Unit 12 Part 2 Reflection and Refraction Notes
Unit 11, Part 2—Reflection and Refraction of Light
Light travels in straight lines at super fast speeds. Remember that the speed of light (and all other em waves) in a vacuum is 3 x 108m/s. Also remember that the speed of any wave depends only on the medium the wave is traveling through. The em (electromagnetic) waves will slow down (slightly) when they travel through any medium that is more dense than a vacuum (like glass or water).
The Law of Reflection
The law of reflection is a simple law that describes how light waves (or other waves) reflect from reflective surfaces. It states that the angle of incidence is equal to the angle of reflection as measured from a “normal” that is drawn perpendicular to the reflecting surface at the point of incidence (the point where the wave hits the surface).
The normal is the dashed (imaginary) line that is perpendicular to the reflecting surface. The angle of incidence and the angle of reflection are the angles each of the rays make with the normal (the dashed line that is perpendicular to the reflecting surface).
Øi = angle of incidence. This is the angle the incident light ray (the original light ray heading into the reflecting surface) makes with the normal. Make sure you know that the angle is NOT the angle the ray makes with the reflecting surface itself.
Ør = angle of reflection. This is the angle the reflected light ray makes with the normal after it is reflected off the reflecting surface. Make sure that you know the angle is NOT the angle the ray makes with the reflecting surface itself.
The Law of Reflection states that
Øi = Ør
Therefore, if the incident ray makes a 300 angle with the normal, the reflected ray will also make a 300 angle with the normal.
Smooth surfaces (like a mirror or a piece of glass) will produce a “regular/smooth reflection”. In “regular/smooth reflection”, if a bunch of incident light rays hit a smooth surface parallel to each other, all the reflected rays will emerge parallel to each other, too. Each individual ray follows the law of reflection.
Rough surfaces (like the surface of a piece of paper), on the other hand, will produce a “diffuse/ irregular reflection” in which all the reflected rays from a bunch of incident, parallel light rays are scattered in many different random directions after they reflect. If incident rays from a beam of light are all parallel to each other as they hit the reflecting surface, they will each individually follow the law of reflection no matter what the surface looks like. Therefore, each individual reflected ray will emerge from the surface at the same angle that each individual incident ray initially hit the surface (as taken from the normal at the point the light ray hits). However, rough surfaces look like this on a microscopic scale:
When all the parallel incident rays strike the surface, they will all have different “normals” as the surface isn’t a perfectly flat/smooth reflecting surface. Remember that all the “normals” are drawn perpendicular to the reflecting surface at the point where the incident ray hits it. And, since the reflecting surface isn’t perfectly flat, the normals will all be at different angles to each other. Since all the incident rays will have different normals, they will all have different angles of incidence and different angles of reflection than each other. The diffuse/irregular reflection from a rough surface will look something like this.
Refraction
When light hits a boundary, some light will be reflected (see above, “Law of Reflection”), but some light will be transmitted into the next medium. Review: When waves (such as light waves) enter a different medium, their speeds will change as it’s the medium that determines the speed that waves travel. Light waves slow down in materials that are more “optically dense” than other mediums, but they speed up in less “optically dense” mediums. The term “optically dense” refers to how dense the materials are to light waves in particular. Ex: Air isn’t very optically dense to light, but diamond is. Scientists long ago assigned practically every transparent/translucent substance that light can move through a particular number (called the index of refraction, more on this later…) to indicate how “optically dense” the particular material is (there’s a chart listing some of them in your book). You will always be provided the optical density/index of refraction information or a chart that you can find it from.
New, super important information: When light enters a new medium and its speed changes, its direction changes, too. The bending of light waves as they cross the boundary between two different mediums is called refraction and is simply due (in a nutshell) to the speed of the light waves changing.
