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Chapter 26: Refraction, Lenses, Optical Instruments • Refraction of light, Snell’s law. Apparent depth • Polarization of light on reflection • Dispersion of light. Prisms, rainbows, sun dogs • Formation of image by lenses, thin lens equation, magnification • Combination of lenses • The human eye – correction for near and farsightedness • Optical instruments • Omit 26.14, lens aberrations Friday, March 9, 2007 1 Refraction of light When light travels from one transparent medium to another, it is in general deflected from its original direction –!this is refraction. The amount by which the light is deflected depends on the refractive index of each medium. Friday, March 9, 2007 2 Refractive index – a measure of the speed of light in a transparent medium Refractive index = n= Speed of light in vacuum Speed of light in medium c c and v = = f λ v n Prob. 26.6: Light has wavelength 340 nm and frequency 5.403"1014 Hz when travelling through some substance. What substance is it? v = c/n = f ! = (5.403"1014 Hz)"(340"10-9m) # = 1.837"108 m/s n = 3"108/1.837"108 = 1.633 – carbon disulphide Friday, March 9, 2007 3 Prob. 26.8: A flat sheet of ice has a thickness of 2 cm. It is on top of a flat sheet of quartz that has a thickness of 1.1 cm. Light strikes the ice perpendicularly and travels through it and then through the quartz. In the time it takes the light to travel through the two sheets, how far would it have travelled in vacuum? Ice: # t1 = l1/v1 = l1 n1/c as v1 = c/n1 Quartz:# t2 = l2/v2 = l2 n2/c Ice n1 = 1.309 l1 = 2 cm Quartz n2 = 1.544 l2 = 1.1 cm In time (t1 + t2), light would travel a distance L = c(t1 + t2) in vacuum. t1 = (0.020 m) " 1.309/c = 0.02618/c t2 = (0.011 m) " 1.544/c = 0.01698/c t1 + t2 = 0.04316/c So L = c(t1 + t2) = 0.0432 m = 4.32 cm Friday, March 9, 2007 4 Refraction and Reflection Snell’s law: n1 sin !1 = n2 sin !2 Light incident from above: If !1 = 30◦ “External reflection” n1 sin !1 n2 1 × sin 30◦ = 1.33 sin !2 = !2 = 22.1◦ Friday, March 9, 2007 5 Snell’s law: n1 sin !1 = n2 sin !2 Light incident from below: If !1 = 30◦ n1 sin !1 n2 1.33 × sin 30◦ = 1 sin !2 = !2 = 41.7◦ “Internal reflection” (inside medium of higher refractive index) Friday, March 9, 2007 6 Prob. 26.C8: Two rays of light converge to a point on a screen. A plane-parallel plate of glass is placed in the path of this converging light and the glass plate is parallel to the screen. Will the point of convergence remain on the screen? If not, will the point move toward the glass of away from it? As n2 > n1, the ray is refracted toward the normal to the surface when it enters the glass. glass n1 = 1 n2 > n1 Friday, March 9, 2007 The converging rays meet farther to the right, behind the screen. Friday, March 9, 2007 7 n1 = 1 n2 > n1 8 Rear view mirror Day setting: $ 100% reflection Night setting: $ 10% reflection Friday, March 9, 2007 9 Apparent Depth At what angle should the searchlight be aimed to illuminate the chest? tan !2 = 2 3.3 !2 = 31.22◦ Snell’s Law: n1 sin !1 = n2 sin !2 sin !1 = 1.33 sin 31.22◦ = 0.6894 1 !1 = 43.6◦ The searchlight is aimed above the chest – the apparent depth of the chest is less than its actual depth. Friday, March 9, 2007 10 Apparent depth – formation of image Apparent position of chest Friday, March 9, 2007 11 Prob. 26.17/19: Find the relationship between true and apparent depth for small angles of incidence. n1 !1 n2 x x d! x tan !2 = d tan !1 = !2 d´ !1 d x apparent depth So, x = d ! tan !1 = d tan !2 Snell’s Law: n1 sin !1 = n2 sin !2 true depth For small angles: sin ! ! tan ! ! ! radians Therefore, n1!1 ! n2!2 and d "!1 ! d!2 So, d ! !2 n1 " " d !1 n2 Friday, March 9, 2007 → d" # dn1 n2 Apparent depth 12 Apparent depth (height) Viewed from below the surface of the water, the object appears further above the surface than it actually is. n2 The same formula holds as for apparent depth: