single-mode fibers- transmitover time due to impurities in the one signal per fiber (used inglass. cable TV and telephones). • Multi-mode fibers- transmit multiple signals per fiber (used in computer networks).
MiragesMirages are caused by the refracting properties of anon-uniform atmosphere.Several examples of mirages include seeing “puddles”ahead on a hot highway or in a desert and the lingeringdaylight after the sun is below the horizon. More Mirages Continued…
Inferior Mirages A person sees a puddle ahead on the hot highway because the road heats the air above it, while the air farther above the road stays cool. Instead of just two layers, hot and cool, there are reallymany layers, each slightly hotter than the layer above it. The cooler air has aslightly higher index of refraction than the warm air beneath it. Rays oflight coming toward the road gradually refract further from the normal,more parallel to the road. (Imagine the wheels and axle: on a light raycoming from the sky, the left wheel is always in slightly warmer air than theright wheel, so the left wheel continually moves faster, bending the axlemore and more toward the observer.) When a ray is bent enough, itsurpasses the critical angle and reflects. The ray continues to refract as itheads toward the observer. The “puddle” is really just an inverted image ofthe sky above. This is an example of an inferior mirage, since the cool are isabove the hot air.
Superior MiragesSuperior mirages occur when alayer of cool air is beneath a layerof warm air. Light rays are bentdownward, which can make anobject seem to be higher in the airand inverted. (Imagine thewheels and axle on a ray comingfrom the boat: the right wheel iscontinually in slightly warmer airthan the left wheel. Thus, the rightwheel moves slightly faster andbends the axle toward theobserver.) When the critical angleis exceeded the ray reflects. Thesemirages usually occur over ice, snow, or cold water. Sometimes superiorimages are produced without reflection. Eric the Red, for example, was able tosee Greenland while it was below the horizon due to the light graduallyrefracting and following the curvature of the Earth.
Sunlight after SunsetLingering daylight after the sunis below the horizon is another Apparenteffect of refraction. Light travels position Observerat a slightly slower speed in of sunEarth’s atmosphere than inspace. As a result, sunlight is Actualrefracted by the atmosphere. In position Earththe morning, this refraction of suncauses sunlight to reach usbefore the sun is actually above Atmospherethe horizon. In the evening, thesunlight is bent above the horizon after the sun has actually set. Sodaylight is extended in the morning and evening because of therefraction of light. Note: the picture greatly exaggerates this effect aswell as the thickness of the atmosphere. Different “shapes” of Sun
Dispersion of LightDispersion is the separation of light into a spectrum by refraction. Theindex of refraction is actually a function of wavelength. For longerwavelengths the index is slightly small. Thus, red light refracts less thanviolet. (The pic is exaggerated.) This effect causes white light to splitinto it spectrum of colors. Red light travels the fastest in glass, has asmaller index of refraction, and bends the least. Violet is slowed downthe most, has the largest index, and bends the most. In other words: thehigher the frequency, the greater the bending. Animation
Atmospheric OpticsThere are many natural occurrences of light optics in our atmosphere. One of the most common of these is the rainbow, which is caused by water droplets dispersing sunlight. Others include arcs, halos, cloud iridescence, and many more. Photo gallery of atmospheric optics.
Rainbows A rainbow is a spectrum formed when sunlight is dispersed by water droplets in the atmosphere. Sunlight incident on a water droplet is refracted. Because of dispersion, each color is refracted at a slightly different angle. At the back surface of the droplet, the light undergoes total internal reflection. On theway out of the droplet, the light is once more refracted and dispersed.Although each droplet produces a complete spectrum, an observer willonly see a certain wavelength of light from each droplet. (The wavelengthdepends on the relative positions of the sun, droplet, and observer.)Because there are millions of droplets in the sky, a complete spectrum isseen. The droplets reflecting red light make an angle of 42 o with respect tothe direction of the sun’s rays; the droplets reflecting violet light make anangle of 40o. Rainbow images
Secondary Secondary Rainbow The secondary rainbow is a rainbow of radius 51°, occasionally visible outside the primary rainbow. It is produced when the light Primary entering a cloud droplet is reflected twice internally and then exits the droplet. The color spectrum is reversed in respect to the primary rainbow, with red appearing on its inner edge.Alexander’sdark region
Supernumerary ArcsSupernumerary arcs are faint arcs of colorjust inside the primary rainbow. Theyoccur when the drops are of uniform size.If two light rays in a raindrop arescattered in the same direction but havetake different paths within the drop, thenthey could interfere with each otherconstructively or destructively. The typeof interference that occurs depends on thedifference in distance traveled by therays. If that difference is nearly zero or amultiple of the wavelength, it isconstructive, and that color is reinforced.If the difference is close to half awavelength, there is destructiveinterference.
