EC560_Fall2013_HW3_Solutions(5)
4
Homework 3 solution 1.Double convex lens. ] 2 ) ( exp[ ] 2 ) ( exp[ ) , ( ) , ( ) , ( 2 2 2 2 1 2 2 1 2 1 f y x jk h f y x jk h y x t y x t y x t o o o o Where 1 1 1 n R f and 1 2 2 n R f and 1 o h , 2 o h ,are constants. Thus ] 2 ) ( exp[ ) , ( 2 2 f y x jk h y x t o o Where 2 1 2 1 1 1 1 1 1 1 R R n f f f and 2 1 o o o h h h is a constant. Note that 2 R is negative 2.Focusing of a plane wave by a thin lens ) exp( ) , ( 1 jkz y x U and ] 2 ) ( exp[ ) , ( 2 2 f y x jk h y x t o Therefore, ) , ( ) , ( ) , ( 1 2 y x t y x U y x U ]} 2 ) ( [ exp{ 2 2 f y x z jk h o The wave fronts of this wave are paraboloids of revolution,defined by f y x z 2 ) ( 2 2 =constant with radius of curvature - f ,i.e., they approximate a spherical wave focused at a point a distance f to the right of the lens.If the incident wave is a plane wave at a small angle , )] ( exp[ ) , ( 1 x z jk y x U , Then ) , ( ) , ( ) , ( 1 2 y x t y x U y x U This is a parabolical wave centered about the point ) , 0 , ( f f ,as illustrated below ]} 2 ) ) (( [ exp{ ]} 2 ) 2 ( [ exp{ ]} 2 ) ( [ exp{ 2 2 2 2 2 2 f y f x z jk h f y x f x z jk h f y x x z jk h o o o
-
Upload
lauren-stevenson -
Category
Documents
-
view
216 -
download
4
description
solutions to photonics problems
Transcript of EC560_Fall2013_HW3_Solutions(5)
Homework 3 solution1.Double convex lens.
]2
)(exp[]2
)(exp[),(),(),(2
22
21
22
121 fyxjkh
fyxjkhyxtyxtyxt o
oo
o
Where1
11 nRf and
12
2 nRf and 1oh , 2oh ,are constants. Thus ]
2)(exp[),(
22
fyxjkhyxt o
o
Where
21
21111111
RR
nfff
and 21 ooo hhh is a constant. Note that 2R is negative
2.Focusing of a plane wave by a thin lens
)exp(),(1 jkzyxU and ]2
)(exp[),(
22
fyxjk
hyxt o
Therefore, ),(),(),( 12 yxtyxUyxU ]}2
)([exp{22
fyxzjkho
The wave fronts of this wave are paraboloids of revolution,defined byfyxz
2)( 22
=constant with radius of
curvature - f ,i.e., they approximate a spherical wave focused at a point a distance f to the right of the lens.If the
incident wave is a plane wave at a small angle , )](exp[),(1 xzjkyxU ,
Then ),(),(),( 12 yxtyxUyxU
This is a parabolical wave centered about the point ),0,( ff ,as illustrated below
]}2
))(([exp{
]}2
)2([exp{
]}2
)([exp{
22
22
22
fyfxzjkh
fyxfxzjkh
fyxxzjkh
o
o
o
3.Diffraction grating
a) )]2cos(1)[2
()( 0
xdxd
)]2cos()2
)(1(exp[)]()1(exp[)exp()( 000000
xdknjhxdknjdjkxt
where )]2
)(1(exp[ 000
dknjh
b) Since t(x) is a periodic function of x with period of A, it can be expended in a Fourier series
)2exp()(
xjqCxt
where qC are the Fourier coefficients. If the incident wave is a plane wave at small angle i i.e.,
)](exp[),( 01 xzjkyxU q , the transmitted wave has amplitude
)](exp[)]2(exp[)()()( 00012 xzjkxqxkzkjxUxtxU qi
Where
qxq
iiq2
. Thus the transmitted wave is composed of plane wave at angles q
4. Transmission through transparent plate.
5.Interference
]cos[)(2 21
2121 IIIII where 2
21
1||
zAI , 2
22
2||
zAI and
)(2
)( 2222
dyx
zyxk
Therefore )](cos[)(2),,(22
21
2121 dyxIIIIdyxI
The locus of constant I are circles 22 yx constant. The function )cos( 2x is plotted in Table A.1-1 on page 920.
It is a sinusoidal function, called the chirp function, whose frequency increase as x increases. This is why the rings in the
interference pattern become closer and closer as 22 yx increases.
6.Interference
]}2)([exp{)(
22
1 zyaxjkjkz
zAU
]}2)([exp{)(
22
2 zyaxjkjkz
zAU
At ),cos(22, 00 IIIdz
where 20 ||
dAI and
daxax
dyaxyax
dk
44]})[(])){[(
2( 2222
Therefore,dawherexII 2)],2cos(1[0
7.Bragg reflection
The phase difference between two reflection is )( 12 ddk .Butsin2dd and
2cos21 dd =
sin2cosd
Therefore, k
sin2sin2)sin()2cos1(
sin2 kddkd
For ,sin,2 kd or 1sin2
d
i.e.,d2
sin
