33. 전자기파 (Electromagnetic waves)optics.hanyang.ac.kr/~choh/degree/general_physics2/Chapter 33-1.pdf · 질문 •At t = 0, z = 0, the electric field of an electro-magnetic

33. 전자기파 (Electromagnetic waves)

33-2. Maxwell의 무지개눈의감도

33-3. 진행하는 전자기파 : 정성적전자기파의 발생기구에 따른 분류

• 장파, 라디오파(방송파) - LC회로: 고전 전자기학 이론• 가시광, x-선, 감마선 - 원자 또는 핵: 양자물리학 이론

라디오파의 발생과 전파

1) LC 발진기 = 교류전류 공급원

2) 안테나(전자기파 방출원) = 진동 전기쌍극자

Question: EM WavesWhich direction should I orient my antenna to

receive a signal from a vertical transmission tower?

1) Vertical 2) Horizontal 3) 45 Degrees

+

-

Direction wave travels

Alternating E field moves charges up and down thru antenna!

진행하는 전자기파 : 정성적Antenna

Electromagnetic Waves

xz

y

• Transverse:횡파 (vs. sound waves – longitudinal: 종파)

• E perpendicular to B and always in phaseE & B increase and decrease at same times

• Can travel in empty space (sound waves can’t!)

• “Speed of light”: v = c = 1/ √(ε0 μ0) = 3 x 108 m/s(300,000 km/second!)

• Frequency: f = v/λ = c/λ

Which of the following are transverse waves?

• sound

• light

• radio

• X-ray

• microwave

• water waves

• “The Wave” (응원파도타기)

All but sound!

33-3. 진행하는 전자기파 : 정성적

안테나가 방출하는 전자기파의 시․공간적 변화

① 관측점 P에서의 시간적 변화

② 진행방향에 따른 공간적 변화

전자기파 속력

전자기장 벡터

증명해 보자

2

2

k

f

πλ

ω π

≡

≡f c

kω λ= =

Er

Br

E B×r r

속력 Cm

m

E cB

=

33-3. 진행하는 전자기파 : 정량적

( )E ds E dE h Eh hdE⋅ = + − =∫r r

Faraday’s Law:

( ){ } ( )Bd d dBB hdx hdxdt dt dtΦ

= =

BdE dsdtΦ

⋅ = −∫r r

dE dBdx dt

= − E Bx t

∂ ∂= −

∂ ∂

{ }

{ }

sin( ) cos( )

sin( ) cos( )

m m

m m

E E kx t kE kx tx xB B kx t B kx tt t

ω ω

ω ω ω

∂ ∂= − = −

∂ ∂∂ ∂

= − = − −∂ ∂

m

m

E cB

=E

cB

=

r

r

Propagation of EM Waves

• Changing B field creates E field• Changing E field creates B field

E = c B

xz

y

If you decrease E, you also decrease B!This is important !

전자기파 속력

0 0EdB ds

dtε μ Φ

⋅ =∫r r

Maxell’s Law:

( )B ds B dB h Bh hdB⋅ = − + + = −∫r r

{ }0 0 0 0 0 0( ) ( )Ed d dEE hdx hdxdt dt dt

ε μ ε μ ε μΦ ⎧ ⎫= = ⎨ ⎬⎩ ⎭

0 0B Ex t

ε μ∂ ∂− =

∂ ∂

0 0cos( ) cos( )m mkB kx t E kx tω ε μ ω ω− − = − −

0 0 0 0

1 1( / )

m

m

E cB k cε μ ω ε μ

= = =

0 0

1cε μ

=

확인문제 1.

(1) B가 증가하고 있을 때,box 양 옆에서의전기장의 방향과 상대적인 크기는?

(2)자기장의 방향과 상대적인 크기는?

cEE+dE

E

BB+dB

질 문

• At t = 0, z = 0, the electric field of an electro-magnetic wave is oriented at an angle θ with respect to the x-axis, as shown.– Which arrow indicates the direction of the

magnetic field at the same location and instant of time?

(a) A (b) B

• This question cannot be answered unless the direction of propagation is specified:

• If the wave propagates in the +z direction, then B-field is along A• If the wave propagates in the –z direction, then B-field is along B

θ

e

x

y

A

B

33-5. 에너지 수송과 Pointing 백터

포인팅 벡터 (Poynting vector)

전자기파에 실려 옮겨지는 단위 시간당, 단위면적당 에너지 흐름의 크기와 방향

[unit : W/m2]

크기:

방향: 전자기파의 진행방향

Er

Br

S E B∝ ×r r r

0

1S Eμ

B≡ ×r r r

확인문제 2. 에너지는 음의 z 축 방향자기장의 방향은?

