Chapter 24Wave Optics

Waves and Optics Physical Optics (1.5 weeks)

Learning Objectives:

1. Interference and diffraction Students should understand the interference and diffraction of waves, so they can:

a) Apply the principles of interference to coherent sources in order to:

(1) Describe the conditions under which the waves reaching an observation point from two or more sources will all interfere constructively, or under which the waves from two sources will interfere destructively. (2) Determine locations of interference maxima or minima for two sources or determine the frequencies or wavelengths that can lead to constructive or destructive interference at a certain point. (3) Relate the amplitude produced by two or more sources that interfere constructively to the amplitude and intensity produced by a single source.

b) Apply the principles of interference and diffraction to waves that pass through a single or double slit or through a diffraction grating, so they can:

(1) Sketch or identify the intensity pattern that results when monochromatic waves pass through a single slit and fall on a distant screen, and describe how this pattern will change if the slit width or the wavelength of the waves is changed. (2) Calculate, for a single-slit pattern, the angles or the positions on a distant screen where the intensity is zero. (3) Sketch or identify the intensity pattern that results when monochromatic waves pass through a double slit, and identify which features of the pattern result from single-slit diffraction and which from two-slit interference. (4) Calculate, for a two-slit interference pattern, the angles or the positions on a distant screen at which intensity maxima or minima occur. (5) Describe or identify the interference pattern formed by a diffraction grating, calculate the location of intensity maxima, and explain qualitatively why a multiple-slit grating is better than a two-slit grating for making accurate determinations of wavelength.

c) Apply the principles of interference to light reflected by thin films, so they can:

(1) State under what conditions a phase reversal occurs when light is reflected from the interface between two media of different indices of refraction. (2) Determine whether rays of monochromatic light reflected perpendicularly from two such interfaces will interfere constructively or destructively, and thereby account for Newton’s rings and similar phenomena, and explain how glass may be coated to minimize reflection of visible light.

2. Dispersion of light and the electromagnetic spectrum Students should understand dispersion and the electromagnetic spectrum, so they can:

a) Relate a variation of index of refraction with frequency to a variation in refraction.

b) Know the names associated with electromagnetic radiation and be able to arrange in order of increasing wavelength the following: visible light of various colors, ultraviolet light, infrared light, radio waves, x-rays, and gamma rays.

Read and take notes on

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Wave Optics

• The wave nature of light is needed to explain various phenomena.– Interference– Diffraction– Polarization

• The particle nature of light was the basis for ray (geometric) optics.

Introduction

Interference

• Light waves interfere with each other much like mechanical waves do.

• All interference associated with light waves arises when the electromagnetic fields that constitute the individual waves combine.

Section 24.1

Conditions for Interference

• For sustained interference between two sources of light to be observed, there are two conditions which must be met.– The sources must be coherent.

• The waves they emit must maintain a constant phase with respect to each other.

– The waves must have identical wavelengths.

Section 24.1

Producing Coherent Sources• Light from a monochromatic source is allowed to

pass through a narrow slit.• The light from the single slit is allowed to fall on a

screen containing two narrow slits.• The first slit is needed to insure the light comes from

a tiny region of the source which is coherent.• Old method

Section 24.1

Producing Coherent Sources, Cont.

• Currently, it is much more common to use a laser as a coherent source.

• The laser produces an intense, coherent, monochromatic beam over a width of several millimeters.

• The laser light can be used to illuminate multiple slits directly.

Section 24.1

Link to a cool YouTube video on Double Slit Experiment

Link to Bright Storm on

Double Slit Experiment

Young’s Double Slit Experiment

• Thomas Young first demonstrated interference in light waves from two sources in 1801.

• Light is incident on a screen with a narrow slit, So

• The light waves emerging from this slit arrive at a second screen that contains two narrow, parallel slits, S1 and S2

Section 24.2

Young’s Double Slit Experiment, Diagram

• The narrow slits, S1 and S2 act as sources of waves.

• The waves emerging from the slits originate from the same wave front and therefore are always in phase.

Section 24.2

Resulting Interference Pattern• The light from the two slits form a visible pattern on

a screen.• The pattern consists of a series of bright and dark

parallel bands called fringes.• Constructive interference occurs where a bright

fringe appears.• Destructive interference results in a dark fringe.

