Chapter 22Reflection and Refraction of Light

Waves and Optics Geometric Optics (1.5 weeks)

Learning Objectives:

1. Reflection and refraction (pgs. 487-493) Students should understand the principles of reflection and refraction, so they can:

a) Determine how the speed and wavelength of light change when light passes from one medium into another. b) Show on a diagram the directions of reflected and refracted rays.

c) Use Snell’s Law to relate the directions of the incident ray and the refracted ray, and the indices of refraction of the media. d) Identify conditions under which total internal reflection will occur.

2. Mirrors(pgs. 446-474) Students should understand image formation by plane or spherical mirrors, so they can:

a) Locate by ray tracing the image of an object formed by a plane mirror, and determine whether the image is real or virtual, upright or inverted, enlarged or reduced in size. b) Relate the focal point of a spherical mirror to its center of curvature.

c) Locate by ray tracing the image of a real object, given a diagram of a mirror with the focal point shown, and determine whether the image is real or virtual, upright or inverted, enlarged or reduced in size. d) Use the mirror 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.

3. Lenses (pgs. 494-511) Students should understand image formation by converging or diverging lenses, so they can:

a) 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. b) 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. c) 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. d) Analyze simple situations in which the image formed by one lens serves as the object for another lens.

Read and take notes on pages 421-426 and 432-

436 in Conceptual Physics Text

If you are not doing the readings from the Conceptual Physics Text then read

and take notes on the following throughout this unit

Reflection: Lesson 1 C and DRefraction: Lesson 1 C-D

Lesson 2 A and BLesson 3 A-C

Read and take notes on

pages 732-735 in College Physics Text

Nature of Light

• Light has a dual nature.– Particle– Wave

• Wave characteristics will be discussed in this chapter.– Reflection– Refraction– These characteristics can be used to understand

mirrors and lenses.

Introduction

A Brief History of Light

• Early models of light– It was proposed that light consisted of tiny particles.

• Newton– Used this particle model to explain reflection and

refraction• Huygens– 1678– Explained many properties of light by proposing light was

wave-like

Section 22.1

A Brief History of Light, Cont.

• Young– 1801– Strong support for wave theory by showing

interference• Maxwell– 1865– Electromagnetic waves travel at the speed of light.

Section 22.1

A Brief History of Light, Final

• Einstein– Particle nature of light– Explained the photoelectric effect– Used Planck’s ideas

Section 22.1

The Particle Nature of Light

• “Particles” of light are called photons.• Each photon has a particular energy.– E = h ƒ– h is Planck’s constant

• h = 6.63 x 10-34 J s

– Encompasses both natures of light• Interacts like a particle• Has a given frequency like a wave

Section 22.1

Dual Nature of Light

• In some experiments light acts as a wave and in others it acts as a particle.– Classical electromagnetic wave theory provides

explanations of light propagation and interference.– Experiments involving the interaction of light with matter

are best explained by assuming light is a particle.

• Light has a number of physical properties, some associated with waves and others with particles.

Section 22.1

EitherLink to Bright Storm on

Reflection and Refraction(Start to minute 3:12)

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Reflection and Refraction

• The processes of reflection and refraction can occur when light traveling in one medium encounters a boundary leading to a second medium.

• In reflection, part of the light bounces off the second medium.

• In refraction, the light passing into the second medium bends.

• Often, both processes occur at the same time.

Section 22.2

Geometric Optics – Using a Ray Approximation

• Light travels in a straight-line path in a homogeneous medium until it encounters a boundary between two different media.

• The ray approximation is used to represent beams of light.

• A ray of light is an imaginary line drawn along the direction of travel of the light beams.

Section 22.2

Ray Approximation• A wave front is a surface

passing through points of a wave that have the same phase and amplitude.

• The rays, corresponding to the direction of the wave motion, are perpendicular to the wave fronts.

Section 22.2

Read and take notes on pages 442-459 in

Conceptual Physics Text

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Reflection

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Specular and Diffuse

Reflection(optional)

Reflection of Light

• A ray of light, the incident ray, travels in a medium.

• When it encounters a boundary with a second medium, part of the incident ray is reflected back into the first medium.– This means it is directed backward into the first

medium.

Section 22.2

Specular Reflection

• Specular reflection is reflection from a smooth surface.

• The reflected rays are parallel to each other.

• All reflection in this text is assumed to be specular.

Diffuse Reflection

• Diffuse reflection is reflection from a rough surface.

• The reflected rays travel in a variety of directions.

• Diffuse reflection makes the dry road easy to see at night.

