Radiation Pressure (1)

Monday, April 7, 2014 Doppler effect

. Laser: narrow frequency ω0 . The frequency of light seen by the car

ω ω (1 + (v/c)) λ = λ0(1 (v/c)) ≈ 0 − (1 + (v/c)) ω = ω0 1 (v/c)2 − . The frequency of reflected light seen by the detector is ω ω (1 + 2(v/c)) ≈ 0

Monday, April 7, 2014 Energy Transport (1)

. Electromagnetic waves carry energy . In the sunlight, we feel warmth … . These phenomena are related to electromagnetic waves emitted from the Sun . Energy generated by nuclear reactions in the Sun is carried by the electromagnetic waves . The rate of energy transported by an electromagnetic wave is

. This quantity is called the Poynting vector after British physicist John Poynting who first discussed its properties

Monday, April 7, 2014 Radiation Pressure (1)

. When you walk out into the sunlight, you feel warmth, but you do not feel any force from the sunlight . EM waves carry energy and momentum! . Sunlight is exerting a pressure on you, but that pressure is small enough that we do not notice it . Let’s calculate the magnitude of the pressure exerted by these radiated electromagnetic waves

Monday, April 7, 2014 Radiation Pressure (2)

. Electromagnetic waves have energy . Electromagnetic waves also have linear momentum . Let’s assume that an electromagnetic wave is incident on an object and that this object totally absorbs the wave over some time interval Δt . The object will then gain energy ΔU over that time interval from the radiated electromagnetic waves and we can relate the change in momentum of the object Δp to the change in energy by E = mv2/2 c.f. ∆E = mv∆v = v∆p . The change in momentum of the object will be in the same direction as the incident electromagnetic wave

Monday, April 7, 2014 Radiation Pressure (3)

. If instead, the object reflects the electromagnetic wave, the object will gain twice the momentum of the incident wave, because of momentum conservation, in the direction of the incident wave

. Newton’s Second Law tells us that

. To get an expression for the pressure exerted by the electromagnetic wave in terms of the intensity of the wave, we remember that • the intensity is power per unit area • power is energy per unit time

Monday, April 7, 2014 Radiation Pressure (4)

. Therefore we can relate the intensity of the wave to the momentum transferred to the object when the electromagnetic wave is absorbed

which gives us an expression for the force exerted by the radiation on the object

. The pressure is defined as force per unit area so we can write the

radiation pressure due to a totally absorbed electromagnetic wave pr as I = F dA Poynting · 48

Monday, April 7, 2014 Radiation Pressure (5)

. If the electromagnetic wave is reflected the radiation pressure is

(for normal impact and normal reflection, specular reflection)

. Diffuse reflection

Monday, April 7, 2014 Radiation Pressure (5)

. If the electromagnetic wave is reflected the radiation pressure is

(for normal impact and normal reflection)

Sunlight: . The intensity of sunlight is at most 1400 W/m2 . The maximum radiation pressure for sunlight that is totally absorbed is

. Thus the radiation pressure from sunlight is very small

Monday, April 7, 2014 Principle of a laser . If I add N waves coherently . If I add N waves incoherently (random phases) . 4 Stot 2

2 0 1 2 3 4 5 6 S = N S0 S = NS0

Monday, April 7, 2014 Principle of a laser . EM power of a wave:

. Let’s add two waves in phase (coherently) E = E sin(ωt kx) 1 0 − E = E sin(ωt kx) 2 0 − E =2E sin(ωt kx) tot 0 − S =4S0

. Adding two waves increased power by 4 times!

Monday, April 7, 2014 E

Monday, April 7, 2014 absorption

hν E

Monday, April 7, 2014 absorption

hν E

Monday, April 7, 2014 absorption emission

hν E

Monday, April 7, 2014 absorption emission

hν E

Monday, April 7, 2014 absorption emission Stimulated emission

hν E

Monday, April 7, 2014 Mirror Population inversion Mirror

Monday, April 7, 2014 Mirror Spontaneous emission Mirror

Monday, April 7, 2014 Mirror Stimulated emission Mirror

Monday, April 7, 2014 Feed-back by the cavity Mirror Mirror

Monday, April 7, 2014 Stimulated emission Mirror Mirror

Monday, April 7, 2014 Feed-back by the cavity Mirror Mirror

Monday, April 7, 2014 After several round trips…

Mirror Mirror

Laser beam Photons with: - same energy : Monochromatic - same direction of propagation : Spatial coherence - all in synchrony: Temporal coherence

Monday, April 7, 2014 20

Monday, April 7, 2014 Polarization consider a single wave

x E k E

y B B k Polarization: direction of oscillations of E-field . plane-polarized wave . plane-polarized wave . in the x direction . in the y direction