Section 24.2

Fringe Pattern• The fringe pattern formed

from a Young’s Double Slit Experiment would look like this.

• The bright areas represent constructive interference.

• The dark areas represent destructive interference.

Section 24.2

Interference Patterns

• Constructive interference occurs at the center point.

• The two waves travel the same distance.– Therefore, they arrive in

phase.

Section 24.2

Interference Patterns, 2• The upper wave has to

travel farther than the lower wave.

• The upper wave travels one wavelength farther.– Therefore, the waves arrive in

phase.• A bright fringe occurs.

Section 24.2

Interference Patterns, 3• The upper wave travels one-

half of a wavelength farther than the lower wave.

• The trough of the bottom wave overlaps the crest of the upper wave.

• This is destructive interference.– A dark fringe occurs.

Section 24.2

Geometry of Young’s Double Slit Experiment

Section 24.2

Interference Equations• The path difference, δ, is

found from the small triangle.

• δ = r2 – r1 = d sin θ– This assumes the paths are

parallel.– Not exactly parallel, but a

very good approximation since L is much greater than d

Section 24.2

Interference Equations, 2

• For a bright fringe, produced by constructive interference, the path difference must be either zero or some integral multiple of the wavelength.

• δ = d sin θbright = m λ– m = 0, ±1, ±2, … – m is called the order number.

• When m = 0, it is the zeroth order maximum.• When m = ±1, it is called the first order maximum.

Section 24.2

Interference Equations, 3

• When destructive interference occurs, a dark fringe is observed.

• This needs a path difference of an odd half wavelength.

• δ = d sin θdark = (m + ½) λ– m = 0, ±1, ±2, …

Section 24.2

Interference Equations, 4• The positions of the fringes can be measured

vertically from the zeroth order maximum.• y = L tan θ L sin θ• Assumptions

– L >> d– d >> λ

• Approximation– θ is small and therefore the approximation tan θ sin θ

can be used.• The approximation is true to three-digit precision only for angles

less than about 4°

Section 24.2

Interference Equations, Final

• For bright fringes

• For dark fringes

Section 24.2

Uses for Young’s Double Slit Experiment

• Young’s Double Slit Experiment provides a method for measuring wavelength of the light.

• This experiment gave the wave model of light a great deal of credibility.– It is inconceivable that particles of light could

cancel each other.

Section 24.2

Since there is no Khan academy videos on wave optics you will watch Walter Lewin from MIT. These videos are appropriate for AP Physics B students. This one lesson is broken down into two parts totaling one hour and twenty minutes. These are long but a must watch!

Link to Walter Lewin on double slit and interference

(start to minute 30:36) Link to sound interference Optional

(30:36-40:20)

EXAMPLE 24.1 Measuring the Wavelength of a Light Source

Goal Show how Young's experiment can be used to measure the wavelength of coherent light. Problem A screen is separated from a double-slit source by 1.20 m. The distance between the two slits is 0.0300 mm. The second-order bright fringe (m = 2) is measured to be 4.50 cm from the centerline. Determine (a) the wavelength of the light and (b) the distance between adjacent bright fringes. Strategy The bright fringe equation relates the positions of the bright fringes to the other variables, including the wavelength of the light. Substitute into this equation and solve for λ. Taking the difference between ym+1 and ym results in a general expression for the distance between bright fringes.

SOLUTION

(a) Determine the wavelength of the light. Solve the bright fringe equation for the wavelength and substitute the values m = 2, y2 = 4.50 10-2 m, L = 1.20 m, and d = 3.00 10-5 m.

λ = y2d

= (4.50 10-2 m)(3.00 10-5 m)

mL 2(1.20 m) = 5.63 10-7 m = 563 nm

(b) Determine the distance between adjacent bright fringes.

Use the bright fringe equation again to find the distance between any adjacent bright fringes (here, those characterized by m and m + 1).

Δy = ym +1 - ym = λL

(m + 1) - λL

m = λL

d d d

= (5.63 10-7 m)(1.20 m) = 2.25

cm 3.00 10-5 m

LEARN MORE

Remarks This calculation depends on the angle θ being small because the small angle approximation was implicitly used. The measurement of the position of the bright fringes yields the wavelength of light, which in turn is a signature of atomic processes, as is discussed in the chapters on modern physics. This kind of measurement therefore helped open the world of the atom. Question Which of the following make the separation between fringes greater in the two slit interference experiment? (Select all that apply.)