- As light rays go from a less optically dense to a more optically dense medium, the waves will slow down and the refracted (transmitted) rays will bend toward the normal.
refraction incidence as light goes from less to more optically dense
where n1 = air (for example)
n2 = water (for example)
- As a light ray moves from a more optically dense medium (for example, water) to a less optically dense medium (ex, air), the transmitted/refracted light ray will speed-up and bend away from the normal.
refraction incidence as light moves from a more optically dense medium to a less optically dense medium
Snell’s Law
Snell’s Law is an equation that enables people (like you) to determine just how much a light ray will bend as it enters a new medium. It states that a ray of light bends in such a way that the ratio of the “sine of the angle of incidence” to the “sine of the angle of refraction” is a constant value. The definition itself is a little confusing, but when you use the equation it gets pretty simple. The equation that we’ll use in class to determine just how much a light ray bends as it enters a new medium is:
ni (sin i ) = nrefr (sin refr)
where:
ni = the “index of refraction” for the incident medium (the medium the ray comes from)
nrefr = the “index of refraction” for the medium the ray will travel into.
The index of refraction in a vacuum is 1.00000 (exactly 1). This represents the smallest index of refraction that exists as light doesn’t travel faster in any other medium. All other index of refractions have values greater than 1.000 and are said to be “more optically dense” than a vacuum. The “index of refraction” is just a number that references how fast waves move in a particular medium. The larger the index of refraction, the slower light will travel in the material. Again, you will never have to memorize particular index of refractions, they will always be given to you (except for the index of refraction in a vacuum…you need to KNOW this!!!). Notice there are no units associated with the index of refraction.
In addition, all the frequencies of light will travel at different speeds from each other in transparent mediums. Red light waves will travel at a different speed in glass, for example, than blue light. Therefore, each of the different frequencies of light will refract (bend) a different amount when they enter a new, transparent medium. This is why prisms can separate light into its individual colors/frequencies. It’s kind of cool if you stop and think about it…
If you ever want to calculate the index of refraction for a particular medium, all you need is the “speed of light constant, c” ( 3 x 108 m/s) and the speed that light travels in the particular medium (v). Then plug everything into the following equation:
nsubstance = c where “n” is the index of refraction
v of light in that substancefor the particular medium
You can also use the above equation to calculate the speed of light in a particular substance if you know the index of refraction (n) for the particular substance.
In order to do the following problems, you will need a table for the indices of refraction for various transparent substances. In the following example problems I’ll provide it for you.
- Light in air is incident upon a piece of crown glass at an angle of 450. What is the angle of refraction in the glass?
nair = 1.0003 (don’t round this number to “1”)
i = 450
refr = ?
ncrown glass = 1.52 (This can be found in the “indices of refraction” table that you will be provided in class, found on the Unit 5 Practice Problems sheets which are also on the website)
ni (sin i ) = nrefr (sin refr)air
The light will bend toward the normal as it’s entering
a more optically dense medium from which it came.
to verify that this is true, do the math and make sure your answercrown glass
is smaller than 450.
sin refr = ni (sin i)
nrefr
When you do the above math, you’re looking for the angle, NOT the “sin of the angle”. So, to do the math: Take the sin of the angle of incidence, multiply it by the index of refraction of the incident medium, and divide the whole thing by the index of refraction for the refracted medium. The number in your calculator is NOT your angle, it’s the “sin” of your angle. To get the actual angle, hit the “inverse sin” button on your calculator. This is just like we did last semester. If you get confused, ask me for help in class.
Plug in your numbers, do the math and you come up with 27.70. This is the angle the refracted ray makes with the normal, and it is definitely smaller than 450. So, our answer “checks”.
- A ray of light passes from air into water at an angle of 300. Find the angle of refraction
ni = 1.0003
nrefr = 1.33ni (sin i ) = nrefr (sin refr)
i = 300
refr = ?
- A ray of light in air is incident upon a diamond at 450. (a) What is the angle of refraction? (b) Does glass or diamond bend light more (compare your answer to question #1)?
ni = 1.0003
nrefr = 2.42
i = 45o
refr = ?ni (sin i ) = nrefr (sin refr)
Notice that diamonds bend light a LOT. This is because of their huge index of refraction….this is also the reason they are so expensive. Diamonds have the ability to bend light so much that they can separate the light into its individual frequencies and you see a brilliant sparkle of a lot of colors.