Section 22.2

Law of Reflection• The normal is a line

perpendicular to the surface.– It is at the point where the

incident ray strikes the surface.

• The incident ray makes an angle of θ1 with the normal.

• The reflected ray makes an angle of θ1’ with the normal.

Section 22.2

Law of Reflection, Cont.

• The angle of reflection is equal to the angle of incidence.

• θ1= θ1’

Section 22.2

EXAMPLE 22.1 The Double Reflecting Light Ray

Mirrors M1 and M2 make an angle of 120° with each other.

Goal Calculate a resultant angle from two reflections. Problem Two mirrors make an angle of 120° with each other, as in the figure. A ray is incident on mirror M1 at an angle of 65° to the normal. Find the angle the ray

makes with the normal to M2 after it is reflected from both mirrors. Strategy Apply the law of reflection twice. Given the incident ray at angle θinc, find the final resultant angle, βref.

SOLUTION

Apply the law of reflection to M1 to find the angle of reflection, θref. θref = θinc = 65°

Find the angle that is the complement of the angle θref. = 90° - θref = 90° - 65° = 25°

Find the unknown angle α in the triangle of M1, M2, and the ray traveling from M1 to M2, using the fact that the three angles sum to 180°.

180° = 25° + 120° + α → α = 35°

The angle α is complementary to the angle of incidence, βinc, for M2.

α + βinc = 90° → βinc = 90° - 35° = 55°

Apply the law of reflection a second time, obtaining βref. βref = βinc = 55°

LEARN MORE

Remarks Notice the heavy reliance on elementary geometry and trigonometry in these reflection problems. Question In general, what is the relationship between the incident angle θinc and the final reflected angle βref when the angle between the mirrors is 90.0°?

θinc - βref = 90.0° θinc + βref = 180.0° θinc + βref = 90.0°

Read and take notes on

pages 736-739 in College Physics Text

Refraction of Light

• When a ray of light traveling through a transparent medium encounters a boundary leading into another transparent medium, part of the ray is reflected and part of the ray enters the second medium.

• The ray that enters the second medium is bent at the boundary.– This bending of the ray is called refraction.

Section 22.2

Refraction of Light, Cont.• The incident ray, the

reflected ray, the refracted ray, and the normal all lie on the same plane.

• The angle of refraction, θ2, depends on the properties of the medium.

Section 22.2

Following the Reflected and Refracted Rays

• Ray is the incident ray.• Ray is the reflected ray.• Ray is refracted into the

Lucite.• Ray is internally reflected

in the Lucite.• Ray is refracted as it

enters the air from the Lucite.

Section 22.2

More About Refraction

• The angle of refraction depends upon the material and the angle of incidence.

• The path of the light through the refracting surface is reversible.

Section 22.2

Refraction Details, 1

• Light may move from a material where its speed is high to a material where is speed is lower.

• The angle of refraction is less than the angle of incidence.– The ray bends toward

the normal.

Section 22.2

Refraction Details, 2

• Light may move from a material where its speed is low to a material where is speed is higher.

• The angle of refraction is greater than the angle of incidence.– The ray bends away from the

normal.

Link to Bright Storm on Index of Refraction

(Minute 3:12- End)

The Index of Refraction

• When light passes from one medium to another, it is refracted because the speed of light is different in the two media.

• The index of refraction, n, of a medium can be defined

Section 22.3

Index of Refraction, Cont.

• Some values of n– For a vacuum, n = 1– For other media, n > 1

• n is a unitless ratio• As the value of n increases, the speed of the

wave decreases.

Section 22.3

EXAMPLE 22.2 Angle of Refraction for Glass

Goal Apply Snell's law to a slab of glass. Problem A light ray of wavelength 589 nm (produced by a sodium lamp) traveling through air is incident on a smooth, flat slab of crown glass at an angle of 30.0° to the normal, as sketched in the figure. Find the angle of refraction, θ2. Strategy Substitute quantities into Snell's law and solve for the unknown angle refraction, θ2.

SOLUTION

Solve Snell's law for sin θ2. (1) sin θ2 = (n1/n2) sin θ1

Using a refraction index, find n1 = 1.00 for air and n2 = 1.52 for crown glass. Substitute these values into Equation (1) and take the inverse sine of both sides. sin θ2 = (1.00 /1.52)(sin 30.0°) = 0.329

θ2 = sin-1 (0.329) = 19.2°

LEARN MORE

Remarks Notice that the light ray bends toward the normal when it enters a material of a higher index of refraction. If the ray left the material following the same path in reverse, it would bend away from the normal. Question If the glass is replaced by a transparent material with smaller index of refraction than that of glass, the refraction angle θ2 will be:

smaller larger unchanged

Frequency Between Media• As light travels from one

medium to another, its frequency does not change.– Both the wave speed and the

wavelength do change.– The wavefronts do not pile

up, nor are created or destroyed at the boundary, so ƒ must stay the same.

