1 / 58

Optics, a fancy word for light

Optics, a fancy word for light. Law of Reflection. This states the angle of incidence equals the angle of reflection. What that means is if I bounce of a light ray at 25 ⁰ relative to the surface, it would leave at 25 ⁰ relavative to the surface. This law is written as θ r = θ i.

williamsvera
Download Presentation

Optics, a fancy word for light

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Optics, a fancy word for light

  2. Law of Reflection • This states the angle of incidence equals the angle of reflection. What that means is if I bounce of a light ray at 25⁰ relative to the surface, it would leave at 25⁰ relavative to the surface. • This law is written as θr = θi

  3. Angle of incidence Angle of reflection i r R Angle of Refraction Normal Line Incident light ray Reflected light ray Refraction of Light Rays air Boundary Refracted Light ray water

  4. Reflection of Light Waves Angle of Incidence = Angle of Reflection Law of Reflection i r

  5. Problems with Reflection • A light hits a mirror at 52⁰ to the normal. The mirror then rotates 35⁰ around the point where the beam strikes the mirror so that t he angle of incidence of the light ray decreases (smaller). The axis of rotation is perpendicular to the plane of the incident and reflected rays. What is the angle of rotation on the reflected ray?

  6. How to solve these kinds of problems: • Draw a picture of what is going on. • Add your light rays into the diagram. • Write out our knowns and unknowns. • Solve for unknowns using appropriate laws/idea’s

  7. Diagram: see board • Known: initial angle is 52⁰, change in mirror rotation, 35⁰ • Unknown: Change in reflection after rotation • Because we’re reducing the angle of incidence we rotate the mirror counter clockwise.

  8. Θi, final= Θi, intial- ∆Θmirror • 52⁰ – 35⁰ = 17⁰, which means our new angle is 17 ⁰ from the new normal (due to rotation) • Look at new diagram on board • Using the law of reflection • Θr, final= Θr, intial • 17⁰ counterclockwise from new normal (in = out) • ∆Θr= 52⁰ + 35⁰ - 17⁰ = 70⁰ clockwise from the original angle. • See final diagram of what this means as I’m sure some of you are slightly confused as the math doesn’t add up

  9. Try some of these type son page 460.

  10. Plane Mirrors • Plane mirrors: A flat smooth surface from which light is reflected in a specular reflection. • Object: The source of light or illuminated image we’re looking at in the mirror. • Diverging rays: rays(light, sight, etc) spreading out as we look at an image. • Virtual image: The image seen by looking at the mirror with diverging rays, always on the opposite side of the mirror. See picture on 461. Use your cell phone to prove my point.

  11. Continued • Some other stuff related to that business from before: • As we get closer to the mirror the closer the image appears to be, the farther away we get the farther away it appears to be. • Anyone want to tell me when you can get tricked by this and its dangerous? • Like here

  12. Height and Position • Again I apologize as I can’t find these images, so lets look at 462 together. • What do you notice? • di = -do • In a plane mirror the distance from the image is the same as original. The negative means that image is virtual. • hi = ho This means the heights of both original and image are the same

  13. Image Orientation • If you have an iphone 4 or newer pull it out and snap a self picture. • If not…what happens to your image when you look in a mirror? • Things go opposite, ie right is left and left if right, so the image is reversed.

  14. Curved Mirrors: This just got serious • A concaved mirror, is exactly as it sounds, a mirror that has a cave in it. • In it we have: • A principal axis: a straight line perpendicular to the middle of the mirror. • Centre(C) and radius(r) of the curvature (seen on next slide) • Focal point(F) and focal length(ƒ) (again on next slide.

  15. Yes from your text via camera phone • You’ll notice here that the focal length (point) and is half the radius so : ƒ = r/2

  16. So what does all this mean? • It means when you point the middle (or principal axis) of a concave mirror towards on object, say like the sun, all the rays get reflected onto a single point (focal point).

  17. Finding the image • First lets watch this little video, instead of my typing mad notes. • Real image: An inverted (because of concave) optical image that is either bigger/smaller because on location?

  18. MOAR! • Here’s an applet that might help

  19. Imagine Image • So here’s the break down: • If the original is outside C (twice the focal point, or more then the radius) you get an inverted smaller image. • If the original is inside C (less then the raduis) you get an inverted image that’s large. • If the original is inside the focal point (f or half the radius) you get a large, upright image. • For diagrams of this see page 465.

  20. And with the math… • So obviously because science rules and it married math, I know right who does that, we can find our image using math. • Mirror equation: • 1 = 1 + 1 f di do The reciprocal of the focal length is equal to the sum of the reciprocals of image and objects(original) position. *This is only approximately correct as light in real life sprays out (diverges) and the mirrors themselves can have nicks and abrasions.

  21. Magnification • How much bigger or smaller did we get, yup there's a formula for that too. • m = hi = -di ho do The magnification of an object by a spherical mirror is defined as the image height divided b the object height, is equal to the negative of the image position divided by the object position.

  22. Formula Sheet • You may wish to add the following to your formula sheet as I manipulated them for you!! • di= fdo/ do – f distance image from mirror • do= fdi / di- f distance object from mirror (original) • f = dido / do + di focal point based on image and original

  23. Concave Example • A concave mirror has a radius of 20 cm. A 2cm tall object is placed 30 cm from the mirror. What is the image position and height? • Draw a ray diagram of the info you know!

