Diffraction by a Slit
As an application of Huygens’s wavelets we will derive the intensity distribution for diffraction
by a slit. Consider a beam of light incident on a slit in a screen of width a as shown. The beam is
assumed to be much wider than the slit and can be modeled as a monochromatic plane wave. The
incident radiation will be assumed to be polarized in a direction parallel to the long axis of the
slit, i.e. along the z-direction.
At any time the wave amplitude at the location of the slit is E(slit, ) = E0 sin (k· 0 - ) =-E0
sin ( ) = -E0 z sin( ) where z is a unit vector in the z direction. For conceptual purposes,
imagine the slit divided into N thin regions as shown, where N is some large integer. Each region
has a width y = a/N. We will number the regions with an index n = 0, 1, 2, ... N-1.
7.1
The center of region n has coordinate yn = -a/2 + (n + ½ ) y where the origin of the y coordinate
is placed at the center of the slit. We will suppose that N is so large that the wave originating in
each region can be considered to be a Huygens wavelet. Ultimately this requires that we take the
limit y 0. The slit width a is assumed to be very small compared to the screen-to-screen
separation D: D>>a. This is called the Fraunhofer limit and the resulting diffraction pattern is
called a Fraunhofer diffraction pattern. Let P be an arbitrary point on the observing screen
characterized by its distance s from the midplane of the slit. In general, we think of D as
measured in meters, s in centimeters (or less) and a in millimeters (or less). The intensity at P at
any instant t will be proportional to E1 +E2 + E3 + ...2 where En is the wave amplitude at P at
time t due to the Huygens wavelet originating at region n. All of these electric fields will point in
the z-direction so we can write the instantaneous intensity as proportional to E1 +E2 + E3 + ...2
where En = En z. The wavelet from region n takes a time rn / c to get to P, where rn is the
distance from the center of region n to P, so that a spherical wave arriving at P from this region
has wave amplitude
(7.1) En(P,t) sin [ (t - rn /c)] / rn sin (k rn - t) / rn
The proportionality constant in (7.1) is determined by the rate at which energy is radiated from
region n. Wider regions radiate more energy than thinner ones and a reasonable estimate for a
thin region is that En y. We can then write (7.1) as
7.2
(7.2)
En(P,t) sin (k rn - t) y / rn
The proportionality constant in (7.2) depends only on the incident field E0 and will not be
needed.
The distance rn from the center of region n to P is given by
rn = [D2 + (s-yn)2 ] = [D2 + s2-2syn + yn2 ]
Let r [D2 + s2 ] = distance from center of slit to P. Then
rn [r2-2syn + yn2 ] = r [1 - 2syn /r2 + yn2/r2 ]1/2
This can be simplified using the approximation that <<1< (1+ )p 1 + p . Here we take
= - 2syn /r2 + yn2/r2 and p = ½. Then
rn r [1 - syn /r2 + yn2/2r2 ] = r - syn / r + yn2/2r .
Note that s /r s /D = sin( ), so that
(7.3)
rn r - yn sin + yn2/2r
The total electric field at P at time t is given by
N-1
(7.4)
E(P,t) sin (k rn - t) y / rn sin (k rn - t) y / r
n=0
where we have replaced rn in the denominator with r. This is permissible because 1/rn is very
close to 1/r with our assumptions about the geometry. This is not true for the numerator where
we expect sin (krn - t) to go through one full cycle of oscillation when rn changes by one
wavelength.
Since r is constant we can take it out of the summation and incorporate it into the overall
proportionality constant and write, using 7.3,
N-1
(7.5)
E(P,t) sin (k rn - t) y
n=0
= sin [k r -kyn sin( ) + kyn2/2r - t] y
This is still a rather complicated summation to perform. The quadratic term in the argument of
the sine function above has maximum value ka 2/2r ka2/2D = a2/ D. This will be very small
when D is large. We define the Fraunhofer condition as
(7.6)
a2 << D.