Real vs. Virtual ImagesReal images are formed by mirrors or lenses when light raysactually converge and pass through the image. Real images will belocated in front of the mirror forming them. A real image can beprojected onto a piece of paper or a screen. If photographic filmwere placed here, a photo could be created.Virtual images occur where light rays only appear to haveoriginated. For example, sometimes rays appear to be coming froma point behind the mirror. Virtual images can’t be projected onpaper, screens, or film since the light rays do not really convergethere.Examples are forthcoming.
Plane Mirror Object Rays emanating from an object at point Pstrike the mirror and are reflected with equalangles of incidence and reflection. After P P’reflection, the rays continue to spread. If weextend the rays backward behind the mirror, Virtualthey will intersect at point P’, which is the Imageimage of point P. To an observer, the raysappear to come from point P’, but no source isthere and no rays actually converging there .For that reason, this image at P’ is a virtualimage. do di O IThe image, I, formed by a plane mirrorof an object, O, appears to be adistance di , behind the mirror, equalto the object distance do. Animation Continued…
Plane Mirror (cont.)Two rays from object P strike the mirror at points B and M. Each ray isreflected such that i = r. Triangles BPM and BP’M are P do B di P’ congruent by ASA (show this), which implies that do= di and h = h’. Thus, the image is the h M h’ same distance behind the mirror Object Image as the object is in front of it, and the image is the same size as the object. object image MirrorWith plane mirrors, the image is reversed left to right (or the front andback of an image ). When you raise your left hand in front of a mirror,your image raises its right hand. Why aren’t top and bottom reversed?
Concave and Convex Mirrors Concave and convex mirrors are curved mirrors similar to portions of a sphere. light rays light raysConcave mirrors reflect light Convex mirrors reflect lightfrom their inner surface, like from their outer surface, like the inside of a spoon. the outside of a spoon.
Concave Mirrors• Concave mirrors are approximately spherical and have a principalaxis that goes through the center, C, of the imagined sphere and endsat the point at the center of the mirror, A. The principal axis isperpendicular to the surface of the mirror at A.• CA is the radius of the sphere,or the radiusof curvature of the mirror, R .• Halfway between C and A is the focalpoint of the mirror, F. This is the pointwhere rays parallel to the principal axis willconverge when reflected off the mirror.• The length of FA is the focal length, f.• The focal length is half of the radius of thesphere (proven on next slide).
r = 2fTo prove that the radius of curvature of a concave mirror istwice its focal length, first construct a tangent line at thepoint of incidence. The normal is perpendicular to thetangent and goes through the center, C. Here, i = r = β. Byalt. int. angles the angle at C is also β, and α = 2 β. s is thearc length from the principle axis to the pt. of incidence.Now imagine a sphere centeredat F with radius f. If the incident tan geray is close to the principle axis, β ntlthe arc length of the new sphere β s ineis about the same as s. From β αs = r θ, we have s = r β and • C • f Fs ≈ f α = 2 f β. Thus, r β ≈ 2 f β,and r = 2 f. r
Focusing Light with Concave Mirrors Light rays parallel to the principal axis will be reflected through the focus (disregarding spherical aberration, explained on next slide.) In reverse, light rays passing through the focus will be reflected parallel to the principal axis, as in a flood light.Concave mirrors can form both real and virtual images, depending onwhere the object is located, as will be shown in upcoming slides.