  24. Known: • Object height = 2cm • Object distance = 30cm • Radius of mirror = 20cm • Known without knowing: • Focal point: 10cm (radius / 2) • Unknown: • Image distance • Image height Again, the image is the thing we’d see, NOT the actual item.

  25. Solution • Using the formula from a few slides ago, trying to find image distance it’s… • di= fdo/ do – f • di= (10cm)(30cm)/(30-10) = 15cm, real image (in front of the mirror, so its real) • Part 2 hi = -di hi= -di x ho ho do do • (-15)(2) / (30) = (-1cm) • Mean its 1cm tall (smaller) and inverted • Lets try 469 problems 12-16

  26. Convex Mirrors • What do you get when you look at a convex mirror? (outside of a spoon). The same thing, right side up, but smaller and wider. • The image we get is a virtual image, so the distance is negative. • Forumla from before is the same for convex mirrors.

  27. Example • A convex mirror, used for secruity purposes, in a bus station has a focal point of -0.50m. A creeper who is 1.7m tall is 6 meters from the mirror. Where would the creeper be, (IE like the virtual image) and how tall would he appear.

  28. Known: • Height of the object: 1.7m • Distance from mirror for object: 6m • Focal point: -0.5m • Unknown: • Image distance • Image height

  29. Solution • di= fdo/ do – f = (-0.5)(6)/(6 – (-0.5)) • This is -0.46m, so virtual image behind the mirror • Part 2 hi = -di hi= -di x ho ho do do • -(-.46)(1.7) / (6)= 0.13m, so smaller and upright • Try 472 17-21

  30. Table 17-1 • Pretty good summary of what just happened.

  31. Snell’s Law n1sin(θ1)=n2sin(θ2) States the relationship between angle of incidence and the angle of refraction Where n = index of refraction This is a measure of the relative speed of light in a given medium. The larger the index of refraction (n) the slower the speed of light in that medium.

  32. Different mediums have different indexice

  33. i Incident medium Refracted medium R Snell’s General Law of Refraction ni x sin i = nr x sin R This equation works for all different media!

  34. Example • A light beam in air hits a sheet of crown glass at an angle of 30 degrees. What is angle of refraction? Index of light is 1.00 and index of crown glass is 1.52. • Hint draw a diagram

  35. Solution • Angle 1 = 30 degree, index 1= 1.00 • Angle 2= ?, index 1.52 • n1sin(θ1)=n2sin(θ2) moving things around I get … • θ2= sin-1(n1 x sin(θ2)) n2 • Putting all the info in I get a solution of 19.2⁰

  36. 23o water Flint glass R Example #1: Snell’s General Law of Refraction What is the angle of refraction?

  37. 0.5197 1.61 Solution: ni x sin i = nr x sin R Snell’s General Law of Refraction 1.33 x sin 23o = 1.61 x sin R 1.33 x 0.3907 = 1.61x sin R = sin R

  38. = sin R 0.3228 Solution (con’t): Snell’s General Law of Refraction Therefore: 18.8o = R (angle of refraction)

  39. Example #2 31.4o Snell’s General Law of Refraction Medium? Quartz (crystal) 55o Determine the incident medium.

  40. 1.2615 0.5201 Solution: ni x sin i = nr x sin R Snell’s General Law of Refraction ni x sin 31.4o = 1.54 x sin 55o ni x 0.5201 = 1.26149 ni =

  41. = 2.42 ni Solution (con’t): Snell’s General Law of Refraction (index of refraction of the incidence medium) Therefore: The incident medium is probably Diamond Try on page 487

  42. C n = Vm Snell’s Law This means that the index of refraction can also be calculated by ratio of speeds: Where: C = 3.00 x 108 m/s (light in vacuum) Vm = speed of light in a medium

  43. C n = Vm 3.00 x 108 m/s 1.87 x 108 m/s Snell’s Law Light travels at 1.87 x108 m/s a new plastic medium. What is the index of refraction for this new plastic? Solution: Example 1: n = n = 1.60

  44. Critical Angle and Total Internal Reflection Snell’s General Law of Refraction Remember the refraction of light lab, when you made the light ray go from water back into air, the refraction stopped at about 49o and you had the light ray bounce off the front of the semi-circular dish. This is called total internal reflection. The angle at which this occurs is called the critical angle.

  45. Total internal Reflection is very important in the design of binoculars and cameras. Snell’s General Law of Refraction Mirrors are only about 80 –90% efficient at returning light energy. This means that 10-20% of the light is lost as heat energy in heating up the mirror. Total Internal Reflection is above 99% efficient in doing the same thing!

  46. Snell’s General Law of Refraction Only occurs when the light ray travels from a higher index of refraction medium to a lower index of refraction medium. For Example: Water into Air or Diamond into glass

  47. Note: At the Critical Angle the angle of refraction is always 90o! Snell’s General Law of Refraction This makes calculating the critical angle very easy! ni x sin i = nr x sin R But R = 90o at the critical Angle

  48. nr sin ic = nc And Sin 90o = 1.00!! Snell’s General Law of Refraction This means that Snell’s Law becomes: nc x sin ic = nr x 1.00 or

  49. air water nr sin ic = nc What is the critical angle for a light ray going from water into air? Snell’s General Law of Refraction

  50. 1.0003 sin ic = 1.33 Snell’s General Law of Refraction sin ic = 0.7521 ic = 49o

More Related