Chapter 4: Basic Physics SEOptics
PART II: Reflection, Refraction, Mirrors and Lenses
The Big Idea
Fermat’s Principle states that light will always take the path of least amount of time (not distance). This principle governs the paths light will take and explains the familiar phenomena of reflection, refraction, lenses and diffraction. Light rarely travels in a straightline path. When photons interact with electrons in matter the time it takes for this interaction determines the path. For example, higher frequency blue light is refracted more than red because blue wavelengths interacts more frequently with electrons than red wavelengths and the path of least time is for blue to bend more then red in order to get out of this ‘slow’ area faster. The rainbows we see are a result of this. Fermat’s Principle explains the many fascinating phenomena of light from rainbows to sunsets to the haloes around the moon.
Key Concepts
 Fermat’s Principle makes the angle of incident light equal to the angle of reflected light. This is the law of reflection.
 When light travels from one type of material (like air) into another (like glass), the speed slows down due to interactions between photons and electrons. If the ray enters the material at an angle Fermat’s Principle dictates that the light also changes the direction of its motion. This is called refraction. See the figure shown to the right, which demonstrates the refraction a light ray experiences as it passes from air into a rectangular piece of glass and out again. Because light travels at slower than usual speed in transparent materials (due to constantly being absorbed and reemitted), the light ray bends in order to get out of this material quicker and satisfy Fermat’s Principle. Note that this means that light doesn’t always travel in a straight line.
 When light is refracted its wavelength and speed change; however, its frequency remains the same as the frequency of the light source. Proof of this is that light retains its original color under water. For example, blue light is still blue under water.
Key Applications
 Total internal reflection occurs when light goes from a slow (high index of refraction) medium to a fast (low index of refraction) medium. With total internal reflection, light refracts so much it actually refracts back into the first medium. This is how fiber optic cables work: no light leaves the wire.
 Lenses, made from curved pieces of glass, focus or defocus light as it passes through. Lenses that focus light are called converging lenses, and these are the ones used to make telescopes and cameras. Lenses that defocus light are called diverging lenses.
 Lenses can be used to make visual representations, called images.
 Mirrors are made from highly reflective metal that is applied to a curved or flat piece of glass. Converging mirrors can be used to focus light – headlights, telescopes, satellite TV receivers, and solar cookers all rely on this principle. Like lenses, mirrors can create images.
 The focal length, \begin{align*}f\end{align*}
f , of a lens or mirror is the distance from the surface of the lens or mirror to the place where the light is focused. This is called the focal point or focus. For diverging lenses or mirrors, the focal length is negative.  When light rays converge in front of a mirror or behind a lens, a real image is formed. Real images are useful in that you can place photographic film at the physical location of the real image, expose the film to the light, and make a twodimensional representation of the world, a photograph.
 When light rays diverge in front of a mirror or behind a lens, a virtual image is formed. A virtual image is a trick, like the person you see “behind” a mirror’s surface when you brush your teeth. Since virtual images aren’t actually “anywhere,” you can’t place photographic film anywhere to capture them.
 Real images are upsidedown, or inverted. You can make a real image of an object by putting it farther from a mirror or lens than the focal length. Virtual images are typically rightsideup. You can make virtual images by moving the mirror or lens closer to the object than the focal length.
 When using the lens makers equation, remember that real things get positive numbers and virtual things get negative numbers. Thus, diverging lenses and virtual images get negative numbers. The object distance is always positive.
Key Equations

\begin{align*}c = \lambda f\end{align*}
c=λf  The product of the wavelength \begin{align*}\lambda\end{align*}λ of the light (in meters) and the frequency \begin{align*}f\end{align*}f of the light (in Hz, or \begin{align*}\frac{1}{sec}\end{align*}1sec ) is always equal to a constant, namely the speed of light \begin{align*}c = 300,000,000 \ m/s\end{align*}c=300,000,000 m/s . 
\begin{align*}n = \frac{c}{v}\end{align*}
n=cv  The index of refraction, \begin{align*}n\end{align*}n , is the ratio of the speed \begin{align*}c\end{align*}c it travels in a vacuum to the slower speed it travels in a material. \begin{align*}n\end{align*}n can depend slightly on wavelength. 
\begin{align*}n_i \sin(\theta_i) = n_r \sin(\theta_r)\end{align*}
nisin(θi)=nrsin(θr)
 \begin{align*}\frac{1}{f}=\frac{1}{d_0}+\frac{1}{d_i}\end{align*}  For lenses, the distance from the center of the lens to the focus is \begin{align*}f\end{align*}. Focal lengths are positive for converging lens and negative for diverging lens. The distance from the center of the lens to the object in question is \begin{align*}d_0\end{align*}, where distances to the left of the lens are positive in sign. The distance from the center of the lens to the image is \begin{align*}d_i\end{align*}. This number is positive for real images (formed to the right of the lens), and negative for virtual images (formed to the left of the lens).
