Electromagnetic Waves

Lecture 13 Electromagnetic Waves Ch. 33

• Cartoon• Opening Demo• Topics

– Electromagnetic waves– Traveling E/M wave - Induced electric and induced magnetic amplitudes– Plane waves and spherical waves– Energy transport Poynting vector– Pressure produced by E/M wave– Polarization– Reflection, refraction,Snell’s Law, Internal reflection– Prisms and chromatic dispersion– Polarization by reflection-Brewsters angle

• Elmo• Polling

Electromagnetic Waves

Eye Sensitivity to Color

Production of Electromagnetic waves

Spherical waves Plane waves

To investigate further the properties of electromagnetic waves we consider the simplest situation of a plane wave. A single wire with variable current generates propagating electric and magnetic fields with cylindrical symmetry around the wire.

If we now stack several wires parallel to each other, and make this stack wide enough (and the wires very close together), we will have a (plane) wave propagating in the z direction, with E-field oriented along x, E = Ex (the current direction) and B-field along y B=By (Transverse waves)

Electromagnetic Wave

How the fieldsvary at a PointP in space as

the wave goesby

Electromagnetic Wave

Electromagnetic Wave Self Generation

• Faraday’s Law ofInduction

• Maxwell’s Law ofInduction

• Changing electric fieldinduces a magneticfield

• Changing magneticfield induces a electricfield

!E.d!s = !

d"B

dx"#

!B.d!s = µ

0!0

d"E

dx"#c =

1

µ0!0

Summary

!S =

dU

Adt

U is the energy carried by a wave

Magnitude of S is like the intensity

S =dU

Adt

B0=E0

c

A point source of light generates a spherical wave. Light isemitted isotropically and the intensity of it falls off as 1/r2

Let P be the power of the sourcein joules per sec. Then the intensity of light at a distance r is

24 r

PI

!=

Lets look at an example

Relation of intensity and power for a Spherical Wave

15 . The maximum electric field at a distance of 10 m from an isotropicpoint light source is 2.0 V/m. Calculate

(a) the maximum value of the magnetic field and

(b) the average intensity of the light there?

(c) What is the power of the source?

(a) The magnetic field amplitude of the wave is

Tc

EB

s

m

m

V

m

m

9

8107.6

10998.2

0.2!

"="

==

(b) The average intensity is

Iavg =E2

m

2µ0c=

2.0V

m( )2

2 4! "10#7T $

m

A( ) 2.998 "108 m

s( )= 5.3"10#3 W

m2

(c) The power of the source is

( ) ( ) WmIrPm

Wavg 7.6103.51044

2

322=!==

"##

Speed of light in Water v =c

n=2.998 !10

8 m

s

1.33= 2.26 !10

8 m

s

Nothing is known to travel faster than light in a vacuum

However, electrons can travel faster than light in water.And when the do the electrons emit light called Cerenkov radiation

Momentum and Radiation Pressure

c

Momentum = p

p=U/c paper

Light beam

1) Black paper absorbs all the light

p=0!p = !U / c

What is the change in momentum of thepaper over some time interval?

2) Suppose light is 100% reflected

!p = p " (" p) = 2p !p = 2!U / c

From this we define the pressure

Radiation Pressure Pr

Want to relate the pressure Pr felt by the paper to the intensity of light

I =Power

Area=!U / !t

A

Pr =F

A=!p / !t

A

!p = !U / c For 100% absorption of light on the paper

Pr=F

A=c!U / !t

A= cI

Pr=F

A=2c!U / !t

A= 2cI For 100% reflection of light

Problem 21What is the radiation pressure 1.5 m away from a 500Watt lightbulb?

Radiation pressure: Light carries momentum

�

Pr

=I

c

This is the force per unit area felt by an object that absorbs light. (Black piece of paper))

�

Pr

=2I

c

This is the force per unit area felt by an object that reflects light backwards. (Aluminum foil)

Another property of light

Polarization of Light

• All we mean by polarization is whichdirection is the electric vector vibrating.