Spherical Aberration F • F • C • C • Spherical Mirror Parabolic MirrorOnly parallel rays close to the principal axis of a spherical mirror willconverge at the focal point. Rays farther away will converge at a pointcloser to the mirror. The image formed by a large spherical mirror will bea disk, not a point. This is known as spherical aberration.Parabolic mirrors don’t have spherical aberration. They are used to focusrays from stars in a telescope. They can also be used in flashlights andheadlights since a light source placed at their focal point will reflect lightin parallel beams. However, perfectly parabolic mirrors are hard to makeand slight errors could lead to spherical aberration. Continued…
Spherical vs. Parabolic MirrorsParallel rays converge at the Parabolic mirrors have nofocal point of a spherical spherical aberration. Themirror only if they are close to mirror focuses all parallel raysthe principal axis. The image at the focal point. That is whyformed in a large spherical they are used in telescopes andmirror is a disk, not a point light beams like flashlights and(spherical aberration). car headlights.
Concave Mirrors: Object beyond Cobject The image formed when an object is placed beyond C is • C • F located between C and F. It is a real, inverted image image that is smaller in size than the object. Animation 1 Animation 2
Concave Mirrors: Object between C and F The image formed object when an object is placed between C and F • C • F is located beyond C. It is a real, inverted image image that is larger in size than the object. Animation 1 Animation 2
Concave Mirrors: Object in front of F The image formed when an object is placed in front of F is object located behind the image mirror. It is a virtual, • C • F upright image that is larger in size than the object. It is virtual since it is formed only Animation where light rays seem to be diverging from.
Concave Mirrors: Object at C or F What happens when an object is placed at C? The image will be formed at C also, but it will be inverted. It will be real and the same size as the object.Animation What happens when an object is placed at F? No image will be formed. All rays will reflect parallel to the principal axis and will never converge. The image is “at infinity.”
Convex Mirrors• A convex mirror has thesame basic properties as a light raysconcave mirror but its focusand center are located behindthe mirror.• This means a convex mirrorhas a negative focal length • Rays parallel to the principal(used later in the mirror axis will reflect as if comingequation). from the focus behind the mirror.• Light rays reflected fromconvex mirrors always • Rays approaching the mirrordiverge, so only virtual on a path toward F will reflectimages will be formed. parallel to the principal axis.
Convex Mirror Diagram The image formed by a convex mirror no matter where theobject object is placed will image be virtual, upright, • F • C and smaller than the object. As the object is moved closer to the mirror, the image will approach the size of the object.
Mirror/Lens Equation Derivation From ∆PCO, β = θ + α, so 2β = 2θ + 2α. From ∆PCO, γ = 2θ + α , so -γ = -2θ - α. P Adding equations yields 2β - γ = α. θ object From s = r θ, we haves θ γ s = r β, s ≈ di α, and β α T • C O s ≈ di α (for rays image close to the principle axis). Thus: s α≈ s β= r d o di γ≈ s do di (cont.)
Mirror/Lens Equation Derivation (cont.) From the last slide, β = s / r, α ≈ s / d0 , γ ≈ s / di , and 2 β - γ = α. Substituting into the last equation yields: P 2s s ss θ θ object r -d = d i o γ β α 2 1 1 T • C O r = di + do image 2 1 1 = d +d 2f i o di 1 1 1 = d +d f i o doThe last equation applies to convex and concave mirrors, as well as tolenses, provided a sign convention is adhered to.
Mirror Sign Convention f = focal length1 1 1 di = image distancef = d i + do do = object distance + for real image di - for virtual image + for concave mirrors f - for convex mirrors
Magnification hi By definition, m = hom = magnificationhi = image height (negative means inverted)ho = object heightMagnification is simply the ratio of image heightto object height. A positive magnification meansan upright image.
hi -di Magnification Identity: m = = ho doTo derive this let’s look at two rays. One hits the mirror on the axis.The incident and reflected rays each make angle θ relative to the axis.A second ray is drawn through the center and is reflected back on topof itself (since a radius is always perpendicular to an tangent line of a circle). The intersection of the reflected rays object determines the location of θ ho the tip of the image. Our • C result follows image, from similar triangles, with the negative sign a height = hi consequence of our sign convention. (In this picture di do hi is negative and di is positive.)
