24.5 The Doppler Effect and Electromagnetic Waves

Electromagnetic (EM) waves also can exhibit a Doppler effect:

1. Increase in observed frequency for source and observer approaching one another

2. Decrease in observed frequency for source and observer receding from one another

1

But Doppler Shift of EM waves differs from that of sound for two reasons: a) Sound waves travels in a medium, whereas electromagnetic waves can

travel in a vacuum (and in some transparent medium)

b) For sound, it is the motion relative to the medium that is important. For electromagnetic waves, only the relative motion of the source and observer is important.

Very much less than

cvc

vff so <<

±= rel

rel if 1

Electromagnetic waves

±=

vvvv

ffs

o

so

1

1

For sound

Not for light

Observed frequency

Source frequency

v

24.5 The Doppler Effect and Electromagnetic Waves Example Radar Guns and Speed Traps. The radar gun of a police car emits an electromagnetic wave with a frequency of 8.0x109Hz. The approach is essentially head on. The wave from the gun reflects from the speeding car and returns to the police car, where on-board equipment measures its frequency to be greater than the emitted wave by 2100 Hz. Find the speed of the car with respect to the highway. The police car is parked.

2

v

24.5 The Doppler Effect and Electromagnetic Waves Example Radar Guns and Speed Traps. The radar gun of a police car emits an electromagnetic wave with a frequency of 8.0x109Hz. The approach is essentially head on. The wave from the gun reflects from the speeding car and returns to the police car, where on-board equipment measures its frequency to be greater than the emitted wave by 2100 Hz. Find the speed of the car with respect to the highway. The police car is parked.

3

v

24.5 The Doppler Effect and Electromagnetic Waves

Example Radar Guns and Speed Traps

The radar gun of a police car emits an electromagnetic wave with a frequency of 8.0x109Hz. The approach is essentially head on. The wave from the gun reflects from the speeding car and returns to the police car, where on-board equipment measures its frequency to be greater than the emitted wave by 2100 Hz. Find the speed of the car with respect to the highway. The police car is parked.

4

1 rel

+=

cvff so

frequency “observed” by speeding car

1 rel

+=′

cvff oo frequency observed

by police car so fff −′=∆

( ) sm39 Hz)100.8(2

Hz2100sm100.32 9

8 =×

×=∆

≈sf

fcv

so fcvf −

+= 1 ss f

cv

cvf −

+

+= 11

−

++= 121 2

2

cv

cvf s

+= 2

2

2cv

cvf s c

vffcv

s21 ≈∆→<<

sffcv

2∆

≈→

ss fcv

cvf −

++= 2

2

21

19.5 Capacitors and Dielectrics

Electrical Energy stored in a capacitor

can be thought of in two different ways:

(1) Electrical potential energy between the excess +q and –q charges spread evenly separated by distance d, RELATIVE to the situation where the two plates are neutral.

(2) Potential energy stored in the electric field (which exists only between the plates, and is zero outside!!!

221Energy CV=

vacuumin

density Energy 2

21

221

VolumeEnergy

E

Eu

o

o

ε

κε

=

===

( ) ( ) )()(Energy 2212

21 volumeuAdEEd

dA

oo ==

= κεκε

Ad= Volume

5 Electric field carries energy of “density”

Formulation done for a parallel plate capacitor but generally applicable

dAC 0κε

=EdV =

22.8 Self Inductance

Magnetic Energy: (1) Inductance Take an IDEAL solenoid of N turns, cross-sectional area A, length h and carrying current I.

N turns

h

Cross-sectional

area A

We now want to increase the current by ∆I over time interval ∆t.

This will cause the magnetic field generated by the solenoid to change, and then induce a back- EMF that tends to prevent the increase in current (Lenz’s Law!).