n1 d! " dn1 n2 As n1 > n2, d´ > d Friday, March 9, 2007 13 Prob. 26.16: A silver medallion is sealed within a transparent block of plastic. An observer in the air, viewing the medallion from directly above, see the medallion at an apparent depth of 1.6 cm beneath the top surface of the block. How far below the top surface would the medallion appear if the observer (not wearing goggles) and the block were under water? Friday, March 9, 2007 14 Prob. 26.20/18: A man in a boat is looking straight down at a fish in the water directly beneath him. The fish is looking right back. They are the same distance from the air-water interface. To the man, the fish appears to be 2 m beneath his eyes. To the fish, how far above its eyes does the man appear to be? Friday, March 9, 2007 15 Chapter 26 so far... Refractive index: n = c/v Snell’s law: n1 sin%1 = n2 sin%2 Apparent depth: d! = dn1/n2 Friday, March 9, 2007 16 Displacement of path of ray of light n1 sin θ1 = n2 sin θ2 = n1 sin θ3 So, θ3 = θ1 Rotating the glass plate changes the amount of displacement – used in some optical instruments to line up an image with reference lines (cross-hairs) Friday, March 9, 2007 17 Prob. 26.19/17: If "1 = 30o, the glass plate is 6 mm thick and n2 = 1.52, what x is the displacement? dt = 6 mm n2 = 1.52 Snell: n1 sin !1 = n2 sin !2 = n3 sin !3 = 30º As n1 = n3, !1 = !3 That is, emerging ray is parallel with incident ray. n1 sin !1 1 × sin 30◦ sin !2 = = → !2 = 19.2◦ n2 1.52 Friday, March 9, 2007 18 26.19 contd θ2 = 19.2 ◦ A B 90º x t = 6 mm BCD: x = L sin(!1 − !2) ABD: L = t/ cos !2 So, x = t sin(!1 − !2) cos !2 C x= L 6 sin(30◦ − 19.2◦) cos 19.2◦ Displacement, x = 1.19 mm !1 − !2 D Friday, March 9, 2007 19 Snell’s Law: a light wave hitting a boundary For any wave: v = f ! For light: v = c/n = f ! So, λ = Rays are perpendicular to wavefronts c 1 , proportional to nf n The incident wave turns into the refracted wave and matches onto it. The wavefronts crumple and tilt to adjust to the new wavelength # the ray is refracted. Friday, March 9, 2007 (small n) (large n) 20 Snell’s Law: a light wave hitting a boundary Rays are perpendicular to wavefronts B 90º – "1 Wavefronts 90º A C n1 ! = c/(nf) for a light wave c = h sin "1 ABC: !1 = n1 f c = h sin "2 ACD: !2 = n2 f c So, = n1 sin !1 = n2 sin !2 hf Snell’s Law! n2 90º D Wavefronts Friday, March 9, 2007 21 Total internal reflection Internal reflection The normal case: both reflected and refracted rays. If the angle of incidence increases, so does the angle of refraction, until... Refraction at the “critical angle for total internal reflection” – the refracted ray is at 90o. Possible only if n2 < n1 Friday, March 9, 2007 n1 sin !c = n2 sin 90◦ sin !c = n2/n1 22 n1 sin θc = n2 sin 90◦ n2 < n1 !1 > !c If the angle of incidence is greater than the critical angle: – according to Snell’s law, the sine of the angle of refraction is greater # than 1.0, so there is no refracted ray – the light undergoes “total internal reflection” Example: n1 = 1.33 (water), n2 = 1 (air): sin "c = 1/1.33, "c = 48.8o 26.13 Friday, March 9, 2007 23 Prob. 26.105/13: A ray of light travels from the coin to the surface of the liquid and is refracted as it enters the air. A person sees the ray as it skims just above the surface of the liquid. How fast is the light travelling in the liquid? • What are the angles of incidence, refraction? Apply Snell’s law, find # the index of refraction of the liquid. 90º " n2 = 1 n1 = ? tan θ = 5/6, θ = 39.8◦ n1 sin " = n2 sin 90º n1 = 1/sin " = 1.56 " Friday, March 9, 2007 v = c/n1 = 1.92"108 m/s 24 Total internal reflection around the bend Friday, March 9, 2007 25 Optical fibre – total internal reflection at the walls steers the light around bends Applications: • Medicine – flexible optical fibres used to look inside the body. # “Keyhole” surgery – add surgical instrument, laser beam to vaporize tissue. • Communications – transmit telephone, radio, TV, internet signals on a # laser beam inside a fibre optic cable – no external interference, much # greater amount of information can be transmitted than with copper cable. 