When a mirror is rotated at an angle the reflected ray moves through double that angle the instrument based on the above principle is?

By the end of this section, you will be able to:

• Explain reflection of light from polished and rough surfaces.

The angle of reflection equals the angle of incidence.

Take a piece of paper and shine a flashlight at an angle at the paper, as shown in Figure 3. Now shine the flashlight at a mirror at an angle. Do your observations confirm the predictions in Figure 3 and Figure 4? Shine the flashlight on various surfaces and determine whether the reflected light is diffuse or not. You can choose a shiny metallic lid of a pot or your skin. Using the mirror and flashlight, can you confirm the law of reflection? You will need to draw lines on a piece of paper showing the incident and reflected rays. (This part works even better if you use a laser pencil.)

• The angle of reflection equals the angle of incidence.
• A mirror has a smooth surface and reflects light at specific angles.
• Light is diffused when it reflects from a rough surface.
• Mirror images can be photographed and videotaped by instruments.

1. Using the law of reflection, explain how powder takes the shine off of a person’s nose. What is the name of the optical effect?

1. Show that when light reflects from two mirrors that meet each other at a right angle, the outgoing ray is parallel to the incoming ray, as illustrated in the following figure.
   
   Figure 7. A corner reflector sends the reflected ray back in a direction parallel to the incident ray, independent of incoming direction.

2. Light shows staged with lasers use moving mirrors to swing beams and create colorful effects. Show that a light ray reflected from a mirror changes direction by 2θ when the mirror is rotated by an angle θ.
3. A flat mirror is neither converging nor diverging. To prove this, consider two rays originating from the same point and diverging at an angle θ. Show that after striking a plane mirror, the angle between their directions remains θ.
   
   Figure 8. A flat mirror neither converges nor diverges light rays. Two rays continue to diverge at the same angle after reflection.

law of reflection: angle of reflection equals the angle of incidence

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Page 2

By the end of this section, you will be able to:

• Determine the index of refraction, given the speed of light in a medium.

The changing of a light ray’s direction (loosely called bending) when it passes through variations in matter is called refraction.

The speed of light c not only affects refraction, it is one of the central concepts of Einstein’s theory of relativity. As the accuracy of the measurements of the speed of light were improved, c was found not to depend on the velocity of the source or the observer. However, the speed of light does vary in a precise manner with the material it traverses. These facts have far-reaching implications, as we will see in the chapter Special Relativity. It makes connections between space and time and alters our expectations that all observers measure the same time for the same event, for example. The speed of light is so important that its value in a vacuum is one of the most fundamental constants in nature as well as being one of the four fundamental SI units.

The speed of light is now known to great precision. In fact, the speed of light in a vacuum c is so important that it is accepted as one of the basic physical quantities and has the fixed value c = 2.9972458 × 108 m/s ≈ 3.00 × 108 m/s, where the approximate value of 3.00 × 108 m/s is used whenever three-digit accuracy is sufficient. The speed of light through matter is less than it is in a vacuum, because light interacts with atoms in a material. The speed of light depends strongly on the type of material, since its interaction with different atoms, crystal lattices, and other substructures varies. We define the index of refraction n of a material to be

n=cvn=\frac{c}{v}\\n=vc​

c = 2.9972458 × 108 m/s ≈ 3.00 × 108 m/s

n=cv\displaystyle{n}=\frac{c}{v}\\n=vc​

Calculate the speed of light in zircon, a material used in jewelry to imitate diamond. The speed of light in a material, v, can be calculated from the index of refraction n of the material using the equation

n=cvn=\frac{c}{v}\\n=vc​

n=cvn=\frac{c}{v}\\n=vc​

v=cnv=\frac{c}{n}\\v=nc​

The index of refraction for zircon is given as 1.923 in Table 1, and c is given in the equation for speed of light. Entering these values in the last expression gives

v=3.00×108 m/s1.923 =1.56×108 m/s\begin{array}{lll}v&=&\frac{3.00\times10^8\text{ m/s}}{1.923}\\\text{ }&=&1.56\times10^8\text{ m/s}\end{array}\\v ​==​1.9233.00×108 m/s​1.56×108 m/s​

Discussion

The amount that a light ray changes its direction depends both on the incident angle and the amount that the speed changes. For a ray at a given incident angle, a large change in speed causes a large change in direction, and thus a large change in angle. The exact mathematical relationship is the law of refraction, or “Snell’s Law,” which is stated in equation form as n1 sinθ1 = n2 sinθ2.

Here n1 and n2 are the indices of refraction for medium 1 and 2, and θ1 and θ2 are the angles between the rays and the perpendicular in medium 1 and 2, as shown in Figure 3. The incoming ray is called the incident ray and the outgoing ray the refracted ray, and the associated angles the incident angle and the refracted angle. The law of refraction is also called Snell’s law after the Dutch mathematician Willebrord Snell (1591–1626), who discovered it in 1621. Snell’s experiments showed that the law of refraction was obeyed and that a characteristic index of refraction n could be assigned to a given medium. Snell was not aware that the speed of light varied in different media, but through experiments he was able to determine indices of refraction from the way light rays changed direction.

A classic observation of refraction occurs when a pencil is placed in a glass half filled with water. Do this and observe the shape of the pencil when you look at the pencil sideways, that is, through air, glass, water. Explain your observations. Draw ray diagrams for the situation.

Find the index of refraction for medium 2 in Figure 3a, assuming medium 1 is air and given the incident angle is 30.0º and the angle of refraction is 22.0º. The index of refraction for air is taken to be 1 in most cases (and up to four significant figures, it is 1.000). Thus n1=1.00 here. From the given information, θ1 = 30.0º and θ2 = 22.0º. With this information, the only unknown in Snell’s law is n2, so that it can be used to find this unknown. Snell’s law is n1 sinθ1 = n2 sinθ2.

Rearranging to isolate n2 gives

n2=n1sin⁡θ1sin⁡θ2\displaystyle{n}_2=n_1\frac{\sin\theta_1}{\sin\theta_2}\\n2​=n1​sinθ2​sinθ1​​

n2=1.00sin⁡30.0∘sin⁡22.0∘=0.5000.375 =1.33\begin{array}{lll}{n}_2&=&1.00\frac{\sin30.0^{\circ}}{\sin22.0^{\circ}}=\frac{0.500}{0.375}\\\text{ }&=&1.33\end{array}\\n2​ ​==​1.00sin22.0∘sin30.0∘​=0.3750.500​1.33​

Discussion

Suppose that in a situation like that in Example 2, light goes from air to diamond and that the incident angle is 30.0º. Calculate the angle of refraction θ2 in the diamond. Again the index of refraction for air is taken to be n1 = 1.00, and we are given θ1 = 30.0º. We can look up the index of refraction for diamond in Table 1, finding n2 = 2.419. The only unknown in Snell’s law is θ2, which we wish to determine. Solving Snell’s law for sin θ2 yields

sin⁡θ2=n1n2sin⁡θ1\displaystyle\sin\theta_2=\frac{n_1}{n_2}\sin\theta_1\\sinθ2​=n2​n1​​sinθ1​

sin⁡θ2=1.002.419sin⁡30.0∘=(0.413)(0.500)=0.207\displaystyle\sin\theta_2=\frac{1.00}{2.419}\sin30.0^{\circ}=\left(0.413\right)\left(0.500\right)=0.207\\sinθ2​=2.4191.00​sin30.0∘=(0.413)(0.500)=0.207

• The changing of a light ray’s direction when it passes through variations in matter is called refraction.

  The speed of light in vacuum c = 2.9972458 × 108 m/s ≈ 3.00 × 108 m/s.

  Index of refraction

  n=cvn=\frac{c}{v}\\n=vc​

  , where v is the speed of light in the material, c is the speed of light in vacuum, and n is the index of refraction.

  Snell’s law, the law of refraction, is stated in equation form as n1 sin θ1 = n2 sinθ2.

