Optics of Vision. MATERIAL TO READ Web: 1.

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1 Optics of Vision MATERIAL TO READ Web: 1. Text: Chap. 3, pp (NB: pg missing) Chap. 5 pp.1-17 Handbook: 1. study guide 3 2. lab 3 Optics of the eye For quiz on study guide 3, you should know: 1. Refraction at a planar surface (Snell s law) 2. Object/image distance relations for refraction at single spherical surface and lenses: lens power 3. Anatomy and function of ocular components 4. Accomodation of the eye 5. Lens type required for correction of eye defects Re: optics in Lab quiz #9, you should know: 1. Items 2, 4 and 5 from list above 2. Familiarity with model eye and purposes of the various lenses used.

2 THE NATURE OF LIGHT (More in SG4) Wave particle duality Electromagnetic waves Photons (fast! earth-sun: 8.3 min.) speed c = lf = m/s of light (in vacuum) λ = nm for light visible to humans Refraction of light Ignore light s wavelike nature for interactions with objects much larger than wavelength, λ of light. Treat light as travelling in straight lines or rays. Refraction: bending of light rays as the pass from one medium to another. Caused by change in speed of light for different media. This change is a result of photonic interactions with molecules in medium.

3 (speed in vacuum) (speed in medium) v = c/n (refractive index: characteristic of material) Take note: 1. n 1 \ v c Speed of light in material is always less than speed in vacuum. 2. FREQUENCY of light, f, DOES NOT CHANGE from one medium to another. 3. Therefore, the wavelength, λ, must be medium dependent: v = l m /n l m = v/f = c/(nn) = l/n wavelength in vacuum

4 Rough Analogy: transverse wave on rope Less optically dense frequency = #jiggles/sec. Same over whole rope More optically dense Refraction incident ray Less reflected optically dense θ 1 ray refractive index : n 1 (Ignore 4%) More optically dense refractive index : n 2 θ 2 assume transparent (until SG5) transmitted ray (for now assume no absorption)

5 Snell s Law of Refraction sinθ n 1 sinq 1 = n 2 sinq 2 sinθ 2 n1 1 = n 2 n 1 if n 2 > n 1 θ 1 then (n 2 /n 1 ) > 1 n 2 so (sinθ 1 /sinθ 2 ) > 1 sinθ 1 > sinθ 2 normal θ 2 θ 1 > θ 2 Therefore as light passes from a medium of low refractive index to one of higher refractive index it bends towards the normal. The opposite occurs if n 2 < n 1 n 1 θ 1 NOW θ 2 > θ 1 θ 2 n 2 (consider case above in reverse direction)

6 Explanation of Refraction d 1 Light beam has nonzero width made of many parallel rays n 1 n 2 d 2 θ 1 Wave front must be perpendicular to direction of beam n 2 > n 1, so v 2 < v 1 θ 2 wave front During the same time t: refracted ray travels d 2 incident ray travels d 1 and d 2 < d 1 because v 2 < v 1. \ to keep wave front to motion, the rays bend Example Why do straight objects in water appear bent up? θ 1 θ 2 < θ 1 Consider: virtual 2 light rays from end of paddle image one straight at eye (misses) one in correct direction to reach eye

7 Total internal reflection Consider a light ray passing from a high index medium (e.g. glass n =1.5 ) to a low index medium (e.g. air n = 1). n 1 > n 2 θ 2 > θ 1 as θ 1 increases eventually θ 2 reaches 90 o n 1 θ 1 θ 1 θ 1 n 2 θ 2 θ 2 θ 2 = 90 o Value of θ 1 is called the critical angle in this case. θ 1 = θ c Using Snell s law: n 1 sinθ 1 = n 2 sinθ 2 we can solve for q c θ 1 = θ c, θ 2 = 90 o n 1 sinθ c = n 2 sin(90 o ) sinq c = n 2 /n 1 e.g.: n 1 = 1.33, n 2 = 1 n 1 = 1.5, n 2 = 1 (water) (air) (glass) (air) θ c = sin -1 (1/1.33) =48.8 o θ c = sin -1 (1/1.5) = 41.8 o