26.29 Friday, March 9, 2007 26 Prob. 26.-/29: The optical fibre shown consists of a core made of flint glass surrounded by a cladding made of crown glass. A beam of light enters the fibre from air at an angle "1 with respect to the normal. What is "1 if the light strikes the core-cladding interface at the critical angle "c? B n1 = 1 A θ2 = 90◦ − θc n3 = 1.523 n2 = 1.667 Friday, March 9, 2007 27 Total internal reflection in a prism 45◦ > !c Glass: n $ 1.5 !c ! sin−1(1/1.5) = 42◦ Prisms “fold” the light path to make the binoculars shorter. Each arm acts as a longer telescope. Friday, March 9, 2007 28 Polarization of light by reflection Light reflected from a surface is in general partially polarized. The reflected light is 100% polarized parallel to the surface when reflection occurs at the “Brewster angle” "B, corresponding to a 90o angle between reflected and refracted rays. θB + θ2 + 90 = 180 ◦ ◦ !2 !B + !2 = 90◦ 90º !B From the triangle: sin !2 = cos !B Snell: n1 sin !B = n2 sin !2 = n2 cos !B So, tan !B = n2 n1 Brewster angle Friday, March 9, 2007 29 Polarization of light by reflection • Sunlight reflected from water – polarized horizontally. "B = 53º • Polaroid type sun glasses reduce glare from reflected sunlight by # filtering out horizontally polarized light. • Digital watches – emitted light is polarized vertically (top to bottom # in the display). • Display turns dark if rotated by 90o when viewed through Polaroid sun # glasses. Friday, March 9, 2007 30 Dispersion by a prism lower n, less refraction Violet is refracted more than red higher n, greater refraction Friday, March 9, 2007 26.38 31 Prob. 26.38: A ray of sunlight is passing from diamond into crown glass; the angle of incidence is 35º. Indices of refraction for red and blue light: Blue: ndiamond = 2.444# # ncrownglass = 1.531 Red: ndiamond = 2.410# # ncrownglass = 1.520 Determine the angle between the refracted red and blue rays in the crown glass. Friday, March 9, 2007 32 Dispersion by rain drops – rainbows Secondary rainbow Primary rainbow http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html Friday, March 9, 2007 33 Dispersion by a raindrop – primary rainbow The refractive index for violet is larger than for red. # violet is refracted through a larger angle # than red Friday, March 9, 2007 34 Dispersion – formation of a rainbow The colours of the rainbow come from raindrops at different height, red from higher up, violet from lower down. Friday, March 9, 2007 35 Violet is refracted through a larger angle than red http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html Friday, March 9, 2007 36 Sun Dogs Friday, March 9, 2007 37 Sun Dogs Refraction by hexagonal ice crystals Violet is refracted through a larger angle than red http://hyperphysics.phy-astr.gsu.edu/hbase/atmos/halo22.html#c3 Friday, March 9, 2007 38 Lenses A positive (converging, convex) lens Focal point Focal point A negative (diverging, concave) lens Friday, March 9, 2007 39 Formation of image by thin lenses Parallel to axis, passes through focal point on right Passes through focal point on left, emerges parallel to axis Passes through centre of lens in a straight line Parallel to axis, ray traced back to axis passes through focal point on left Heads toward focal point on right, emerges parallel to axis Passes through centre of lens in a straight line Friday, March 9, 2007 40 Formation of a real image by a converging lens Light appears to originate from image Object is outside the focal point A “real” image – can be seen on a screen placed at the position of the image. Image is inverted. Friday, March 9, 2007 41 Formation of a virtual image by a converging lens A “virtual” image – cannot be formed on a screen. Image is upright, and magnified. Light appears to originate from image Object is inside the focal point Friday, March 9, 2007 42 Formation of virtual image by a diverging lens Light appears to originate from image Image is virtual, upright, diminished Friday, March 9, 2007 43 Thin lens equation 1 3 ! ho ! ! ! 