Announcements  Today’s material WILL NOT be covered on Exam 3! Material for Exam 3: Chapters 8, 9, 10  Reminder: Special Homework #9 is due tomorrow. (download and print if you need it)  Friday of this week is the last day to withdraw.  Exam 3 is one week from tomorrow. I will need to know by Wednesday of any students who have special needs different than for exam 2.

Today’s agenda: Introduction to Light. You must develop a general understanding of what light is and how it behaves. Reflection and Refraction (Snell’s Law). You must be able to determine the path of light rays using the laws of reflection and refraction. Total Internal Reflection and Fiber Optics. You must be able to determine the conditions under which total internal reflection occurs, and apply total internal reflection to fiber optic and similar materials. Dispersion. You must understand that the index of refraction of a material is wavelength-dependent.

Light How many physicists does it take to change a light bulb? Normally, “light” refers to the type of electromagnetic wave that stimulates the retina of our eyes.light Light acts like a wave except when it acts like particles.

Light—Waves or Particles?* (quantum007.jpg) *Light is both! Take Physics 107 for enlightenment!

Light is a type of electromagnetic wave and travels with the speed c = x10 8 m/s in a vacuum. (Just use 3x10 8 !) The Speed of Light How many physicists does it take to change a light bulb? Eleven. One to do it and ten to co-author the paper.

Visible light is a small part of the electromagnetic spectrum.

News Flash! Light slows down when it passes from a vacuum into matter! Details later!

Although light is actually an electromagnetic wave, it generally travels in straight lines (like particles do!). We can describe many properties of light by assuming that it travels in straight-line paths in the form of rays. A ray is a straight line along which light is propagated. In other contexts, the definition of ray might be extended to include bent or curved lines. Geometric Optics (nz341.jpg)

A light ray is an infinitely thin beam of light. Of course, there really isn’t such a thing, but the concept helps us visualize properties of light. (nz342.jpg) thankfully, there really isn’t such a thing

Light rays from some external source strike an object and reflect off it in all directions. We only see those light rays that reflect in the direction of our eyes. If you can see something, it must be a source of light! Zillions* of rays are simultaneously reflected in all directions from any point of an object. Later, when we study mirrors and lenses, we won’t try do draw them all! Just enough representative ones to understand what the light is doing. *one zillion = 10 a big number

Today’s agenda: Introduction to Light. You must develop a general understanding of what light is and how it behaves. Reflection and Refraction (Snell’s Law). You must be able to determine the path of light rays using the laws of reflection and refraction. Total Internal Reflection and Fiber Optics. You must be able to determine the conditions under which total internal reflection occurs, and apply total internal reflection to fiber optic and similar materials. Dispersion. You must understand that the index of refraction of a material is wavelength-dependent.

Light striking a surface may be reflected, transmitted, or absorbed. Reflected light leaves the surface at the same angle it was incident on the surface: ii rr Reflection Real Important Note: the angles are measured relative to the surface normal. Possible homework hint: the sum of the angles in a 4-sided polygon is 360 .

Reflection from a smooth surface is specular (mirror- like). Reflection from a rough surface is diffuse (not mirror-like).

Light travels in a straight line except when it is reflected or when it moves from one medium to another. Refraction—the “bending,” or change of direction of light rays when light moves from one medium to a different one—takes place because light travels with different speeds in different media. Refraction

The speed of light in a vacuum is c = 3x10 8 m/s. The index of refraction of a material is defined by where c is the speed of light in a vacuum and v is the speed of light in the material. The speed and wavelength of light change when it passes from one medium to another, but not the frequency, so If you study light in advanced classes, you’ll find it is more complex than this.

Because light never travels faster than c, n  1.* For water, n = 1.33 and for glass, n  1.5. Indices of refraction for several materials are listed in your text. Example: calculate the speed of light in diamond (n = 2.42). *Actually, not true but don’t worry about it unless you take advanced courses in optics.

Because n  1, we see this: If n < 1, we would see this: Thanks to Dr. Xiaodong Yang for the images.

When light moves from one medium into another, some is reflected at the boundary, and some is transmitted. “Toy”“Toy” The transmitted light is refracted (“bent”).  a is the angle of incidence, and  b is the angle of refraction. Snell’s Law air (n a ) water (n b ) aa bb n b >n a incident ray refracted ray air (n b ) water (n a ) bb aa n a >n b incident ray refracted ray

air (n a ) water (n b ) aa bb n b >n a incident ray refracted ray Light passing from air (n  1) into water (n  1.33). Light “bends” towards the normal to the surface as it slows down in water.

air (n b ) water (n a ) bb aa n a >n b incident ray refracted ray Light passing from water (n  1.33) into air (n  1). Light “bends” away from the normal to the surface as it speeds up in air.

Snell’s law, also called the law of refraction, gives the relationship between angles and indices of refraction: aa bb water (n b ) air (n a ) aa bb water (n b ) air (n a )  is the angle the ray makes with the normal! You are free to choose which is “a” and which is “b.”