Narrower slits. Smaller separation of the two slits. Wider slits.Larger separation of the two slits.

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Lloyd’s Mirror• An arrangement for

producing an interference pattern with a single light source

• Waves reach point P either by a direct path or by reflection.

• The reflected ray can be treated as a ray from the source S’ behind the mirror.

Section 24.3

Interference Pattern from the Lloyd’s Mirror

• An interference pattern is formed.• The positions of the dark and bright fringes

are reversed relative to pattern of two real sources.

• This is because there is a 180° phase change produced by the reflection.

Section 24.3

Phase Changes Due To Reflection

• An electromagnetic wave undergoes a phase change of 180° upon reflection from a medium of higher index of refraction than the one in which it was traveling.– Analogous to a reflected pulse on a string

Section 24.3

Phase Changes Due To Reflection, Cont.

• There is no phase change when the wave is reflected from a boundary leading to a medium of lower index of refraction.– Analogous to a pulse in a string reflecting from a free support

Section 24.3

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Interference in Thin Films

• Interference effects are commonly observed in thin films.– Examples are soap bubbles and oil on water

• The interference is due to the interaction of the waves reflected from both surfaces of the film.

Section 24.4

Interference in Thin Films, 2

• Facts to remember– An electromagnetic wave traveling from a medium

of index of refraction n1 toward a medium of index of refraction n2 undergoes a 180° phase change on reflection when n2 > n1

• There is no phase change in the reflected wave if n2 < n1

– The wavelength of light λn in a medium with index of refraction n is λn = λ/n where λ is the wavelength of light in vacuum.

Section 24.4

Interference in Thin Films, 3• Ray 1 undergoes a phase

change of 180° with respect to the incident ray.

• Ray 2, which is reflected from the lower surface, undergoes no phase change with respect to the incident wave.

Section 24.4

Interference in Thin Films, 4

• Ray 2 also travels an additional distance of 2t before the waves recombine.

• For constructive interference– 2 n t = (m + ½ ) λ m = 0, 1, 2 …

• This takes into account both the difference in optical path length for the two rays and the 180° phase change

• For destructive interference– 2 n t = m λ m = 0, 1, 2 …

Section 24.4

Interference in Thin Films, 5• Two factors influence interference.

– Possible phase reversals on reflection– Differences in travel distance

• The conditions are valid if the medium above the top surface is the same as the medium below the bottom surface.

• If the thin film is between two different media, one of lower index than the film and one of higher index, the conditions for constructive and destructive interference are reversed.

Section 24.4

Interference in Thin Films, Final

• Be sure to include two effects when analyzing the interference pattern from a thin film.– Path length– Phase change

Section 24.4

Newton’s Rings• Another method for viewing interference is to

place a planoconvex lens on top of a flat glass surface.

• The air film between the glass surfaces varies in thickness from zero at the point of contact to some thickness t.

• A pattern of light and dark rings is observed.– These rings are called Newton’s Rings.– The particle model of light could not explain the origin of the rings.

• Newton’s Rings can be used to test optical lenses.

Newton’s Rings, Diagram

Section 24.4

Problem Solving Strategy with Thin Films, 1

• Identify the thin film causing the interference.• Determine the indices of refraction in the film

and the media on either side of it.• Determine the number of phase reversals:

zero, one or two.

Section 24.4

Problem Solving with Thin Films, 2

• The interference is constructive if the path difference is an integral multiple of λ and destructive if the path difference is an odd half multiple of λ.– The conditions are reversed if one of the waves undergoes

a phase change on reflection.• Substitute values in the appropriate equation.• Solve and check.