Sr

Br

단위면적당 일률을나타내는 벡터량

Intensity (I or S) = Power/Area

• Energy (U) in box:U = u x Volume

= u (AL)• Power (P):

L

A

L=ct

P = U/t= U (c/L)= u A c

• Intensity (I or S):S = P/A

= uc

U = Energyu = Energy Density (Energy/Volume)A = Cross section Area of lightL = Length of box

전자기파의 세기 (intensity)

I ≡ Savg =

전자기파에서의 전기 및 자기 에너지 밀도 비교

점원으로부터 거리에 따른 전자기파의 세기 변화

22( ) (4 ) ( )

4s

sPP I r r I rr

ππ

= ⋅ → =

( )2

222 2

0 0 000 0

1 1 1 12 2 2 2E B

Bu E cB B uε ε εμε μ

⎛ ⎞= = = = =⎜ ⎟⎜ ⎟

⎝ ⎠

E Bu u=

전기장과 자기장이 가지고 있는 에너지는 항상 동일하다.

33-6. 복사압 (Radiation pressure)

Power incident on the Earth's surface due to radiation from the sun is about 1370 W/m2

Radiation Pressure at Earth's surface is 1370/c = 4.5710-6N/m2 or 4.57mPa

PRUp Qc

ΔΔ =

Newton 제 2 법칙으로부터

전자기파가 물체에 조사되었을 때,운동량 (p = U/c) 변화량

pFt

Δ=

Δ

U IA tΔ = Δ

PRIAF Qc

= r PRF IP QA c

= =

보기문제 33-2.

33-7. 편광 (Polarization)

편광방향 : 전기장(E)이 진동하는 방향

전자기파의 E 진동방향이일정할 때 (혹은, 일정하게 변화할 때)

편광된 전자기파(Polarized EM wave)

전자기파의 E 진동방향이시간에 따라 마구잡이로 변할 때

막편광, 비편광된 전자기파(Unolarized EM wave)

y

x

x yE E x E y= +r ) )

편광판 (Polarizer)어느 한 방향의 선편광 성분을 선택적으로 투과시키는 판(sheet)

2 2 2 20 0cos cosyI E E Iθ θ= = =

2cosI S θ= ∝ (Malus 법칙)

선 편광판 (Linear Polarizers)

• Linear Polarizers absorb all electric fieldsperpendicular to their transmission axis.

Unpolarized Light on Linear Polarizer

• Most light comes from electrons accelerating in random directions and is unpolarized.

• Averaging over all directions: Stransmitted= ½ Sincident

Always true for unpolarized light!

Linearly Polarized Light on Linear Polarizer (Law of Malus)

Etranmitted = Eincident cos(θ)

Stransmitted = Sincident cos2(θ)

TA

θ

θ is the angle between the incoming light’s polarization, and the transmission axis

θ

Transmission axisIncident E

ETransmitted

Eabsorbed

=Eincidentcos(θ)

Question

Unpolarized light (like the light from the sun) passes through a polarizing sunglass (a linear polarizer). The intensity of the light when it emerges is

1. zero 2. ½ what it was before 3. ¼ what it was before 4. ⅓ what it was before 5. need more information

Now, horizontally polarized light passes through the same glasses (which arevertically polarized). The intensity of the light when it emerges is

1. zero2. ½ what it was before3. ¼ what it was before4. ⅓ what it was before5. Need more information

90°

TA

TA

S1

S2

S0

60°°

TATA

S1

S2

S0

60°°

Law of Malus

A B

1) S2A > S2

B 2) S2A = S2

B 3) S2A < S2

B

S1= S0cos2(60)

S2= S1cos2(30)= S0 cos2(60) cos2(30)

S1= S0cos2(60)

S2= S1cos2(60) = S0 cos4(60)

E0E0

unpolarized light

E1

45°

I = I0

TATA

90°

TA

E0

I3

B1

unpolarized light

E1

45°

I = I0

TATA

90°

TA

E0

I3

B1

Law of Malus – 3 Polarizers

I2= I1cos2(45)

2) Light transmitted through first polarizer is vertically polarized. Angle between it and second polarizer is θ=45º. I2 = I1 cos2 (45º) = ½ I0 cos2 (45º)