• If there is no preferred direction the wave isunpolarized

• If the preferred direction is vertical, then wesay the wave is vertically polarized

Pass though a polarizing sheet aligned to pass only the y-component

A polarizing sheet or polaroid filter is specialmaterial made up of rows ofmolecules that only allow light to pass when the electricvector is in one direction.

Resolved into its y and z-components The sum of the y-components and z components are equal

Unpolarized light

Pass though a polarizing sheet aligned to pass only the y-componentMalus’s Law

Intensity I0

One Half RuleHalf the intensity out

I =I0

2

I2= I

1cos

2!

I0

I1=I0

2

I ! Ey

2

Ey= E cos"

E2

y= E

2cos

2"

I2= I

1cos

2"

y

Ey= E cos!

Polarizer P1

Analyzer P2

Sunglasses are polarized vertically.Light reflected from sky is partially polarizedand light reflected from car hoods is polarizedin the plane of the hood

Sunglass 1

Sunglass 2

Rotate sun glass 90 deg and no light gets through because cos 90 = 0

35. In the figure, initially unpolarized light is sent through threepolarizing sheets whose polarizing directions make angles of θ1 = 40o,θ2 = 20o, and θ3 = 40o with the direction of the y axis. What percentageof the light’s initial intensity is transmitted by the system? (Hint: Becareful with the angles.)

Let Io be the intensity of the unpolarized light thatis incident on the first polarizing sheet. Thetransmitted intensity of is I1 = (1/2)I0, and thedirection of polarization of the transmitted light is θ1 = 40o counterclockwise from the y axis in thediagram. The polarizing direction of the secondsheet is θ2 = 20o clockwise from the y axis, so theangle between the direction of polarization that isincident on that sheet and the the polarizingdirection of the sheet is 40o + 20o = 60o. Thetransmitted intensity is

,60cos2

160cos

2

0

2

12

ooIII ==

and the direction of polarization of the transmitted light is 20o clockwisefrom the y axis.

I1

I2

I3

I0

35. In the figure, initially unpolarized light is sent through threepolarizing sheets whose polarizing directions make angles of θ1 = 40o,θ2 = 20o, and θ3 = 40o with the direction of the y axis. What percentageof the light’s initial intensity is transmitted by the system? (Hint: Becareful with the angles.)

The polarizing direction of the third sheet is θ3= 40o counterclockwise from the y axis.Consequently, the angle between thedirection of polarization of the light incident onthat sheet and the polarizing direction of thesheet is 20o + 40o = 60o. The transmittedintensity is

I3= I

2cos

260

o

I2= I

1cos

260

o

I1=1

2I0

I3

I0

=1

2cos

460 = 3.125 !10

"2

Thus, 3.1% of the light’s

initial intensity is transmitted.

Chapter 33 Problem 34In Figure 33-42, initially unpolarized light is sent into a system ofthree polarizing sheets whose polarizing directions make angles ofθ1 = θ2 = θ3 = 16° with the direction of the y axis. Whatpercentage of the initial intensity is transmitted by the system?

Fig. 33-42

Chapter 33 Problem 49

In Figure 33-51, a 2.00 m long vertical pole extends from thebottom of a swimming pool to a point 90.0 cm above thewater. Sunlight is incident at angle θ = 55.0°. What is thelength of the shadow of the pole on the level bottom of thepool?

Fig. 33-51

Chapter 33 Problem 62 Suppose the prism of Figure 33-55 has apex angle ϕ = 69.0° and index of

refraction n = 1.59.

Fig. 33-55(a) What is the smallest angle of incidence θ for which a ray can enter the

left face of the prism and exit the right face?(b) What angle of incidence θ is required for the ray to exit the prism with

an identical angle θ for its refraction, as it does in Figure 33-55?

In Figure 33-51, a 2.00 m long vertical pole extends from the bottom of aswimming pool to a point 90.0 cm above the water. Sunlight isincident at angle θ = 55.0°. What is the length of the shadow of thepole on the level bottom of the pool?