Section 24.4

Problem Solving with Thin Films, 3

Equationm = 0, 1, 2, …

1 phase reversal

0 or 2 phase reversals

2nt = (m + ½) l constructive destructive

2nt = m l destructive constructive

Section 24.4

Interference in Thin Films, Example

• An example of different indices of refraction

• A coating on a solar cell• There are two phase

changes

Section 24.4

Link to Walter Lewin on Thin Film

(From minute 40:20 - 53:00 then 56:00 – 117:00 )

EXAMPLE 24.2 Interference in a Soap Film

Goal Study constructive interference effects in a thin film. Problem (a) Calculate the minimum thickness of a soap-bubble film (n = 1.33) that will result in constructive interference in the reflected light if the film is illuminated by light with wavelength 602 nm in free space. (b) Recalculate the minimum thickness for constructive interference when the soap-bubble film is on top of a glass slide with n = 1.50. Strategy In part (a) there is only one inversion, so the condition for constructive interference is 2nt = (m + 1/2)λ. The minimum film thickness for constructive interference corresponds to m = 0 in this equation. Part (b) involves two inversions, so 2nt = mλ is required

SOLUTION

(a) Calculate the minimum thickness of the soap bubble film that will result in constructive interference. Solve 2nt = λ/2 for the thickness t and substitute.

t = λ

= 602 nm

= 113 nm 4n 4(1.33)

(b) Find the minimum soap film thickness when the film is on top of a glass slide with n = 1.33.

Write the condition for constructive interference, when two inversions take place.

2nt = mλ

Solve for t and substitute.

t = mλ

= 1·(602 nm)

= 226 nm 2n 2(1.33)

LEARN MORE

Remarks The swirling colors in a soap bubble result from the thickness of the soap layer varying from one place to another. Question A soap film looks red in one area and violet in a nearby area. In which area is the soap film thicker?

In the red region, because red light has a shorter wavelength than violet light.

In the red region, because red light has a longer wavelength than violet light.

In the violet region, because red light has a shorter wavelength than violet light.

In the violet region, because red light has a longer wavelength than violet light.

EXAMPLE 24.3 Nonreflective Coatings for Solar Cells and Optical Lenses

Reflective losses from a silicon solar cell are minimized by coating it with a thin film of silicon monoxide (SiO). Goal Study destructive interference effects in a thin film when there are two inversions. Problem Semiconductors such as silicon are used to fabricate solar cells, devices that generate electric energy when exposed to sunlight. Solar cells are often coated with a transparent thin film, such as silicon monoxide (SiO; n = 1.45), to minimize reflective losses. A silicon solar cell (n = 3.50) is coated with a thin film of silicon monoxide for this

purpose. Assuming normal incidence, determine the minimum thickness of the film that will produce the least reflection at a wavelength of 552 nm. Strategy Reflection is least when rays 1 and 2 in the figure meet the condition for destructive interference. Note that both rays undergo 180° phase changes on reflection. The condition for a reflection minimum is therefore 2nt = λ/2.

SOLUTION

Solve 2nt = λ/2 for t, the required thickness.

t = λ

= 552 nm

= 95.2 nm 4n 4(1.45)

LEARN MORE

Remarks Typically, such coatings reduce the reflective loss from 30% (with no coating) to 10% (with a coating), thereby increasing the cells efficiency because more light is available to create charge carriers in the cell. In reality the coating is never perfectly nonreflecting because the required thickness is wavelength dependent and the incident light covers a wide range of wavelengths. Question To minimize reflection of light with a smaller wavelength, should the thickness of the coating be thicker or thinner?

thicker thinner

EXAMPLE 24.4 Interference in a Wedge-Shaped Film

Interference bands in reflected light can be observed by illuminating a wedge-shaped film with monochromatic light. The dark areas in the interference pattern correspond to positions of destructive interference. Goal Calculate interference effects when the film has variable thickness.

Problem A pair of glass slides 10.0 cm long and with n = 1.52 are separated on one end by a hair, forming a triangular wedge of air, as illustrated in the figure. When coherent light from a helium—neon laser with wavelength 633 nm is incident on the film from above, 15.0 dark fringes per centimeter are observed. How thick is the hair? Strategy The interference pattern is created by the thin film of air having variable thickness. The pattern is a series of alternating bright and dark parallel bands. A dark band corresponds to destructive interference, and there is one phase reversal, so 2nt = mλ should be used. We can also use the similar triangles in the figure to obtain the relation t/x = D/L. We can find the thickness for any m, and if the position x can also be found, this last equation gives the diameter of the hair, D.