8 What happens if q 1 > q c?? n 1 sinθ 1 = n 2 sinθ 2 sinθ 2 = (n 1 /n 2 )sinθ 1 θ 1 > θ c sin(θ 1 ) > sin(θ c ) = n 2 /n 1 sin(θ1) > n 2 /n 1 sinθ 2 > (n 1 /n 2 )(n 2 /n 1 ) = 1 THIS HAS NO sinθ 2 > 1 SOLUTION!! e.g.: sin -1 (1.2) = ERROR MESSAGE THERE IS NO TRANSMISSION OF LIGHT INTO MEDIUM 2. ALL LIGHT IS INTERNALLY REFLECTED IN MEDIUM 1. n 1 θ 1 > θ C θ r = θ 1 Angle of Reflection, θ r n 2 θ 2 has NO SOLUTION EQUALS Angle of incidence Application: FIBER OPTICS

9 Anatomy and function of ocular components Lateral cross-section of human eye. cornea: front surface: most refraction here iris: covering of lens, controls size of pupil. pupil: opening in iris size determines how much light enters eye lens: provides additional fine focussing of light, accomodation.

10 ciliary muscles (or radial ligaments): control curvature of lens, which determines accomodation. Aqueous and Vitreous humour: Fluids (mostly water) filling eye cavities (n ~ 1.33). Retina: contains photoreceptors - special cells which detect light and convert it into nerve signals. Optic Nerve: Carries nerve signals to brain. Blind spot: Where optic nerve attaches to retina (See lab#3 page 2) fovea: A small (0.3 mm dia.) depression in retina where images are focussed for acute vision and colour vision. Pigment epithelium: A layer of pigmented cels below retina which absorb stray and scattered light.

11 From Peter Gray's Psychology (3rd ed.) 1999 Worth Publishers Notes: To focus on an object, eyes rotate so that image forms on fovea Retina is backwards. Light sensitive parts are positioned deeper (pointing away from light) than non-light sensitive parts.

12 Photoreceptors: Light triggers biochemical change of certain proteins in the retina. Convert light into electrical nerve impulse going to brain For more detail see Section 3 of Chapter 5 in text. DETAILED KNOWLEDGE OF THE BIOCHEMISTRY NOT REQUIRED FOR QUIZ There are two types of photoreceptors 1) cones: Sensitive to daylight and colour vision. Occur mainly in fovea 2) rods: sensitive to lower light intensities than cones and

13 Refraction at a single Spherical Surface AH VH Index of refraction (n) Air 1 Cornea aqueous humour 1.34 Lens 1.41 Lens vitreous humuor 1.34 Largest change in n occurs at air-cornea interface most (80%) bending of the light occurs there. Can use simplified Reduced eye where the VH, lens, AH and cornea are approximated by a single refracting surface. n = 1 n= 1.34 more on this later Clearly we will need to apply Snell s law to a curved (spherical) surface. Consider a portion of a spherical surface: All lines through C cross C Center Surface at right angle of curvature

14 Measuring the angles made by a light ray w.r.t. these normal (perpendicular) lines. θ 1 refracted ray center of θ 2 curvature incident rays focal length n 1 n 2 We can apply Snell s law: n 1 sin(θ 1 ) = n 2 sin(θ 2 ) NB: if θ 1 = 90 o θ 2 = 90 o no bending! Consider an object placed in front of surface: n 1 θ 1 n 2 (SEE TEXT pg. 3-11) h θ 2 α γ β r p q Assume all angles are small and that lens is thin: Then: tan(α) = h/p α tan(β) = h/q β tan(γ) = h/r γ

