Ch 26 Light Refraction: Lenses and Optical Instruments

$Page 1: Ch 26 Light Refraction: Lenses and Optical Instruments$
Chapter 26

The Refraction of Light:

Lenses and Optical Instruments

AP Learning Objectives

Geometric optics Reflection and refraction

Students should understand the principles of reflection and refraction, so they can:– Determine how the speed and wavelength of light change

when light passes from one medium into another.– Show on a diagram the directions of refracted rays.– Use Snell’s Law to relate the directions of the incident ray

and the refracted ray, and the indices of refraction of the media.

– Identify conditions under which total internal reflection will occur.

AP Learning Objectives

Geometric optics Lenses

Students should understand image formation by converging or diverging lenses, so they can:

– Determine whether the focal length of a lens is increased or decreased as a result of a change in the curvature of its surfaces, or in the index of refraction of the material of which the lens is made, or the medium in which it is immersed.

– Determine by ray tracing the location of the image of a real object located inside or outside the focal point of the lens, and state whether the resulting image is upright or inverted, real or virtual.

– Use the thin lens equation to relate the object distance, image distance, and focal length for a lens, and determine the image size in terms of the object size.

– Analyze simple situations in which the image formed by one lens serves as the object for another lens.

Table Of Contents

1. The Index of Refraction2. Snell’s Law and the Refraction of Light3. Total Internal Reflection4. Polarization and the Refraction of Light (AP?)5. The Dispersion of Light: Prisms and Rainbows (AP?)6. Lenses7. The Formation of Images by Lenses8. The Thin-Lens Equation and the Magnification Equation9. Lenses in Combinations10. The Human Eye (AP?)11. Angular Magnification and the Magnifying Glass (AP?)12. The Compound Microscope13. The Telescope14. Lens Aberrations

Chapter 26:The Refraction of Light:


Section 1:

The Index of Refraction

Boundary Effects

Back in Chapter 16 we discussed two important effects a

boundary can have on a wave

– Reflection (Chapter 25)

Topic for this chapter

Occurs when a wave strikes a medium boundary and

“bounces back” into the original medium

– Refraction

will be discussed in this Chapter

Transmission of wave from one medium to the next.

Light travels through a vacuum at a speed

Light travels through materials at a speed less than its speed

in a vacuum.

DEFINITION OF THE INDEX OF REFRACTION

– The index of refraction of a material is the ratio of the

speed of light in a vacuum to the speed of light in the

material:

sm1000.3 8c

v

cn

material in thelight of Speed

in vacuumlight of Speed

Index of Refraction


26.1.1. In which one of the following substances does light have the largest speed?

a) diamond

b) benzene

c) carbon dioxide

d) water

e) None of the above. The speed of light has the same value everywhere in the Universe.

26.1.2. Through experiment, the speed of light passing through material A is 1.4 times greater than when the same light passes through material B. What is the ratio of the refractive index of material A to that of material B?

a) 0.71

b) 0.84

c) 1.0

d) 1.2

e) 1.4

26.1.3. If the index of refraction for a given material is 1.5, what is the speed of light in that material relative to the speed of light in a vacuum, c?

a) No material can have an index greater than 1 since light cannot travel faster than c.

b) c/2

c) 2c/3

d) c

e) 1.5c



Section 2:

Snell’s Law and the Refraction of Light

SNELL’S LAW OF REFRACTION

When light travels from a material withone index of refraction to a material witha different index of refraction, the angleof incidence is related to the angle ofrefraction by

2211 sinsin nn

SNELL’S LAW

Example 1 Determining the Angle of RefractionA light ray strikes an air/water surface at anangle of 46 degrees with respect to thenormal. Find the angle of refraction whenthe direction of the ray is (a) from air towater and (b) from water to air.

2

112

sinsin

n

n

(a)

(b)

332

2

112

sinsin

n

n

742

33.1

46sin00.1

54.0

2211 sinsin nn

00.1

46sin33.1

96.0

APPARENT DEPTH

Example 2 Finding a Sunken Chest

The searchlight on a yacht is being used to illuminate a sunkenchest. At what angle of incidence should the light be aimed?