SOLUTION

Solve the destructive-interference equation for the thickness of the film, t, with n = 1 for air. t = mλ/2

If d is the distance from one dark band to the next, then the x-coordinate of the mth band is a multiple of d. x = md

By dimensional analysis, d is just the inverse of the number of bands per centimeter. d = (15.0 bands/cm)-1 = 6.67 10-2 cm/band

Now use similar triangles and substitute all the information. t

= mλ/2

= λ

= D

x md 2d L Solve for D and substitute given values.

D = λL

= (6.33 10-7 m)(0.100 m)

= 4.75 10-5 m 2d 2(6.67 10-4 m)

LEARN MORE

Remarks Some may be concerned about interference caused by light bouncing off the top and bottom of, say, the upper glass slide. It's unlikely, however, that the thickness of the slide will be half an integer multiple of the wavelength of the helium neon laser (for some very large value of m). In addition, in contrast to the air wedge, the thickness of the glass doesn't vary. Question If the air wedge is filled with water, how are the wavelength and the distance between dark bands affected? (Select all that apply.)

The wavelength decreases. The interference bands get closer

together. The interference bands get farther apart. The wavelength increases.

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CD’s and DVD’s• Data is stored digitally.

– A series of ones and zeros read by laser light reflected from the disk

• Strong reflections correspond to constructive interference.– These reflections are chosen to represent zeros.

• Weak reflections correspond to destructive interference.– These reflections are chosen to represent ones.

Section 24.5

CD’s and Thin Film Interference

• A CD has multiple tracks.– The tracks consist of a sequence of pits of varying

length formed in a reflecting information layer.• The pits appear as bumps to the laser beam.

– The laser beam shines on the metallic layer through a clear plastic coating.

Section 24.5

Reading a CD• As the disk rotates, the laser

reflects off the sequence of bumps and lower areas into a photodector.– The photodector converts the

fluctuating reflected light intensity into an electrical string of zeros and ones.

• The pit depth is made equal to one-quarter of the wavelength of the light.

Section 24.5

Reading a CD, Cont.• When the laser beam hits a rising or falling bump

edge, part of the beam reflects from the top of the bump and part from the lower adjacent area.– This ensures destructive interference and very low

intensity when the reflected beams combine at the detector.

• The bump edges are read as ones.• The flat bump tops and intervening flat plains are

read as zeros.

Section 24.5

DVD’s

• DVD’s use shorter wavelength lasers.– The track separation, pit depth and minimum pit

length are all smaller.– Therefore, the DVD can store about 30 times more

information than a CD.

Section 24.5

EXAMPLE 24.5 Pit Depth in a CD

Cross section of a CD showing metallic pits of depth t and a laser beam detecting the edge of a pit. Goal Apply interference principles to a CD. Problem Find the pit depth in a CD that has a plastic transparent layer with index of refraction of 1.60 and is designed for use in a CD player using a laser with a wavelength of 7.80 102 nm in air. Strategy Rays 1 and 2 (see figure) both reflect from the metal layer, which acts like a mirror, so there is no phase difference due to reflection between those rays. There is, however, the usual phase difference caused by the extra distance 2t traveled by ray 2. The wavelength is λ/n, where n is the index of refraction in the substance.

SOLUTION

Use the appropriate condition for destructive interference in a thin film. 2t = λ/2n

Solve for the thickness t and substitute.

t = λ

= 7.80 102 nm

= 1.22 102 nm 4n (4)(1.60)

LEARN MORE

Remarks Different CD systems have different tolerances for scratches. Anything that changes the reflective properties of the disk can affect the readability of the disk. Question Given two plastics with different indices of refraction, the material with the larger index of refraction will have:

a larger pit depth. a smaller pit depth. the same pit depth.

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Diffraction• Huygen’s principle requires

that the waves spread out after they pass through slits.

• This spreading out of light from its initial line of travel is called diffraction.– In general, diffraction occurs

when waves pass through small openings, around obstacles or by sharp edges.

Section 24.6

Diffraction, 2

• A single slit placed between a distant light source and a screen produces a diffraction pattern.– It will have a broad, intense central band.– The central band will be flanked by a series of narrower,

less intense secondary bands.• Called secondary maxima

– The central band will also be flanked by a series of dark bands.

• Called minima

Section 24.6

Diffraction, 3• The results of the single slit

cannot be explained by geometric optics.– Geometric optics would say

that light rays traveling in straight lines should cast a sharp image of the slit on the screen.