1. Diffusion by reflection from a rough surface is described in this chapter. Light can also be diffused by refraction. Describe how this occurs in a specific situation, such as light interacting with crushed ice.
2. Why is the index of refraction always greater than or equal to 1?
3. Does the fact that the light flash from lightning reaches you before its sound prove that the speed of light is extremely large or simply that it is greater than the speed of sound? Discuss how you could use this effect to get an estimate of the speed of light.
4. Will light change direction toward or away from the perpendicular when it goes from air to water? Water to glass? Glass to air?
5. Explain why an object in water always appears to be at a depth shallower than it actually is? Why do people sometimes sustain neck and spinal injuries when diving into unfamiliar ponds or waters?
6. Explain why a person’s legs appear very short when wading in a pool. Justify your explanation with a ray diagram showing the path of rays from the feet to the eye of an observer who is out of the water.
7. Why is the front surface of a thermometer curved as shown?
   
   Figure 4. The curved surface of the thermometer serves a purpose.

8. Suppose light were incident from air onto a material that had a negative index of refraction, say –1.3; where does the refracted light ray go?

1. What is the speed of light in water? In glycerine?
2. What is the speed of light in air? In crown glass?
3. Calculate the index of refraction for a medium in which the speed of light is 2.012 × 108 m/s, and identify the most likely substance based on Table 1.
4. In what substance in Table 1 is the speed of light 2.290 × 108 m/s?
5. There was a major collision of an asteroid with the Moon in medieval times. It was described by monks at Canterbury Cathedral in England as a red glow on and around the Moon. How long after the asteroid hit the Moon, which is 3.84 × 105 km away, would the light first arrive on Earth?
6. A scuba diver training in a pool looks at his instructor as shown in Figure 5. What angle does the ray from the instructor’s face make with the perpendicular to the water at the point where the ray enters? The angle between the ray in the water and the perpendicular to the water is 25.0º.
   
   Figure 5. A scuba diver in a pool and his trainer look at each other.

7. Components of some computers communicate with each other through optical fibers having an index of refraction n = 1.55. What time in nanoseconds is required for a signal to travel 0.200 m through such a fiber?
8. (a) Using information in Figure 5, find the height of the instructor’s head above the water, noting that you will first have to calculate the angle of incidence. (b) Find the apparent depth of the diver’s head below water as seen by the instructor.
9. Suppose you have an unknown clear substance immersed in water, and you wish to identify it by finding its index of refraction. You arrange to have a beam of light enter it at an angle of 45.0º, and you observe the angle of refraction to be 40.3º. What is the index of refraction of the substance and its likely identity?
10. On the Moon’s surface, lunar astronauts placed a corner reflector, off which a laser beam is periodically reflected. The distance to the Moon is calculated from the round-trip time. What percent correction is needed to account for the delay in time due to the slowing of light in Earth’s atmosphere? Assume the distance to the Moon is precisely 3.84 × 108 m, and Earth’s atmosphere (which varies in density with altitude) is equivalent to a layer 30.0 km thick with a constant index of refraction n = 1.000293.
11. Suppose Figure 6 represents a ray of light going from air through crown glass into water, such as going into a fish tank. Calculate the amount the ray is displaced by the glass (Δx), given that the incident angle is 40.0º and the glass is 1.00 cm thick.
    
    Figure 6. A ray of light passes from one medium to a third by traveling through a second. The final direction is the same as if the second medium were not present, but the ray is displaced by Δx (shown exaggerated).

12. Figure 6 shows a ray of light passing from one medium into a second and then a third. Show that θ3 is the same as it would be if the second medium were not present (provided total internal reflection does not occur).
13. Unreasonable Results. Suppose light travels from water to another substance, with an angle of incidence of 10.0º and an angle of refraction of 14.9º. (a) What is the index of refraction of the other substance? (b) What is unreasonable about this result? (c) Which assumptions are unreasonable or inconsistent?
14. Construct Your Own Problem. Consider sunlight entering the Earth’s atmosphere at sunrise and sunset—that is, at a 90º incident angle. Taking the boundary between nearly empty space and the atmosphere to be sudden, calculate the angle of refraction for sunlight. This lengthens the time the Sun appears to be above the horizon, both at sunrise and sunset. Now construct a problem in which you determine the angle of refraction for different models of the atmosphere, such as various layers of varying density. Your instructor may wish to guide you on the level of complexity to consider and on how the index of refraction varies with air density.
15. Unreasonable Results. Light traveling from water to a gemstone strikes the surface at an angle of 80.0º and has an angle of refraction of 15.2º. (a) What is the speed of light in the gemstone? (b) What is unreasonable about this result? (c) Which assumptions are unreasonable or inconsistent?

index of refraction: for a material, the ratio of the speed of light in vacuum to that in the material

1. 2.25 × 108 m/s in water; 2.04 × 108 m/s in glycerine 3. 1.490, polystyrene 5. 1.28 s 7. 1.03 ns

9. n = 1.46, fused quartz

13. (a) 0.898; (b) Can’t have n < 1.00 since this would imply a speed greater than c; (c) Refracted angle is too big relative to the angle of incidence.

15. (a)

c5.00\frac{c}{5.00}\\5.00c​

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Page 3

By the end of this section, you will be able to:

• Explain the phenomenon of total internal reflection.
• Describe the workings and uses of fiber optics.
• Analyze the reason for the sparkle of diamonds.

Consider what happens when a ray of light strikes the surface between two materials, such as is shown in Figure 1a. Part of the light crosses the boundary and is refracted; the rest is reflected. If, as shown in the figure, the index of refraction for the second medium is less than for the first, the ray bends away from the perpendicular. (Since n1 > n2, the angle of refraction is greater than the angle of incidence—that is, θ1 > θ2.) Now imagine what happens as the incident angle is increased. This causes θ2 to increase also. The largest the angle of refraction θ2 can be is 90º, as shown in Figure 1b.The critical angle θc for a combination of materials is defined to be the incident angle θ1 that produces an angle of refraction of 90º. That is, θc is the incident angle for which θ2 = 90º. If the incident angle θ1 is greater than the critical angle, as shown in Figure 1c, then all of the light is reflected back into medium 1, a condition called total internal reflection.

The incident angle θ1 that produces an angle of refraction of 90º is called the critical angle, θc.

n1 sin θ1 = n2 sin θ2.

n1 sin θ1 = n2.

θc=sin⁡−1(n2n1)\theta_{c}=\sin^{-1}\left(\frac{n_2}{n_1}\right)\\θc​=sin−1(n1​n2​​)

What is the critical angle for light traveling in a polystyrene (a type of plastic) pipe surrounded by air? The index of refraction for polystyrene is found to be 1.49 in Figure 2, and the index of refraction of air can be taken to be 1.00, as before. Thus, the condition that the second medium (air) has an index of refraction less than the first (plastic) is satisfied, and the equation

θc=sin⁡−1(n2n1)\theta_{c}=\sin^{-1}\left(\frac{n_2}{n_1}\right)\\θc​=sin−1(n1​n2​​)

θc=sin⁡−1(n2n1)\theta_{c}=\sin^{-1}\left(\frac{n_2}{n_1}\right)\\θc​=sin−1(n1​n2​​)

θc=sin⁡−1(1.001.49) =sin⁡−1(0.671) =42.2∘\begin{array}{lll}\theta_{c}&=&\sin^{-1}\left(\frac{1.00}{1.49}\right)\\\text{ }&=&\sin^{-1}\left(0.671\right)\\\text{ }&=&42.2^{\circ}\end{array}\\θc​  ​===​sin−1(1.491.00​)sin−1(0.671)42.2∘​

Discussion

Fiber optics is one application of total internal reflection that is in wide use. In communications, it is used to transmit telephone, internet, and cable TV signals. Fiber optics employs the transmission of light down fibers of plastic or glass. Because the fibers are thin, light entering one is likely to strike the inside surface at an angle greater than the critical angle and, thus, be totally reflected (See Figure 2.) The index of refraction outside the fiber must be smaller than inside, a condition that is easily satisfied by coating the outside of the fiber with a material having an appropriate refractive index. In fact, most fibers have a varying refractive index to allow more light to be guided along the fiber through total internal refraction. Rays are reflected around corners as shown, making the fibers into tiny light pipes.

Bundles of fibers can be used to transmit an image without a lens, as illustrated in Figure 3. The output of a device called an endoscope is shown in Figure 3b. Endoscopes are used to explore the body through various orifices or minor incisions. Light is transmitted down one fiber bundle to illuminate internal parts, and the reflected light is transmitted back out through another to be observed. Surgery can be performed, such as arthroscopic surgery on the knee joint, employing cutting tools attached to and observed with the endoscope. Samples can also be obtained, such as by lassoing an intestinal polyp for external examination.