Mirror Equation Sample Problem Suppose AllStar, who is 3 and a half feet tall, stands 27 feet in front of a concave mirror with a radius of curvature of • C • F 20 feet. Where will his image be reflected and what will its size be? di = 15.88 feet hi = -2.06 feet
Mirror Equation Sample Problem 2 Casey decides to join in the fun and she finds a convex mirror to stand in front of. She sees her image reflected 7 feet behind the mirror which • F • C has a focal length of 11 feet. Her image is 1 foot tall. Where is she standing and how tall is she? d =19.25 feet o ho = 2.75 feet
LensesLenses are made of transparent Convex (Converging)materials, like glass or plastic, that Lenstypically have an index of refractiongreater than that of air. Each of a lens’two faces is part of a sphere and can beconvex or concave (or one face may beflat). If a lens is thicker at the centerthan the edges, it is a convex, or Concave (Diverging)converging, lens since parallel rays will Lensbe converged to meet at the focus. Alens which is thinner in the center thanthe edges is a concave, or diverging,lens since rays going through it will bespread out.
Lenses: Focal Length• Like mirrors, lenses have a principal axis perpendicular to theirsurface and passing through their midpoint.• Lenses also have a vertical axis, or principal plane, through theirmiddle.• They have a focal point, F, and the focal length is the distance fromthe vertical axis to F.• There is no real center of curvature, so 2F is used to denote twicethe focal length.
Ray Diagrams For LensesWhen light rays travel through a lens, they refract at both surfaces ofthe lens, upon entering and upon leaving the lens. At each interfacethe bends toward the normal. (Imagine the wheels and axle.) Tosimplify ray diagrams, we often pretend that all refraction occurs atthe vertical axis. This simplification works well for thin lenses andprovides the same results as refracting the light rays twice. • • 2F F • 2F F • • • 2F F • 2F F • Reality Approximation
Convex LensesRays traveling parallel to the principalaxis of a convex lens will refract toward • • 2F F • 2F F •the focus. Rays traveling from the focus will • F 2F • • 2F F • refract parallel to the principal axis.Rays traveling directly through thecenter of a convex lens will leave the • • 2F F • 2F F •lens traveling in the exact samedirection.
Convex Lens: Object Beyond 2F The image formed when an object isobject placed beyond 2F is located behind • • • • the lens between F 2F F F 2F and 2F. It is a real, image inverted image which is smaller than the object Experiment with this diagram itself.
Convex Lens: Object Between 2F and F The image formedobject when an object is placed between 2F and F is•2F • F • F • 2F located beyond 2F behind the lens. It is a real, image inverted image, larger than the object.
Convex Lens: Object within F The image formed when an object is placed in front of F is located somewhereimage beyond F on the same side of the lens as the object. It is a virtual, upright image • 2F • F • F • 2F which is larger than the object object. This is how a magnifying glass works. When the object is brought close to the lens, it will be convex lens used magnified greatly. as a magnifier
Concave Lenses Rays traveling parallel to the principal axis of a concave lens will•2 • F • F • 2 refract as if coming from the focus.F F Rays traveling toward the 2F • • F • 2 F • focus will refract parallel to the principal axis. F Rays traveling directly through the2F •• F • 2 F • center of a concave lens will leave the lens traveling in the exact same F direction, just as with a convex lens.
Concave Lens Diagram No matter where the object is placed, theobject image will be on the same side as the • 2F • F • F • 2F object. The image is image virtual, upright, and smaller than the object with a concave lens. Experiment with this diagram
Lens Sign Convention f = focal length1 1 1f = d +d di = image distance i o do = object distance + for real image di - for virtual image + for convex lenses f - for concave lenses
Lens / Mirror Sign Convention The general rule for lenses and mirrors is this: + for real image di - for virtual imageand if the lens or mirror has the ability to converge light,f is positive. Otherwise, f must be treated as negative forthe mirror/lens equation to work correctly.
Lens Sample Problem Tooter, who stands 4 feet tall (counting his snorkel), finds himself 24 feet in front of a convex lens and he sees his image reflected 35 feet•2F • F • F • 2F behind the lens. What is the focal length of the lens and how tall is his image? f = 14.24 feet hi = -5.83 feet
Lens and Mirror AppletThis application shows where images will be formedwith concave and convex mirrors and lenses. You canchange between lenses and mirrors at the top. Changingthe focal length to negative will change betweenconcave and convex lenses and mirrors. You can alsomove the object or the lens/mirror by clicking anddragging on them. If you click with the right mousebutton, the object will move with the mirror/lens. Thefocal length can be changed by clicking and dragging atthe top or bottom of the lens/mirror. Object distance,image distance, focal length, and magnification can alsobe changed by typing in values at the top. Lens and Mirror Diagrams
Convex Lens in Water Glass Glass H2O AirBecause glass has a higher index of refraction that water the convexlens at the left will still converge light, but it will converge at agreater distance from the lens that it normally would in air. This isdue to the fact that the difference in index of refraction betweenwater and glass is small compared to that of air and glass. A largedifference in index of refraction means a greater change in speed oflight at the interface and, hence, a more dramatic change ofdirection.