Section 24.6

Fraunhofer Diffraction• Fraunhofer Diffraction

occurs when the rays leave the diffracting object in parallel directions.– Screen very far from the slit– Converging lens (shown)

• A bright fringe is seen along the axis (θ = 0) with alternating bright and dark fringes on each side.

Section 24.6

Single Slit Diffraction• According to Huygen’s

principle, each portion of the slit acts as a source of waves.

• The light from one portion of the slit can interfere with light from another portion.

• The resultant intensity on the screen depends on the direction θ

Section 24.7

Single Slit Diffraction, 2

• All the waves that originate at the slit are in phase.• Wave 1 travels farther than wave 3 by an amount

equal to the path difference (a/2) sin θ – a is the width of the slit

• If this path difference is exactly half of a wavelength, the two waves cancel each other and destructive interference results.

Section 24.7

Single Slit Diffraction, 3

• In general, destructive interference occurs for a single slit of width a when sin θdark = mλ / a– m = 1, 2, 3, …

• Doesn’t give any information about the variations in intensity along the screen

Section 24.7

Single Slit Diffraction, 4• The general features of the

intensity distribution are shown.

• A broad central bright fringe is flanked by much weaker bright fringes alternating with dark fringes.

• The points of constructive interference lie approximately halfway between the dark fringes.

Section 24.7

Diffraction Grating

• The diffracting grating consists of many equally spaced parallel slits.– A typical grating contains several thousand lines

per centimeter.• The intensity of the pattern on the screen is

the result of the combined effects of interference and diffraction.

Section 24.8

EXAMPLE 24.6 A Single-Slit Experiment

Goal Find the positions of the dark fringes in single-slit diffraction. Problem Light of wavelength 5.80 102 nm is incident on a slit of width 0.300 mm. The observing screen is placed 2.00 m from the slit. Find the positions of the first dark fringes and the width of the central bright fringe. Strategy This problem requires substitution to find the sines of the angles of the first dark fringes. The positions can then be found with the tangent function because for small angles sin θ ≈ tan θ. The extent of the central maximum is defined by these two dark fringes.

SOLUTION

The first dark fringes that flank the central bright fringe correspond to m = ±1.

sin θ = ± λ

= 5.80 10-7 m

= ±1.93 10-3 a 3.00 10-4 m

Relate the position of the fringe to the tangent function. tan θ = y1/L

Because θ is very small, we can use the approximation sin θ ≈ tan θ and then solve for y1. sin θ ≈ tan θ ≈ y1/ L

y1 ≈ L sin θ = (2.00 m)(±1.93 10-3) = ±3.86 10-3 m

Compute the distance between the positive and negative first—order maxima, which is the width w of the central maximum. w = +3.86 10-3 m - (-3.86 10-3 m) = 7.72 10-3 m

LEARN MORE

Remarks Note that this value of w is much greater than the width of the slit. As the width of the slit is increased, however, the diffraction pattern narrows, corresponding to smaller values of θ. In fact, for large values of a, the maxima and minima are so closely spaced that the only observable pattern is a large central bright area resembling the geometric image of the slit. Because the width of the geometric image increases as the slit width increases, the narrowest image occurs when the geometric and diffraction widths are equal. Question Suppose the entire apparatus is immersed in water. If the same wavelength of light (in air) is incident on the slit immersed in water, is the resulting central maximum larger or smaller? Explain. (Select all that apply.)

The wavelength has become smaller. The central maximum width is

smaller. The wavelength has become larger. The wavelength has remained

the same. The central maximum is wider. The central maximum width is the same.

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Diffraction Grating, Cont.• The condition for maxima is

– d sin θbright = m λ• m = 0, ±1, ±2, …

• The integer m is the order number of the diffraction pattern.

• If the incident radiation contains several wavelengths, each wavelength deviates through a specific angle.

Section 24.8

Diffraction Grating, Final• All the wavelengths are

focused at m = 0– This is called the zeroth order

maximum• The first order maximum

corresponds to m = 1• Note the sharpness of the

principle maxima and the broad range of the dark area.– This is in contrast to the

broad, bright fringes characteristic of the two-slit interference pattern.