The cladding prevents light from being transmitted between fibers in a bundle.

Most telephone conversations and Internet communications are now carried by laser signals along optical fibers. Extensive optical fiber cables have been placed on the ocean floor and underground to enable optical communications. Optical fiber communication systems offer several advantages over electrical (copper) based systems, particularly for long distances. The fibers can be made so transparent that light can travel many kilometers before it becomes dim enough to require amplification—much superior to copper conductors. This property of optical fibers is called low loss. Lasers emit light with characteristics that allow far more conversations in one fiber than are possible with electric signals on a single conductor. This property of optical fibers is called high bandwidth. Optical signals in one fiber do not produce undesirable effects in other adjacent fibers. This property of optical fibers is called reduced crosstalk. We shall explore the unique characteristics of laser radiation in a later chapter.

Total internal reflection, coupled with a large index of refraction, explains why diamonds sparkle more than other materials. The critical angle for a diamond-to-air surface is only 24.4º, and so when light enters a diamond, it has trouble getting back out. (See Figure 7.) Although light freely enters the diamond, it can exit only if it makes an angle less than 24.4º. Facets on diamonds are specifically intended to make this unlikely, so that the light can exit only in certain places. Good diamonds are very clear, so that the light makes many internal reflections and is concentrated at the few places it can exit—hence the sparkle. (Zircon is a natural gemstone that has an exceptionally large index of refraction, but not as large as diamond, so it is not as highly prized. Cubic zirconia is manufactured and has an even higher index of refraction (≈2.17), but still less than that of diamond.) The colors you see emerging from a sparkling diamond are not due to the diamond’s color, which is usually nearly colorless. Those colors result from dispersion, the topic of Dispersion: The Rainbow and Prisms. Colored diamonds get their color from structural defects of the crystal lattice and the inclusion of minute quantities of graphite and other materials. The Argyle Mine in Western Australia produces around 90% of the world’s pink, red, champagne, and cognac diamonds, while around 50% of the world’s clear diamonds come from central and southern Africa.

Explore bending of light between two media with different indices of refraction. See how changing from air to water to glass changes the bending angle. Play with prisms of different shapes and make rainbows.

• The incident angle that produces an angle of refraction of 90º is called critical angle.
• Total internal reflection is a phenomenon that occurs at the boundary between two mediums, such that if the incident angle in the first medium is greater than the critical angle, then all the light is reflected back into that medium.
• Fiber optics involves the transmission of light down fibers of plastic or glass, applying the principle of total internal reflection.
• Endoscopes are used to explore the body through various orifices or minor incisions, based on the transmission of light through optical fibers.
• Cladding prevents light from being transmitted between fibers in a bundle.
• Diamonds sparkle due to total internal reflection coupled with a large index of refraction.

1. A ring with a colorless gemstone is dropped into water. The gemstone becomes invisible when submerged. Can it be a diamond? Explain.
2. A high-quality diamond may be quite clear and colorless, transmitting all visible wavelengths with little absorption. Explain how it can sparkle with flashes of brilliant color when illuminated by white light.
3. Is it possible that total internal reflection plays a role in rainbows? Explain in terms of indices of refraction and angles, perhaps referring to Figure 8. Some of us have seen the formation of a double rainbow. Is it physically possible to observe a triple rainbow?
   
   Figure 8. Double rainbows are not a very common observance. (credit: InvictusOU812, Flickr)

4. The most common type of mirage is an illusion that light from faraway objects is reflected by a pool of water that is not really there. Mirages are generally observed in deserts, when there is a hot layer of air near the ground. Given that the refractive index of air is lower for air at higher temperatures, explain how mirages can be formed.

1. Verify that the critical angle for light going from water to air is 48.6º, as discussed at the end of Example 1, regarding the critical angle for light traveling in a polystyrene (a type of plastic) pipe surrounded by air.
2. (a) At the end of Example 1, it was stated that the critical angle for light going from diamond to air is 24.4º. Verify this. (b) What is the critical angle for light going from zircon to air?
3. An optical fiber uses flint glass clad with crown glass. What is the critical angle?
4. At what minimum angle will you get total internal reflection of light traveling in water and reflected from ice?
5. Suppose you are using total internal reflection to make an efficient corner reflector. If there is air outside and the incident angle is 45.0º, what must be the minimum index of refraction of the material from which the reflector is made?
6. You can determine the index of refraction of a substance by determining its critical angle. (a) What is the index of refraction of a substance that has a critical angle of 68.4º when submerged in water? What is the substance, based on Figure 9? (b) What would the critical angle be for this substance in air?
   
   Figure 9. A light ray inside a liquid strikes the surface at the critical angle and undergoes total internal reflection.

7. A ray of light, emitted beneath the surface of an unknown liquid with air above it, undergoes total internal reflection as shown in Figure 9. What is the index of refraction for the liquid and its likely identification?
8. A light ray entering an optical fiber surrounded by air is first refracted and then reflected as shown in Figure 10. Show that if the fiber is made from crown glass, any incident ray will be totally internally reflected.
   
   Figure 10. A light ray enters the end of a fiber, the surface of which is perpendicular to its sides. Examine the conditions under which it may be totally internally reflected.

fiber optics: transmission of light down fibers of plastic or glass, applying the principle of total internal reflection

corner reflector: an object consisting of two mutually perpendicular reflecting surfaces, so that the light that enters is reflected back exactly parallel to the direction from which it came

zircon: natural gemstone with a large index of refraction

3. 66.3º 5. > 1.414 7. 1.50, benzene

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Page 4

By the end of this section, you will be able to:

• Explain the phenomenon of dispersion and discuss its advantages and disadvantages.

We see about six colors in a rainbow—red, orange, yellow, green, blue, and violet; sometimes indigo is listed, too. Those colors are associated with different wavelengths of light, as shown in Figure 2. When our eye receives pure-wavelength light, we tend to see only one of the six colors, depending on wavelength. The thousands of other hues we can sense in other situations are our eye’s response to various mixtures of wavelengths. White light, in particular, is a fairly uniform mixture of all visible wavelengths. Sunlight, considered to be white, actually appears to be a bit yellow because of its mixture of wavelengths, but it does contain all visible wavelengths. The sequence of colors in rainbows is the same sequence as the colors plotted versus wavelength in Figure 2. What this implies is that white light is spread out according to wavelength in a rainbow. Dispersion is defined as the spreading of white light into its full spectrum of wavelengths. More technically, dispersion occurs whenever there is a process that changes the direction of light in a manner that depends on wavelength. Dispersion, as a general phenomenon, can occur for any type of wave and always involves wavelength-dependent processes.

Dispersion is defined to be the spreading of white light into its full spectrum of wavelengths.

Any type of wave can exhibit dispersion. Sound waves, all types of electromagnetic waves, and water waves can be dispersed according to wavelength. Dispersion occurs whenever the speed of propagation depends on wavelength, thus separating and spreading out various wavelengths. Dispersion may require special circumstances and can result in spectacular displays such as in the production of a rainbow. This is also true for sound, since all frequencies ordinarily travel at the same speed. If you listen to sound through a long tube, such as a vacuum cleaner hose, you can easily hear it is dispersed by interaction with the tube. Dispersion, in fact, can reveal a great deal about what the wave has encountered that disperses its wavelengths. The dispersion of electromagnetic radiation from outer space, for example, has revealed much about what exists between the stars—the so-called empty space.

Rainbows are produced by a combination of refraction and reflection.

How does a lens form an image? See how light rays are refracted by a lens. Watch how the image changes when you adjust the focal length of the lens, move the object, move the lens, or move the screen.

• The spreading of white light into its full spectrum of wavelengths is called dispersion.
• Rainbows are produced by a combination of refraction and reflection and involve the dispersion of sunlight into a continuous distribution of colors.
• Dispersion produces beautiful rainbows but also causes problems in certain optical systems.