Convex Lens Made of Water Glass Since water has a higher index of refraction than air, a convex lens made of water will converge light just as a glass lens of the same shape. However, theAir glass lens will have a smaller focal length n = 1.5 than the water lens (provided the lenses are of same shape) because glass has an index of refraction greater than that of water. Since there is a bigger differenceH2O in refractive index at the air-glass interface than at the air-water interface, the glass lens will bend light more than the water lens.Air n = 1.33
Air & Water Lenses On the left is depicted a concave lens filled with water, and light rays entering it from an air-filled environment. Water has a higher index than air, so the rays diverge just like Air they do with a glass lens.Concave lens made of H2OTo the right is an air-filled convex lenssubmerged in water. Instead ofconverging the light, the rays divergebecause air has a lower index than water. H2O Convex lens made of Air What would be the situation with a concave lens made of air submerged in water?
Chromatic AberrationAs in a raindrop or a prism, different wave-lengths of light are refracted at differentangles (higher frequency ↔ greaterbending). The light passing through a lensis slightly dispersed, so objects viewedthrough lenses will be ringed with color.This is known as chromatic aberration andit will always be present when a single lens Chromatic Aberrationis used. Chromatic aberration can begreatly reduced when a convex lens iscombined with a concave lens with adifferent index of refraction. Thedispersion caused by the convex lens willbe almost canceled by the dispersioncaused by the concave lens. Lenses such asthis are called achromatic lenses and are Achromatic Lensused in all precision optical instruments. Examples
Human eyeThe human eye is a fluid-filled object thatfocuses images of objects on the retina. Thecornea, with an index of refraction of about1.38, is where most of the refraction occurs.Some of this light will then passes throughthe pupil opening into the lens, with an indexof refraction of about 1.44. The lens is flexi- Human eye w/raysble and the ciliary muscles contract or relax to change its shape andfocal length. When the muscles relax, the lens flattens and the focallength becomes longer so that distant objects can be focused on theretina. When the muscles contract, the lens is pushed into a moreconvex shape and the focal length is shortened so that close objectscan be focused on the retina. The retina contains rods and cones todetect the intensity and frequency of the light and send impulses to thebrain along the optic nerve.
Hyperopia The first eye shown suffers from farsightedness, which is also known as hyperopia. This is due to a focal length that is too long, causing the image to be focused behind the retina, making it difficult for the person to see close up things. Formation of image behind The second eye is being helped with a the retina in a hyperopic eye. convex lens. The convex lens helps the eye refract the light and decrease the image distance so it is once again focused on the retina. Hyperopia usually occurs among adults due to weakened ciliary Convex lens correction muscles or decreased lens flexibility. for hyperopic eye.Farsighted means “can see far” and the rays focus too far from the lens.
The first eye suffers from Myopia nearsightedness, or myopia. This is a result of a focal length that is too short, causing the images of distant objects to be focused in front of the retina. The second eye’s vision is being Formation of image in front corrected with a concave lens. The of the retina in a myopic eye. concave lens diverges the light rays, increasing the image distance so that it is focused on the retina. Nearsightedness is common among young people, sometimes the result of a bulging cornea (which will Concave lens correction refract light more than normal) or an for myopic eye. elongated eyeball.Nearsighted means “can see near” and the rays focus too near the lens.
