1. 2 Today’s agendum: Electromagnetic Waves. Energy Carried by Electromagnetic Waves. Momentum and Radiation Pressure of an Electromagnetic Wave

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2

Today’s agendum:

Electromagnetic Waves.

Energy Carried by Electromagnetic Waves.

Momentum and Radiation Pressure of an Electromagnetic Wave.

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We began this course by studying fields that didn’t vary with time—the electric field due to static charges, and the magnetic field due to a constant current.

In case you didn’t notice—about five lectures ago things started moving!

We found that changing magnetic field gives rise to an electric field. Also a changing electric field gives rise to a magnetic field.

These time-varying electric and magnetic fields can propagate through space.

4

Electromagnetic Waves

These four equations provide a complete description of electromagnetism.

0

E

02

1 dEB= +μ J

c dt

B 0

dB×E=-

dt

Maxwell’s Equations

dt

dsdE B

o

enclosedqAdE

0AdB

dt

dI sdB E

oenclo

5

Production of Electromagnetic Waves

Apply a sinusoidal voltage to an antenna.

Charged particles in the antenna oscillate sinusoidally.

The accelerated charges produce sinusoidally varying electric and magnetic fields, which extend throughout space.

The fields do not instantaneously permeate all space, but propagate at the speed of light.

direction ofpropagation

y

z

x

6

http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=35

This static image doesn’t show how the wave propagates.

Here are a couple of animations, available on-line:


y

z

x

http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=35

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Electromagnetic waves are transverse waves, but are not mechanical waves (they need no medium to vibrate in).


Therefore, electromagnetic waves can propagate in free space.

At any point, the magnitudes of E and B (of the wave shown) depend only upon x and t, and not on y or z. A collection of such waves is called a plane wave.

y

z

x

8

Manipulation of Maxwell’s equations leads to the following plane wave equations for E and B:

These equations have solutions:

You can verify this by direct substitution.

2 2y y

0 02 2

E E (x,t)=

x t

2 2z z

0 02 2

B B (x,t)=

x t

y maxE =E sin kx- t

z maxB =B sin kx- t

where , ,

2

k= =2 f and f = =c.k

Emax and Bmax in these notes are sometimes written by others as E0 and B0.

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You can also show that

At every instant, the ratio of the magnitude of the electric field to the magnitude of the magnetic field in an electromagnetic wave equals the speed of light.

y zE B

=-x t

max maxE k cos kx- t =B cos kx- t

.

max

max 0 0

E E 1= = =c=

B B k

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Summary of Important Properties of Electromagnetic WavesThe solutions of Maxwell’s equations are wave-like with both E and B satisfying a wave equation.

y maxE =E sin kx- t

z maxB =B sin kx- t

Electromagnetic waves travel through empty space with the speed of light c = 1/(00)½.

Emax and Bmax are the electric and magnetic field amplitudes.

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Summary of Important Properties of Electromagnetic WavesThe components of the electric and magnetic fields of plane EM waves are perpendicular to each other and perpendicular to the direction of wave propagation. The latter property says that EM waves are transverse waves.

The magnitudes of E and B in empty space are related byE/B = c.

max

max

E E= = =c

B B k


y

z

x

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Today’s agendum:




13

The magnitude S represents the rate at which energy flows through a unit surface area perpendicular to the direction of wave propagation.

Energy Carried by Electromagnetic Waves

Electromagnetic waves carry energy, and as they propagate through space they can transfer energy to objects in their path. The rate of flow of energy in an electromagnetic wave is described by a vector S, called the Poynting vector.*

0

1S= E B

Thus, S represents power per unit area. The direction of S is along the direction of wave propagation. The units of S are J/(s·m2) =W/m2.

*J. H. Poynting, 1884.

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z

x

y

c

SE

B Because B = E/c we can write

These equations for S apply at any instant of time and represent the instantaneous rate at which energy is passing through a unit area.

0

1S= E B

For an EM wave

so

E B =EB

.0

EBS=

.

2 2

0 0

E cBS= =

c

15

The time average of sin2(kx - t) is ½, so

2 2

0 0 0

EB E cBS= = =

c

EM waves are sinusoidal.

The average of S over one or more cycles is called the wave intensity I.

2 2max max max max

average0 0 0

E B E cBI =S = S = = =

2 2 c 2

y maxE =E sin kx- t

z maxB =B sin kx- t

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The energy densities (energy per unit volume) associated with electric field and magnetic fields are:

Using B = E/c and c = 1/(00)½ we can write

Energy Density

2E 0

1u = E

2

2

B0

1Bu =

2

222

20 0B 0

0 0 0

EE1B 1 1 1cu = = = = E

2 2 2 2

22

B E 00

1 1Bu =u = E =

2 2

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For an electromagnetic wave, the instantaneous energy density associated with the magnetic field equals the instantaneous energy density associated with the electric field.

22

B E 00

1 1Bu =u = E =

2 2

22

B E 00

Bu=u +u = E =

Hence, in a given volume the energy is equally shared by the two fields. The total energy density is equal to the sum of the energy densities associated with the electric and magnetic fields:

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When we average this instantaneous energy density over one or more cycles of an electromagnetic wave, we again get a factor of ½ from the time average of sin2(kx - t).

so we see thatRecall

The intensity of an electromagnetic wave equals the average energy density multiplied by the speed of light.

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B E 00

Bu=u +u = E =

22 max

0 max0

B1 1u = E =

2 2

2 2max max

average0 0

E cB1 1S = S = =

2 c 2.S =c u

, 2E 0 max

1u = E

4,

2max

B0

B1u =

4and

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Example: a radio station on the surface of the earth radiates a sinusoidal wave with an average total power of 50 kW. Assuming the wave is radiated equally in all directions above the ground, find the amplitude of the electric and magnetic fields detected by a satellite 100 km from the antenna.

R

Station

SatelliteAll the radiated power passes through the hemispherical surface* so the average power per unit area (the intensity) is

4

-7 222 5

average

5.00 10 Wpower PI = = = =7.96 10 W m

area 2 R 2 1.00 10 m

*In problems like this you need to ask whether the power is radiated into all space or into just part of

space.

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R

Station

Satellite

2max

0

E1I = S =

2 c

max 0E = 2 cI

-7 8 -7= 2 4 10 3 10 7.96 10

-2 V=2.45 10 m

-2

-11maxmax 8

V2.45 10E mB = = =8.17 10 Tc 3 10 m s

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Example: for the radio station in the example on the previous two slides, calculate the average energy densities associated with the electric and magnetic field.