1. (a) What is the ratio of the speed of red light to violet light in diamond, based on [link]? (b) What is this ratio in polystyrene? (c) Which is more dispersive?
2. A beam of white light goes from air into water at an incident angle of 75.0º. At what angles are the red (660 nm) and violet (410 nm) parts of the light refracted?
3. By how much do the critical angles for red (660 nm) and violet (410 nm) light differ in a diamond surrounded by air?
4. (a) A narrow beam of light containing yellow (580 nm) and green (550 nm) wavelengths goes from polystyrene to air, striking the surface at a 30.0º incident angle. What is the angle between the colors when they emerge? (b) How far would they have to travel to be separated by 1.00 mm?
5. A parallel beam of light containing orange (610 nm) and violet (410 nm) wavelengths goes from fused quartz to water, striking the surface between them at a 60.0º incident angle. What is the angle between the two colors in water?
6. A ray of 610 nm light goes from air into fused quartz at an incident angle of 55.0º. At what incident angle must 470 nm light enter flint glass to have the same angle of refraction?
7. A narrow beam of light containing red (660 nm) and blue (470 nm) wavelengths travels from air through a 1.00 cm thick flat piece of crown glass and back to air again. The beam strikes at a 30.0º incident angle. (a) At what angles do the two colors emerge? (b) By what distance are the red and blue separated when they emerge?
8. A narrow beam of white light enters a prism made of crown glass at a 45.0º incident angle, as shown in Figure 6. At what angles, θR and θV, do the red (660 nm) and violet (410 nm) components of the light emerge from the prism?
   
   Figure 6. This prism will disperse the white light into a rainbow of colors. The incident angle is 45.0º, and the angles at which the red and violet light emerge are θR and θV.

rainbow: dispersion of sunlight into a continuous distribution of colors according to wavelength, produced by the refraction and reflection of sunlight by water droplets in the sky

2. 46.5º, red; 46.0º, violet 4. (a) 0.043º; (b) 1.33 m 6. 71.3º 8. 53.5º, red; 55.2º, violet

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Page 5

By the end of this section, you will be able to:

• List the rules for ray tracking for thin lenses.
• Illustrate the formation of images using the technique of ray tracking.
• Determine power of a lens given the focal length.

The word lens derives from the Latin word for a lentil bean, the shape of which is similar to the convex lens in Figure 1. The convex lens shown has been shaped so that all light rays that enter it parallel to its axis cross one another at a single point on the opposite side of the lens. (The axis is defined to be a line normal to the lens at its center, as shown in Figure 1.) Such a lens is called a converging (or convex) lens for the converging effect it has on light rays. An expanded view of the path of one ray through the lens is shown, to illustrate how the ray changes direction both as it enters and as it leaves the lens. Since the index of refraction of the lens is greater than that of air, the ray moves towards the perpendicular as it enters and away from the perpendicular as it leaves. (This is in accordance with the law of refraction.) Due to the lens’s shape, light is thus bent toward the axis at both surfaces. The point at which the rays cross is defined to be the focal point F of the lens. The distance from the center of the lens to its focal point is defined to be the focal length f of the lens. Figure 2 shows how a converging lens, such as that in a magnifying glass, can converge the nearly parallel light rays from the sun to a small spot.

The lens in which light rays that enter it parallel to its axis cross one another at a single point on the opposite side with a converging effect is called converging lens.

The point at which the light rays cross is called the focal point F of the lens.

The distance from the center of the lens to its focal point is called focal length f.

The greater effect a lens has on light rays, the more powerful it is said to be. For example, a powerful converging lens will focus parallel light rays closer to itself and will have a smaller focal length than a weak lens. The light will also focus into a smaller and more intense spot for a more powerful lens. The power P of a lens is defined to be the inverse of its focal length. In equation form, this is

P=1fP=\frac{1}{f}\\P=f1​

The power P of a lens is defined to be the inverse of its focal length. In equation form, this is 

P=1fP=\frac{1}{f}\\P=f1​

1D=1m, or 1m−11\text{D}=\frac{1}{\text{m}}\text{, or }1\text{m}^{-1}\\1D=m1​, or 1m−1

Suppose you take a magnifying glass out on a sunny day and you find that it concentrates sunlight to a small spot 8.00 cm away from the lens. What are the focal length and power of the lens? The situation here is the same as those shown in Figure 1 and Figure 2. The Sun is so far away that the Sun’s rays are nearly parallel when they reach Earth. The magnifying glass is a convex (or converging) lens, focusing the nearly parallel rays of sunlight. Thus the focal length of the lens is the distance from the lens to the spot, and its power is the inverse of this distance (in m). The focal length of the lens is the distance from the center of the lens to the spot, given to be 8.00 cm. Thus,

f = 8.00 cm.

P=1f=10.0800 m=12.5 DP=\frac{1}{f}=\frac{1}{0.0800\text{ m}}=12.5\text{ D}\\P=f1​=0.0800 m1​=12.5 D

Discussion

Figure 3 shows a concave lens and the effect it has on rays of light that enter it parallel to its axis (the path taken by ray 2 in the Figure is the axis of the lens). The concave lens is a diverging lens, because it causes the light rays to bend away (diverge) from its axis. In this case, the lens has been shaped so that all light rays entering it parallel to its axis appear to originate from the same point, F, defined to be the focal point of a diverging lens. The distance from the center of the lens to the focal point is again called the focal length f of the lens. Note that the focal length and power of a diverging lens are defined to be negative.

For example, if the distance to F in Figure 3 is 5.00 cm, then the focal length is f = –5.00 cm and the power of the lens is P = –20 D. An expanded view of the path of one ray through the lens is shown in the Figure to illustrate how the shape of the lens, together with the law of refraction, causes the ray to follow its particular path and be diverged.

A lens that causes the light rays to bend away from its axis is called a diverging lens.

Ray tracing is the technique of determining or following (tracing) the paths that light rays take. For rays passing through matter, the law of refraction is used to trace the paths. Here we use ray tracing to help us understand the action of lenses in situations ranging from forming images on film to magnifying small print to correcting nearsightedness. While ray tracing for complicated lenses, such as those found in sophisticated cameras, may require computer techniques, there is a set of simple rules for tracing rays through thin lenses.

A thin lens is defined to be one whose thickness allows rays to refract, as illustrated in Figure 1, but does not allow properties such as dispersion and aberrations. An ideal thin lens has two refracting surfaces but the lens is thin enough to assume that light rays bend only once. A thin symmetrical lens has two focal points, one on either side and both at the same distance from the lens. (See Figure 6.)

A thin lens is defined to be one whose thickness allows rays to refract but does not allow properties such as dispersion and aberrations.

Look through your eyeglasses (or those of a friend) backward and forward and comment on whether they act like thin lenses.

1. A ray entering a converging lens parallel to its axis passes through the focal point F of the lens on the other side. (See rays 1 and 3 in Figure 1.)
2. A ray entering a diverging lens parallel to its axis seems to come from the focal point F. (See rays 1 and 3 in Figure 2.)
3. A ray passing through the center of either a converging or a diverging lens does not change direction. (See Figure 5, and see ray 2 in Figure 1 and Figure 2.)
4. A ray entering a converging lens through its focal point exits parallel to its axis. (The reverse of rays 1 and 3 in Figure 1.)
5. A ray that enters a diverging lens by heading toward the focal point on the opposite side exits parallel to the axis. (The reverse of rays 1 and 3 in Figure 2.)

1. A ray entering a converging lens parallel to its axis passes through the focal point F of the lens on the other side.
2. A ray entering a diverging lens parallel to its axis seems to come from the focal point F.
3. A ray passing through the center of either a converging or a diverging lens does not change direction.
4. A ray entering a converging lens through its focal point exits parallel to its axis.
5. A ray that enters a diverging lens by heading toward the focal point on the opposite side exits parallel to the axis.

The image formed in Figure 7 is a real image, meaning that it can be projected. That is, light rays from one point on the object actually cross at the location of the image and can be projected onto a screen, a piece of film, or the retina of an eye, for example. Figure 8 shows how such an image would be projected onto film by a camera lens. This Figure also shows how a real image is projected onto the retina by the lens of an eye. Note that the image is there whether it is projected onto a screen or not.

The image in which light rays from one point on the object actually cross at the location of the image and can be projected onto a screen, a piece of film, or the retina of an eye is called a real image.