When this condition is met we can write
7.3
N-1
(7.7)
E(P,t) sin [(kr - t) - kyn sin ] y
n=0
Now take the limit N . In this limit yn becomes a continuous variable y and the sum becomes
an integral
a/2
(7.8)
E(P,t) , sin[(kr - t) - kysin ] dy
-a/2
y = a/2
= (k sin )-1 cos[kr - t - kysin ] y = -a/2
= (k sin )-1 { cos[kr - t - ½ kasin ] - cos[kr - t + ½ kasin ] }
Define a new quantity
(7.9)
u ½ k a sin = a sin /
so that
(7.10)
E(P,t) (k sin )-1 {cos(kr - t - u) - cos(kr - t + u)}
Use identity (A.6) from appendix A
cos ( - ) - cos( + ) = 2 sin sin
(7.11) E(P,t) (k sin )-1 sin(kr - t) sin(u) sin(kr - t) sin(u) / u
The instantaneous intensity is given by
(7.12)
I(P,t) E(P,t)2 sin2(kr - t) sin2(u) / u2
The average intensity at P is given by averaging the above expression over a full period of
oscillation. Since sin2( X - t ) = ½ for any quantity X we have
(7.13)
I(P) = I0 sin2(u) / u2
where I0 is an overall proportionality constant. Equation (7.13) is the final result of our
calculation. A frequently occurring function in optics is sinc(x) sin(x) / x, so a more compact
way of writing (7.25) is I(P) = I0 sinc2(u). A plot of sinc2(u) is shown below.
7.4
sinc2(u)
This intensity distribution should look familiar.
The pattern has a central bright spot
surrounded by secondary dimmer bright
regions. The intensity zeros occur where u =
±m ( m = 1, 2, 3, ...). The angular extent of
the central bright region is - /a < sin < /a.
When /a is a small number, this can be
written as - /a < < /a. I.e. the angular width
of the central bright region is 2 /a. All of the
secondary bright regions have an angular width
/a. The central bright region always occurs at
= 0 while the brightest part of a satellite
bright region lies approximately at the location
of the maximum of sin(u), i.e. approximately
at the angular values sin = ±(m + ½ ) /a
where m = 1, 2, 3, ...
Problem: Why is the word “approximately” used in the above sentence? Refer to eq. (7.13)
Problem: Two lasers of different
wavelengths (632.8nm and blue light of an
unknown wavelength) produce beams that
fall on a beam splitter as shown. A beam
splitter is a device such as a lightly silvered
piece of glass that reflects a portion of the
incident light falling upon it (from either
side) and transmits the rest. The combined
beam falls on a thin slit and produces a
Fraunhofer diffraction pattern on a screen
1.6m away.
A. What laser safety consideration exists for
this experiment that was not present for the
single-slit experiment performed in class?
B. If the center of the third satellite red
bright region lies at the same position as the intensity zero separating the fourth and fifth blue
satellite bright regions, what is the unknown wavelength?
7.5
Diffraction by a Circular Aperture
The diffraction pattern depends on the geometry of the diffracting aperture. Consider a circular
aperture of diameter a cut in an absorbing screen. The screen occupies the xy plane and a parallel
observing screen is located at z = D where D >> a . A point on the aperture can be specified by
the circular coordinates , where x = cos , y = sin . Again we divide up the aperture into
notional regions, each so small that Huygens’s principle can be applied to find the wave
amplitude radiated by each region. It is convenient to divide the aperture into thin annuli of
thickness
and into thin segments of angular width . The area of a piece of aperture with
radius lying between and + and angle between and + is A =
.
