 OPTICS 04 OPENING QUESTIONS: Review: What is retroreflection? What shapes are used to create situations where retroreflection occurs? How does retroreflection happen in the human eye? Why does retroreflection result in red eye? What equation did we use to calculate the speed of light in water? ═══════════════════════════ FORMULAE OBJECTUS: θi = θr Angle of incidence = angle of reflection c/v = n Index of refraction Snell's Law: n1sin Θ1= n2/sinΘ2 LEARNING TARGET: I will be able to recognize and utilize Snell's Law after today's class. WORDS O' THE DAY: medium (material through which light is traveling) Reflection ("Light waves bouncing off an object") Emission ("Light is given off by an object") Absorption ("Some wavelengths of light are absorbed by an object") Refraction ("Light bending as it changes medium") Angle of Incidence ("The angle a light ray takes as it moves toward a reflecting surface") Angle of Reflection ("The angle a light ray takes as it moves away from a reflecting object") Retroreflection ("incident light waves are 'double bounced' back to the observer, regardless of the angle of incidence") Snell's Law: WORK O' THE DAY: ═══════════════════════════ Let us now please take a look at another aspect of refraction-- The INDEX of refraction. We noted yesterday that the speed of light (c) is variable in that it depends on the *medium* through which it passes. As light passes from one medium to another, it either speeds up (when passing into a less dense medium) or slows down (when passing into a more dense medium). The obvious caveat here is that light must be able to move through BOTH media. Visible light doesn't pass from air through wood, for example so a room made entirely of wood is dark... But I digress... There is a specific measurement that indicates how much light speeds up or slows down when passing from a vacuum into that medium, we call that the INDEX of refraction. NOTE: As I noted in yesterday's work, the speed of light in air is VERY NEARLY the same as the speed of light in a vacuum. You should already know that the index of refraction is the ratio of the speed of light in a vacuum compared to the speed of light in that particular medium: c/v = n Where c is the speed of light in a vacuum, v is the speed of light in some other medium (such as glass or water) and n is the index of refraction in that medium ═══════════════════════════ Let's take a break to have some fun with numbers: What do you suppose the index of refraction of air might be? (if the index of refraction in a vacuum is 1.0000000 etc) INDEED! The index of refraction is just a small bit above that since air is not particularly dense. At Standard Temperatures and Pressures (if you've had chem you know that as STP, otherwise don't worry), the index of refraction is 1.00027, so let's just KISS and keep it at 1.00   You know how to calculate the index of refraction You also know about the relationship between the angles a light ray makes to the speed of light across two different mediums? media's? Yesterday, I asked you to work with that information to derive an equation that relates the incident angle, the refracted angle and the index of refraction of both mediums ═══════════════════════════ x x x x x x Step 0: Snell's Law as written: n1sin Θ1= n2/sinΘ2 Step 2: Rewrite Snell's Law: n1/n2= sinΘ2/sin Θ1   Step 3: The definition of index of refraction: c/v = n Step 4: This part is a bit tricky. Remember, we're dealing with an incident light ray going through one medium at one speed (we'll call that initial velocity v1) and entering another medium and either speeding up or slowing down as it does (we'll call that new velocity v2): n1/n2= sinΘ2/sin Θ1 substitute for n1 and n2 using the definition of the index of refraction and using v1 for the initial speed of light through the first medium and v2 for the final speed of light as it travels through the second medium (c/v1)/(c/v2)= sinΘ2/sin Θ1 Step 5: Cancel out the 'c' terms (c/v1)/(c/v2)= sinΘ2/sin Θ1 Step 6: Rewrite: (v2)/(v1)= sinΘ2/sin Θ1 Step 7: Recognize that we ALREADY have a Simi relationship (see step #2 above) sinΘ2/sin Θ1= n1/n2 Therefore: sinΘ2/sin Θ1= n1/n2 = v2/v1   WHOOOOOAAAAA... how totally cool is THAT!   ═══════════════════════════ Practice problems: 1) The index of refraction of isopropyl alcohol is 1.38. If a light ray in air is incident at 27 degrees at the surface of isopropyl alcohol (I'll do this one for you!). Sketch that situation [Notice we always measure both the angle of incidence (incoming light ray) and the angle of refraction (outgoing light ray) to a vertical reference line that we draw as part of our solution] Will the refracted light ray bend towards the vertical or away from it? It will refract towards the vertical centerline Why? The light ray will refract (bend) TOWARDS the center line because the index of refraction for isopropyl alcohol is greater than the index of refraction for air and the light ray will slow down (review the barrel example in yesterday's lesson plan!) Calculate the resulting angle of refraction write the easy form of Snell's Law: sinΘ2/sin Θ1= n1/n2 Isolate sinΘ2: sinΘ2= sin Θ1(n1/n2) Isolate Θ2: Θ2= sin-1[sin Θ1(n1/n2)] Substitute given values: Θ2= sin-1[sin 27o(1.00/1.38)] Solve: Θ2= sin-1[sin 27o(1.00/1.38)] Answer: 19.2o Which makes sense. The index of refraction for air (1.00) is much less than the index of refraction for isopropyl alcohol. Therefore the speed of light in alcohol is much less and the light ray bends towards the vertical as expected. 2) A flashlight is shown upwards towards the surface of water (n = 1.33) at an angle of 18 degrees to the normal line. Sketch that situation Will the refracted light ray bend towards the vertical or away from it? Why? Calculate the angle of the light ray in air? 3) A flash light is shown through air upon an unknown liquid at an incident angle of 21 degrees to the normal. The resulting angle of incidence tells you the substance is benzene with an index of refraction of 1.50. Sketch that situation Will the refracted light ray bend towards the vertical or away from it? Why? Calculate the angle of refraction in benzene   4) Consider problem #3. A light ray shown upwards from benzene into air would bend... towards the normal (perpendicular) line? Away from the normal (perpendicular) line? Be fully reflected back into the benzene? (this is kindofa trick question) 5) Calculate the speed of light in isopropyl alcohol in problem #1   BE PREPARED TO DISCUSS THESE PROBLEMS ON MONDAY