313.30.2tan 12

1

221

sinsin

n

n

441

00.1

31sin33.1

69.0

1

2

n

ndd

Apparent depth,observer directlyabove object

Apparent Depth

Conceptual Example 4 On the Inside Looking Out

A swimmer is under water and looking up at the surface. Someoneholds a coin in the air, directly above the swimmer’s eyes. To theswimmer, the coin appears to be at a certain height above the water. Is the apparent height of the coin greater, less than, or the same as its actual height?

1

2

n

nddSince:

and n2 (air) is less than n1 (water), the object will appear farther away

THE DISPLACEMENT OF LIGHT BY A SLAB OF MATERIAL

Refraction will occur at both surfaces, with each surfacebending according to Snell’s Law

THE DERIVATION OF SNELL’S LAW

26.2.1. A spherical ball is submerged in a shallow pan of water as shown. As we look directly at the ball, it appears spherical, what does the ball look like if you are looking at the rays emerging from above the surface of the water at Point P?

P

26.2.2. One dark evening, you are standing at the edge of a swimming pool that is 1.21 m deep. Directing a flashlight down into the water, you notice a shiny metal plate at the bottom and hold the light there. The light from the flashlight enters the pool at an incident angle of 50.0 at a point 0.85 m from the edge you are standing near. How far from this edge is the metal plate?

a) 0.85 m

b) 1.1 m

c) 1.5 m

d) 1.7 m

e) 1.9 m

26.2.3. You are standing directly above a fish in an aquarium. The actual depth of the fish is one-half the distance from the surface to the bottom. As you look down at it, where does the fish appear to be?

a) I see it at its actual depth.

b) I see it at the surface of the water.

c) I see it below its actual depth, but above the bottom of the aquarium.

d) I see it above its actual depth, but below the surface of the water.

e) I see it at the bottom of the aquarium.

26.2.4. A laser beam is directed at the left side of a plastic block as shown. Through which side and at what angle does the light leave the block?

a) top side, 40

b) right side, 40

c) top side, 68.8

d) right side, 68.8

e) bottom side, 40

26.2.5. Is light bent more, less, or not at all when entering a medium with a larger index of refraction than that of the incident medium when the angle of incidence is greater than 0?

a) more

b) less

c) not at all

26.2.6. Is light bent more, less, or not at all when entering a medium with a smaller index of refraction than that of the incident medium when the angle of incidence is greater than 0?

a) more

b) less

c) not at all

26.2.7. Blue light travels from air into glass that has an index of refraction of 1.54. How does the wavelength of the light in air compare with the wavelength of the light in the glass?

a) The wavelength of the light is the same in both media, only the frequency changes upon entering the glass.

b) The wavelength of the light becomes shorter in the glass.

c) The wavelength of the light becomes longer in the glass.

26.2.8. A single light beam is split into two equal beams, denoted A and B. Beam A travels through a medium with a higher index of refraction than the medium that beam B travel through. When both beams exit their media back into air, how do their wavelengths compare?

a) Beam A has a longer wavelength.

b) Beam B has a longer wavelength.

c) Both beams have the same wavelength.

26.2.9. In which of the following cases, if any, does the angle of reflection not equal to the angle of incidence?

a) Light travels from a material with a lower index of refraction and reflects from the interface with a materials of higher index of refraction.

b) Light travels from a materials with a higher index of refraction and reflects from an interface with a material of lower index of refraction.

c) None of the light incident on a surface is refracted into the material.

d) Most, but not all, of the light incident on a surface is refracted into the material.

e) In all of the above cases, the angle of reflection will equal the angle of incidence.



Section 3:

Total Internal Reflection

When light passes from a medium of larger refractive index into oneof smaller refractive index, the refracted ray bends away from the normal.

Critical angle 211

2 sin nnn

nc

Total Internal Reflection

Example 5 Total Internal Reflection

A beam of light is propagating through diamond and strikes the diamond-air interface at an angle of incidence of 28 degrees. (a) Will part of the beam enter the air or will there be total internal reflection? (b) Repeat part (a) assuming that the diamond is surrounded by water.