Section 22.3

Index of Refraction Extended

• The frequency stays the same as the wave travels from one medium to the other.

• v = ƒ λ• The ratio of the indices of refraction of the two

media can be expressed as various ratios.

Section 22.3

Some Indices of Refraction

Section 22.3

EXAMPLE 22.3 Light in Fused Quartz

Goal Use the index of refraction to determine the effect of a medium on light's speed and wavelength. Problem Light of wavelength 589 nm in vacuum passes through a piece of fused quartz of index of refraction n = 1.458. (a) Find the speed of light in fused quartz. (b) What is the wavelength of this light in fused quartz? (c) What is the frequency of the light in fused quartz? Strategy Substitute values into the speed of light in media equation and the wavelength in media equations.

SOLUTION

(a) Find the speed of light in fused quartz. Obtain the speed from the speed of light in medium equation.

v = c

= 3.00 108 m/s

= 2.06 108 m/s n 1.458

(b) What is the wavelength of this light in fused quartz?

Use the relevant equation to calculate the wavelength:

λn = λ0

= 589 nm

= 404 nm n 1.458

(c) What is the frequency of the light in fused quartz?

The frequency in quartz is the same as in vacuum. Solve c = fλ for the frequency.

f = c

= 3.00 108 m/s

= 5.09 1014 Hz λ 5.89 10-7 m

LEARN MORE

Remarks It's interesting to note that the speed of light in vacuum, 3.00 108 m/s, is an upper limit for the speed of material objects. In studying relativity, we will find that this upper limit is consistent with experimental observations. However, it's possible for a particle moving in a medium to have a speed that exceeds the speed of light in that medium. For example, it's theoretically possible for a particle to travel through fused quartz at a speed greater than 2.06 108 m/s, but it must still have a speed less than 3.00 108 m/s. Question If light with wavelength λ in glass passes into water with index nw, the new wavelength of the light is:

λ/nw λ λng/nw λnw/ng

Either

Link to Bright Storm on Snell’s Law

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Snell’s Law of Refraction

• n1 sin θ1 = n2 sin θ2 – θ1 is the angle of incidence

– θ2 is the angle of refraction

• The experimental discovery of this relationship is usually credited to Willebrørd Snell (1591 – 1626).

Section 22.3

Khan Academy on example problems of Snell’s law

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EXAMPLE 22.4 Light Passing Through a Slab

When light passes through a flat slab of material, the emerging beam is parallel to the incident beam; therefore, θ1 = θ3. Goal Apply Snell's law when a ray passes into and out of another medium Problem A light beam traveling through a transparent medium of index of refraction n1 passes through a thick transparent slab with parallel faces and index of refraction n2.

Show that the emerging beam is parallel to the incident beam. Strategy Apply Snell's law twice, once at the upper surface and once at the lower surface. The two equations will be related because the angle of refraction at the upper surface equals the angle of incidence at the lower surface. The ray passing through the slab makes equal angles with the normal at the entry and exit points. This procedure will enable us to compare angles θ1 and θ3.

SOLUTION

Apply Snell's law to the upper surface.

(1) sin θ2 = n1

sin θ1 n2

Apply Snell's law to the lower surface.

(2) sin θ3 = n2

sin θ2 n1

Substitute Equation (1) into Equation (2).

sin θ3 = n2 (

n1 sin θ1 ) = sin θ1 n1 n2

Take the inverse sine of both sides, noting that the angles are positive and less than 90°. θ3 = θ1

LEARN MORE

Remarks The preceding result proves that the slab doesn't alter the direction of the beam. It does, however, produce a lateral displacement of the beam, as shown in the figure. Question Suppose an additional slab with index n3 were placed below the slab of glass. Would the exit angle at the bottom surface of this second slab still equal the incident angle at the upper surface of the first slab?

No, never. Sometimes yes and sometimes no, depending on the refractive

indices of the media. They are equal only if the incident light is perpendicular

to the first surface. Yes, always.

EXAMPLE 22.5 Refraction of Laser Light in a Digital Videodisc (DVD)

(a) A micrograph of a DVD surface showing tracks and pits along each track. (b) Cross section of a cone-shaped laser beam used to read a DVD.

