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Single Slit
As shown in …gure 1, consider a single slit of width d: Assume that a coherent
plane wave is approaching the slit from the left side. Now consider a screen
located at a distance L to the right of the slit where L >> as well as
L >> d: A point on the screen is de…ned as a distance y above the axis
that de…nes
p the center of the slit so that the distance to this point is given
by r = L2 + y 2 ; and the angle that de…nes this position on the screen is
sin = y=r:
ρ
r
x
y
θ
L
Figure 1: Thin Slit
Now consider a source of a Huygen’s wavelet that is a distance x above
the center of the slit. The distance to this point on the screen is given by
q
= L2 + (y x)2 :
Thus the electric …eld of the Huygen’s spherical wavelet that propagates out
from this point is
sin (k
!t)
Ewav /
:
For a plane wave incident upon the slit, all of the wavelets are in phase and
have the same …eld strength. This enables us to sum the contribution of all
1
of the Huygen’s wavelets inside the slit by simply performing the integral
Etot /
Z
d=2
sin (k
!t)
dx;
d=2
where k = 2 = : We are interested in the total …eld strength at the screen.
This enables us to make several approximations. Since L >> d we can make
the approximation
p
p
y
= L2 + y 2 2xy + x2 = r2 2xy = r
x = r x sin :
r
With this approximation the integral for Etot becomes
Etot /
Z
Etot /
1
r
Etot
1
/
r
d=2
d=2
Z
Z
sin (kr !t kx sin )
dx
r x sin
d=2
sin (kr
!t) cos (kx sin ) cos (kr
1 x sin =r
!t) sin (kx sin )
sin (kr
!t) cos (kx sin )
!t) sin (kx sin ) dx:
d=2
d=2
cos (kr
dx
d=2
Consistent with our approximations we ignore the term that is proportional
to x=r as it is much less than 1. However, kd is of the order of 1, and
the e¤ects of the phase changes can be signi…cant, hence these terms are not
ignored. It is useful to note that the second term in the integral, sin (kx sin ) ;
is odd in x while the integration has symmetric limits so that the contribution
from this term vanishes. Our integral now becomes
Etot
Etot
Z
d=2
sin (kr
/
r
!t)
sin (kr
/
r
!t) sin (kx sin )
k sin
Etot /
d
sin (kr
r
!t)
cos (kx sin ) dx;
d=2
sin =2
;
=2
where we have de…ned
= kd sin :
2
d=2
;
d=2
The intensity of the light on the screen is then proportional to
I /
I /
2
Etot
1
/
2
d
r
2
sin2 =2
;
( =2)2
sin2 =2
:
( =2)2
This is the desired result. It is useful to note that we could have obtained
the same result if we asssumed the Huygen wavelets were of the form
Ewav /
cos (k
!t)
;
or any arbitrary phase relative to the sine function as long as they are all in
phase.
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