4.2442.2

00.1sinsin 1

1

21

n

nc(a)

(b) 3.3342.2

33.1sinsin 1

1

21

n

nc

Conceptual Example 6 The Sparkle of a Diamond

The diamond is famous for its sparkle because the light coming fromit glitters as the diamond is moved about. Why does a diamond exhibit such brilliance? Why does it lose much of its brilliance whenplaced under water?

Another use for Total Internal Reflection

T.I.R. in Fiber Optics

26.3.1. A laser beam is directed at the left side of a plastic block (n = 1.53) as shown at an angle . The beam then undergoes total internal reflection as the light strikes the top interface at the critical angle for the block and the surrounding air. What is the value of the angle ? Note: The drawing is not drawn to scale and the angles shown are not the actual angles.

a) 40.8

b) 49.2

c) 63.1

d) 68.7

e) This is not possible since no angle will satisfy this situation.

26.3.2. If total internal reflection is to occur, what must be incident angle be relative to the critical angle?

a) It must be equal to the critical angle.

b) It must be larger than or equal to the critical angle.

c) It must be smaller than or equal to the critical angle.

d) It must be smaller than the critical angle.

e) Total internal reflection only depends on the indices of refraction of the two materials, so it doesn’t matter what the incident angle is.



Section 4:

Polarization and the Refraction of Light

1

2tann

nB Brewster’s law

Polarized Light & Reflection

26.4.1. A beam of unpolarized light is directed at a liquid within a transparent container. When the light strikes the air-liquid interface, Jason observes that the reflected ray and the refracted ray are perpendicular to one another. Investigating, Jason places a polarizer in the path of the reflected ray. What does Jason observe when the transmission axis of the polarizer is perpendicular to the surface of the water?

a) No light is transmitted through the polarizer.

b) About one quarter of the light is transmitted through the polarizer.

c) About one half of the light is transmitted through the polarizer.

d) About three quarters of the light is transmitted through the polarizer.

e) All of the light is transmitted through the polarizer.

26.4.2. Unpolarized light from a laser is directed toward a horizontal, flat sheet of glass mounted on a stand as shown. Some of the light that reflects from the sheet passes through a polarizer with a vertical transmission axis. Any light that passes through the polarizer may be observed on the wall behind the polarizer. Janet holds the polarizer as shown and observes the wall as Andrew varies the angle of incidence, which equals the angle of reflection , of the laser light. What will Janet observe as Andrew gradually varies from 80 to 40?

a) The intensity of the light will increase uniformly as the angle is decreased.

b) The intensity of the light will decrease uniformly as the angle is decreased.

c) The intensity of the light will decrease to zero near 60, but otherwise some light will reach the wall.

d) The light on the wall will appear to be at constant intensity.

e) No light will pass through the polarizer sheet at any of these angles.



Section 5:

The Dispersion of Light: Prisms and Rainbows

The net effect of a prism is to change the direction of a light ray.

Light rays corresponding to different colors bend by different amounts.

Dispersion Of Light


Rainbows

26.5.1. Which one of the following statements concerning refraction of light by a prism is true?

a) Light of higher frequency is refracted more than light of lower frequency.

b) Light of lower frequency is refracted more than light of higher frequency.

c) Light of longer wavelength is refracted more than light of shorter wavelength.

d) Orange light is refracted more than blue light.

e) All light is refracted the same amount, it only depends on the index of refraction of the prism.

26.5.2. Usually when we see a rainbow, we see a single bow with red at the top and violet at the bottom. The process of rainbow formation is described in the text. On rare occasions, we may see a “double rainbow,” which is one rainbow below another. The lower one appears to be the usual rainbow, while other has the colors reversed with red at the bottom and violet at the top. Which of the following provides the best explanation of a double rainbow?

a) The upper rainbow is formed from water droplets that are higher than those that form the lower rainbow; and it is formed in the same way as the lower rainbow.

b) The upper rainbow is formed from water droplets that are higher than those that form the lower rainbow; and it is formed by light that is reflected two times within multiple water droplets, unlike the single reflections within multiple droplets that form the lower rainbow.

c) The upper rainbow is formed from the water droplets that form the lower rainbow, but it is formed by light that is reflected two times within multiple water droplets, in addition to the single reflections within the same droplets that form the lower rainbow.

d) The upper rainbow is formed by the light that is transmitted out the each water droplet at the point where the light is also reflected within each droplet. This transmitted light is then reflected from other droplets to our eyes.