- A block of unknown material is submerged in water. Light in the water is incident on the block at an angle of 310. The angle of refraction in the block is 270. What is the index of refraction of the unknown material?
i = 310
refr = 270
ni = 1.33 (the ray is in water)
nrefr = ?ni (sin i ) = nrefr (sin refr)
- What is the speed of light in ethanol. quartz, and flint glass? (Use you index of refraction table for this…)
Vethanol = ?
Vquartz = ?
Vflint glass = ?
Look up the index of refraction’s for all the above substances in the index of refraction chart you will be given in class (one is provided for you on the Unit 5 Practice Problems sheets which can be accessed on the website). In order to calculate the speed that light travels in each of the above substances (Vsubstance) , you need the index of refraction for the particular medium (nsubstance) and the speed of light in a vacuum (c ). You can then use the following formula:
nsubstance = c
v of light in that substance
- The speed of light in plastic is 2 x 108 m/s. What is the index of refraction of plastic?
Vplastic = 2 x 108 m/s
C = 3 x 108 m/s
nplastic = ?nsubstance = c
v of light in that substance
Total Internal Reflection (why fiber optics is such a great thing)
We now have the ability to transmit thousands of signals (information) simultaneously through a super thin cable over very long distances without any appreciable loss of energy (and information) due to something called “total internal reflection”. A few reminders on waves is needed first: (1) Remember that all waves transmit energy from one place to another and (2) em waves also have the ability to transmit information encoded within their energy/frequency.
The whole reason why we can use fiber optic cables to transmit information without any appreciable loss of energy or interference from other waves is due to the “index of refraction” of the material the cable is made of and the “index of refraction” of the material surrounding the fiber optic cable.
Remember that when a ray passes from a more optically dense medium (higher “n”) to a less optically dense medium (lower “n”), the ray bends away from the normal (so refr > i ). As the angle of incidence increases, the angle of refraction will increase, too (see fig. #1 below and look back over this information in the notes on the previous pages if you need a refresher). Eventually, as the angle of incidence increases, there will be a point that the angle of refraction will make a 900 angle with the normal and will lie directly along the surface of the boundary between the two mediums (see fig. #2). The angle of incidence that will produce an angle of refraction of 900 is called the “critical angle”. Any angle of incidence greater than the “critical angle” will not be able to produce a refracted ray and the ray won’t be able to cross the boundary into the next medium from the incident medium. Instead, the ray will “reflect” and follow the law of reflection (see fig. #3). Remember that the law of reflection states that the angle of incidence is equal to the angle of reflection.
Fig#1Fig#2
i = critical
(critical
angle)
WaterWater
(high n) i
refr = 900
Air refrAir
(low n)
(i > critical)
Fig #3 reflection
Water
Air
Any ray that reaches the boundary (this only works for em waves moving from high n to low n) between two media with an angle of incidence larger than the critical angle (critical) won’t be able to cross the boundary and enter the new media. It will instead be reflected according to the Law of Reflection. When this happens, the angle of incidence will equal the angle of reflection. This is the whole reason fiber optics works. The material the cable is made of has and index of refraction that is larger than the material coating the cable itself. A light ray (with encoded information in it) is then sent through the cable. If the original/incident ray enters the cable at any angle that is larger than the critical angle (scientists have the ability to determine that information), the ray will be reflected when it hits the side of the cable wall. It won’t be able to escape the internal confines of the cable the whole way down the length of the cable. It will look something like this:
In order to calculate the “critical angle” (the angle that will produce an angle of refraction of 900) between two different substances, you can use the same equation you used earlier:
ni (sin i ) = nrefr (sin refr)
where i = “the critical angle” (critical)
ni = the index of refraction for the incident media
refr = 900
nrefr = index of refraction for the “media across the boundary”
IMPORTANT: Remember in the above equation, the critical angle (critical) is the angle of incidence that will produce an angle of refraction of 900. Any angle greater than the critical angle will not allow the ray to escape, it will reflect instead.
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Unit 11 Notes, Part 2: Reflection and Refraction