2E 0 max

1u = E

4

2 -12 -2

E

1u = 8.85 10 2.45 10

4

-15E 3

Ju =1.33 10

m

2max

B0

B1u =

4

2

-11

B -7

8.17 101u =

4 4 10

-15B 3

Ju =1.33 10

m

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Today’s agendum:




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Momentum and Radiation Pressure

EM waves carry linear momentum as well as energy. When this momentum is absorbed at a surface pressure is exerted on that surface.

If we assume that EM radiation is incident on an object for a time t and that the radiation is entirely absorbed by the object, then the object gains energy U in time t.Maxwell showed that the momentum change of the object is then:

The direction of the momentum change of the object is in the direction of the incident radiation.

incident

Up = (total absorption)

c

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If instead of being totally absorbed the radiation is totally reflected by the object, and the reflection is along the incident path, then the magnitude of the momentum change of the object is twice that for total absorption.

The direction of the momentum change of the object is again in the direction of the incident radiation.

2 Up = (total reflection along incident path)

c

incident

reflected

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The radiation pressure on the object is defined as the force per unit area:

From Newton’s 2nd Law (F = dp/dt) we have:

For total absorption,

So

Radiation Pressure

FP=

A

F 1 dpP= =

A A dt

Up=

c

dU1 dp 1 d U 1 SdtP= = = =A dt A dt c c A c

incident

(Equations on this slide involve magnitudes of vector quantities.)

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This is the instantaneous radiation pressure in the case of total absorption:

SP=

c

For the average radiation pressure, replace S by <S>=Savg=I:

averagerad

S IP = =

c c

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Following similar arguments it can be shown that:

rad

IP = (total absorption)

c

rad

2IP = (total reflection)

c

incident

incident

reflected

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Example: a satellite orbiting the earth has solar energy collection panels with a total area of 4.0 m2. If the sun’s radiation is incident perpendicular to the panels and is completely absorbed find the average solar power absorbed and the average force associated with the radiation pressure. The intensity (I or Saverage) of sunlight prior to passing through the earth’s atmosphere is 1.4 kW/m2.

3 2 32

WPower=IA= 1.4 10 4.0 m =5.6 10 W=5.6 kWm

Assuming total absorption of the radiation:

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average -6rad 8

W1.4 10S I mP = = = =4.7 10 Pamc c 3 10 s

-6 2 -52rad

NF=P A= 4.7 10 4.0 m =1.9 10 Nm

Caution! The letter P (or p) has been

used in this lecture for power,

pressure, and momentum!

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Today’s agendum:

Introduction to Light.You must develop a general understanding of what light is and how it behaves.

Reflection and Refraction (Snell’s “Law”).You must be able to determine the path of light rays using the laws of reflection and refraction.

Total Internal Reflection and Fiber Optics.You must be able to determine the conditions under which total internal reflection occurs, and apply total internal reflection to fiber optic and similar materials.

Dispersion.You must understand that the index of refraction of a material is wavelength-dependent.

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Light

Normally, “light” refers to the type of electromagnetic wave that stimulates the retina of our eyes.

Light acts like a wave except when it acts like particles.

http://www.physicsclassroom.com/Class/light/U12L2a.html

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*Light—Waves or Particles?

http://www.nearingzero.net (quantum007.jpg)

*Both! Take Physics 203 for further enlightenment!

http://www.nearingzero.net/

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Light is a type of electromagnetic wave and travels with the speed c = 2.9979x108 m/s in a vacuum. (Just use 3x108!)

The Speed of Light

How many physicists does it take to change a light bulb? Eleven. One to do it and ten to co-author the paper.

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Visible light is a small part of the electromagnetic spectrum.

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Although light is actually an electromagnetic wave, it generally travels in straight lines (like particles do!).

We can describe many properties of light by assuming that it travels in straight-line paths in the form of rays.

A ray is a straight line along which light is propagated. In other contexts, the definition of ray might be extended to include bent or curved lines.

Geometric Optics

http://www.nearingzero.net (ray.jpg)

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A light ray is an infinitely thin beam of light. Of course, there really isn’t such a thing, but the concept helps us visualize properties of light.

http://www.nearingzero.net (rays.jpg)

there really isn’t such a thing

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Light rays from some external source strike an object and reflect off it in all directions.

We only see those light rays that reflect in the direction of our eyes.

If you can see something, it must be reflecting light!Zillions* of rays are simultaneously reflected in all directions from any point of an object. Later, when we study mirrors and lenses, we won’t try do draw them all! Just enough representative ones to understand what the light is doing.

*one zillion = 10a big number

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Today’s agendum:





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Light striking a surface may be reflected, transmitted, or absorbed. Reflected light leaves the surface at the same angle it was incident on the surface:

i r

Reflection

Real Important Note: the angles are measured relative to the surface normal.

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Reflection from a smooth surface is specular (mirror- like). Reflection from a rough surface is diffuse (not mirror-like).

http://acept.la.asu.edu/PiN/rdg/reflection/reflection.shtml

http://www.mic-d.com/java/specular/

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Light travels in a straight line except when it is reflected or when it moves from one medium to another.

Refraction—the “bending” of light rays when light moves from one medium to a different one—takes place because light travels with different speeds in different media.

Refraction

http://id.mind.net/~zona/mstm/physics/light/rayOptics/refraction/refraction1.html

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The speed of light in a vacuum is c = 3x108 m/s. The index of refraction of a material is defined by

cn = ,

v

where c is the speed of light in a vacuum and v is the speed of light in the material.

The speed and wavelength of light change when it passes from one medium to another, but not the frequency, so

.

n

cv = and =

n n

If you study light in advanced classes, you’ll find it is more complex than this.

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Because light never travels faster than c, n 1. For water, n = 1.33 and for glass, n 1.5. Indices of refraction for several materials are listed in your text.

cv =

n

83×10 m/sv =

2.42

8v = 1.24×10 m/s

Example: calculate the speed of light in diamond (n = 2.42).

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When light moves from one medium into another, some is reflected at the boundary, and some is transmitted.

The transmitted light is refracted (“bent”).

a is the angle of incidence, and b is the angle of refraction.

Snell’s “Law”

air (na)

water (nb)

a

b

nb>na

incident ray

refracted ray

air (nb)

water (na)

b

a

na>nb

incid

ent ray

refracted ray

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air (na)

water (nb)

a

b

nb>na

incident ray

refracted ray

Light passing from air (n 1) into water (n 1.33).

Light “bends” towards the normal to the surface as it slows down in water.