26.5.3. Which of the following is an example of dispersion?

a) A rainbow seems to connect two mountain peaks.

b) A magnifying glass is used to focus light to start a fire.

c) A spoon in a glass of water looks bent.

d) Snell’s law is a direct result from chromatic dispersion.

e) Laser light is used to accurately cut parts for an automobile.



Section 6:

Lenses

Lenses

Lenses refract light in such a way that an image of the light source is formed.

With a converging lens, paraxial rays that are parallel to the principal axis converge to the focal point.

With a diverging lens, paraxial rays that are parallel to the principal axis appear to originate from the focal point.

Examples of Lenses

26.6.1. Parallel rays of blue light enter a transparent plastic block at 0 and pass through without being diverted. The rays then pass through a converging lens and are focused at a point P. If the block is then rotated so that the parallel rays approach the block at an angle of 45, as shown, without moving the lens, were will the light be focused relative to the point P?

a) above point P

b) below point P

c) to the left of point P

d) to the right of point P

e) at point P

26.6.2. Parallel rays of red light that are directed at a converging lens are focused at a point P on the principle axis to the right of the lens when the lens is surrounded by air. If the lens is surrounded by water instead of air, where will the red parallel rays be focused relative to point P?

a) above point P

b) below point P

c) to the left of point P

d) to the right of point P

e) at point P



Section 7:

The Formation of Images by Lenses

Ray Diagrams

Ray tracing is also used for lenses. We use the same principal rays we used for mirrors.

– the p-ray, which travels parallel to the principal axis, then refracts through focus.

– the f-ray, which travels through focus, then refracts parallel to the principal axis.

– the c-ray, which travels through center and continues without bending.

You must draw 2 of the 3 principal rays. Assume light bends at the middle of the lens (vertical axis)

IMAGE FORMATION BY A CONVERGING LENS

In this example, when the object is placed further thantwice the focal length from the lens, the real image is inverted and smaller than the object.

When the object is placed between F and 2F, the real image is inverted and larger than the object.


When the object is placed between F and the lens, the virtual image is upright and larger than the object.


IMAGE FORMATION BY A DIVERGING LENS

A diverging lens always forms an upright, virtual, diminished image.

26.7.1. The knight from a chess set is placed at the focal point of a diverging lens as shown. By carefully constructing a ray diagram, determine where the image of the knight will appear?

a) no image is formed

b) the image is at a distance f to the left of the lens, but it is inverted

c) the image is at a distance f to the right of the lens, but it is upright

d) the image is at a distance f/2 to the right of the lens, but it is inverted

e) the image is at a distance f/2 to the left of the lens, but it is upright

26.7.2. The knight from a chess set is placed at the focal point of a converging lens as shown. By carefully constructing a ray diagram, determine where the image of the knight will appear?

a) no image is formed

b) the image is at a distance greater than f to the left of the lens, but it is inverted

c) the image is at a distance greater than f to the right of the lens, but it is upright

d) the image is at a distance f/2 to the right of the lens, but it is inverted

e) the image is at a distance f/2 to the left of the lens, but it is upright

26.7.3. An object is located 25 cm to the left of a converging lens that has a focal length of 12 cm. Consider a carefully drawn ray diagram for this situation. A real image is produced by this configuration. If you wanted to produce a larger real image without changing the relative distance of the object and lens, which of the following choices would produce the desired result?

a) Replace the lens with a diverging lens with a focal length of 4 cm.

b) Replace the lens with a converging lens with a focal length of 4 cm.

c) Replace the lens with a diverging lens with a focal length of 12 cm.

d) Replace the lens with a converging lens with a focal length of 20 cm.

e) Replace the lens with a diverging lens with a focal length of 20 cm.