For mirrors, the same equation holds! However, the object and image distances are both positive for real images formed to the left of the mirror. For virtual images formed to the right of the mirror, the image distance is negative.
Light Problem Set
 What’s the difference between diffuse reflection and specular reflection? The size of the irregularities on the surface of a material will affect how the light reflects. Do a little research online and diagram light reflecting from a) a rough asphalt road and b) the same road made smooth by filling in the bumps with water. Explain why this makes it difficult to drive at night in the rain.
 If you’re standing on the side of a lake spearfishing, do you aim right at the fish you’re trying to hit, above it, or below it? Explain and include a diagram. If you’re hitting the fish with a laser instead where would you aim?
 Consider the following table, which states the indices of refraction for a number of materials.
Material  \begin{align*}n\end{align*} 

vacuum  1.00000 
air  1.00029 
water  1.33 
typical glass  1.52 
cooking oil  1.53 
heavy flint glass  1.65 
sapphire  1.77 
diamond  2.42 

 For which of these materials is the speed of light slowest?
 Which two materials have the most similar indices of refraction?
 What is the speed of light in cooking oil?
 A certain light wave has a frequency of \begin{align*}4.29 \times 10^{14} \ Hz\end{align*}. What is the wavelength of this wave in empty space? In water?
 A light ray bounces off a fish in your aquarium. It travels through the water, into the glass side of the aquarium, and then into air. Draw a sketch of the situation, being careful to indicate how the light will change directions when it refracts at each interface. Include a brief discussion of why this occurs.
 In the “disappearing test tube” demo, a test tube filled with vegetable oil vanishes when placed in a beaker full of the same oil. How is this possible? Would a diamond tube filled with water and placed in water have the same effect?
 Imagine a thread of diamond wire immersed in water. Can such an object demonstrate total internal reflection? Draw a picture along with your calculations.
 Explain with a diagram how an optical fiber works. Do a little online research – what are optical fibers used for?
 Explain (include a diagram) how dispersion and total internal reflection combine to create the sparkly colors seen in a diamond. What’s the importance of the high index of refraction of diamond to this effect?
 The figure above is immersed in water. Draw the light ray as it enters the Gas medium through its exit back into the water medium at the bottom. Be very careful drawing your light rays, making sure that it is clear when (and by how much) the angle is increasing, when decreasing and when staying the same. You may want to calculate the refracted angles to check your light ray drawings.
 Use your ruler to draw the path of the ray as it goes through the following stack of materials. At each boundary draw the normal line.
 Who can see whom in the plane mirror below? Draw the lines with a ruler to prove your results.
 Does the size of your image in a mirror as seen by you depend upon your distance from the mirror? Put another way, do you need a bigger mirror to see your entire face if you’re close to the mirror, but smaller if you’re far away? Try this after you shower: rub out a section of your bathroom mirror just large enough to contain your face. Then move closer, then farther away. What happens? Discuss briefly.