Goal Apply Snell's law together with geometric constraints. Problem A DVD is a video recording consisting of a spiral track

about 1.0 µm wide with digital information. (See Fig. a.) The digital information consists of a series of pits that are "read" by a laser beam sharply focused on a track

in the information layer. If the width a of the beam at the information layer must

equal 1.0 µm to distinguish individual tracks and the width w of the beam as it enters the plastic is 0.7000 mm, find the angle θ1 at which the conical beam should

enter the plastic. (See Fig. b.) Assume the plastic has a thickness t = 1.20 mm and

an index of refraction n = 1.55. Note that this system is relatively immune to small dust particles degrading the video quality because particles would have to be as large as 0.700 mm to obscure the beam at the point where it enters the plastic. Strategy Use right-triangle trigonometry to determine the angle θ2 and then apply Snell's law to obtain the angle θ1.

SOLUTION

From the top and bottom of figure b, obtain an equation relating w, b, and a. w = 2b + a

Solve this equation for b and substitute given values.

b = w − a

= 7.000 10-4 m − 1.0 10-6 m

= 349.5 µm 2 2

Now use the tangent function to find θ2.

tan θ2 = b

= 349.5 µm

→ θ2 = 16.2° t 1.20 103 µm

Finally, use Snell's law to find θ1. n1 sin θ1 = n2 sin θ2

sin θ1 = n2 sin θ2

= 1.55 sin 16.2°

= 0.433 n1 1.00

θ1 = sin-1 (0.433) = 25.7°

LEARN MORE

Remarks Despite its apparent complexity, the problem isn't that different from calculating the refraction of light by glass. Question Suppose the plastic were replaced by a material with a higher index of refraction. How would the width of the beam at the information layer be affected?

It would decrease. It would increase. It would remain the same.

Read and take notes on

pages 742-746 in College Physics Text

Dispersion

• The index of refraction in anything except a vacuum depends on the wavelength of the light.

• This dependence of n on λ is called dispersion.• Snell’s Law indicates that the angle of refraction

made when light enters a material depends on the wavelength of the light.

Section 22.4

Variation of Index of Refraction with Wavelength

• The index of refraction for a material usually decreases with increasing wavelength.

• Violet light refracts more than red light when passing from air into a material.

Section 22.4

Refraction in a Prism• The amount the ray is bent

away from its original direction is called the angle of deviation, δ

• Since all the colors have different angles of deviation, they will spread out into a spectrum.– Violet deviates the most.– Red deviates the least.

Section 22.4

Prism Spectrometer

• A prism spectrometer uses a prism to cause the wavelengths to separate.

• The instrument is commonly used to study wavelengths emitted by a light source.

Section 22.4

Using Spectra to Identify Gases

• All hot, low pressure gases emit their own characteristic spectra

• The particular wavelengths emitted by a gas serve as “fingerprints” of that gas.

• Some uses of spectral analysis– Identification of molecules– Identification of elements in distant stars– Identification of minerals

Section 22.4

The Rainbow

• A ray of light strikes a drop of water in the atmosphere.

• It undergoes both reflection and refraction.– First refraction at the front of the drop• Violet light will deviate the most.• Red light will deviate the least.

Section 22.5

The Rainbow, 2• At the back surface the light

is reflected. • It is refracted again as it

returns to the front surface and moves into the air.

• The rays leave the drop at various angles.– The angle between the white

light and the violet ray is 40°– The angle between the white

light and the red ray is 42°

Section 22.5

Observing the Rainbow• If a raindrop high in the sky

is observed, the red ray is seen.

• A drop lower in the sky would direct violet light to the observer.

• The other colors of the spectra lie in between the red and the violet.

Section 22.5

Read and take notes on

pages 746-749 in College Physics Text

Christian Huygens

• 1629 – 1695• Best known for

contributions to fields of optics and dynamics

• Deduced the laws of reflection and refraction

• Explained double refraction

Section 22.6

Huygen’s Principle

• Huygen assumed that light is a form of wave motion rather than a stream of particles.

• Huygen’s Principle is a geometric construction for determining the position of a new wave at some point based on the knowledge of the wave front that preceded it.

Section 22.6

Huygen’s Principle, Cont.

• All points on a given wave front are taken as point sources for the production of spherical secondary waves, called wavelets, which propagate in the forward direction with speeds characteristic of waves in that medium.– After some time has elapsed, the new position of

the wave front is the surface tangent to the wavelets.

Section 22.6

Huygen’s Construction for a Plane Wave

• At t = 0, the wave front is indicated by the plane AA’

• The points are representative sources for the wavelets.