26.7.4. A real object is placed to the right of a converging lens at a distance 2f. The focal length of the lens is f. By carefully drawing a ray diagram for this situation, determine which of the following statements best describes the image formed.

a) An observer on the left side of the lens would see a real image to the left of the lens that is inverted and larger than the object.

b) An observer on the left side of the lens would see a real image to the left of the lens that is inverted and smaller than the object.

c) An observer on the right side of the lens would see a real image to the left of the lens that is upright and larger than the object.

d) An observer on the right side of the lens would see a real image to the right of the lens that is upright and smaller than the object.

e) An observer on the left side of the lens would see a virtual image to the right of the lens that is inverted and smaller than the object.

26.7.5. Consider the two rays drawn from the top of an object. One of the rays crosses the principle axis at point A as shown. Using the information provided, which one of the following statements best describes the location of the focal point on the right side of the converging lens.

a) The focal point is a short distance to the left of point A.

b) The focal point is a large distance to the right of point A, where the rays converge.

c) The focal point is a short distance to the right of point A.

d) The focal point is at point A.

26.7.6. An object is placed to the left of a converging lens. The distance from the object to the lens is one and one-half times the focal length of the lens. A screen is used to view the image of the object on the right side of the lens. If the top half of the lens is covered with black electrical tape such that no light can enter the lens on that half, what would you see on the screen?

a) The top half of the image is no longer seen.

b) The bottom half of the image is no longer seen.

c) The upper most portion and lower most portion of the image are no longer seen.

d) The entire image is seen just as before.

e) The entire image is seen, but about one half as bright.

26.7.7. Which one of the following phrases best describes images formed by diverging lenses?

a) always smaller than the object

b) always larger than the object

c) always inverted

d) always virtual

e) always real

26.7.8. A physics student desires to create a beam of light that consists of parallel rays. Which one of the following arrangements would allow her to accomplish this task?

a) A light bulb is placed at the focal point of a convex mirror.

b) A light bulb is placed at the focal point of a diverging lens.

c) A light bulb is placed at the focal point of a converging lens.

d) A light bulb is located at twice the focal length from a concave mirror.

e) A light bulb is located at twice the focal length from a converging lens.



Section 8:

The Thin-Lens Equation and the Magnification Equation

fdd io

111

o

i

o

i

d

d

h

hm

Lens Equations

Summary of Sign Conventions for Lenses

lens. converging afor is f

lens. diverging afor is f

lens. theofleft the toisobject theif is od

lens. theofright the toisobject theif is od

image). (real lens theofright the toformed imagean for is id

image). (virtual lens theofleft the toformed imagean for is id

image.upright an for is m

image. invertedan for is m

Example 9 The Real Image Formed by a Camera Lens

A 1.70-m tall person is standing 2.50 m in front of a camera. Thecamera uses a converging lens whose focal length is 0.0500 m. (a)Find the image distance and determine whether the image isreal or virtual. (b) Find the magnification and height of the imageon the film.

1m 6.19m 50.2

1

m 0500.0

1111 oi dfd

(a)

m 0510.0id real image

(b) 0204.0m 50.2

m 0510.0

o

i

d

dm

m 0347.0m 50.20204.0 oi mhh

26.8.1. An object is placed at a distance 2f to the left of a converging lens with a focal length f. Using the thin lens equation and the magnification equation, determine the location and magnification of the image formed by this configuration.

a) The image is formed at a distance f to the right of the lens and it has a magnification of 1/2.

b) The image is formed at a distance 2f to the right of the lens and it has a magnification of 1/2.

c) The image is formed at a distance f to the right of the lens and it has a magnification of 1.

d) The image is formed at a distance 2f to the right of the lens and it has a magnification of 1.

e) The image is formed at a distance f/2 to the right of the lens and it has a magnification of 1/4.

26.8.2. An object is placed at a distance 5.0 cm to the left of a converging lens with a focal length 2.5 cm. Using the thin lens equation and the magnification equation, determine the location and magnification of the image formed by this configuration.

a) The image is formed 2.5 cm to the right of the lens and it has a magnification of 1/2.

b) The image is formed 5.0 cm to the right of the lens and it has a magnification of 1/2.

c) The image is formed 2.5 cm to the right of the lens and it has a magnification of 1.

d) The image is formed 5.0 cm to the right of the lens and it has a magnification of 1.

e) The image is formed 1.25 cm to the right of the lens and it has a magnification of 1/4.