3) Light transmitted through second polarizer is polarized 45º from vertical. Angle between it and third polarizer is θ=45º. I3 = I2 cos2 (45º) = ½ I0 cos4 (45º)

I1= ½ I0

33-8. 반사 (reflection), 굴절 (refraction)

Light incident on an object• Absorption

Everything true for wavelengths << object size

• Reflection (bounces)– See it– Mirrors

• Refraction (bends)– Lenses

• Often some of each

33-8. 반사 (reflection), 굴절 (refraction)

반사 법칙 : '1 1θ θ=

1 1 2 2sin sinn nθ θ= =일정굴절 법칙 (Snell 법칙) :

굴절률 (index of refraction) :

cnv

⎛ ⎞≡ = ⎜ ⎟

⎝ ⎠

진공중에서의속력

매질내에서의속력

12 nn >

12 nn <

1n

2n

n2

n1

d

d′

′ d = d n2

n1

Apparent depth:

Apparent Depth

50

actual fish

apparent fish

굴절률 (refractive index )

cnv

⎛ ⎞≡ = ⎜ ⎟

⎝ ⎠

진공중에서의속력

매질내에서의속력

με1

=v(ε, μ) 매질:0μ μ≈0ε κε≡

0 0

cnv

μεκ

μ ε= = =

굴절시 전자기파의 진동수는 변하지 않는다

11 λfv = 22 λfv =Wavefronts

Wavefronts

θ1

θ1

θ2

θ2

l

1 2

1 2

v vλ λ

=

1 1 2 2n nλ λ= =일정 0

nλλ =

색의 분산 (Color dispersion)

파장 λ = 589 nm에서의 굴절률

빛의 파장에 따라 굴절률이 다름 굴절률이 다름 빛이 퍼짐 “분산”

무지개 (Rainbow)

쌍무지개

In second rainbow

pattern is reversed

33-9. 내부전반사 (Total internal reflection)빛이 굴절율이 높은 매질에서 낮은 매질로 들어갈 때 ( n1 > n2),그 경계면에서 빛이 모두 반사되는 현상

{ }1 21 2 2 1 2

1

sin sin 90 sin cnn n n n nn

θ θ −= = ⇒ = >oc

임계각 (critical angle) : θc

광섬유

프리즘

Fiber OpticsAt each contact w/ the glass air interface, if the light hits atgreater than the critical angle, it undergoes total internal reflection and stays in the fiber.

We can be certain that ncladding < ninside

ninside

ncladding

noutside

Add “cladding” so outside material doesn’t matter!

33-10. 반사에 의한 편광

브루스터 법칙(Brewster’s Law)

반사광선과 굴절광선이 90도를 이룰 때, 편광방향이 입사평면에 대해 나란한 빛은 전혀 반사되지 않는 현상(따라서 이 때 반사되는 빛은 편광방향이 입사평면에대해 수직)

브루스터 각 (Brewster angle) : θB

90 θ θ+ = o1 2 일때,

( )1 2

2

2

sin sin

sin 90

cos

n n

n

n

θ θ

θ

θ

=

= −

=

o

1 2

1

1

2

1

tan nn

θ =1

1 2

1

tan nn

θ − ⎛ ⎞≡ ⎜ ⎟

⎝ ⎠B

1θ

2θ

Question: Brewster’s AngleWhen a polarizer is placed between the light source and the surface with transmission axis aligned as shown, the intensity of the reflected light:

(1) Increases (2) Unchanged (3) Decreases

T.A.

Flat Lens (Window)

n1n2

Incident ray is displaced, but its direction is not changed.

←t→

θ1

θ3If θ1 is not large, and if t is small, the displacement, d, will be quite small.

d

2 11

2

n nd tn

θ⎧ ⎫−

= ⎨ ⎬⎩ ⎭

n1

3 1θ θ=

33. Summary : EM waves

Er

Br

E B×r r

속력 CE cB=

0 0

1cε μ

=

[unit : W/m2]

0

1S E Bμ

≡ ×r r r

'1 1θ θ=

1 1 2 2sin sinn nθ θ= =일정

cnv

⎛ ⎞≡ = ⎜ ⎟

⎝ ⎠

진공중에서의속력

매질내에서의속력 { }1 21 2

1

sin cn n nn

θ −= >

1 2

1

tan nn

θ − ⎛ ⎞≡ ⎜ ⎟

⎝ ⎠B