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air (nb)

water (na)

b

a

na>nb

incid

ent ray

refracted ray

Light passing from water (n 1.33) into air (n 1).

Light “bends” away from the normal to the surface as it speeds up in air.

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Snell’s “Law”, also called the law of refraction, gives the relationship between angles and indices of refraction: a a b bn sin θ = n sin θ .

a

b

water (nb)

air (na)

a

b

water (nb)

air (na)

is the angle the ray makes with the normal!

You are free to choose which is “a” and which is “b.”

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Today’s agendum:





48

Suppose n2<n1. The largest possible value of sin(2) is 1 (when 2 = 90). The largest possible value of sin(1) is

1 1 2 2n sin θ = n sin θ

21 2

1

nsin θ = sin θ

n

21,max

1

nsin θ = .

n

This value of is called the critical angle, C. For any angle of incidence larger than C, all of the light incident at an interface is reflected, and none is transmitted.

Total Internal Reflection; Fiber Optics

For 1 larger than this, Snell’s Law cannot be satisfied!

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1 < C 1 close to C

1 > C

1

Another visualization here.

50

n2

n1>n2

Ray incident normal to surface is not “bent.”

Ray incident normal to surface is not “bent.” Some is reflected,Ray incident normal to surface is not “bent.” Some is reflected, some is transmitted.

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n2

n1>n2

Increasing angle of incidence…

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n2

n1>n2

Increasing angle of incidence…more…

53

n2

n1>n2

Increasing angle of incidence…more…critical angle reached…Increasing angle of incidence…more…critical angle reached… some of incident energy is reflected, some is “transmitted along the boundary layer.

54

n2

n1>n2

Light incident at any angle beyond C is totally internally reflected.

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application: fiber optics

http://laser.physics.sunysb.edu/~wise/wise187/janfeb2001/reports/andrea/report.html

56

i

f

ni=1 (air)

nf>1

Light is incident at an angle i on a transparent fiber.

The light refracts at an angle f.

i i ffn sin θ = n sin θ

i ffsin θ = n sin θ

Example: determine the incident angle i for which light strikes the inner surface of a fiber optic cable at the critical angle.

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i

f90-f

ni=1 (air)

nf>1

Light strikes the fiber wall an an angle of90-f normal to the surface.

At the critical angle, instead of exiting the fiber, the refracted light travels along the fiber-air boundary. In this case, 90-f is the critical angle.

ff f c an sin 90-θ =n sin θ =n sin 90 1

ff

1sin 90-θ =

n

90

Solve the above for f and use to solve for i.

i ffsin θ = n sin θ

58

application: swimming underwater

If you are looking up from underwater, if your angle of sight (relative to the normal to the surface) is too large, you see an underwater reflection instead of what’s above the water.

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application: perfect mirrors

(used in binoculars)

application: diamonds

60

Today’s agendum:





61

Dispersion

We’ve treated the index of refraction of a material as if it had a single value for all wavelengths of light.

In fact, the index of refraction is generally wavelength- (or color-) dependent. When white light passes from airinto glass, the different colours are refracted by different angles, and therefore spread out, or are dispersed.

It is observed that the shorter the wavelength of the light, the greater is the refraction.

62Picture from the Exploratorium (http://www.exploratorium.edu/).

63

Today’s agendum:

Plane Mirrors.You must be able to draw ray diagrams for plane mirrors, and be able to calculate image and object heights, distances, and magnifications.

Spherical Mirrors: concave and convex mirrors.You must understand the differences between these two kinds of mirrors, be able to draw ray diagrams for both kinds of mirrors, and be able to solve the mirror equation for both kinds of mirrors.

64

Mirrors

Plane mirrors form virtual images; no light actually comes from the image. The solid red rays show the actual light path after reflection; the dashed black rays show the perceived light path.

Images Formed by Plane Mirrors

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*The object distance and image distance are equal: s=-s’.The object height and image height are equal: y=y’. The magnification of a plane mirror is therefore one.

s s’

y’y

The image is upright and virtual.

The image is reversed front-to-back relative to the object.

*The – sign is needed because of sign conventions—see later.

66

s s’

Example: how tall must a full-length mirror be?

A light ray from the top of your head reflects directly back from the top of the mirror.

67

s s’

y/2

y/2

To reach your eye, a light ray from your foot must reflect halfway up the mirror (because I = R = ).

68

s s’

y/2

y/2

The mirror needs to be only half as tall as you.

This calculation assumed your eyes are at the top of your head.

69

Today’s agendum:


Spherical Mirrors: concave and convex mirrors.You must understand the differences between these two kinds of mirrors, be able to draw ray diagrams for both kinds of mirrors, and be able to solve the mirror equation for both kinds of mirrors.

70

Images Formed by Spherical Mirrors

Spherical mirrors are made from polished sections cut from a spherical surface.

The center of curvature, C, is the center of the sphere, of which the mirror is a section.

Of course, you don’t really make these mirrors by cutting out part of a sphere of glass.

C

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The radius of curvature, R, is the radius of the sphere, or the distance from V to C.

C VR

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The principal axis (or optical axis) is the line that passes through the center of curvature and the center of the mirror.

C VR

Principal or Optical Axis

The center of the mirror is often called the vertex of the mirror.

73

Paraxial rays are parallel to the principal axis of the mirror (from an object infinitely far away). Reflected paraxial rays pass through a common point known as the focal point F.

C VF

74

The focal length f is the distance from P to F. Your text shows that f = R/2.

C PF

f

R

75

Reality check: paraxial rays don’t really pass exactly through the focal point of a spherical mirror (“spherical aberration”).

C VF

76

If the mirror is small compared to its radius of curvature, or the object being imaged is close to the principal axis, then the rays essentially all focus at a single point.

C VF

We will assume mirrors with large radii of curvature and objects close to the principal axis.

77

In “real life” you would minimize aberration by using a parabolic mirror.

C VF

78

Today’s agendum:


Spherical Mirrors: concave and convex mirrors.You must understand the differences between these two kinds of mirrors, be able to draw ray diagrams for both kinds of mirrors, and be able to solve the lens equation for both kinds of mirrors.

79

Concave and Convex Mirrors

There are two kinds of spherical mirrors: concave and convex.

Fconcave convex

80

Ray Diagrams for Mirrors

We can use three “principal rays” to construct images. In this example, the object is “outside” of F.

Ray 1 is parallel to the axis and reflects through F.Ray 2 passes through F before reflecting parallel to the axis.

