Physics 102: Lecture 19 Lenses and your EYE Ciliary Muscles 1

3 Cases for Converging Lenses Object Image Inverted Reduced Real Past 2F Image Object Inverted Enlarged Real Between F & 2F Demo 71 Image Object Upright Enlarged Virtual Inside F

Only 1 Case for Diverging Lenses Example F P.A. Image Object F Image is always virtual, upright, and reduced.

Review of Lenses Preflight 18.8 Focal point determined by geometry and Snell’s Law: n1 sin(q1) = n2 sin(q2) P.A. n1<n2 F Fat in middle = Converging Thin in middle = Diverging Larger n2/n1 = more bending, shorter focal length. Smaller n2/n1 = less bending, longer focal length. n1 = n2 => No Bending, f = infinity Lens in water has ________focal length since n2/n1 is _________!

Preflight 19.1 A converging lens is used to project a real image onto a screen. A piece of black tape is then placed over the upper half of the lens.

Same as mirror equation Lens Equation Same as mirror equation do F P.A. Image Object f F do = distance object is from lens: Positive: object in front of lens Negative: object behind lens di Example di = distance image is from lens: Positive: real image (behind lens) Negative: virtual image (in front of lens) Just like for mirrors f = focal length lens: Positive: converging lens Negative: diverging lens

Lens Equation Example If d0 larger, di is smaller, |m| smaller do F P.A. Image Object f F di Example Just like for mirrors. Make d0 larger. Then di smaller, m smaller Make d0 smaller. Then di larger, m larger If d0 larger, di is smaller, |m| smaller If d0 smaller, di is larger, |m| larger

Multiple Lenses Example Image from lens 1 becomes object for lens 2 1 Lens 1 creates a real, inverted and enlarged image of the object. Lens 2 creates a real, inverted and reduced image of the image from lens 1. The combination gives a real, upright, enlarged image of the object.

Multiple Lenses: Magnification do = 15 cm 1 2 f1 f2 f1 = 10 cm f2 = 5 cm Example di = 30 cm First find image from lens 1.

Multiple Lenses: Magnification 1 2 do = 15 cm L = 42 cm di = 8.6 cm f1 f2 f1 = 10 cm f2 = 5 cm Example di = 30 cm do=12 cm Now find image from lens 2.

Multiple Lenses: Magnification 1 2 do = 15 cm L = 42 cm di = 8.6 cm f1 f2 f1 = 10 cm f2 = 5 cm Example di = 30 cm do=12 cm Net magnification: mnet = m1 m2 40

The Eye One of first organs to develop. ~100 million Receptors ~200,000 /mm2 Sensitive to single photon! Candle from 12 miles Ciliary Muscles Numbers in red are for a digital camera Macula/fovea is super sensitive part only sees about 15 degrees http://hyperphysics.phy-astr.gsu.edu/hbase/vision/retina.html#c2

ACT: Focusing and the Eye Ciliary Muscles Cornea n= 1.38 Lens n = 1.4 Vitreous n = 1.33 Which part of the eye does most of the light bending? 1) Lens 2) Cornea 3) Retina 4) Cones

Eye (Relaxed) Example 25 mm Determine the focal length of your eye when looking at an object far away. 25 mm ~ 1 inch Object is far away: Want image at retina:

Eye (Tensed) Example 250 mm 25 mm Determine the focal length of your eye when looking at an object up close (25 cm). Tensed means more curvature, more bending, shorter focal length Object is up close: Want image at retina:

Preflight 19.3 A person with normal vision (near point at 26 cm) is standing in front of a plane mirror. What is the closest distance to the mirror where the person can stand and still see himself in focus? 26cm 13cm 1) 13 cm 2) 26 cm 3) 52 cm

Near Point, Far Point Eye’s lens changes shape (changes f ) Object at any do should have image be at retina (di = approx. 25 mm) Can only change shape so much “Near Point” Closest do where image can be at retina Normally, ~25 cm (if far-sighted then further) “Far Point” Furthest do where image can be at retina Normally, infinity (if near-sighted then closer)

If you are nearsighted... Example (far point is too close) do dfar Too far for near-sighted eye to focus dfar Near-sighted eye can focus on this! Contacts form virtual image at far point – becomes object for eye. do flens = - dfar Want to have (virtual) image of distant object, do = , at the far point, di = -dfar.

Refractive Power of Lens Diopter = 1/f where f is focal length of lens in meters. Example: My prescription reads -6.5 dipoters flens = -1/6.5 = -0.154 m = -15.4 cm (a diverging lens) dfar = 15.4 cm (!) My eyes: 6.5 diopters, implies f=-0.15 m (my far point is 0.15 m, or 6”, or 15 cm) Big # means smaller f, more bending (worse eyesight) flens = - dfar

If you are farsighted... Example (near point is too far) Too close for far-sighted eye to focus dnear Far-sighted eye can focus on this! do Contacts form virtual image at near point – becomes object for eye. Example My near point with far correction: 65 cm Want near point at 25 cm Need 1/f=1/.25-1/.65 1/f=2.5 diopters Want the near point to be at do. When object is at do, lens must create an (virtual) image at -dnear.

Farsightedness Near point dnear > 25 cm To correct, produce virtual image of object at d0 = 25 cm to the near point (di = dnear) Example: My near prescription reads +2.5 dipoters flens = +1/2.5 = 0.4 m = 40 cm therefore dnear = 67 cm (with my far correction)

ACT/Preflight 19.4 Two people who wear glasses are camping. One of them is nearsighted and the other is farsighted. Which person’s glasses will be useful in starting a fire with the sun’s rays? nearsighted farsighted