diffraction by n-slits. optical disturbance due to n slits
Post on 14-Dec-2015
229 Views
Preview:
TRANSCRIPT
/ 2 / 2
/ 2 / 2
2 / 2 ( 1) / 2
2 / 2 ( 1) / 2
( ) ( )
( ) ........ ( )
( ) sin[ ( sin )]
b a b
b a b
a b N a b
a b N a b
E C F z dz F z dz
F z dz F z dz
F z t k R z
Optical disturbance due to N slits
Contribution of N slits
1
0
1( ) 2
0
sinIm ( )
sin sinsin[ ( 1) ]
N
j
Ni t kR i j
j
E E
bc e e
NbC t kR N
2 2
0
20
sin sin( ) ( ) ( )*
sin
(0)
NI E E I
I N I
Irradiance due to N-slits
I0 = Irradiance by single slit at =0
For minima
sin0
sin2 ( 1) ( 1)
for , ,....., , ,..
N
N N
N N N N
Between consecutive principal maxima, there will be N-1 minima
Subsidiary maxima
3 5 for , ,.....
2 2N N
(N-2) subsidiary maxima between consecutive principle maxima
The nominal track separation on a CD is 1.6 micrometers, corresponding to about 625 tracks per millimeter. This is in the range of ordinary laboratory diffraction gratings. For red light of wavelength 600 nm, this would give a first order diffraction maximum at about 22° .
ProblemConsider a opaque screen with 5 equally spaced narrow slits (spacing between them is d) and with monochromatic plane wave (wavelength ) incident normally.Draw a sketch of the transmitted intensity vs. angle to the normal for = 0 to = 1/5 radian. Take Sin = over this range and assume d/ =10. What is the ratio of least intense to the most intense peak?What is the angular distance of the first intense peak away from = 0.
top related