Ray 3 passes through C and reflects back on itself.

FC

We’ll also use three for convex mirrors, but there will be a different version of ray 2.

81

We use three “principal rays” to construct images.

C

An image is formed where the rays converge.

The image from a concave mirror, object outside the focal point, is real, inverted, and smaller than the object.

“Real” image: you could put a camera there and detect the image.

F

Ray Diagrams for Concave Mirrors

Two rays would be enough to show us where the image is. We include the third ray for “safety.” You don’t have to use principal rays, but they are easiest to trace.

82

C

The image from a concave mirror, object inside the focal point, is virtual, upright, and larger than the object.Ray 1: parallel to the axis then through F.

Ray 2: “through” F then parallel to the axis.

Ray 3: “through” C.

F

With this size object, there was a bit of spherical aberration present, and I had to “cheat” my C a bit to the left to make the diagram look “nice.”

83

C

You could show that if an object is placed at the focal point, reflected rays all emerge parallel, and *no image is formed.Ray 1: parallel to the axis then through F.

Ray 2: “through” F then parallel to the axis. Can’t do!

Ray 3: through C.

F

no image

*Actually, the image is formed at infinity.Worth thinking about: what if the object is placed between F and C?

84

With a bit of geometry, you can show that

C

The magnification is the ratio of the image to the object height:

F

The Mirror Equation

1 1 1+ =

s s' f

f

s

s’

y' s'm= = -

y s

y

y’

85

Sign conventions for the mirror equation:

C F

1 1 1+ =

s s' f

f

s

s’

y' s'm= = -

y s

y

y’

When the object, image, or focal point is on the reflecting side of the mirror, the distance is positive.When the object, image, or focal point is “behind” the mirror, the distance is negative.

The image height is positive if the image is upright, and negative if the image is inverted relative to the object.

86

Example: a dime (height is 1.8 cm) is placed 100 cm away from a concave mirror. The image height is 0.9 cm and the image is inverted. What is the focal length of the mirror.

1 1 1+ =

s s' fy' s'

m= = -y s

C F

f

s

s’

y

y’

s’, s, or f on reflecting side are +y is – if image is inverted

87

Applications of concave mirrors.

Shaving mirrors.

Makeup mirrors.

Solar cookers.

Flashlights, headlamps, stove reflectors.

Satellite dishes (when used with electromagnetic radiation).

88

Today’s agendum:


Spherical Mirrors: concave and convex mirrors.You must understand the differences between these two kinds of mirrors, be able to draw ray diagrams for both kinds of mirrors, and be able to solve the lens equation for both kinds of mirrors.

89

CF

Ray Diagrams for Convex Mirrors

Ray 1: parallel to the axis then through F.

Ray 2: “through” Vertex.

Ray 3: “through” C.

The image is virtual, upright, and smaller than the object.

90

CF

Instead of sending ray 2 “through” V, we could have sent it “through” F. The ray is reflected parallel to the principal axis.

Your text talks about all four of the “principal rays” we have used.

91

CF

Because they are on the “other” side of the mirror from the object, s’ and f are negative.

The mirror equation still works for convex mirrors.

1 1 1+ =

s s' fy' s'

m= = -y s

s

fs’y

y’

92

The ray diagram looks like the one on the previous slide, but with the object much further away (difficult to draw).

On reflecting sidepositive.

Not on reflecting sidenegative.

Example: a convex rearview car mirror has a radius of curvature of 40 cm. Determine the location of the image and its magnification for an object 10 m from the mirror.

1 1 1+ =

s s' f

1 1 1 1 1

= = s' f s -0.2 m 10 m

93

1 1 1

= s' -0.2 m 10 m

…algebra…

s'= -0.196 m= -19.6 cm

s' -0.196 m 1m= - = - =

s 10 m 51

Remind me… what does it say on passenger side rear view mirrors?

94

Applications of convex mirrors.

Passenger side rear-view mirrors.

Grocery store aisle mirrors.

Anti-shoplifting (surveillance) mirrors.

Christmas tree ornaments.

Railroad crossing mirrors.

95

Sign Conventions Introduced Today

When the object, image, or focal point is on the reflecting side of the mirror, the distance is positive.

When the object, image, or focal point is “behind” the mirror, the distance is negative.


96

Summary of Sign Conventions

Object Distance. When the object is on the same side as the incoming light, the object distance is positive (otherwise is negative).

Here’s a compact way of expressing mirror and lens (coming soon) sign conventions all at once.

Image Distance. When the image is on the same side as the outgoing light, the image distance is positive (otherwise is negative).

Radius of Curvature. When the center of curvature C is on the same side as the outgoing light, R is positive (otherwise is negative).

97

Today’s agendum:

Refraction at Spherical Surfaces.You must be able to calculate properties of images formed by refraction at spherical surfaces.

Thin Lenses: Concave and Convex Lenses, Ray Diagrams, Solving the Lens Equation.You must understand the differences between these two kinds of lenses, be able to draw ray diagrams for both kinds of lenses, and be able to solve the lens equation for both kinds of lenses.

If Time Allows: Lens Combinations, Optical Instruments.You should be aware of this useful information.

98

Refraction at Spherical Surfaces

Convex surface:

C Faxis

R

f

1

2

na

nb>n

a

Geometry: a light ray parallel to the axis passes through F.

b

b a

nf = R>R

n -n

99

An extended object will form an image inside the nb medium.

C Faxis

R

f

1

2

na

nb>n

a

This image is real and inverted.

ss’

Ray 1: parallel to the axis, through F.Ray 3: through C.

100

Concave surface:

CF

R

nb>n

a

Geometry: a light ray parallel to the axis seems to have come from F.

b

b a

nf = R>R

n -n

axis

na

f

101

CF

R

nb>n

a

The image is virtual and upright.

axis

na

f

An extended object will form an image inside the na medium.Ray 1: parallel to the axis, through F.Ray 3: through C.

There are three different places to put the object. The different images formed are always virtual and upright.

102

We can use geometry to derive an equation relating the image and source positions, and an equation for the magnification.

C Faxis R

f

na

nb

ss’

a b b an n n -n+ =

s s' Ra

b

n s'y'm= =-

y n s

103

CF

R

nb

axis

na

f

The same equations work for concave surfaces.

s

s’

a b b an n n -n+ =

s s' Ra

b

n s'y'm= =-

y n s

104

Approximations Were Used!

The equations in this section are excellent approximations if both the angles of incidence and refraction are small.

105

Sign Conventions

R is positive when it is in the medium into which the light propagates. R is negative when it is in the medium from which the light radiates.

