! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! Solution PS#1

Q1:!

a) The threshold of power: Pth=Nhv=3.6x10-17

W

b) Red side: Er= =1.6ev

Blue side: = =3.27ev

c) The rest-mass energy of an electron is

Ee=mec2=8.19x10

-14J, me=9.1x10

-31kg

The energy of photon Ep=Ee=hc

! thus =2.43x10

-12m

Momentum P=h

!=2.73x10

-22Js/m

d) Irradiance I(r)=2

100

4 r! u(r)=

2

100

4

I

c c r!=

Energy =1.51.5 1.5

2 2 2

0 0 0

100( ) 8

4 ( )cubic

u r dV dxdydzc x y z!

=+ +

" " " "

= dxdy

=6.11x10-7

J (numerically)

e) The irradiance of the beam

I=2

2

c E!=

3 2

2

c B!! ! thus B=

3

2I

c !=2.89x10

-6T

For infinite long straight wire, B=2

i

r

µ

!, where i is the current

Thus i=2 rB!

µ=14.45mA

Q3, b) from the lorenze model of collision electrons

2

2 2

2 2

0

( ) 1p

n n iki

!

! ! "!= + = +

# #! (1)

In the medal, the electron is considered as free-electron, thus the restoring force is

zero, the resonate frequency is zero, if without collision, that is

Equation (1) turns to be

2

2

21

pn

!

!= "! , only when 2 2

p! !> , the medal is not

transparent to the EM waves, and there is a penetration depth, here the plasma

frequency dominate the penetration. For 2 2

p! !< , since 2

0, is realn n>! ! , the

metal is a transparent to the EM wave, the penetration depth is infinite….

For

2314 , 349 , 0.235, 0.485

penetration depth 114.582

p nm nm n n ik i

d nmk

! !

!

"

= = # $ = =

= #

! !

For

2314 , 314 , 0, 0

penetration depth 2

p nm nm n n ik

dk

! !

!

"

= = # = =

= # $

! !

For

2314 , 289 , 0, 0, 0

the wave can propagate in the medal infinate long, the medal is transparent medium to the light

p nm nm n n n ik! != = > = > =! !

"#$%&%'(! )*! $%! +#,-).%'! /0%! +#11)-)#,! #*! %1%+/'#,-(! %234/)#,! 567! /3',-! /#! 8%!

2 2

2 2

2

( / )( ) 1 1

1 ( / )

p pn n ik

i i

! ! !

! "! " != + = # = #

+ +!

If / 1! " >> (!/0%,!

2

2 2( / )

( ) 1( / )

pn n ik

i

! !

" != + # $!

2 2 22

2 2 1/ 21 / 1( / )

1 2 ( )( / ) 2

ppn k nk k

! " !! !

" !

+ ## = = $ =

Penetration depth 2

dk

!

"= (!,#!94//%'!/0%!*'%23%,+:!)-!14';%'!#'!-9411%'!/04,!/0%!<14-94!

*'%23%,+:(!/0%'%!)-!41$4:-!4!-=),>.%</0!),!9%/41!

!

?*! / 1! " << (!$0)+0!)-!/0%!+4-%!1)-/%.!*#'!/0%!/0'%%!$4&%1%,;/0!),!/0)-!23%-/)#,(!%234/)#,!567!

/3',-!#3/!/#!8%! !

!

2

2 2 2( / )

( ) 1 1 ( / ) (1 ( / ))1 ( / )

11 for 1

1

p

pn n ik ii

x xx

! !! ! " !

" != + = # $ # #

+

$ # <<+

!

Thus

2

2 2 2

3

2 2 2 2 4 6

1/ 2

1 ( / ) 2

( / ) 1 [1 ( / ) ] / ( )

2

p

p

p p p

n k nk

k

! "! !

!

! ! ! ! " ! !

# = # =

# + # +

$ =

Substitute the value for wavelength into the above equation, one can get the depth

for waves can propagate inside the medal is

! 2( / ) 1p

! ! " 2 4 6/

p! " " 2

k

2d

k

!

"=

349nm 0.235 5.27x10-6

0.235 719.9nm

314nm 0 2.77x10-6

0.832x10-3

10.88µm

289nm -0.153 1.69x10-6

2.7x10-5

55µm

Q5: the value is

The maximum difference in the real part of refraction index is 1.84x10-4

The percentage difference of group and phase velocity is

At 7.0x1014

Hz" 0.035%

At 9.0x1014

Hz, 0.003%