1 3 f Object distance tan ! = So, ho hi = do di hi di di − f = = ho do f That is: Friday, March 9, 2007 1 1 1 + = do di f Image distance ho hi tan ! = = f di − f Divide by di : Thin lens equation 1 1 1 = − do f di Applicable to diverging lenses too 44 Linear magnification ! ho ! ! ! f Linear magnification, m = (height of image)/(height of object) hi hi di (or from tan !) m= Similar triangles: = ho ho do Sign convention, image is inverted, so: m = − di do Friday, March 9, 2007 45 Conventions for the thin lens equation Draw ray diagrams with rays travelling from left to right. Normal situation: Object # Lens # Image Real object to left of lens, object distance, do is positive Real image to right of lens, image distance, di is positive Virtual object to right of lens, do is negative (2 or more lenses) Virtual image to left of lens, di is negative Converging (positive) lens, focal length f is positive Diverging (negative) lens, focal length f is negative Ex8 Friday, March 9, 2007 46 Converging lens, f > 0 f>0 do (> 0) di (> 0) Object # Lens # Image Diverging lens, f < 0 do (> 0) f<0 di (< 0) Friday, March 9, 2007 47 Example: A 1.7 m tall person stands 2.5 m in front of a camera. The focal length of the lens is 0.05 m. a) Find the image distance b) Find the magnification and the height of the image on the film. Friday, March 9, 2007 48 Prob. 26.50/104: To focus a camera on objects at different distances, the converging lens is moved toward or away from the film, so a sharp image always falls on the film. A camera with a lens of focal length f = 200 mm is to be focussed on an object located at a distance of 3.5 m and then at 50 m. Over what distance must the lens be movable? Friday, March 9, 2007 49 Thin lens equation 1 ho 3 ! ! ! ! 1 3 f Object distance Thin lens equation: Image distance 1 1 1 + = do di f Linear magnification: m = Friday, March 9, 2007 hi di =− ho do 50 Combinations of lenses – microscope • Find the location of the image formed by the first lens as if the second # lens did not exist. • Use that image as an object (source of light) for the second lens using the # sign convention for real and virtual objects. 1 3´ 2 3 2 1´ Lens 2: light appears to come from intermediate image A microscope producing a virtual, inverted and magnified final image. The eyepiece acts as a magnifying glass. Friday, March 9, 2007 51 Prob. 26.59/60: Two identical diverging lenses are separated by 16 cm. The focal length of each lens is –8 cm. An object is located 4 cm to the left of the lens that is on the left. Determine the final image distance relative to the lens on the right. Friday, March 9, 2007 52 Thin lens equation 1 ho 3 ! ! ! ! 1 3 f Object distance Image distance 1 1 1 Thin lens equation: + = do di f Linear magnification: m = hi di =− ho do Object to left: do > 0 Image to right: di > 0 Positive lens: f > 0 (converging, convex) Negative lens: f < 0 (diverging, concave) Friday, March 9, 2007 53 Combinations of lenses – microscope • Find the location of the image formed by the first lens as if the second # lens did not exist. • Use that image as an object (source of light) for the second lens using the # sign convention for real and virtual objects. 