The image distance is positive when the image is in the medium into which the light propagates, and negative if it is in the medium from which the light radiates (virtual image).

The object distance is positive when the object is in the medium from which the light radiates (the usual case—a real object), and negative if on the side opposite to the light source (a virtual object).

These are really “the same” as for mirrors.

106

Example: a Jurassic mosquito is discovered embedded in an amber sphere which has an index of refraction of 1.6. The radius of curvature of the sphere is 3.0 mm. The mosquito is located on the principal axis and appears to be imbedded 5.0 mm into the amber. How deep is the mosquito really?

R

s

s’

a b b an n n -n+ =

s s' R na=1.6nb=1

The object is in the amber, so na=1.6 and nb=1.

The image is in the medium from which the light radiates so s’=-5.0 mm. Notice the reversed orientation. . .

107

R

s

s’

na=1.6nb=1

R is negative because it is in the medium from which the light radiates. R=-3.0 mm.

1.6 1 1-1.6+ =

s -5.0 -3

s= 4 mm

108

Today’s agendum:




109

Thin Lenses

A lens in this section is taken to be a single object made of transparent material of refractive material n>1.There are two surface boundaries. Light from an object incident on the first surface forms an image, which becomes the object for the second surface.A thin lens is one for which the distance from the object to each of the two surfaces is the “same” (and the distance from the image to each surface is the “same”).

This would NOT qualify as a thin lens.

110

Until I figure out how to use Powerpoint to fill in the lens color, I will make my lenses look “hollow,” like this.

There are several surface combinations from which we can make lenses. Here are three (there are more).

111

Thin lenses can be converging or diverging.

Converging and Diverging Lenses

The converging lens is thicker in the center. The diverging lens is thicker at the edges.

There are focal points on both sides of each lens. The focal length is the same whether light passes from left to right or right to left.

112

There are two surfaces at which light refracts. Our equations (provided later) “automatically” take care of this.

In your diagrams, simply draw the incident ray up to the center of the lens, then draw the refracted ray in its final direction.

113

Today’s agendum:




114

Ray Diagrams for Converging Lenses

Ray 1 is parallel to the axis and refracts through F.

Ray 2 passes through F’ before refracting parallel to the axis.Ray 3 passes straight through the center of the lens.

The image is real and inverted. In this case, it is larger than the object.

O F’ FI

115

Ray Diagrams for Diverging Lenses

Ray 1 is parallel to the axis and refracts as if through F.Ray 2 heads towards F’ before refracting parallel to the axis.Ray 3 passes straight through the center of the lens.

The image is virtual and upright. It is smaller than the object.

O F F’I

116

Converging and Diverging Lenses

The image formed by a diverging lens is always virtual, upright, and smaller than the object. See this web page.

The image formed by a converging lens may be real, inverted, and either smaller or larger than the object. It may also be virtual, upright, and larger than the object. See this web page.

Do these lens properties remind you of anything you’ve studied recently?

117

Today’s agendum:




118

The Lensmaker’s Equation

a b

1 1 1= n-1

f R R

1 1 1+ =

s s' f

a b

1 1 1 1+ = n-1

s s' R R

The Lensmaker’s Equation

s s

s’

s’

y' s'M= = -

y s

119

Sign Conventions for The Lens Equation

1 1 1+ =

s s' fy' s'

M= = -y s

The object distance s is positive if the object is on the side of the lens from which the light is coming; otherwise s is negative.

The image distance s’ is positive if the image is on the opposite side of the lens from where the light is coming; otherwise s’ is negative.

The focal length f is positive for converging lenses and negative for diverging lenses.

The image height y’ is positive if the image is upright and negative if the image is inverted relative to the object.

120

Example: an object is located 5 cm in front of a converging lens of 10 cm focal length. Find the image distance and magnification. Is the image real or virtual?

O FF’

1 1 1 1 1 1= - = - =-

s' f s +10 +5 10

Image distance is 10 cm, image is on side of lens light is coming from, so image is virtual. M=2 so image is upright.

s' -10M= - = - = 2

s 5

It’s just a coincidence that the image is located at F’.

121

Today’s agendum:




122

Lens Combinations

To determine the image formed by a combination of two lenses, simply...

…calculate the image formed by the first lens…

…then use the first lens image as the source (object) for the second lens.

There is no homework on lens combinations.

123

Optical Instruments

A Simple Magnifier

h O

h´

Magnifier

FIq

p

25 cm (near point)

O

124

Refracting Telescope

tane

h

f

tan o o

o

h

f

e o

o eo

hf f

Mh f

f

For viewing very far objects. Object distance taken as infinity.

125

Terrestrial Telescopes

For producing upright images:

Galilean telescope Field-lens telescope

126

Reflecting Telescope

Newtonian-focus reflecting telescope

127

Compound Microscope

Again has objective and eyepiece, but because it is for viewing very near objects it is very different from the telescope.

.25e

e

Mf

Eyepiece magnification:

i eo

o

h l fqm

h p p

Objective magnification:

Overall magnification:

.25 .25ee o

e e o

l f lM M m

f p f f

128


When the object, image, or focal point is on the reflecting side of the mirror, the distance is positive.

When the object, image, or focal point is “behind” the mirror, the distance is negative.


Mirrors

The object distance s is positive if the object is on the side of the lens from which the light is coming; otherwise s is negative (and the object is virtual).The image distance s’ and radius of curvature R are positive if the image is on the side of the lens into which the light is going; otherwise negative.

The focal length f is positive for converging lenses and negative for diverging lenses.


Lenses

129


Object Distance. When the object is on the same side as the incoming light, the object distance is positive (otherwise is negative).

Here’s a more compact way of expressing the sign conventions all at once.

Image Distance. When the image is on the same side as the outgoing light, the image distance is positive (otherwise is negative).

Radius of Curvature. When the center of curvature C is on the same side as the outgoing light, R is positive (otherwise is negative).

130

Today’s agendum:

Review of Waves.You are expected to recall facts about waves from Physics 103.

Young’s Double Slit Experiment.You must understand how the double slit experiment produces an interference pattern.

Conditions for Interference in the Double Slit Experiment.You must be able to calculate the conditions for constructive and destructive interference in the double slit experiment.

Intensity in the Double Slit Experiment.You must be able to calculate intensities in the double slit experiment.

131

Interference

Review of Waves

This section is a review of material you learned in your previous physics course (perhaps Physics 103).

Consider a wave described by

y(x,t) A sin (kx ωt) .