Several important distances appear in Figure 7. We define do to be the object distance, the distance of an object from the center of a lens. Image distance di is defined to be the distance of the image from the center of a lens. The height of the object and height of the image are given the symbols ho and hi, respectively. Images that appear upright relative to the object have heights that are positive and those that are inverted have negative heights. Using the rules of ray tracing and making a scale drawing with paper and pencil, like that in Figure 7, we can accurately describe the location and size of an image. But the real benefit of ray tracing is in visualizing how images are formed in a variety of situations. To obtain numerical information, we use a pair of equations that can be derived from a geometric analysis of ray tracing for thin lenses. The thin lens equations are

1do+1di=1f\frac{1}{d_\text{o}}+\frac{1}{d_{\text{i}}}=\frac{1}{f}\\do​1​+di​1​=f1​

hiho=dido=m\frac{h_{\text{i}}}{h_\text{o}}=\frac{d_{\text{i}}}{d_{\text{o}}}=m\\ho​hi​​=do​di​​=m

(hiho)\left(\frac{h_{\text{i}}}{h_\text{o}}\right)\\(ho​hi​​)

The distance of the image from the center of the lens is called image distance.

1do+1di=1f\displaystyle\frac{1}{d_\text{o}}+\frac{1}{d_{\text{i}}}=\frac{1}{f}\\do​1​+di​1​=f1​

hiho=dido=m\displaystyle\frac{h_{\text{i}}}{h_\text{o}}=\frac{d_{\text{i}}}{d_{\text{o}}}=m\\ho​hi​​=do​di​​=m

A clear glass light bulb is placed 0.750 m from a convex lens having a 0.500 m focal length, as shown in Figure 9. Use ray tracing to get an approximate location for the image. Then use the thin lens equations to calculate both the location of the image and its magnification. Verify that ray tracing and the thin lens equations produce consistent results.

The thin lens equations can be used to find di from the given information:

1do+1di=1f\frac{1}{d_\text{o}}+\frac{1}{d_{\text{i}}}=\frac{1}{f}\\do​1​+di​1​=f1​

1di=1f−1do\frac{1}{d_{\text{i}}}=\frac{1}{f}-\frac{1}{d_\text{o}}\\di​1​=f1​−do​1​

1di\frac{1}{d_{\text{i}}}\\di​1​

1di=10.500 m−10.750 m=0.667m\frac{1}{d_{\text{i}}}=\frac{1}{0.500\text{ m}}-\frac{1}{0.750\text{ m}}=\frac{0.667}{\text{m}}\\di​1​=0.500 m1​−0.750 m1​=m0.667​

di=m0.667=1.50 md_{\text{i}}=\frac{\text{m}}{0.667}=1.50\text{ m}\\di​=0.667m​=1.50 m

1di=1f−1do\frac{1}{d_\text{i}}=\frac{1}{f}-\frac{1}{d_{\text{o}}}\\di​1​=f1​−do​1​

di=fdodo−fd_{\text{i}}=\frac{fd_{\text{o}}}{d_{\text{o}}-f}\\di​=do​−ffdo​​

The thin lens equations can be used to find the magnification m, since both di and do are known. Entering their values gives

m=−dido=−1.50 m0.750 m=−2.00\displaystyle{m}=-\frac{d_{\text{i}}}{d_{\text{o}}}=-\frac{1.50\text{ m}}{0.750\text{ m}}=-2.00\\m=−do​di​​=−0.750 m1.50 m​=−2.00

Discussion

A different type of image is formed when an object, such as a person's face, is held close to a convex lens. The image is upright and larger than the object, as seen in Figure 10b, and so the lens is called a magnifier. If you slowly pull the magnifier away from the face, you will see that the magnification steadily increases until the image begins to blur. Pulling the magnifier even farther away produces an inverted image as seen in Figure 10a. The distance at which the image blurs, and beyond which it inverts, is the focal length of the lens. To use a convex lens as a magnifier, the object must be closer to the converging lens than its focal length. This is called a case 2 image. A case 2 image is formed when do < f and f is positive.

Figure 11 uses ray tracing to show how an image is formed when an object is held closer to a converging lens than its focal length. Rays coming from a common point on the object continue to diverge after passing through the lens, but all appear to originate from a point at the location of the image. The image is on the same side of the lens as the object and is farther away from the lens than the object. This image, like all case 2 images, cannot be projected and, hence, is called a virtual image.

An image that is on the same side of the lens as the object and cannot be projected on a screen is called a virtual image.

Suppose the book page in Figure 11a is held 7.50 cm from a convex lens of focal length 10.0 cm, such as a typical magnifying glass might have. What magnification is produced? We are given that do = 7.50 cm and f = 10.0 cm, so we have a situation where the object is placed closer to the lens than its focal length. We therefore expect to get a case 2 virtual image with a positive magnification that is greater than 1. Ray tracing produces an image like that shown in Figure 11, but we will use the thin lens equations to get numerical solutions in this example. To find the magnification m, we try to use magnification equation,

m=−didom=-\frac{d_{\text{i}}}{d_{\text{o}}}\\m=−do​di​​

1di=1f−1do\frac{1}{d_\text{i}}=\frac{1}{f}-\frac{1}{d_{\text{o}}}\\di​1​=f1​−do​1​

1di\frac{1}{d_\text{i}}\\di​1​

1di=110.0 cm−17.50 cm=−0.0333cm\frac{1}{d_\text{i}}=\frac{1}{10.0\text{ cm}}-\frac{1}{7.50\text{ cm}}=\frac{-0.0333}{\text{cm}}\\di​1​=10.0 cm1​−7.50 cm1​=cm−0.0333​

di=−cm0.0333=−30.0 cmd_{\text{i}}=-\frac{\text{cm}}{0.0333}=-30.0\text{ cm}\\di​=−0.0333cm​=−30.0 cm

m=−dido=−−30.0 cm10.0 cm=3.00\displaystyle{m}=-\frac{d_{\text{i}}}{d_{\text{o}}}=-\frac{-30.0\text{ cm}}{10.0\text{ cm}}=3.00\\m=−do​di​​=−10.0 cm−30.0 cm​=3.00

Discussion

A third type of image is formed by a diverging or concave lens. Try looking through eyeglasses meant to correct nearsightedness. (See Figure 12.) You will see an image that is upright but smaller than the object. This means that the magnification is positive but less than 1. The ray diagram in Figure 13 shows that the image is on the same side of the lens as the object and, hence, cannot be projected—it is a virtual image. Note that the image is closer to the lens than the object. This is a case 3 image, formed for any object by a negative focal length or diverging lens.

Suppose an object such as a book page is held 7.50 cm from a concave lens of focal length –10.0 cm. Such a lens could be used in eyeglasses to correct pronounced nearsightedness. What magnification is produced? This example is identical to the preceding one, except that the focal length is negative for a concave or diverging lens. The method of solution is thus the same, but the results are different in important ways. To find the magnification m, we must first find the image distance di using thin lens equation 

1di=1f−1do\frac{1}{d_\text{i}}=\frac{1}{f}-\frac{1}{d_{\text{o}}}\\di​1​=f1​−do​1​

di=fdodo−f{d_\text{i}}=\frac{fd_{\text{o}}}{d_{\text{o}}-f}\\di​=do​−ffdo​​

We are given that f = –10.0 cm and do = 7.50 cm. Entering these yields a value for

1di\frac{1}{d_{\text{i}}}\\di​1​

1di=1−10.0 cm−17.50 cm=−0.2333cm\displaystyle\frac{1}{d_{\text{i}}}=\frac{1}{-10.0\text{ cm}}-\frac{1}{7.50\text{ cm}}=\frac{-0.2333}{\text{cm}}\\di​1​=−10.0 cm1​−7.50 cm1​=cm−0.2333​

di=−cm0.2333=−4.29 cm\displaystyle{d_\text{i}}=-\frac{\text{cm}}{0.2333}=-4.29\text{ cm}\\di​=−0.2333cm​=−4.29 cm

di=(7.5)(−10)(7.5−(−10))=−7517.5=−4.29 cm\displaystyle{d}_{\text{i}}=\frac{\left(7.5\right)\left(-10\right)}{\left(7.5-\left(-10\right)\right)}=-\frac{75}{17.5}=-4.29\text{ cm}\\di​=(7.5−(−10))(7.5)(−10)​=−17.575​=−4.29 cm

m=−dido=−−4.29 cm7.50 cm=0.571\displaystyle{m}=-\frac{d_{\text{i}}}{d_{\text{o}}}=-\frac{-4.29\text{ cm}}{7.50\text{ cm}}=0.571\\m=−do​di​​=−7.50 cm−4.29 cm​=0.571

Discussion

Find several lenses and determine whether they are converging or diverging. In general those that are thicker near the edges are diverging and those that are thicker near the center are converging. On a bright sunny day take the converging lenses outside and try focusing the sunlight onto a piece of paper. Determine the focal lengths of the lenses. Be careful because the paper may start to burn, depending on the type of lens you have selected.