1 3´ 2 3 2 1´ Lens 2: light appears to come from intermediate image A microscope producing a virtual, inverted and magnified final image. The eyepiece acts as a magnifying glass. Friday, March 9, 2007 54 26.66: Two converging lenses (f1 = 9 cm, f2 = 6 cm) are separated by 18 cm. The lens on the left has the longer focal length. An object stands 12 cm to the left of the left-hand lens. a) Locate the final image relative to the lens on the right. b) Obtain the overall magnification. c) Is the final image real or virtual, upright or inverted, larger or smaller than the object? Friday, March 9, 2007 55 Prob. 26.64: A coin is located 20 cm to the left of a converging lens (f = 16 cm). A second, identical, lens is placed to the right of the first lens, such that the image formed by the combination has the same size and orientation as the original coin. Find the separation between the lenses. Friday, March 9, 2007 56 The Human Eye n = 1.33-1.34 n = 1.38 Most of the refraction occurs at the cornea n = 1.41-1.45 n = 1.34 Sharpest image, best colour discrimination Biomedical Applications of Introductory Physics, Tuszynski & Dixon Friday, March 9, 2007 57 The human eye The eye focuses an image onto the retina by adjusting the focal length of the eye lens. This is known as accommodation. Eye lens has its longest focal length Ciliary muscle, relaxed Greatest distance at which eye can focus: “far point” Normal value: infinity Eye lens compressed, focal length decreased Closest distance at which eye can focus: “near point” Normal value: N = 25 cm Friday, March 9, 2007 58 Near and far points Near point: closest distance at which unaided eye can focus, normal value, N = 25 cm Far point: greatest distance at which unaided eye can focus, normal value infinity Accommodation: the ability of the eye to adjust its focal length to focus on objects at different distances. Nearsighted eye: far point less than infinity " distant objects are blurred Farsighted eye: near point greater than 25 cm " objects close by are blurred Friday, March 9, 2007 59 Nearsightedness Objects in focus The eye lens forms an image of a distant object in front of the retina # blurred image on the retina Correction is with a diverging lens that moves the image back onto the retina. The corrective lens forms a virtual image in front of the eye that is close enough for the eye to focus on. Friday, March 9, 2007 60 Farsightedness Objects in focus The eye lens forms an image of a nearby object behind the retina # blurred image on the retina Correct with a converging lens that forms a virtual image far enough away for the eye to focus on. Friday, March 9, 2007 61 Correction of near and farsightedness Use a corrective lens to form a virtual image at a distance at which the eye can focus. Nearsighted: • The corrective lens forms an image of a distant object at the person’s far point, or closer. Farsighted: • The corrective lens forms an image of an object at the person’s near point, or further. Power of a lens: • Power is 1/f, focal length in metres, power in diopters. # Example, f = –10 cm, power = 1/(– 0.1) = –10 diopters. Friday, March 9, 2007 62 Prob. 26.107/71: A nearsighted person cannot read a sign that is more than 5.2 m from his eyes. He wears contact lenses that do not correct his vision completely, but do allow him to read signs located up to distances of 12 m from his eyes. What is the focal length of the contacts? Friday, March 9, 2007 63 Prob. 26.67: A farsighted person has a near point that is 67 cm from her eyes. She wears eyeglasses that are designed to enable her to read a newspaper held at a distance of 25 cm from her eyes. Find the focal length of the eyeglasses, assuming that they are worn – a) 2.2 cm from the eyes, b) 3.3 cm from the eyes. Friday, March 9, 2007 64 Angular size, magnification ho Angular size of the object: ! = (near point, closest eye can focus at) N (small angle, in radians, object viewed by unaided eye at near point, N) Angular size of image formed by the magnifying glass: θ! = ho do Angular magnification, M: !! ho N N = × = ! do ho do The magnifying glass lets the user view the object closer than the near point M= Friday, March 9, 2007 65 Magnifying Glass Angular magnification: M = !! N = ! do di = – ! Two cases: ! 