Electromagnetic Waves. Lecture 13 Electromagnetic Waves Ch. 33 Cartoon Opening Demo Topics –Electromagnetic waves –Traveling E/M wave - Induced electric

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

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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 fields vary at a Point P in space as

the wave goes by



Electromagnetic Wave

Electromagnetic Wave Self Generation

• Faraday’s Law of Induction

• Maxwell’s Law of Induction

• Changing electric field induces a magnetic field

• Changing magnetic field induces a electric field

rE.d

rs =−

dφB

dx—∫

rB.d

rs =μ0ε0

dφE

dx—∫c =

1μ0ε0



Summary



rS =

dUAdt

U is the energy carried by a wave

Magnitude of S is like the intensity



S =dUAdt



B0 =E0

c

A point source of light generates a spherical wave. Light is emitted 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 rPIπ

=

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 isotropic point 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

TcE

Bsm

mV

mm

98

107.610998.2

0.2 −×=×

==

(b) The average intensity is

Iavg =E2

m

2μ0c=

2.0 Vm( )

2

2 4π ×10−7T ⋅mA( ) 2.998 ×108 m

s( )=5.3×10−3 W

m2

(c) The power of the source is

( ) ( ) WmIrPmW

avg 7.6103.51044 2

322 =×== −ππ





Speed of light in Water v =cn=2.998 ×108 m

s

1.33=2.26 ×108 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=0Δp = ΔU / c

What is the change in momentum of the

paper over some time interval?

2) Suppose light is 100% reflected

Δp = p − (−p) = 2p Δp = 2ΔU / 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 =PowerArea

=ΔU / Δt

A

Pr =FA=Δp / Δt

A

Δp = ΔU / c For 100% absorption of light on the paper

Pr =FA=

cΔU / ΔtA

=cI

Pr =FA=2cΔU / Δt

A=2cI For 100% reflection of light

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

Radiation pressure: Light carries momentum

€

Pr =I

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

€

Pr =2I

cThis 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 which direction is the electric vector vibrating.

• If there is no preferred direction the wave is unpolarized

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

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

A polarizing sheet or polaroid filter is special material made up of rows of molecules that only allow light to pass when the electric vector 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 =I 02

I2 =I1 cos2θ

I0

I1 =I 02

I ∝ Ey2

Ey =Ecosθ

E2y =E2 cos2θ

I 2 =I1 cos2θ

y

Ey =Ecosθ

Polarizer P1

Analyzer P2

Sunglasses are polarized vertically. Light reflected from sky is partially polarized and light reflected from car hoods is polarized in 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 three polarizing sheets whose polarizing directions make angles of θ1 = 40o, θ2 = 20o, and θ3 = 40o with the direction of the y axis. What percentage of the light’s initial intensity is transmitted by the system? (Hint: Be careful with the angles.)

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

,60cos2

160cos 2

02

12oo III ==

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

I1

I2

I3

I0

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

The polarizing direction of the third sheet is θ3 = 40o counterclockwise from the y axis. Consequently, the angle between the direction of polarization of the light incident on that sheet and the polarizing direction of the sheet is 20o + 40o = 60o. The transmitted intensity is

I3 =I 2 cos2 60o

I 2 =I1 cos2 60o

I1 =12

I 0

I 3I 0

=12

cos4 60 =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 of three polarizing sheets whose polarizing directions make angles of θ1 = θ2 = θ3 = 16° with the direction of the y axis. What percentage 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 the bottom of a swimming pool to a point 90.0 cm above the water. Sunlight is incident at angle θ = 55.0°. What is the length of the shadow of the pole on the level bottom of the pool?

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 a swimming pool to a point 90.0 cm above the water. Sunlight is incident at angle θ = 55.0°. What is the length of the shadow of the pole on the level bottom of the pool?

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Electromagnetic Waves. Lecture 13 Electromagnetic Waves Ch. 33 Cartoon Opening Demo Topics –Electromagnetic waves –Traveling E/M wave - Induced electric