Example: a 45  -45  -90  glass (n=1.50) prism is surrounded by water (n=1.33). Light is incident at a 23  angle, as shown in the diagram. What angle does the light make when it exits the prism? 45  ii n g =1.50 n w =1.33

Example: a 45  -45  -90  glass (n=1.50) prism is surrounded by water (n=1.33). Light is incident at a 23  angle, as shown in the diagram. What angle does the light make when it exits the prism? 45  ii n g =1.50 n w =1.33 gg keep an extra digit to reduce roundoff error

Example: a 45  -45  -90  glass (n=1.50) prism is surrounded by water (n=1.33). Light is incident at a 23  angle, as shown in the diagram. What angle does the light make when it exits the prism? 45  n w =1.33 Trig… 23   90   =   180   - 45  =   90   =  There are a variety of ways to get this.

Example: a 45  -45  -90  glass (n=1.50) prism is surrounded by water (n=1.33). Light is incident at a 23  angle, as shown in the diagram. What angle does the light make when it exits the prism? 45  n w =  

Example: a 45  -45  -90  glass (n=1.50) prism is surrounded by water (n=1.33). Light is incident at a 23  angle, as shown in the diagram. What angle does the light make when it exits the prism? 45  n w =1.33  

Interactive applet: Fun with Snell’s Law.Fun Same applet as slide 20, try glass into water.

Suppose n 2 <n 1. The largest possible value of sin(  2 ) is 1 (when  2 = 90). The largest possible value of sin(  1 ) is Example: n 1 =1.5 and n 2 =1 then sin  1 =1/1.5 sin  2 =0.667 sin  2. If you set  1 =45 , then n 1 sin  1 =n 2 sin  2 has no solution. A ray coming in at a 45  angle cannot pass through the interface and be refracted. Total Internal Reflection; Fiber Optics For  1 larger than this, Snell’s Law cannot be satisfied! This value of  is called the critical angle,  C. For any angle of incidence larger than  C, all of the light incident at an interface is reflected, and none is transmitted.

 1 <  C  1 close to  C  1 >  C 11 11 11

n2n2 n 1 >n 2 Ray incident normal to surface is not “bent.”  Ray incident normal to surface is not “bent.” Some is reflected,Ray incident normal to surface is not “bent.” Some is reflected, some is transmitted.

n2n2 n 1 >n 2  Increasing angle of incidence…

n2n2 n 1 >n 2  Increasing angle of incidence…more…

n2n2 n 1 >n 2  Increasing angle of incidence…more…critical angle reached…Increasing angle of incidence…more…critical angle reached… some of incident energy is reflected, some is “transmitted along the boundary layer.

n2n2 n 1 >n 2  Light incident at any angle beyond  C is totally internally reflected.

application: fiber optics ea/report.html

ii ff n i =1 (air) n f >1 Light is incident at an angle  i on a transparent fiber. The light refracts at an angle  f. Example: determine the incident angle  i for which light strikes the inner surface of a fiber optic cable at the critical angle.

ii ff 90  -  f n i =1 (air) n f >1 Light strikes the fiber wall an an angle of 90-  f normal to the surface. At the critical angle, instead of exiting the fiber, the refracted light travels along the fiber-air boundary. In this case, 90  -  f is the critical angle. 90  Solve the above for  f and use to solve for  i.

ii ff 90  -  f n i =1 (air) n f >1 Numerical example: what  i will result in the critical angle if n f =1.4? 90  Begin the analysis at the fiber-into-air interface: keep an extra digit to reduce roundoff error

ii ff 90  -  f n i =1 (air) n f >1 Numerical example: what  i will result in the critical angle if n f =1.4? 90  Next consider the air-into-fiber interface. This is a very large angle of incidence! If you want the incident light to be nearly parallel to the fiber axis, you must surround the fiber with a coating with n outside <n coating <n fiber.

application: swimming underwaterswimming underwater If you are looking up from underwater, if your angle of sight (relative to the normal to the surface) is too large, you see an underwater reflection instead of what’s above the water.

application: perfect mirrorsperfect mirrors (used in binoculars) application: diamondsdiamonds

Today’s agenda: Introduction to Light. You must develop a general understanding of what light is and how it behaves. Reflection and Refraction (Snell’s Law). You must be able to determine the path of light rays using the laws of reflection and refraction. Total Internal Reflection and Fiber Optics. You must be able to determine the conditions under which total internal reflection occurs, and apply total internal reflection to fiber optic and similar materials. Dispersion. You must understand that the index of refraction of a material is wavelength-dependent.

Picture from the Exploratorium (

Dispersion We’ve treated the index of refraction of a material as if it had a single value for all wavelengths of light. In fact, the speed of light in a material substance depends on the wavelength, so the index of refraction is generally wavelength- (or color-) dependent.