+ −

Increasing current

tBNA

tN

∆∆

−=∆∆Φ

−=E

tN

∆∆Φ

−=E

“−” sign indicates “back”-emf

hNInIB 0

0µµ ==

tI

hNNA

hNI

tNA

∆∆

−=

∆

∆−= 001 µµ

E

hANL

tIL

20 , µ

=∆∆

−=E

Where L is defined to be the (self-) inductance of the solenoid

(Henry) H 1As1V =⋅SI Unit of L

6

22.8 Self Inductance

(2) The energy stored in an inductor h

ANLtIL

20 , µ

=∆∆

−=EThe work required to increase the current from I by a small ∆I is given by:

( ) ILItItIL

tIqW

∆=∆

∆∆

=

∆−=∆−=∆

)(

)(EE

LII

W=

∆∆ I

W∆

∆

( ) ILIW ∆=∆I∆

I

Total magnetic potential energy (MPE) stored is then equal to the total work required to ramp up the current from 0 to I, which is given by the triangular area under the line I

W∆

∆

I∆

I

LII

W=

∆∆

LIIW =∆∆

2

21))((

21

LIILI

MPE

=

=

)Volume(2

12

121

21

2

02

2220

0

22

02

=

=

==

BAhh

IN

Ih

ANLIMPE

µµ

µ

µ

7

24.4 The Energy Carried by Electromagnetic Waves

The total energy density carried by an electro-magnetic wave (which is itself a function of location and time)

[ ] [ ]22 ),(2

1),(21

Volume)( Total),( txBtxEMPEEPEtxu

oo µ

ε +=+

=

8

2

21

Volume Total EEPEu oE ε==

From Ch. 20 electric energy density

From Ch. 22 magnetic energy density

2

21

Volume Total BMPEu

oB µ

==

+

=

λππ

µλππε xftBxftEtxu o

ooo

22sin2

122sin21),( 2222

Averaging over space and time: 2122sin2 =

λππ xft

+

=→ 2

02

0 21

21

21

21 BEu

oo µ

ε

20

20 4

141 BEu

oo µ

ε += But 00

0 Ec

EB ooµε== 20

20 EB ooµε=→ 2

0

20 EB

oo

εµ

=→

202

1 Eu oε=2

021 Bu

oµ=and

EPE = MPE

24.4 The Energy Carried by Electromagnetic Waves

A related quantity: Intensity of EM waves (How bright is the light hitting you?) Analogous to sound intensity (how loud is the sound wave arriving at you)?

9

y

x

z area A

tS

∆==

(Area)energy EM Total

AreaPower

t∆=

(Area)olume)density)(Venergy (EM

We compute the intensity by finding the energy contained the gray box (of length c∆t, and area A) passing through the yellow frame (area A) in time ∆t.

tcx ∆=∆ ( ) uctA

AtcuS ⋅=∆

⋅∆⋅=

Intensity: S (unit: W/m2)

ucS ⋅=→

202

1 EcS oε⋅=2

021 BcS

oµ⋅=and

Average intensity of an electromagnetic wave:

10

Example: Sunlight enters the top of the earth's atmosphere with an electric field whose RMS value is Erms = 720 N/C. Find

(a) the average intensity of this electromagnetic wave and (b) the rms value of the sunlight's magnetic field At the Earth, orbiting the Sun at 1.00 a.u. (c) Repeat (a) and (b) for Venus, which orbits the Sun at a radius of 0.723 a.u. *** a.u. = astronomical unit.

11

Example: Sunlight enters the top of the earth's atmosphere with an electric field whose RMS value is Erms = 720 N/C. Find

(a) the average intensity of this electromagnetic wave and (b) the rms value of the sunlight's magnetic field At the Earth, orbiting the Sun at 1.00 a.u. (c) Repeat (a) and (b) for Venus, which orbits the Sun at a radius of 0.723 a.u. *** a.u. = astronomical unit.

23

2

2

212822

0 mW 10 38.1

CN 720

mNC 1085.8

sm 1000.3

21 )a( ×=

⋅

×

×=== −

rmsoo EcEcS εε

T 1040.2 m/s 1000.3

N/C 720 )b( 68

00

−×=×

==→=c

EBc

EB rmsrms

723.0a.u. 00.1a.u. 723.0 ,

4 and

4 )c(

2

2

2

2

2

==

==→==

−

E

V

E

V

V

E

E

V

E

SUNE

V

SUNV r

rrr

rr

SS

rPS

rPS

ππ

( )28

2

8

2

2

W/m 1063.2 (0.723)

W/m 1038.1/

×=×

==

=

−

EV

EE

E

VV rr

SSrrS

2202

1rms

oo

BcBcSµµ

=⋅=

T 1032.3 m/s 1000.3

)W/m 10m/A)(2.63T 104( 6-8

237

×=×

×⋅×==→

−πµcSB o

rms

24.6 Polarization

In (linearly) polarized light, the electric field oscillates along a single transverse direction.