The phase of this wave is

θ(x,t) kx ωt .

Also dθ dxk ω .

dt dt

y

x

132

If is constant with time (i.e., d/dt=0), then we are moving with the wave, and

dx .

dt k

The phase velocity, vp, is given by

p

ωv .

k

Imagine yourself riding on any point on this wave. The point you are riding moves to the right. The velocity it moves at is vp.

If the wave is moving from left to right then /k must be positive.

y

x

133

When waves of the same nature travel past some point at the same time, the amplitude at that point is the sum of the amplitudes of all the waves

The amplitude of the electric field at a point is found by adding the instantaneous amplitudes, including the phase, of all electric waves at that point.

In Physics 103 you may have learned that power (or intensity) is proportional to amplitude squared. The intensity of the superposed waves is proportional to the square of theamplitude of the resulting sum of waves.

Superposition—a Characteristic of All Waves

134

Constructive Interference: If the waves are in phase, they reinforce to produce a wave of greater amplitude.

Destructive Interference: If the waves are out of phase, they reinforce to produce a wave of reduced amplitude.

Interference—a Result of the Superposition of Waves

135

Today’s agendum:





136

This experiment demonstrates the wave nature of light.

Consider a single light source, and two slits. Each slit acts as a secondary source of light.

Light waves from secondary slits interfere to produce alternating maxima and minima in the intensity.

Reference and “toys:” fsu magnet lab, colorado.

Young’s Double Slit Experiment

Interesting reading: the double slit experiment and quantum mechanics.

137

How does this work?

Light waves from the two slits arriving at the detection screen in phase will interfere constructively and light waves arriving out of phase will interfere destructively.

In phase—constructive.

Out of phase—destructive.

138

Today’s agendum:





139

Conditions for Interference

Sources must be monochromatic-of a single wavelength.

Sources must be coherent-- must maintain a constant phase with respect to each other.

Here’s the geometry I will use in succeeding

diagrams.

140

dL2

L1

L = L2 –L1 = d sin

For an infinitely distant* screen:

P

S2

S1

L

R

y

L1

L2

d

tany

R

*so that all the angles labeled are approximately equal

141

Destructive Interference:

Constructive Interference:

The parameter m is called the order of the interference fringe. The central bright fringe at = 0 (m = 0) is known as the zeroth-order maximum. The first maximum on either side (m = ±1) is called the first-order maximum.

dL2

L1

L = L2 –L1 = d sin

, , , L d sin m m=0 1 2...

, , ,

1L d sin m+ m=0 1 2...

2

142

P

S2

S1

L

R

y

L1

L2

d

tany

R

y R tan R sin

Bright fringes:

m d sin

ym d

R

Ry m

d

Do not use the small-angle approximation unless it is valid!

143

P

S2

S1

L

R

y

L1

L2

d

tany

R

y R tan R sin

Dark fringes:

1m d sin

2

1 ym d

2 R

R 1y m

d 2

Do not use the small-angle approximation unless it is valid!

144

Example: a viewing screen is separated from the double-slit source by 1.2 m. The distance between the two slits is 0.030 mm. The second-order bright fringe (m = 2) is 4.5 cm from the center line. Determine the wavelength of the light.

y R tan R sin

Bright fringes:

m d sin

ym d

R

yd

Rm

-2 -5

74.5 10 m 3.0 10 m

5.6 10 m 560 nm1.2 m 2

P

S2

S1

L

R

yL1

L2

tany

R

145

Example: a viewing screen is separated from the double-slit source by 1.2 m. The distance between the two slits is 0.030 mm. The second-order bright fringe (m = 2) is 4.5 cm from the center line. Find the distance between adjacent bright fringes.

y R tan R sin

Bright fringes:

m d sin

y

m d R

Ry m

d

72

m+1 m -5

5.6 10 m 1.2 mR R Ry -y m 1 m 2.2 10- m 2.2 cm

d d d 3.0 10 m

P

S2

S1

L

R

yL1

L2

tany

R

146

Example: a viewing screen is separated from the double-slit source by 1.2 m. The distance between the two slits is 0.030 mm. The second-order bright fringe (m = 2) is 4.5 cm from the center line. Find the width of the bright fringes.Define the bright fringe width to be the distance between two adjacent destructive minima.

darky1m d sin d

2 R

dark

R 1y m

d 2

7

dark,m+1 dark,m -5

5.6 10 m 1.2 my -y 2.2 cm

3.0 10 m

P

S2

S1

L

R

yL1

L2

tany

R

dark,m+1 dark,m

R 1 R 1 Ry -y m 1 m

d 2 d 2 d

147

Today’s agendum:





148

Intensity in the Double Slit Experiment

Our equations for the minima and maxima intensity positions are for the centers of the fringes.

In this section, we calculate distribution of light intensity in the double-slit interference pattern.

149

The derivation of the double-slit intensity equation is not particularly difficult, so study it if you find derivations helpful for your understanding.

A path length difference L= corresponds to a phase difference of =2.

A path length difference L=m corresponds to a phase difference of =2m.In general, for non-integral m, the phase difference at P between the waves from S1 and S2 is

φ π φ

π λ λ λ

L d sin 2= = = d sin

2 x x x

150

Your text writes the equation for the intensity distribution in the double-slit experiment in terms of the phase difference on the previous slide.

Your starting equation for the intensity is

φ

20I =I cos

where I0 is 4 times the peak intensity of either of the two interfering waves:

0 single waveI =4I

Why did my previous diagrams show this?

151

Today’s agendum: Interference Due to Reflection.

Phase Change Due to Reflection.You must be able to determine whether or not a phase change occurs when a wave is reflected.

Thin Film Interference.You must be able to calculate thin film thicknesses for constructive or destructive interference.

Examples.You must be able to solve problems similar to these examples.

152

Light undergoes a phase change of 180° ( radians) upon reflection from a medium that has a higher index of refraction than the one in which the wave is traveling.

Interference from Reflection

Phase Change Due to Reflection

153





154

Thin film interference is caused by…

Thin Film Interference

…phase difference of reflected waves due to reflection off a higher-n material, and…

…phase difference of reflected waves due to path length differences.

http://www.photographyblog.com/gallery/showphoto.php?photo=5545

155

No phase change

Air

Air

Filmt nAir < nFilm

Do the reflected rays interfere destructively or constructively?Caution! The wavelength in the film is different than in air.

180° phase change

Thin Film Interference

Ray undergoes a phase change on reflection.

Ray has a phase change due to the path difference.