15 From the diagram : α + γ + (180 o - θ 1 ) = 180 o α + γ = θ 1 β + θ 2 + (180 o - γ) = 180 o θ 2 = γ - β Snell s law: n 1 sin(θ 1 ) = n 2 sin(θ 2 ) n 1 θ 1 = n 2 θ 2 due to small angles n 1 (α + γ) = n 2 (γ - β) Using expressions from previous page: n 1 (h/p + h/r) = n 2 (h/r h/q) n 1 /p + n 1 /r = n 2 /r n 2 /q h cancels n 1 /p + n 2 /q = (n 2 - n 1 )/r = P where: p = object distance q = image distance P = refractive POWER of surface (P constant for each surface: need n 2,n 1 r) P dimensions: distance -1 : unit = m -1 = diopter

16 Sign convention: +ve p travelling against light from surface to object +ve q travelling with light from surface to image +ve r travelling with light from surface to center of curvature +ve Magnification if object has same orientation as image M = Magnification: ratio of image height to object height M = h /h n 1 n 2 h θ 1 θ 2 h p q Assume all angles are small and that lens is thin:

17 n 1 sinθ 1 = n 2 sin(θ 2 ) n 1 tanθ 1 = n 2 tan(θ 2 ) n 1 (h/p) = n 2 (-h )/q - n 1 q/(n 2 p) = h /h = M Example: optical power of reduced eye m P = (n 2 -n 1 )/r = (1.34 1)/0.008 n = 1 =43 diopters n= 1.34 a 0.05 m object 25 cm in front of reduced eye wouldbe focussed at a distance: P = n 1 /p + n 2 /q 43 = 1/ /q q = 1/39 = 3.44 cm (avg. eyeball diameter = 2.5 cm) and have a height: h = Mh = h[-n 1 q/(n 2 p)] = -5[-1(0.034)/(1.34(.25))] = cm

18 Examples for using sign convention I O q < 0 r > 0 light p > 0 I O r < 0 q > 0 light p > 0 -ve p possible for systems with several optical surfaces

19 Real vs. virtual images Real Image n 1 n 2 REAL IMAGE h p q 1) Light rays actually pass through image points 2) Light rays converge after passing thru surface 3) Image visible on screen at image position 4) Image is downstream from surface (q > 0) 5) Image is inverted (M < 0) Virtual Image h n 1 n 2 rays must be continued rays from object backwards to intersect p do not intersect q

20 1) Light rays do not pass thru surface to image points 2) Rays diverge after passing thru surface as if they came from I. 3) Image is not visible on screen at image position but can be seen by observer looking upstream. 4) Image is upstream from surface (q < 0) 5) Image is upright (m > 0) Example Object r = 0.1 m p = 0.2 m q =? Image Find q. n 1 = 1 n 2 = 1.34 First, find power: P = (n 2 n 1 )/r = (1.34-1)/0.1 = 3.4 d then use: n 1 /p + n 2 /q = P (NB: use n 2 with n 2 /q = P n 1 /p q even though n 2 /q = 3.4 1/(.2) image appears n 2 /q = -1.6 m -1 in medium 1) Therefore, q = n 2 /(-1.6) = 1.34/(-1.6) = 0.84 m

21 Also, find magnification: M = -(n 1 q)/(n 2 p) = 0.84/( ) = 3.13 (+ve > 1, \ image larger than object) Another example of a virtual image is a reflection in a mirror. Here s a 3 rd example: Example- Object in a fishbowl (Closely related to paperweight problem: self-test II, #2) Where does the 10 cm fish appear to be? 5 cm The back surface of the bowl can be ignored. A simplified diagram results: r = [n 1 ] 10 cm [n 2 ] light What is the image p = 5 cm distance, q?

22 We will also ignore the difference between the refractive index between the water and the bowl. Therefore: n 1 = n water = 1.33 n 2 = n air = 1.00 The radius of the fishbowl is 10 cm, r = m [NB negative] 1 st calculate the power of the surface: P = (n 2 - n 1 )/r = ( )/(-0.10) = 3.3 d Next calculate the image distance: P = n 1 /p + n 2 /q 3.3 = 1.33/ /q q = m Magnification: - (n 2 q)/(n 1 p) = ( )/0.05 = 1.14 Image is larger and between object and surface if redo with p > r then 4.3 cm image further than object