1) Final image is at infinity: so do = f Then: M = N N = do f 1 1 1 = − do f −∞ " $ minimum magnification 2) Final image is at the near point, so di = – N Thin lens equation: Then, M = Friday, March 9, 2007 1 1 1 f +N = − = do f −N fN N f +N N = = 1+ do f f $ maximum usable magnification 66 Magnification Markings Lenses are sometimes marked with the magnification they produce when an image is formed at infinity. For example, “10"”. This means that, " N = 10, " M= f with N = 25 cm, the normal near point. So, f = N 25 = = 2.5 cm 10 10 Friday, March 9, 2007 67 Prob. 26.112: A stamp collector is viewing a stamp with a magnifying glass held next to her eye. Her near point is 25 cm from her eye. a) What is the refractive power of a magnifying glass that has an angular magnification of 6 when the image of the stamp is located at the near point? b) What is the angular magnification when the image of the stamp is 45 cm from the eye? Friday, March 9, 2007 68 Astronomical Telescope fo fe di1 Formation of intermediate image by the objective. hi hi (small angles) != ! di1 fo The eyepiece acts as a magnifying do2 glass to produce a magnified final image. hi hi !! = − #− do2 fe ! ! hi fo fo M= "− × =− ! fe hi fe (exact when object and final image are at infinity) Friday, March 9, 2007 69 Astronomical Telescope di1 Object at infinity: • first image at focal point of objective # di1 = fo Final image at infinity: • first image at focal point of eyepiece as well # do2 = fe # distance between lenses is: " L = fo + fe do2 and di1 fo M =− =− exactly do2 fe # Friday, March 9, 2007 70 Prob. 26.92/90: An astronomical telescope has an angular magnification of –132 and uses an objective with a refractive power of 1.5 diopters. What is the refractive power of the eyepiece? Angular magnification, M =− Refractive power, P = so, M =− 1 f fo = −132 fe Pe = −132 Po Therefore, Pe = 132 # Po = 132 # 1.5 = 198 diopters. fe = 1 = 0.00505 m = 5.1 mm Pe Friday, March 9, 2007 71 Opera Glasses Like an astronomical telescope but with an eyepiece that is a diverging (negative) lens. The distance between the lenses is still: L = fo + fe and the angular magnification is M = – fo/fe But, as fe < 0: • the length is less than an astronomical telescope of the same # magnification • the image is the right way up (M > 0) Friday, March 9, 2007 72 Compound Microscope !! fo fe Distance between lenses = L Magnification: compare angular size of final image, "%, to angular size, ", of object at near point viewed with the naked eye. With: • object just outside the focal point of the objective, so do1 $ fo • first image at focal point of eyepiece (# final image at infinity) Angular magnification, M = θ! (L − fe)N "− θ fo fe N = near point Friday, March 9, 2007 73 Compound Microscope ho1 do1 ! fo do2 ! fe di1 hi1 fo !! Final image fe ! " ! " L − fe di1 × ho1 # − × ho1 hi1 = m1ho1 = − d01 fo hi1 hi1 " do2 f " !e L − fe ! × ho1 So, ! " − fo fe !! = ho1 With the object at the near point and no microscope: ! = N " ! ! ! L − fe ×N Therefore, M = " − ! fo fe Friday, March 9, 2007 74 Prob. 26.86/98: A microscope for viewing blood cells has an objective with a focal length of 0.5 cm and an eyepiece with a focal length of 2.5 cm. The distance between the two lenses is 14 cm. If a blood cell subtends an angle of 2.1 " 10-5 rad when viewed with the naked eye at a near point of 25 cm, what angle does it subtend when viewed through the microscope? Friday, March 9, 2007 75 Summary of Chapter 26 • Snell’s Law: ## # # n1 sin "1 = n2 sin "2, # v = c/n • Apparent depth: ## # d! = d n1/n2 • Total internal reflection: #n1 sin "c = n2, "" " " " " " " " " " " " • Lens equation: # # # 1/do + 1/di = 1/f "c = critical angle for total " internal reflection • Linear magnification: # m = – di/do " Two lenses:# # # # m = m1 m2 • Angular magnification:# M = "!/" • Magnifying glass:# # # # # # # # Friday, March 9, 2007 M = N/f # # (image at infinity, N = near point) M = N/f + 1# (image at near point) 76