Dark lines in drawings are there to help you see the boundaries, and are not a separate medium.

156

We will get destructive interference when the path difference is an integral number of wavelengths:

Assume the incident light is nearly perpendicular to the film surface.

No phase change

Air

Air

Filmt nAir < nFilm

180° phase changeThe path length difference is approximately 2t.

There is a 180 phase difference (½ of a wavelength) due to the first reflection.

film

film

=n

, , ,

film filmfilm

2t=m =m 2n t=m m=0 1 2...n

157

We will get constructive interference when the path difference is a half-integral number of wavelengths:

Assume the incident light is nearly perpendicular to the film surface.

No phase change

Air

Air

Filmt nAir < nFilm

180° phase change

We get constructive interference when the path difference is film/2, 3film/2, 5film/2, etc.

film

film

=n

1 1 1 , , ,

2 2 2

film filmfilm

2t= m+ = m+ 2n t= m+ m=0 1 2...n

158

No phase change

Air

Air

Filmt nAir < nFilm

180° phase change

film

film

=n

1, , ,

2

film2n t= m+ m=0 1 2...

, , , film2n t=m m=0 1 2...

You need to apply the reasoning used in deriving them to each of your thin film interference problems.

These are only true when the film is surrounded by a medium with lower index of refraction than the film!

The equations below are not on your starting equation sheet.

159

No phase change

Air

Air

Filmt nAir < nFilm

180° phase change

film

film

=n

1

2

film2n t= m+

film2n t=m

These are valid when the light is incident almost perpendicular to the film:

Caution!

The incident ray in the diagram clearly does not qualify visually as “almost perpendicular.” That’s because the angle relative to the normal is exaggerated for viewing convenience.

160

No phase change

Air

Air

Filmt nAir < nFilm

180° phase change

film

film

=n

Caution!

For truly non-perpendicular incidence, you have to take into account the extra path length of the ray reflected at the air-film interface (as well as the extra path length inside the film).

161

Thin Film Interference Problem Solving Tips

Identify the thin film causing the interference.

Determine the phase relationship between the portion of the wave reflected at the upper surface and the portion reflected at the lower surface.

Phase differences have two causes: (1) path differences and (2) phase changes upon reflection.

When the total phase change is an integer multiple of the wavelength (, 2, 3, etc.) the interference is constructive, and when it is a half-integer multiple of the wavelength (/2, 3/2, 5/2, etc.) it is destructive.

162





163

Example: a glass lens is coated on one side with a thin film of MgF2 to reduce reflection from the lens surface. The index of refraction for MgF2 is 1.38 and for glass is 1.50. What is the minimum thickness of MgF2 that eliminates reflection of light of wavelength λ = 550 nm? Assume approximately perpendicular angle of incidence for the light.

Both rays and experience a 180 phase shift on reflection so the total phase difference is due to the path difference of the two rays.

Air

MgF2

tn= 1.38

nAir = 1.00

180° phase change

180° phasechange

glass, ng =1.50

164

The minimum thickness is for m=0.

The reflected light is minimum when the two light rays meet the condition for destructive interference: the path length difference is a half-integral multiple of the light wavelength in MgF2.

1, , ,

2

MgF2

2t= m+ m=0 1 2...n

min

MgF2

2t =2n

minMgF2

550 nmt = = =99.6 nm

4n 4 1.38

Air

MgF2

tn= 1.38

nAir = 1.00

180° phase change

180° phasechange

glass, ng =1.50

165

Example: two glass plates 10 cm long are in contact on one side and separated by a piece of paper 0.02 mm thick on the other side. What is the spacing between the interference fringes? Assume monochromatic light with a wavelength in air of λ = 500 nm incident perpendicular to the slides.

Ray is not phase shifted on reflection. Ray is shifted 180 on reflection.

H

tx

L = 10 cmH = 2x10-5 m

For destructive interference

The light that is partly reflected at the bottom of the first glass surface and partly transmitted is responsible for the interference fringes.*

2t=m m=0 1 2...

*This reference explains why there is no visible interference due to the relatively thick glass plates themselves.

166

H

tx

L = 10 cmH = 2x10-5 m

2t=m m=0 1 2...

t H Hx

= t=x L L

-5

0.1 m 500 nmHx L2 =m x =m =m =m 1.25 mm

L 2H 2 2 10 m

Successive dark fringes are separated by 1.25 mm.

x is the distance from the contact point to where destructive interference takes place.

167

H

tx

L = 10 cmH = 2x10-5 m

t H Hx

= t=x L L

1 1

2 2

Hx L2 = m+ x = m+

L 2H

Successive bright fringes occur for m+½ and (m+1)+½.

1

2

3

2

Hx2 = m+1 +

L

Lx = m+

2H

For constructive interference

1

2

2t= m+ m=0 1 2...

168

H

tx

L = 10 cmH = 2x10-5 m

Successive bright fringes are separated by 1.25 mm.

169

Example: suppose the glass plates have ng = 1.50 and the space between them contains water (nw = 1.33). What happens now?

Ray is not phase shifted on reflection. Ray is shifted 180 on reflection. Both are the same as before.

H

tx

L = 10 cmH = 2x10-5 m

For destructive interference

2t=m m=0 1 2...

But the path difference now occurs in water, where the light will have a wavelength

watern

Repeat the calculation, using water.

170

H

tx

L = 10 cmH = 2x10-5 m

water

-5

0.1 m 500 nm/1.33Lx =m =m =m 0.94 mm

2H 2 2 10 m

Successive dark fringes are separated by 0.94 mm.

For destructive interference, we now have

171

Two lectures ago I showed you these two plots of the intensity distribution in the double-slit experiment:

Which is correct?

Peak intensity varies with angle.

Peak intensity independent of angle.

172

Diffraction

Light is an electromagnetic wave, and like all waves, “bends” around obstacles.

d

<<d d >>d

This bending, which is most noticeable when the dimension of the obstacle is close to the wavelength of the light, is called “diffraction.” Only waves diffract.

173

Diffraction pattern from a penny positioned halfway between a light source and a screen.The shadow of the penny is the circular dark spot.

Notice the circular bright and dark fringes.

The central bright spot is not a defect in the picture. It is a result of light “bending” around the edges of the penny and interfering constructively in the exact center of the shadow.

174

Single Slit Diffraction

In the previous chapter we calculated the interference pattern from a pair of slits.

One of the assumptions in the calculation was that the slit width was very small compared with the wavelength of the light.Now we consider the effect of finite slit width. We start with a single slit.

a

Each part of the slit acts as a source of light rays, and these different light rays interfere.