26.8.3. An object is placed at a distance 5.0 cm to the left of a diverging lens with a focal length 2.5 cm. Using the thin lens equation and the magnification equation, determine the location and magnification of the image formed by this configuration.

a) The image is formed 1.7 cm to the left of the lens and it has a magnification of +1/3.

b) The image is formed 0.6 cm to the left of the lens and it has a magnification of +3/25.

c) No image is formed in this configuration.

d) The image is formed 0.6 cm to the right of the lens and it has a magnification of 3/25.

e) The image is formed 1.7 cm to the right of the lens and it has a magnification of 1/3.



Section 9:

Lenses in Combinations

The image produced by one lens serves asthe object for the nextlens.

Lens Combos

26.9.1. The compound microscope described in the text is made from two lenses. Which one of the following statements is true concerning the operation of this microscope?

a) Both lenses form real images.

b) Both lenses form virtual images.

c) Only the lens closest to the eye forms an image.

d) The lens closest to the object forms a real image; the other lens forms a virtual image.

e) The lens closest to the object forms a virtual image; the other lens forms a real image.

26.9.2. In her biology class, Chris examines an insect wing under a compound microscope that has an objective lens with a focal length of 0.70 cm, an eyepiece with a focal length of 3.0 cm, and a lens separation distance of 16.00 cm. Chris has a near point distance of 22.5 cm. What is the approximate angular magnification of the microscope as Chris views the insect wing?

a) 75

b) 110

c) 140

d) 190

e) 250



Section 10:

The Human Eye

ANATOMY

OPTICS

The lens only contributes about 20-25% of the refraction, but its functionis important.

NEARSIGNTEDNESS

The lens creates an image of the distance object at the far pointof the nearsighted eye.

Example 12 Eyeglasses for the Nearsighted Person

A nearsighted person has a far point located only 521 cm from theeye. Assuming that eyeglasses are to be worn 2 cm in front of the eye, find the focal length needed for the diverging lens of the glassesso the person can see distant objects.

cm 519

11111

io ddf

cm 519f

FARSIGNTEDNESS

The lens creates an image of the close object at the near pointof the farsighted eye.



Section 11:

Angular Magnification and the Magnifying Glass

The size of the image on the retina determines how largean object appears to be.

Angular Magnification

o

o

d

h sizeAngular radiansin


Example 14 A Penny and the Moon

Compare the angular size of a penny held at arms length with that ofthe moon.

rad 027.0cm 71

cm 9.1

o

o

d

hPenny

Moonrad 0090.0

m 103.9

m 105.3

8

6

o

o

d

h

Angular magnification

M

Ndf

Mi

11

Angular magnificationof a magnifying glass




Section 12:

The Compound Microscope

THE REFRACTIVE POWER OF A LENS – THE DIOPTER

metersin

1diopters)(in power Refractive

f

Optometrists who prescribe correctional lenses and the opticianswho make the lenses do not specify the focal length. Insteadthey use the concept of refractive power.

To increase the angular magnification beyond that possible with a magnifyingglass, an additional converging lenscan be included to “premagnify” the object.

Angular magnification ofa compound microscope

eo

e

ff

NfLM

Compound Microscope



Section 13:

The Telescope

Angular magnification ofan astronomical telescope e

o

f

fM

Telescope



Section 14:

Lens Aberrations

In a converging lens, sphericalaberration prevents light raysparallel to the principal axis fromconverging at a single point.

Spherical aberration can be reducedby using a variable-aperture diaphragm.

Lens Aberration

Chromatic aberration arises when different colors are focused atdifferent points along the principal axis.

Chromatic Aberration

26.14.1. Which one of the following statements best explains why chromatic aberration occurs in lenses, but not in mirrors?

a) The shape of the mirror prevents chromatic aberration.

b) The thickness of a lens varies from top to bottom.

c) The frequency of light changes when it passes through glass.

d) The angle of incidence varies over the surface of a lens for incident parallel rays of light.

e) Different colors of light are refracted by different amounts as the light passes through a lens.


Documents

Ch 26 Light Refraction: Lenses and Optical Instruments