Step 1. Examine the situation to determine that image formation by a lens is involved. Step 2. Determine whether ray tracing, the thin lens equations, or both are to be employed. A sketch is very useful even if ray tracing is not specifically required by the problem. Write symbols and values on the sketch. Step 3. Identify exactly what needs to be determined in the problem (identify the unknowns). Step 4. Make alist of what is given or can be inferred from the problem as stated (identify the knowns). It is helpful to determine whether the situation involves a case 1, 2, or 3 image. While these are just names for types of images, they have certain characteristics (given in Table 1) that can be of great use in solving problems. Step 5. If ray tracing is required, use the ray tracing rules listed near the beginning of this section. Step 6. Most quantitative problems require the use of the thin lens equations. These are solved in the usual manner by substituting knowns and solving for unknowns. Several worked examples serve as guides.

Step 7. Check to see if the answer is reasonable: Does it make sense? If you have identified the type of image (case 1, 2, or 3), you should assess whether your answer is consistent with the type of image, magnification, and so on.

We do not realize that light rays are coming from every part of the object, passing through every part of the lens, and all can be used to form the final image. We generally feel the entire lens, or mirror, is needed to form an image. Actually, half a lens will form the same, though a fainter, image.

• Light rays entering a converging lens parallel to its axis cross one another at a single point on the opposite side.
• For a converging lens, the focal point is the point at which converging light rays cross; for a diverging lens, the focal point is the point from which diverging light rays appear to originate.
• The distance from the center of the lens to its focal point is called the focal length f.

  Power P of a lens is defined to be the inverse of its focal length,

  P=1fP=\frac{1}{f}\\P=f1​

  . A lens that causes the light rays to bend away from its axis is called a diverging lens. Ray tracing is the technique of graphically determining the paths that light rays take.

  The image in which light rays from one point on the object actually cross at the location of the image and can be projected onto a screen, a piece of film, or the retina of an eye is called a real image.

• Thin lens equations are

1do+1di=1f\frac{1}{{d}_{\text{o}}}+\frac{1}{{d}_{\text{i}}}=\frac{1}{f}do​1​+di​1​=f1​

hiho=−dido=m\frac{{h}_{\text{i}}}{{h}_{\text{o}}}=-\frac{{d}_{\text{i}}}{{d}_{\text{o}}}=m\\ho​hi​​=−do​di​​=m

• The distance of the image from the center of the lens is called image distance.

  An image that is on the same side of the lens as the object and cannot be projected on a screen is called a virtual image.

1. It can be argued that a flat piece of glass, such as in a window, is like a lens with an infinite focal length. If so, where does it form an image? That is, how are di and do related?
2. You can often see a reflection when looking at a sheet of glass, particularly if it is darker on the other side. Explain why you can often see a double image in such circumstances.
3. When you focus a camera, you adjust the distance of the lens from the film. If the camera lens acts like a thin lens, why can it not be a fixed distance from the film for both near and distant objects?
4. A thin lens has two focal points, one on either side, at equal distances from its center, and should behave the same for light entering from either side. Look through your eyeglasses (or those of a friend) backward and forward and comment on whether they are thin lenses.
5. Will the focal length of a lens change when it is submerged in water? Explain.

1. What is the power in diopters of a camera lens that has a 50.0 mm focal length?
2. Your camera’s zoom lens has an adjustable focal length ranging from 80.0 to 200 mm. What is its range of powers?
3. What is the focal length of 1.75 D reading glasses found on the rack in a pharmacy?
4. You note that your prescription for new eyeglasses is –4.50 D. What will their focal length be?
5. How far from the lens must the film in a camera be, if the lens has a 35.0 mm focal length and is being used to photograph a flower 75.0 cm away? Explicitly show how you follow the steps in the Problem-Solving Strategy for lenses.
6. A certain slide projector has a 100 mm focal length lens. (a) How far away is the screen, if a slide is placed 103 mm from the lens and produces a sharp image? (b) If the slide is 24.0 by 36.0 mm, what are the dimensions of the image? Explicitly show how you follow the steps in the Problem-Solving Strategy for Lenses (above).
7. A doctor examines a mole with a 15.0 cm focal length magnifying glass held 13.5 cm from the mole (a) Where is the image? (b) What is its magnification? (c) How big is the image of a 5.00 mm diameter mole?
8. How far from a piece of paper must you hold your father’s 2.25 D reading glasses to try to burn a hole in the paper with sunlight?
9. A camera with a 50.0 mm focal length lens is being used to photograph a person standing 3.00 m away. (a) How far from the lens must the film be? (b) If the film is 36.0 mm high, what fraction of a 1.75 m tall person will fit on it? (c) Discuss how reasonable this seems, based on your experience in taking or posing for photographs.
10. A camera lens used for taking close-up photographs has a focal length of 22.0 mm. The farthest it can be placed from the film is 33.0 mm. (a) What is the closest object that can be photographed? (b) What is the magnification of this closest object?
11. Suppose your 50.0 mm focal length camera lens is 51.0 mm away from the film in the camera. (a) How far away is an object that is in focus? (b) What is the height of the object if its image is 2.00 cm high?
12. (a) What is the focal length of a magnifying glass that produces a magnification of 3.00 when held 5.00 cm from an object, such as a rare coin? (b) Calculate the power of the magnifier in diopters. (c) Discuss how this power compares to those for store-bought reading glasses (typically 1.0 to 4.0 D). Is the magnifier’s power greater, and should it be?
13. What magnification will be produced by a lens of power –4.00 D (such as might be used to correct myopia) if an object is held 25.0 cm away?
14. In Example 3, the magnification of a book held 7.50 cm from a 10.0 cm focal length lens was found to be 3.00. (a) Find the magnification for the book when it is held 8.50 cm from the magnifier. (b) Do the same for when it is held 9.50 cm from the magnifier. (c) Comment on the trend in m as the object distance increases as in these two calculations.
15. Suppose a 200 mm focal length telephoto lens is being used to photograph mountains 10.0 km away. (a) Where is the image? (b) What is the height of the image of a 1000 m high cliff on one of the mountains?
16. A camera with a 100 mm focal length lens is used to photograph the sun and moon. What is the height of the image of the sun on the film, given the sun is 1.40 × 106 km in diameter and is 1.50 × 108 km away?
17. Combine thin lens equations to show that the magnification for a thin lens is determined by its focal length and the object distance and is given by

    m=f(f−do)m=\frac{f}{\left(f-d_{\text{o}}\right)}\\m=(f−do​)f​

    .

diverging lens: a concave lens in which light rays that enter it parallel to its axis bend away (diverge) from its axis

focal point: for a converging lens or mirror, the point at which converging light rays cross; for a diverging lens or mirror, the point from which diverging light rays appear to originate

focal length: distance from the center of a lens or curved mirror to its focal point

magnification: ratio of image height to object height

power: inverse of focal length

real image: image that can be projected

virtual image: image that cannot be projected

2. 5.00 to 12.5 D 4. –0.222 m 6. (a) 3.43 m; (b) 0.800 by 1.20 m 7. (a) −1.35 m (on the object side of the lens); (b) +10.0; (c) 5.00 cm 8. 44.4 cm 10. (a) 6.60 cm; (b) –0.333 12. (a) +7.50 cm; (b) 13.3 D; (c) Much greater 14. (a) +6.67; (b) +20.0; (c) The magnification increases without limit (to infinity) as the object distance increases to the limit of the focal distance. 16. −0.933 mm

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Page 6

By the end of this section, you will be able to:

• Illustrate image formation in a flat mirror.
• Explain with ray diagrams the formation of an image using spherical mirrors.
• Determine focal length and magnification given radius of curvature, distance of object and image.