• After the wavelets have moved a distance cΔt, a new plane BB’ can be drawn tangent to the wavefronts.

Section 22.6

Huygen’s Construction for a Spherical Wave

• The inner arc represents part of the spherical wave.• The points are representative points where wavelets are

propagated.• The new wavefront is tangent at each point to the wavelet.

Section 22.6

Huygen’s Principle and the Law of Reflection

• The Law of Reflection can be derived from Huygen’s Principle.• AA’ is a wave front of incident light.• The reflected wave front is CD.

Section 22.6

Huygen’s Principle and the Law of Reflection, Cont.

• Triangle ADC is congruent to triangle AA’C.• θ1 = θ1’• This is the Law of Reflection.

Section 22.6

Huygen’s Principle and the Law of Refraction

• In time Δt, ray 1 moves from A to B and ray 2 moves from A’ to C.

• From triangles AA’C and ACB, all the ratios in the Law of Refraction can be found.– n1 sin θ1 = n2 sin θ2

Section 22.6

Khan Academy on Total Internal Reflection

Total Internal Reflection• Total internal reflection can

occur when light attempts to move from a medium with a higher index of refraction to one with a lower index of refraction.– Ray 5 shows internal

reflection

Section 22.7

Critical Angle

• A particular angle of incidence will result in an angle of refraction of 90°– This angle of incidence is

called the critical angle.

Section 22.7

Critical Angle, Cont.

• For angles of incidence greater than the critical angle, the beam is entirely reflected at the boundary.– This ray obeys the Law of Reflection at the boundary.

• Total internal reflection occurs only when light is incident on the boundary of a medium having a lower index of refraction than the medium in which it is traveling.

Section 22.7

Fiber Optics• An application of internal

reflection• Plastic or glass rods are

used to “pipe” light from one place to another.

• Applications include– Medical use of fiber optic

cables for diagnosis and correction of medical problems

– Telecommunications

EXAMPLE 22.7

Goal Apply the concept of total internal reflection. Problem (a) Find the critical angle for a water-air boundary. (b) Use the result of part (a) to predict what a fish will see if it looks up toward the water surface at angles of 40.0°, 48.6°, and 60.0°.

Strategy After finding the critical angle by substitution, use that the path of a light ray is reversible: at a given angle, wherever a light beam can go is also where a light beam can come from, along the same path

SOLUTION

(a) Find the critical angle for a water-air boundary. Substitute to find the critical angle.

sin θc = n2

= 1.00

= 0.750 n1 1.333

θc = sin-1 (0.750) = 48.6°

(b) Predict what a fish will see if it looks up toward the water surface at angles of 40.0°, 48.6°, and 60.0°. At an angle of 40.0°, a beam of light from underwater will be refracted at the surface and enter the air above. Because the path of a light ray is reversible (Snell's law works both going and coming), light from above can follow the same path and be perceived by the fish. At an angle of 48.6°, the critical angle for water, light from underwater is bent so that it travels along the surface. So light following the same path in reverse can reach the fish only by skimming along the water surface before being refracted toward the fish's eye. At angles greater than the critical angle of 48.6°, a beam of light shot toward the surface will be completely reflected down toward the bottom of the pool. Reversing the path, the fish sees a reflection of some object on the bottom.

LEARN MORE

Question If the water is replaced by a transparent fluid with a higher index of refraction, the critical angle of the fluid-air boundary is:

larger. the same as for water. smaller.

Link to Webassign Unit 4 B Reflection and Refraction Discussion Questions

Link to Webassign Unit

4 B Reflection and Refraction Homework 1-7

Grading Rubric for Unit 4B Reflection and Refraction

Name: ______________________ Conceptual Physics Text Pgs 421-426 ---------------------------------------------------_____

Pgs 432-336 ---------------------------------------------------_____ Pgs 442-459 ---------------------------------------------------_____

Advanced notes from text book:

Pgs 732-735 --------------------------------------------------_____ Pgs 736-739 ---------------------------------------------------_____ Pgs 742-746 ---------------------------------------------------_____ Pgs 746-749 --------------------------------------------------_____

Example Problems:

22.1 (a) ----------------------------------------------------------_____ 22.2 (a) ----------------------------------------------------------_____ 22.3 (a-c) --------------------------------------------------------_____ 22.4 (a) ----------------------------------------------------------_____ 22.5 (a) ----------------------------------------------------------_____ 22.7 (a-b) -------------------------------------------------------_____ Web Assign 1 (a-d), 2 (a-c), 3 (a-d), 4, 5, 6, 7 (a-f) -----------------------____