Monday, April 7, 2014 Polarization

Two possible polarizations

E B E

B k k

Monday, April 7, 2014 . Any E E B E = + B k k k

Monday, April 7, 2014 Circular polarization

. Let take two linearly polarized E B waves and add them with a E phase shift of 90o B k k . E = E sin(ωt kx) x 0 − E = E sin(ωt kx π/2) y 0 − −

http://webphysics.davidson.edu/physlet_resources/dav_optics/

Examples/polarization.html 24

Monday, April 7, 2014 . Two polarizations: • two linear • two circular . circular = sum of linear . linear = sum of circular

Monday, April 7, 2014 Add many waves

. Possibility 1: Each wave has its electric field vector oscillating in a different plane . This light is called unpolarized light

Monday, April 7, 2014 Add many waves

. Possibility 2: Waves have a preferred direction of oscillations . This light is called partially polarized light .

Monday, April 7, 2014 Add many waves

. Waves have one direction of oscillations (of E-field) . polarized light .

Monday, April 7, 2014 Polarizer (1) . We can change unpolarized light to polarized light by passing it through a polarizer

. A polarizer allows only one component of the polarization of the light to pass through

. One way to make a polarizer is to produce a material that consists of long parallel chains of molecules that effectively only let light pass if it has a certain direction of polarization and block it if it has a polarization in the perpendicular direction

Monday, April 7, 2014 Polarizer (3)

. The components of the unpolarized light that have the same polarization as the polarizer are transmitted but those with polarization perpendicular to the polarizer are absorbed

. If light with polarization parallel to the polarizing angle is incident on the polarizer, all the light passes through

. If light with polarization perpendicular to the polarizing angle is incident on the polarizer, none of the light is transmitted

Monday, April 7, 2014 Polarizer (2) . We will discuss polarizers without taking into account the details of the molecular structure . Instead we will characterize each polarizer with a polarizing direction . Unpolarized light passing through a polarizer will emerge polarized in the polarizing direction

Monday, April 7, 2014 Intensity of polarized light Case 1: coherent addition

E E B E = + B k k k

E = E sin(ωt kx) x 0 − E = E sin(ωt kx) y 0 − If phases are equal, the sum remains 100% polarized

Monday, April 7, 2014 Intensity of polarized light

Case 2: incoherent addition

. If phases are random, the sum is unpolarized

= + k k

In most applications phases are random

Monday, April 7, 2014 Polarizer (4)

. Now let’s consider the intensity of the light that passes through a polarizer

. We begin with unpolarized light with intensity I0 . Unpolarized light has equal components of polarization in the y and z directions . After passing through a vertical polarizer only the y component of the polarization remains . The intensity I of the light passing through the polarizer is given by

because the unpolarized light had equal contribution from the y and z components and only the y components are transmitted by the vertical polarizer . This factor of one half only applies to the case of unpolarized light passing through a polarizer

Monday, April 7, 2014 Intensity after Polarizer (1)

. Now let’s assume that polarized light passes through a polarizer and that this light has a polarization that is not parallel or perpendicular to the polarizing direction of the polarizer . The angle between the incident polarization is θ . The component of the electric field E of the light that is transmitted is given by

where E0 is the electric field of the incident polarized light

. The intensity of the light I0 before the polarizer is given by

Monday, April 7, 2014 Intensity after Polarizer (2)

. After the light passes through the polarizer, the intensity I is given by

. The transmitted intensity in terms of the initial intensity is

. This result is called the Law of Malus . This equation only applies to the case of polarized light incident on a polarizer

Monday, April 7, 2014 iClicker . Unpolarized light first passes through one polarizer, and then through another tilted by 45

degrees. Intensity of unpolarized light is I0. The intensity of the passed through light is

A: I0/2 3/2 B: I0/ 2

C: I0/4

D: I0/8 E: 0

Monday, April 7, 2014 iClicker

. The intensity of the unpolarized light is I0 . The intensity of the light passing through the first polarizer is

The intensity of the light passing the second polarizer is

= I0/4 .

Monday, April 7, 2014 iClicker Unpolarized light first passes through one polarizer, and then through another tilted by 45 degrees and then through another tilted by 45 degrees with respect to second. Intensity of unpolarized light is I0. The intensity of the passed through light is

A: I0/2

B: I0/4 5/2 . C: I0/2

D: I0/8 E: 0

Monday, April 7, 2014 iClicker: Three Polarizers (3) . The intensity of the light passing the second polarizer is

. The intensity of the light passing the third polarizer is

. The fact that 1/8th of the intensity of the light is transmitted is somewhat surprising because polarizers 1 and 3 have polarizing angles that are perpendicular to each other . The fact that polarizer 2 is in between these two polarizers allows light to pass through

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Monday, April 7, 2014