12

POLARIZED ELECTROMAGNETIC WAVES

In unpolarized light, the electric field oscillates along random transverse direction at a given moment in time.

θ E0

E0’=E0cosθ

24.6 Polarization

Polarized light may be produced from unpolarized light with the aid of polarizing material.

13

2002

1 EcS ε=

( ) ( )200

200 cos

21'

21' θεε EcEcS ==

θcos' 00 EE =

Average Intensity of light:

Absorption and retransmission by antennae in polarizer

θθε 22200 coscos

21' SEcS =

=

Malus’ Law For unpolarized incident light Random Input polarization:

21coscos 22 =→ θθ

( ) SSS21cos' 2 == θ

Primed = after polarizer Unprimed = before polarizer

24.6 Polarization

MALUS’ LAW (general case)

θ2cosoSS =

intensity before analyzer

intensity after analyzer 14 The textbook’s

notational convention

E0

E0cosθ

θ = angle between polarization before and after analyzer

24.6 Polarization

Example Using Polarizers and Analyzers What value of θ should be used so the average intensity of the polarized light reaching the photocell is one-tenth the average intensity of the unpolarized light?

15

( ) θ221

101 cosoo SS =

θ251 cos=

51cos =θ

4.63=θ

24.6 Polarization

THE OCCURANCE OF POLARIZED LIGHT IN NATURE

16

Chapter 25

The Reflection of Light: Mirrors

17

We will start to treat light in terms of “rays” -- technical for the poetic “beam” of light. This works as long as the “width” of our rays are significantly larger than the wavelength of the light

25.1 Wave Fronts and Rays A hemispherical view of a sound wave emitted by a small pulsating sphere (i.e. a point-like source).

18

At large distances from the source, the wave fronts become less and less curved.

−=

λππψψ xfttx 22sin),( 0

The rays are perpendicular to the wave fronts (locations where the wave is at the same phase angle, e.g. at the maxima or compressions.

They become plane-waves described by:

25.2 The Reflection of Light

LAW OF REFLECTION

The incident ray, the reflected ray, and the normal to the surface all lie in the same plane, and the angle of incidence equals the angle of reflection.

19

2 types of reflecting surfaces: (a) smooth surface

Specular reflection: the reflected rays are parallel to each other.

(b) rough surface

Diffuse reflection

Plane mirror

Law of reflection still applies locally, where the normal is the direction perpendicular to the tangent plane

B

B

25.4 Spherical Mirrors

Just as it does for a plane mirror, the Law of reflection still applies locally, where the normal is the direction perpendicular to the tangent plane

20

Mirrors cut from segments of a sphere (Spherical mirrors) are used in many optical instruments.

If the inside surface of the spherical mirror is silvered or polished, it is a concave mirror.

If the outside surface is silvered or polished, it is a convex mirror.

The principal axis of the mirror is a straight line drawn through the center (C) and the midpoint of the mirror (B).

21

Applications of spherical mirrors

Wide-angle rear-view convex mirror found on many vehicles.

8.2 m diameter (largest single piece) reflector (concave mirror***) of the SUBARU Telescope National Astronomical Observatory of Japan (Mauna Kea Hawaii)

*** actually the SUBARU main mirror is closer to an hyperboloid

Large radius-of-curvature (R) concave mirrors to enlarge nearby object.

e.g. compact mirror

25.4 Spherical Mirrors

Light rays near and parallel to the principal axis are reflected from the concave mirror and converge at the focal point.

22

The focal point of a spherical concave mirror is (approximately) halfway between the center of curvature of the mirror (C) and the middle of the mirror at B.

In Class Demo The focal length, f , is the distance between the focal point and the middle of the mirror.

Rf 21=

25.4 Spherical Mirrors

The effect is called spherical aberration.

23

Rays that lie close to the principal axis are called paraxial rays.

Rays that are far from the principal axis do not converge to a single point.

When paraxial light rays that are parallel to the principal axis strike a convex mirror, the rays appear to originate from the focal point behind the mirror.

Rf 21−=

Since the light rays do not really converge to this point, it is a virtual focus

The SIGN convention is to assign a negative value to virtual quantities

In Class Demo