Now let us consider the focal length of a mirror—for example, the concave spherical mirrors in Figure 2. Rays of light that strike the surface follow the law of reflection. For a mirror that is large compared with its radius of curvature, as in Figure 2a, we see that the reflected rays do not cross at the same point, and the mirror does not have a well-defined focal point. If the mirror had the shape of a parabola, the rays would all cross at a single point, and the mirror would have a well-defined focal point. But parabolic mirrors are much more expensive to make than spherical mirrors. The solution is to use a mirror that is small compared with its radius of curvature, as shown in Figure 2b. (This is the mirror equivalent of the thin lens approximation.) To a very good approximation, this mirror has a well-defined focal point at F that is the focal distance f from the center of the mirror. The focal length f of a concave mirror is positive, since it is a converging mirror.

Just as for lenses, the shorter the focal length, the more powerful the mirror; thus,

P=1fP=\frac{1}{f}\\P=f1​

f=R2f=\frac{R}{2}\\f=2R​

The convex mirror shown in Figure 3 also has a focal point. Parallel rays of light reflected from the mirror seem to originate from the point F at the focal distance f behind the mirror. The focal length and power of a convex mirror are negative, since it is a diverging mirror.

1. A ray approaching a concave converging mirror parallel to its axis is reflected through the focal point F of the mirror on the same side. (See rays 1 and 3 in Figure 2b.)
2. A ray approaching a convex diverging mirror parallel to its axis is reflected so that it seems to come from the focal point F behind the mirror. (See rays 1 and 3 in Figure 3.)
3. Any ray striking the center of a mirror is followed by applying the law of reflection; it makes the same angle with the axis when leaving as when approaching. (See ray 2 in Figure 4.)
4. A ray approaching a concave converging mirror through its focal point is reflected parallel to its axis. (The reverse of rays 1 and 3 in Figure 2.)
5. A ray approaching a convex diverging mirror by heading toward its focal point on the opposite side is reflected parallel to the axis. (The reverse of rays 1 and 3 in Figure 3.)

Consider the situation shown in Figure 4, concave spherical mirror reflection, in which an object is placed farther from a concave (converging) mirror than its focal length. That is, f is positive and do > f, so that we may expect an image similar to the case 1 real image formed by a converging lens. Ray tracing in Figure 4 shows that the rays from a common point on the object all cross at a point on the same side of the mirror as the object. Thus a real image can be projected onto a screen placed at this location. The image distance is positive, and the image is inverted, so its magnification is negative. This is a case 1 image for mirrors. It differs from the case 1 image for lenses only in that the image is on the same side of the mirror as the object. It is otherwise identical.

Electric room heaters use a concave mirror to reflect infrared (IR) radiation from hot coils. Note that IR follows the same law of reflection as visible light. Given that the mirror has a radius of curvature of 50.0 cm and produces an image of the coils 3.00 m away from the mirror, where are the coils? We are given that the concave mirror projects a real image of the coils at an image distance di=3.00 m. The coils are the object, and we are asked to find their location—that is, to find the object distance do. We are also given the radius of curvature of the mirror, so that its focal length is

f=R2=25.0 cmf=\frac{R}{2}=25.0\text{ cm}\\f=2R​=25.0 cm

1do+1di=1f\frac{1}{d_{\text{o}}}+\frac{1}{d_{\text{i}}}=\frac{1}{f}\\do​1​+di​1​=f1​

Rearranging to isolate do gives 

1do=1f−1di\frac{1}{d_{\text{o}}}=\frac{1}{f}-\frac{1}{d_{\text{i}}}\\do​1​=f1​−di​1​

Entering known quantities gives a value for

1do\frac{1}{d_{\text{o}}}\\do​1​

1do=10.250 m−13.00 m=3.667m\frac{1}{d_{\text{o}}}=\frac{1}{0.250\text{ m}}-\frac{1}{3.00\text{ m}}=\frac{3.667}{\text{m}}\\do​1​=0.250 m1​−3.00 m1​=m3.667​

This must be inverted to find do:

do=1 m3.667=27.3 cmd_{\text{o}}=\frac{1\text{ m}}{3.667}=27.3\text{ cm}\\do​=3.6671 m​=27.3 cm

One of the solar technologies used today for generating electricity is a device (called a parabolic trough or concentrating collector) that concentrates the sunlight onto a blackened pipe that contains a fluid. This heated fluid is pumped to a heat exchanger, where its heat energy is transferred to another system that is used to generate steam—and so generate electricity through a conventional steam cycle. Figure 5 shows such a working system in southern California. Concave mirrors are used to concentrate the sunlight onto the pipe. The mirror has the approximate shape of a section of a cylinder. For the problem, assume that the mirror is exactly one-quarter of a full cylinder.

1. If we wish to place the fluid-carrying pipe 40.0 cm from the concave mirror at the mirror’s focal point, what will be the radius of curvature of the mirror?
2. Per meter of pipe, what will be the amount of sunlight concentrated onto the pipe, assuming the insolation (incident solar radiation) is 0.900 k W/m2?
3. If the fluid-carrying pipe has a 2.00-cm diameter, what will be the temperature increase of the fluid per meter of pipe over a period of one minute? Assume all the solar radiation incident on the reflector is absorbed by the pipe, and that the fluid is mineral oil.

A=14(2πR)L\text{A}=\frac{1}{4}\left(2\pi\text{R}\right)\text{L}\\A=41​(2πR)L

A=π2R(1.00 m)=(3.14)2(0.800 m)(1.00 m)=1.262\displaystyle\text{A}=\frac{\pi}{2}R\left(1.00\text{ m}\right)=\frac{\left(3.14\right)}{2}\left(0.800\text{ m}\right)\left(1.00\text{ m}\right)=1.26^2\\A=2π​R(1.00 m)=2(3.14)​(0.800 m)(1.00 m)=1.262

(9.00×102Wm2)(1.26 m2)=1130 W\displaystyle\left(9.00\times10^{2}\frac{\text{W}}{\text{m}^2}\right)\left(1.26\text{ m}^2\right)=1130\text{ W}\\(9.00×102m2W​)(1.26 m2)=1130 W

Solution to Part 3

m=ρV=ρπ(d2)2(1.00 m) =(8.00×102 kg/m3)(3.14)(0.0100 m)2(1.00 m) =0.251 kg\begin{array}{lll}{m}&=&\rho\text{V}=\rho\pi\left(\frac{d}{2}\right)^2\left(1.00\text{ m}\right)\\\text{ }&=&\left(8.00\times10^2\text{ kg/m}^3\right)\left(3.14\right)\left(0.0100\text{ m}\right)^2\left(1.00\text{ m}\right)\\\text{ }&=&0.251\text{ kg}\end{array}\\m  ​===​ρV=ρπ(2d​)2(1.00 m)(8.00×102 kg/m3)(3.14)(0.0100 m)2(1.00 m)0.251 kg​

ΔT=Qmc =(1130 W)(60.0 s)(0.251 kg)(1670 J⋅ kg/∘ C) =162∘C\begin{array}{lll}\Delta{T}&=&\frac{Q}{mc}\\\text{ }&=&\frac{\left(1130\text{ W}\right)\left(60.0\text{ s}\right)}{\left(0.251\text{ kg}\right)\left(1670\text{ J}\cdot\text{ kg/}^{\circ}\text{ C}\right)}\\\text{ }&=&162^{\circ}\text{C}\end{array}\\ΔT  ​===​mcQ​(0.251 kg)(1670 J⋅ kg/∘ C)(1130 W)(60.0 s)​162∘C​

Discussion for Part 3

What happens if an object is closer to a concave mirror than its focal length? This is analogous to a case 2 image for lenses ( do <f and f positive), which is a magnifier. In fact, this is how makeup mirrors act as magnifiers. Figure 6a uses ray tracing to locate the image of an object placed close to a concave mirror. Rays from a common point on the object are reflected in such a manner that they appear to be coming from behind the mirror, meaning that the image is virtual and cannot be projected. As with a magnifying glass, the image is upright and larger than the object. This is a case 2 image for mirrors and is exactly analogous to that for lenses.