 Marjan is looking at herself in a mirror (a normal flat mirror). Assume that her eyes are 10 cm below the top of her head, and that she stands 180 cm tall. Calculate the minimum length flat mirror that Marjan would need to see her body from eye level all the way down to her feet. Sketch at least 3 ray traces from her eyes showing the topmost, bottommost, and middle rays. In the following six problems, you will do a careful ray tracing with a ruler (including the extrapolation of rays for virtual images). It is best if you can use different colors for the three different ray tracings. When sketching diverging rays, you should use dotted lines for the extrapolated lines behind a mirror or in front of a lens in order to produce the virtual image. When comparing measured distances and heights to calculated distances and heights, values within 10% are considered “good.” Use the following cheat sheet as your guide. CONVERGING (CONCAVE) MIRRORS Ray #1: Leaves tip of candle, travels parallel to optic axis, reflects back through focus. Ray #2: Leaves tip, travels through focus, reflects back parallel to optic axis. Ray #3: Leaves tip, reflects off center of mirror with an angle of reflection equal to the angle of incidence. DIVERGING (CONVEX) MIRRORS Ray #1: Leaves tip, travels parallel to optic axis, reflects OUTWARD by lining up with focus on the OPPOSITE side as the candle. Ray #2: Leaves tip, heads toward the focus on the OPPOSITE side, and emerges parallel to the optic axis. Ray #3: Leaves tip, heads straight for the mirror center, and reflects at an equal angle. CONVERGING (CONVEX) LENSES Ray #1: Leaves tip, travels parallel to optic axis, refracts and travels to through to focus. Ray #2: Leaves tip, travels through focus on same side, travels through lens, and exits lens parallel to optic axis on opposite side. Ray #3: Leaves tip, passes straight through center of lens and exits without bending. DIVERGING (CONCAVE) LENSES Ray #1: Leaves tip, travels parallel to optic axis, refracts OUTWARD by lining up with focus on the SAME side as the candle. Ray #2: Leaves tip, heads toward the focus on the OPPOSITE side, and emerges parallel from the lens. Ray #3: Leaves tip, passes straight through the center of lens and exits without bending.
 Consider a concave mirror with a focal length equal to two units, as shown below.
 Carefully trace three rays coming off the top of the object in order to form the image.
 Measure \begin{align*}d_o\end{align*} and \begin{align*}d_i\end{align*} and record their values below
 Use the mirror/lens equation to calculate \begin{align*}d_i\end{align*}.
 Find the percent difference between your measured \begin{align*}d_i\end{align*} and your calculated \begin{align*}d_i\end{align*}.
 Measure the magnification \begin{align*}M\end{align*} in your ray tracing above.
 Consider a concave mirror with unknown focal length that produces a virtual image six units behind the mirror.
 Calculate the focal length of the mirror and draw an \begin{align*}\times\end{align*} at the position of the focus.
 Carefully trace three rays coming off the top of the object and show how they converge to form the image.
 Does your image appear bigger or smaller than the object?
 Consider a concave mirror with a focal length equal to six units.
 Carefully trace three rays coming off the top of the object and form the image.
 Measure \begin{align*}d_o\end{align*} and \begin{align*}d_i\end{align*}.
 Is it an virtual image or a real one?
 Use the mirror/lens equation to calculate \begin{align*}d_i\end{align*}.
 Find the percent difference between your measured \begin{align*}d_i\end{align*} and your calculated \begin{align*}d_i\end{align*}.
 Measure the magnification \begin{align*}M\end{align*} from your drawing above.
 Consider a converging lens with a focal length equal to three units.
 Carefully trace three rays coming off the top of the object and form the image. Remember to treat the center of the lens as the place where the light ray bends.
 Measure \begin{align*}d_o\end{align*} and \begin{align*}d_i\end{align*}.
 Use the mirror/lens equation to calculate \begin{align*}d_i\end{align*}.
 Find the percent difference between your measured \begin{align*}d_i\end{align*} and your calculated \begin{align*}d_i\end{align*}.
 Measure the magnification \begin{align*}M\end{align*} from your drawing above.
 Consider a converging lens with a focal length equal to three units.
 Carefully trace three rays coming off the top of the object and form the image. Remember to treat the center of the lens as the place where the light ray bends.
 Measure \begin{align*}d_o\end{align*} and \begin{align*}d_i\end{align*}.
 Use the mirror/lens equation to calculate \begin{align*}d_i\end{align*}.
 Find the percent difference between your measured \begin{align*}d_i\end{align*} and your calculated \begin{align*}d_i\end{align*}.
 Measure the magnification \begin{align*}M\end{align*} from your drawing above.
 Consider a diverging lens with a focal length equal to four units.
 Carefully trace three rays coming off the top of the object and show where they converge to form the image.
 Measure \begin{align*}d_o\end{align*} and \begin{align*}d_i\end{align*}.
 Use the mirror/lens equation to calculate \begin{align*}d_i\end{align*}.
 Find the percent difference between your measured \begin{align*}d_i\end{align*} and your calculated \begin{align*}d_i\end{align*}.
 Measure the magnification \begin{align*}M\end{align*} from your drawing above.