175

a/2

a/2a

sin2

a

Divide the slit in half.

Ray travels farther* than ray by (a/2)sin. Likewise for rays and .If this path difference is exactly half a wavelength (corresponding to a phase difference of 180°) then the two waves will cancel each other and destructive interference results.Destructive interference:

a

sin =2 2

*All rays from the slit are converging at a point P very far to the right and out of the picture.

176

a/2

a/2a

sin2

a

Destructive interferenc

e:

a

sin =2 2

a sin =

If you divide the slit into 4 equal parts, destructive

interference occurs when .

2

sin =a

sin =

a

.

3

sin =a

If you divide the slit into 6 equal parts, destructive

interference occurs when

177

a/2

a/2a

sin2

a

In general, destructive interference occurs when

, , , ,

sin =m m=1 2 3 ...a

The above equation gives the positions of the dark fringes. The bright fringes are approximately halfway

in between.

178

a

O

x

y

Use this geometry for tomorrow’s single-slit homework problems.

If is small,* then it is valid to use the approximation sin .( must be expressed in radians.)

*The approximation is quite good for angles of 10 or less, and not bad for even larger angles.

179

I won’t derive the intensity distribution for the single slit. The general features of that distribution are shown below.

Most of the intensity is in the central maximum. It is twice the width of the other (secondary) maxima.

Single Slit Diffraction Intensity

180

New starting equations for single-slit intensity:

2

= a sin

2

0

sin /2I =I

/2

“Toy”

181

Example: 633 nm laser light is passed through a narrow slit and a diffraction pattern is observed on a screen 6.0 m away. The distance on the screen between the centers of the first minima outside the central bright fringe is 32 mm. What is the slit width?

y1 = (32 mm)/2 tan = y1/L tan sin for small

1 1

Lsin = a=

a sin y /L y

-9

-3

6.0 m 633 10 ma=

16 10 m

-4a=2.37 10 m

182

The ability of optical systems to distinguish closely spaced objects is limited because of the wave nature of light.

If the sources are far enough apart so that their central maxima do not overlap, their images can be distinguished and they are said to be resolved.

Resolution of Single Slit (and Circular Aperture)

183

When the central maximum of one image falls on the first minimum of the other image the images are said to be just resolved. This limiting condition of resolution is called Rayleigh’s criterion.

184

From Rayleigh’s criterion we can determine the minimum angular separation of the sources at the slit for which the images are resolved.

For a slit of width a:

For a circular aperture of diameter D:

Resolution is wavelength limited!

=

a

1.22

=D

a=sin

These come from

the small angle approximation, and geometry.

185

If a single slit diffracts, what about a double slit?

Remember the double-slit interference pattern from the chapter on interference?

2max

d sinI =I cos

If the slit width (not the spacing between slits) is small (i.e., comparable to the wavelength of the light), you must account for diffraction.

interference only

186

S2

S1

d

a

P

L

y

r1

r2

Double Slit Diffraction

187

A diffraction grating consists of a large number of equally spaced parallel slits.

d

= d sin

The path difference between rays from any two adjacent slits is = dsin .

Interference maxima occur for

Diffraction Gratings

If is equal to some integer multiple of the wavelength then waves from all slits will arrive in phase at a point on a distant screen.

, , , , d sin =m m=1 2 3 ...

188

Ok what’s with this equation monkey business?

, , , , a sin =m m=1 2 3 ...

, , , , d sin =m m=1 2 3 ...

, , , , d sin =m m=1 2 3 ... double-slit interferenceconstructive

single-slit diffractiondestructive!

diffraction gratingconstructive

189

d

= d sin The intensity maxima are brighter and sharper than for the two slit case.

Interference Maxima:

Diffraction Grating Intensity Distribution

d sin =m

190

191

Application: spectroscopy

You can view the atomic spectra for each of the elements here.

visible light

hydrogen

helium

mercury

192

Interference Maxima:

First-order violet:

Example: the wavelengths of visible light are from approximately 400 nm (violet) to 700 nm (red). Find the angular width of the first-order visible spectrum produced by a plane grating with 600 slits per millimeter when white light falls normally on the grating.

d sin =m

-61d = =1.67 10 m

600 slits/mm

-9

VV -6

1 400 10 msin =m = =0.240

d 1.67 10 mV =13.9

193

First-order red:

-9

RR -6

1 700 10 msin =m = =0.419

d 1.67 10 mR =24.8

R V =24.8 -13.9 =10.9

visible light

10.9

194

No matter what the grating spacing, d, the largest angle for the 2nd order spectrum (for the red end) is always greater than the smallest angle for the 3rd

order spectrum (for the violet end), so 2nd and 3rd orders always overlap.

Example: for this diffraction grating show that the violet end of the third-order spectrum overlaps the red end of the second-order spectrum.

Third-order violet:

-9 -6

VV

3 400 10 m 1.20 10 msin =m = =

d d d

Second-order red:

-9 -6

VR

2 700 10 m 1.40 10 msin =m = =

d d d

R2 V3sin >sin

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Diffraction gratings let us measure wavelengths by spreading apart the diffraction maxima associated with different wavelengths. In order to distinguish two nearly equal wavelengths the diffraction must have sufficient resolving power, R.

Consider two wavelengths λ1 and λ2 that are nearly equal.

The average wavelength is and the difference is

The resolving power is defined as

Diffraction Grating Resolving Power

1 2

avg

+=

2 2 1= - .

avgR= .

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For a grating with N lines illuminated it can be shown that the resolving power in the mth order diffraction is

avgR=

R=Nm.

Dispersion

mercury

Spectroscopic instruments need to resolve spectral lines of nearly the same wavelength.

The greater the angular dispersion, the better a spectrometer is at resolving nearby lines.

angular dispersion=

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Example: Light from mercury vapor lamps contain several wavelengths in the visible region of the spectrum including two yellow lines at 577 and 579 nm. What must be the resolving power of a grating to distinguish these two lines?

mercury

avg 578 nmR= = =289

2 nm

avg

577 nm + 579 nm= = 578 nm

2

=579 nm - 577 nm= 2 nm

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Example: how many lines of the grating must be illuminated if these two wavelengths are to be resolved in the first-order spectrum?

mercury

R=289

R 289

R=Nm N= = =289m 1

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Documents

1. 2 Today’s agendum: Electromagnetic Waves. Energy Carried by Electromagnetic Waves. Momentum and Radiation Pressure of an Electromagnetic Wave