A convex mirror is a diverging mirror (f is negative) and forms only one type of image. It is a case 3 image—one that is upright and smaller than the object, just as for diverging lenses. Figure 7a uses ray tracing to illustrate the location and size of the case 3 image for mirrors. Since the image is behind the mirror, it cannot be projected and is thus a virtual image. It is also seen to be smaller than the object.

A keratometer is a device used to measure the curvature of the cornea, particularly for fitting contact lenses. Light is reflected from the cornea, which acts like a convex mirror, and the keratometer measures the magnification of the image. The smaller the magnification, the smaller the radius of curvature of the cornea. If the light source is 12.0 cm from the cornea and the image’s magnification is 0.0320, what is the cornea’s radius of curvature? If we can find the focal length of the convex mirror formed by the cornea, we can find its radius of curvature (the radius of curvature is twice the focal length of a spherical mirror). We are given that the object distance is do = 12.0 cm and that m = 0.0320. We first solve for the image distance di, and then for f.

m=−didom=-\frac{d_{\text{i}}}{d_{\text{o}}}\\m=−do​di​​

Entering known values yields di = –(0.0320)(12.0 cm) = –0.384 cm.

1f=1do+1di\frac{1}{f}=\frac{1}{d_{\text{o}}}+\frac{1}{d_{\text{i}}}\\f1​=do​1​+di​1​

1f=112.0 cm+1−0.384 cm=−2.52cm\frac{1}{f}=\frac{1}{12.0\text{ cm}}+\frac{1}{-0.384\text{ cm}}=\frac{-2.52}{\text{cm}}\\f1​=12.0 cm1​+−0.384 cm1​=cm−2.52​

This must be inverted to find f:

f=cm−2.52=−0.400 cmf=\frac{\text{cm}}{-2.52}=-0.400\text{ cm}\\f=−2.52cm​=−0.400 cm

The radius of curvature is twice the focal length, so that R = 2|f| = 0.800 cm.

The three types of images formed by mirrors (cases 1, 2, and 3) are exactly analogous to those formed by lenses, as summarized in the table at the end of Image Formation by Lenses. It is easiest to concentrate on only three types of images—then remember that concave mirrors act like convex lenses, whereas convex mirrors act like concave lenses.

Find a flashlight and identify the curved mirror used in it. Find another flashlight and shine the first flashlight onto the second one, which is turned off. Estimate the focal length of the mirror. You might try shining a flashlight on the curved mirror behind the headlight of a car, keeping the headlight switched off, and determine its focal length.

Step 1. Examine the situation to determine that image formation by a mirror is involved.

Step 2. Refer to the Problem-Solving Strategies for Lenses. The same strategies are valid for mirrors as for lenses with one qualification—use the ray tracing rules for mirrors listed earlier in this section.

• The characteristics of an image formed by a flat mirror are: (a) The image and object are the same distance from the mirror, (b) The image is a virtual image, and (c) The image is situated behind the mirror.
• Image length is half the radius of curvature: 

  f=R2f=\frac{R}{2}\\f=2R​

• A convex mirror is a diverging mirror and forms only one type of image, namely a virtual image.

1. What are the differences between real and virtual images? How can you tell (by looking) whether an image formed by a single lens or mirror is real or virtual?
2. Can you see a virtual image? Can you photograph one? Can one be projected onto a screen with additional lenses or mirrors? Explain your responses.
3. Is it necessary to project a real image onto a screen for it to exist?
4. At what distance is an image always located—at do, di, or f?
5. Under what circumstances will an image be located at the focal point of a lens or mirror?
6. What is meant by a negative magnification? What is meant by a magnification that is less than 1 in magnitude?
7. Can a case 1 image be larger than the object even though its magnification is always negative? Explain.
8. Figure 8 shows a light bulb between two mirrors. One mirror produces a beam of light with parallel rays; the other keeps light from escaping without being put into the beam. Where is the filament of the light in relation to the focal point or radius of curvature of each mirror?
   
   Figure 8. The two mirrors trap most of the bulb’s light and form a directional beam as in a headlight.

9. The two mirrors trap most of the bulb’s light and form a directional beam as in a headlight.
10. Two concave mirrors of different sizes are placed facing one another. A filament bulb is placed at the focus of the larger mirror. The rays after reflection from the larger mirror travel parallel to one another. The rays falling on the smaller mirror retrace their paths.
11. Devise an arrangement of mirrors allowing you to see the back of your head. What is the minimum number of mirrors needed for this task?
12. If you wish to see your entire body in a flat mirror (from head to toe), how tall should the mirror be? Does its size depend upon your distance away from the mirror? Provide a sketch.
13. It can be argued that a flat mirror has an infinite focal length. If so, where does it form an image? That is, how are di and do related?
14. Why are diverging mirrors often used for rear-view mirrors in vehicles? What is the main disadvantage of using such a mirror compared with a flat one?

1. What is the focal length of a makeup mirror that has a power of 1.50 D?
2. Some telephoto cameras use a mirror rather than a lens. What radius of curvature mirror is needed to replace a 800 mm focal length telephoto lens?
3. (a) Calculate the focal length of the mirror formed by the shiny back of a spoon that has a 3.00 cm radius of curvature. (b) What is its power in diopters?
4. Find the magnification of the heater element in Example 1. Note that its large magnitude helps spread out the reflected energy.
5. What is the focal length of a makeup mirror that produces a magnification of 1.50 when a person’s face is 12.0 cm away?
6. A shopper standing 3.00 m from a convex security mirror sees his image with a magnification of 0.250. (a) Where is his image? (b) What is the focal length of the mirror? (c) What is its radius of curvature?
7. An object 1.50 cm high is held 3.00 cm from a person’s cornea, and its reflected image is measured to be 0.167 cm high. (a) What is the magnification? (b) Where is the image? (c) Find the radius of curvature of the convex mirror formed by the cornea. (Note that this technique is used by optometrists to measure the curvature of the cornea for contact lens fitting. The instrument used is called a keratometer, or curve measurer.)
8. Ray tracing for a flat mirror shows that the image is located a distance behind the mirror equal to the distance of the object from the mirror. This is stated di = −do, since this is a negative image distance (it is a virtual image). (a) What is the focal length of a flat mirror? (b) What is its power?
9. Show that for a flat mirror hi = ho, knowing that the image is a distance behind the mirror equal in magnitude to the distance of the object from the mirror.
10. Use the law of reflection to prove that the focal length of a mirror is half its radius of curvature. That is, prove that

    f=R2f=\frac{R}{2}\\f=2R​

    . Note this is true for a spherical mirror only if its diameter is small compared with its radius of curvature.
11. Referring to the electric room heater considered in the first example in this section, calculate the intensity of IR radiation in W/m2 projected by the concave mirror on a person 3.00 m away. Assume that the heating element radiates 1500 W and has an area of 100 cm2, and that half of the radiated power is reflected and focused by the mirror.
12. Consider a 250-W heat lamp fixed to the ceiling in a bathroom. If the filament in one light burns out then the remaining three still work. Construct a problem in which you determine the resistance of each filament in order to obtain a certain intensity projected on the bathroom floor. The ceiling is 3.0 m high. The problem will need to involve concave mirrors behind the filaments. Your instructor may wish to guide you on the level of complexity to consider in the electrical components.

diverging mirror: a convex mirror in which light rays that strike it parallel to its axis bend away (diverge) from its axis

law of reflection: angle of reflection equals the angle of incidence

1. +0.667 m

3. (a) −1.5 × 10−2 m; (b) −66.7 D

9. 

m=hiho=−dido=−−dodo=dodo=1⇒hi=hom=\frac{{h}_{\text{i}}}{{h}_{\text{o}}}=-\frac{{d}_{\text{i}}}{{d}_{\text{o}}}=-\frac{-{d}_{\text{o}}}{{d}_{\text{o}}}=\frac{{d}_{\text{o}}}{{d}_{\text{o}}}=1\Rightarrow {h}_{\text{i}}={h}_{\text{o}}\\m=ho​hi​​=−do​di​​=−do​−do​​=do​do​​=1⇒hi​=ho​

11. 6.82 kW/m2

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