 A piece of transparent goo falls on your paper. You notice that the letters on your page appear smaller than they really are. Is the goo acting as a converging lens or a diverging lens? Explain. Is the image you see real or virtual? Explain.
 An object is placed 30 mm in front of a lens. An image of the object is formed 90 mm behind the lens.
 Is the lens converging or diverging? Explain your reasoning.
 (optional) What is the focal length of the lens?
 To the right is a diagram showing how to make a “ghost light bulb.” The real light bulb is below the box and it forms an image of the exact same size right above it. The image looks very real until you try to touch it. What is the focal length of the concave mirror?
 Little Red Riding Hood (aka RHood) gets to her grandmother’s house only to find the Big Bad Wolf (akaBBW) in her place. RHood notices that BBW is wearing her grandmother’s glasses and it makes the wolf’s eyes look magnified (bigger).
 Are these glasses for nearsighted or farsighted people? For full credit, explain your answer thoroughly. You may need to consult some resources online.
 Create a diagram below of how these glasses correct a person’s vision.
 Draw ray diagrams and explain why some eyes are nearsighted and others farsighted – what’s the problem with the eye in each case? Draw a second set of diagrams and explain what kind of lens is used to correct each problem.
 Do a little online research and explain below how LASIK surgery can correct vision.
 Explain the fundamental differences between how a camera focuses and how the eye focuses. You may need to do a little research here. While you’re researching, look up how a whale’s eye focuses. Is it more like the human eye or like a camera?
Answers:

 Diamond
 vacuum & air
 \begin{align*}1.96 \times 10^8 \ m/s\end{align*}
 \begin{align*}6.99 \times 10^{7} \ m; \ 5.26 \times 10^{7} \ m\end{align*}
 85 cm
 (c) 4 units (e) 1 (same size, but inverted)
 (a) 6 units (c) bigger (about twice as big)
 (b) \begin{align*}d_o = 2 \ units, \ d_i 3 \ units\end{align*}, virtual image (c) 3 units (e) Image should be about 1.5 times as big
 (c) \begin{align*}\frac{21}{4} \ units\end{align*} (e) About \begin{align*}\frac{3}{4}\end{align*} as big
 (c) 6 (so 6 units on left side) (e) 3 times bigger
 (c) \begin{align*}\frac{28}{11} \ units\end{align*}.
 (b) 22.5 mm
 32 cm
OPTIONAL PROBLEMS USING SNELL’S LAW
 Nisha stands at the edge of an aquarium 3.0 m deep. She shines a laser at a height of 1.7 m that hits the water of the pool 8.1 m from the edge.
 Draw a diagram of this situation. Label all known lengths.
 How far from the edge of the pool will the light hit bottom?
 If her friend, James, were at the bottom and shined a light back, hitting the same spot as Nisha’s, how far from the edge would he have to be so that the light never leaves the water?
 A light source sits in a tank of water, as shown.
 If one of the light rays coming from inside the tank of water hits the surface at \begin{align*}35.0^\circ\end{align*}, as measured from the normal to the surface, at what angle will it enter the air?
 Now suppose the incident angle in the water is \begin{align*}80^\circ\end{align*} as measured from the normal. What is the refracted angle? What problem arises?
 Find the critical angle for the waterair interface. This is the incident angle that corresponds to the largest possible refracted angle, \begin{align*}90^\circ\end{align*}.
 Consider a convex mirror with a focal length equal to two units.
 Carefully trace three rays coming off the top of the object and form the image.
 Measure \begin{align*}d_o\end{align*} and \begin{align*}d_i\end{align*}.
 Use the mirror/lens equation to calculate \begin{align*}d_i\end{align*}.
 Find the percent difference between your measured \begin{align*}d_i\end{align*} and your calculated \begin{align*}d_i\end{align*}.
 Measure the magnification \begin{align*}M\end{align*} from your drawing above.
Answers to Optional Problems:
 (b) 11.4 m (c) 11.5 m

 \begin{align*}49.7^\circ\end{align*}
 no such angle, it reflects
 \begin{align*}48.8^\circ\end{align*}
 (c) 1.5 units (e) About \begin{align*}\frac{2}{3}\end{align*} as big
Chapter Summary
Image Attributions
To add resources, you must be the owner of the FlexBook® textbook. Please Customize the FlexBook® textbook.