Refracting TelescopesRefracting telescopes are comprised of two convex lenses. The objectivelens collects light from a distant source, converging it to a focus andforming a real, inverted image inside the telescope. The objective lensneeds to be fairly large in order to have enough light-gathering power sothat the final image is bright enough to see. An eyepiece lens is situatedbeyond this focal point by a distance equal to its own focal length. Thus,each lens has a focal point at F. The rays exiting the eyepiece are nearlyparallel, resulting in a magnified, inverted, virtual image. Besidesmagnification, a good telescope also needs resolving power, which is itsability to distinguish objects with very small angular separations. F
Reflecting TelescopesGalileo was the first to use a refracting telescope for astronomy. It isdifficult to make large refracting telescopes, though, because theobjective lens becomes so heavy that it is distorted by its own weight. In1668 Newton invented a reflecting telescope. Instead of an objectivelens, it uses a concave objective mirror, which focuses incoming parallelrays. A small plane mirror is placed at this focal point to shoot the lightup to an eyepiece lens (perpendicular to incoming rays) on the side ofthe telescope. The mirror serves to gather as much light as possible,while the eyepiece lens, as in the refracting scope, is responsible for themagnification.
Huygens’ PrincipleChristiaan Huygens, a contemporary of Newton, wasan advocate of the wave theory of light. (Newtonfavored the particle view.) Huygens’ principle statesthat a wave crest can be thought of as a series ofequally-spaced point sources that produce waveletsthat travel at the same speed as the original wave.These wavelets superimpose with one another.Constructive interference occurs along a line parallelto the original wave at a distance of one wavelengthfrom it. This principle explains diffraction well:When light passes through a very small slit, it is as ifonly one of these point sources is allowed through. ChristiaanSince there are no other sources to interfere with it, Huygenscircular wavefronts radiate outwards in all directions. Applet showing reflect • • • • •
screen PDiffraction: Single SlitLight enters an opening of width a and isdiffracted onto a distant screen. All points at theopening act as individual point sources of light.These point sources interfere with each other, bothconstructively and destructively, at different pointson the screen, producing alternating bands oflight and dark. To find the first dark spot, let’sconsider two point sources: one at the left edge,and one in the middle of the slit. Light from the leftpoint source must travel a greater distance to pointP on the screen than light from the middle pointsource. If this extra distance Extrais a half a wavelength, λ/2, distancedestructive interference willoccur at P and there willbe a dark spot there. a/2 applet a Continued…
Single Slit (cont.)Let’s zoom in on the small triangle in the last slide. Since a / 2 isextremely small compared to the distanced to the screen, the twoarrows pointing to P are essentially parallel. The extra distance isfound by drawing segment AC perpendicular to BC. This means thatangle A in the triangle is also θ. Since AB is the hypotenuse of aright triangle, the extra distance is given by (a / 2) sinθ. Thus, using (a / 2) sinθ = λ/2, or equivalently, P a sinθ = λ, we can locate the first dark i nt po C spot on the screen. Other dark spots can To be located by dividing the slit further. e c an P istθ θ int ad po tr θ Ex ToB a/2 A
screen P Diffraction: Double Slit Light passes through two openings, each of which acts as a point source. Here a is the distance between the openings rather than the width of a particular opening. As before, if d1 - d2 = n λ (a multiple of the wavelength), light from the two sources will be in phase and there will a bright d1 spot at P for that wavelength. By the d2 Pythagorean theorem, the exact difference L in distance is d1 - d2 = [ L2 + (x + a / 2)2 ] ½ - [ L2 + (x - a / 2)2 ] ½Approximation on next slide. Link 1 Link 2 a x
Double Slit (cont.) screen PIn practice, L is far greater than a, meaningthat segments measuring d1 and d2 arevirtually parallel. Thus, both rays make anangle θ relative to the vertical, and thebottom right angle of the triangle is also θ(just like in the single slit case). This meansthe extra distance traveled is given by a sinθ. d1Therefore, the required condition for a bright d2spot at P is that there exists a natural number, Ln, such that: a sinθ = n λ θ θIf white light is shone at theslits, different colors will bein phase at different angles.Electron diffraction a
Diffraction GratingsA different grating has numerous tiny slits, equally spaced. Itseparates white light into its component colors just as a double slitwould. When a sinθ = n λ, light of wavelength λ will be reinforcedat an angle of θ. Since different colors have different wavelengths,different colors will be reinforced at different angles, and a prism-likespectrum can be produced. Note, though, that prisms separate light viarefraction rather than diffraction. The pic on the left shows red lightshone through a grating. The CD acts as a diffraction grating since thetracks are very close together (about 625/mm).