OPTICS 04 

OPENING QUESTIONS: Review:
═══════════════════════════ FORMULAE OBJECTUS:
LEARNING TARGET: I will be able to recognize and utilize Snell's Law after today's class. WORDS O' THE DAY:
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
Step 0: Snell's Law as written: n_{1}sin Θ_{1}= n_{2}/sinΘ_{2} Step 2: Rewrite Snell's Law: n_{1}/n_{2}= sinΘ_{2}/sin Θ_{1}
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 v_{1}) and entering another medium and either speeding up or slowing down as it does (we'll call that new velocity v_{2}): n_{1}/n_{2}= sinΘ_{2}/sin Θ_{1} 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/v_{1})/(c/v_{2})= sinΘ_{2}/sin Θ_{1} Step 5: Cancel out the 'c' terms ( Step 6: Rewrite: (v_{2})/(v_{1})= sinΘ_{2}/sin Θ_{1} Step 7: Recognize that we ALREADY have a Simi relationship (see step #2 above) sinΘ_{2}/sin Θ_{1}= n_{1}/n_{2} Therefore: sinΘ_{2}/sin Θ_{1}= n_{1}/n_{2} = v_{2}/v_{1}
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!).
sinΘ_{2}/sin Θ_{1}= n_{1}/n_{2}
2) A flashlight is shown upwards towards the surface of water (n = 1.33) at an angle of 18 degrees to the normal line.
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.
4) Consider problem #3. A light ray shown upwards from benzene into air would bend...
5)