Section 24.8

Diffraction Grating in CD Tracking• A diffraction grating can be

used in a three-beam method to keep the beam on a CD on track.

• The central maximum of the diffraction pattern is used to read the information on the CD.

• The two first-order maxima are used for steering.

Section 24.8

EXAMPLE 24.7 A Diffraction Grating

Goal Calculate different—order principal maxima for a diffraction grating. Problem Monochromatic light from a helium—neon laser (λ = 632.8 nm) is incident normally on a diffraction grating containing 6.00 103 lines/cm. Find the angles at which one would observe the first order maximum, the second-order maximum, and so forth. Strategy Find the slit separation by inverting the number of lines per centimeter, then substitute values into the interference pattern maxima equation.

SOLUTION

Invert the number of lines per centimeter to obtain the slit separation.

d = 1

= 1.67 10-4 cm = 1.67 103 nm 6.00 103 cm-1

Substitute m = 1 to find the sine of the angle corresponding to the first order maximum.

sin θ1 = λ

= 632.8 nm

= 0.379 d 1.67 103 nm

Take the inverse sine of the preceding result to find θ1 θ1 = sin-1 0.379 = 22.3°

Repeat the calculation for m = 2.

sin θ2 = 2λ

= 2(632.8 nm)

= 0.758 d 1.67 103 nm

θ2 = 49.3°

Repeat the calculation for m = 3.

sin θ3 = 3λ

= 3(632.8 nm)

= 1.14 d 1.67 103 nm

Because sin θ can't exceed 1, there is no solution for θ3.

LEARN MORE

Remarks The foregoing calculation shows that there can only be a finite number of principal maxima. In this case only zeroth-, first-, and second-order maxima would be observed. Question If a diffraction grating has more lines per centimeter, how does that affect the spacing of the lines on the grating and the separation between adjacent principal maxima? (Select all that apply.)

The separation between adjacent maxima increases. The spacing of the

lines on the grating increases. The spacing of the lines on the grating

decreases. The separation between adjacent maxima decreases. The spacing of the lines on the grating remains the same.

Link to WebassignDiscussion Questions

Unit 4C Physics /Wave Optics

Link to Webassign

Unit 4 C Physical / Wave Optics Homework

Chapter 21Electromagnetic Waves

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Link to Bright Storm on Electromagnetic Waves

Electromagnetic Waves Produced by an Antenna

• When a charged particle undergoes an acceleration, it must radiate energy.– If currents in an ac circuit change rapidly, some energy is

lost in the form of EM waves.– EM waves are radiated by any circuit carrying alternating

current.• An alternating voltage applied to the wires of an

antenna forces the electric charges in the antenna to oscillate.

Section 21.10

EM Waves by an Antenna, Cont.

• Two rods are connected to an ac source, charges oscillate between the rods (a).

• As oscillations continue, the rods become less charged, the field near the charges decreases and the field produced at t = 0 moves away from the rod (b).

• The charges and field reverse (c).• The oscillations continue (d).

Section 21.10

EM Waves by an Antenna, Final• Because the oscillating

charges in the rod produce a current, there is also a magnetic field generated.

• As the current changes, the magnetic field spreads out from the antenna.

• The magnetic field is perpendicular to the electric field.

Section 21.10

Charges and Fields, Summary

• Stationary charges produce only electric fields.• Charges in uniform motion (constant velocity)

produce electric and magnetic fields.• Charges that are accelerated produce electric and

magnetic fields and electromagnetic waves.• An accelerating charge also radiates energy.

Section 21.10

Electromagnetic Waves, Summary

• A changing magnetic field produces an electric field.

• A changing electric field produces a magnetic field.

• These fields are in phase.– At any point, both fields reach their maximum

value at the same time.

Section 21.10

Electromagnetic Waves are Transverse Waves

• The and fields are perpendicular to each other.

• Both fields are perpendicular to the direction of motion.– Therefore, EM waves are

transverse waves.

Section 21.10

Properties of EM Waves• Electromagnetic waves are transverse waves.• Electromagnetic waves travel at the speed of light.

– Because EM waves travel at a speed that is precisely the speed of light, light is an electromagnetic wave.

Section 21.11

Properties of EM Waves, 2• The ratio of the electric field to the magnetic field is

equal to the speed of light.

• Electromagnetic waves carry energy as they travel through space, and this energy can be transferred to objects placed in their path.

Section 21.11

Properties of EM Waves, 3

• Energy carried by EM waves is shared equally by the electric and magnetic fields.

– Intensity (I) is average power per unit area.

Section 21.11

Properties of EM Waves, Final

• Electromagnetic waves transport linear momentum as well as energy.– For complete absorption of energy U, p=U/c– For complete reflection of energy U, p=(2U)/c

• Radiation pressures can be determined experimentally.

Section 21.11

Determining Radiation Pressure• This is an apparatus for

measuring radiation pressure.

• In practice, the system is contained in a vacuum.

• The pressure is determined by the angle at which equilibrium occurs.

Section 21.11

Properties of EM Waves, Summary

• EM waves travel at the speed of light.• EM waves are transverse waves because the electric

and magnetic fields are perpendicular to the direction of propagation of the wave and to each other.

• The ratio of the electric field to the magnetic field in an EM wave equals the speed of light.

• EM waves carry both energy and momentum, which can be delivered to a surface.

Section 21.11

Read and take notes on

pgs. 720-723in College Physics

Text

Link to Bright Storm on Electromagnetic Spectrum

The Spectrum of EM Waves

• Forms of electromagnetic waves exist that are distinguished by their frequencies and wavelengths.– c = ƒλ

• Wavelengths for visible light range from 400 nm to 700 nm.

• There is no sharp division between one kind of EM wave and the next.

Section 21.12

The EM Spectrum• Note the overlap between

types of waves.• Visible light is a small

portion of the spectrum.• Types are distinguished by

frequency or wavelength.

Section 21.12

Notes on The EM Spectrum

• Radio Waves– Used in radio and television communication

systems• Microwaves

– Wavelengths from about 1 mm to 30 cm– Well suited for radar systems– Microwave ovens are an application

Section 21.12

Notes on the EM Spectrum, 2• Infrared waves

– Incorrectly called “heat waves”– Produced by hot objects and molecules– Wavelengths range from about 1 mm to 700 nm– Readily absorbed by most materials

• Visible light– Part of the spectrum detected by the human eye– Wavelengths range from 400 nm to 700 nm– Most sensitive at about 560 nm (yellow-green)

Section 21.12

Notes on the EM Spectrum, 3• Ultraviolet light

– Covers about 400 nm to 0.6 nm– The Sun is an important source of uv light.– Most uv light from the sun is absorbed in the stratosphere

by ozone.• X-rays

– Wavelengths range from about 10 nm to 10-4 nm– Most common source is acceleration of high-energy

electrons striking a metal target– Used as a diagnostic tool in medicine

Section 21.12

Notes on the EM Spectrum, Final

• Gamma rays– Wavelengths from about 10-10 m to 10-14 m – Emitted by radioactive nuclei– Highly penetrating and cause serious damage

when absorbed by living tissue• Looking at objects in different portions of the

spectrum can produce different information.

Section 21.12

Crab Nebula in Various Wavelengths

Section 21.12

Grading Rubric for Unit 4C Physical and Wave Optics Name: ______________________ Conceptual Physics Text Pgs 480-490---------------------------------------------------_____

Pgs 490-494---------------------------------------------------_____ Advanced notes from text book:

Pgs 790-794 --------------------------------------------------_____ Pgs 795-798 --------------------------------------------------_____ Pgs 800-801 --------------------------------------------------_____ Pgs 802-804 --------------------------------------------------_____ Pgs 805-807 --------------------------------------------------_____ Pgs 714-717 --------------------------------------------------_____ Pgs 720-723 --------------------------------------------------_____

Example Problems: 24.1 a,b-------------------------------------------------------------_____ 24.2 a -------------------------------------------------------------_____ 24.3 a -------------------------------------------------------------_____ 24.4 a -------------------------------------------------------------_____ 24.5 a -------------------------------------------------------------_____ 24.6 a -------------------------------------------------------------_____ 24.7 a -------------------------------------------------------------_____ Web Assign 1 (a), 2 (a-c), 3 (a-b), 4 (a-b), 5 (a), 6 (a4), 7 (a), 8 (a), 9 (a3-b3) 10 (a) ------____