We proceed as in the previous section but for the sake of definiteness we now assume that the
particular point P of interest on the observing screen has x-coordinate zero. There is no loss of
generality in requiring this, since the entire diffraction pattern has circular symmetry. With the
coordinates of P being (x, y, z) = (0, s, D) and the coordinates of the center of the nth region
being ( cos , sin , 0), the distance between these two points is
(7.14) rn = [D2 + ( cos )2 + (s - sin )2 ]1/2
= [D2 + 2 cos2 + s2 -2s sin + 2 sin2 ]1/2 = [D2 + s2 -2s sin +
2 1/2
]
again, define r = [D2 + s2 ]1/2 = distance from center of aperture to P, so that
(7.15) rn = [r2 -2s sin + 2 ]1/2 = r [1 - 2s sin / r2 + 2 / r2] ½
r [1 - s sin / r2 + 2 /2r2 ] = r - s sin /r + 2 /2r = r - sin sin + 2 /2r
where we have again written s/r = sin . The wave amplitude at P due to this region is given by
an expression identical to that of (7.2)
(7.16)
En(P,t) sin [ (t - rn /c)] / rn sin (k rn - t) / rn
The proportionality constant in (7.16) depends on the intensity of the light leaving the nth region.
For small regions, this will be proportional to the area A of that region, so we have
(7.17)
En(P,t) sin (k rn - t) A / rn
7.6
The net wave amplitude at P is
(7.18)
E(P,t) sin (k rn - t) A / rn sin (k rn - t) A / r
where we have again approximated rn in the denominator by r, which can be removed from the
summation and absorbed into the overall proportionality constant. Taking the limit A 0, the
summation in (7.18) becomes the double integral
= a /2
(7.19) E(P,t) ,
=0
=2
,
= a /2
,
=
=0
sin (k rn - t) d d
=0
=2
,
sin (k r - k sin sin + k
2
/2r - t) d d
=0
where (7.15) has been used. This integral simplifies in the Fraunhofer limit a2 << D. Assuming
this condition we have
= a /2
E(P,t) =2
,
,
=0
=0
sin [(k r - t) - k sin sin ]
d d
Perform the integration first. Expand the sine function using identity (A.2), sin (A-B) = sin(A)
cos(B) -cos(A) sin(B)
(7.20) sin [(k r - t) - k sin sin ]
= sin (k r - t) cos( k sin sin ) - cos (k r - t) sin(k sin sin )
The second term in (7.34) makes no contribution to E(P,t) since
=2
,
sin [ C sin ] d = 0 for any constant C (why is this true?)
=0
so that we need only evaluate the quantity
2
sin (k r - t) ,cos( k sin sin ) d
0
This integral cannot be expressed in terms of elementary functions. It can be expressed in terms
of quantities called Bessel functions. For any number x we have
2
(7.21)
,cos(x sin ) d = 2 J0 (x)
0
where J0 (x) is the Bessel function of order zero and is illustrated below.
7.7
The Bessel Function J0(x)
Problem: Explain the general features of the above graph using the integral (7.21)
At this point in the derivation we have
a/2
(7.22) E(P,t) sin (k r - t) , J0(k sin ) d
0
The Bessel function of order one, J1(x), satisfies
, J0(x) x dx = x J1(x) and is illustrated below
The Bessel Function J1(x)
7.8
therefore
(7.23)
E(P,t) sin (k r - t) J1( ½ ka sin ) / ka sin sin (k r - t) J1( u) / u
where
(7.24)
u = ½ ka sin = ( a/ ) sin
The light intensity at P is proportional to the time-averaged value of E(P,t)2 so that finally,
(7.25)
I(P) = I0 [ J1( u) / u ]2
A graph of this intensity distribution is shown below
[ J1 (u) / u ]2
This intensity distribution is zero at the values of u satisfying J1(u) = 0. The first few zeros of
J1(u) are at u ±3.832, ±7.016, ±10.173. The central maximum of the intensity pattern falls in
the angular range -3.832 ( a/ ) sin +3.832. This angular range is usually so small that it is a
good approximation to replace sin with . To a good approximation, the angular width of the
central diffraction maximum is therefore
(7.26)
2.43 ( / a)
Problem: Compare (7.26) with the corresponding result for diffraction by a slit.
Diffraction is often the factor that limits the resolution of an optical system. Each point on an
extended luminous object acts as a source of outgoing spherical waves and when these waves
reach a distant imaging system such as a lens or aperture they are effectively plane waves. As
such, they form a diffraction pattern like the one discussed in this section.
7.9
Problem: Why are we not usually aware of this fact when using lenses?
Rayleigh’s criterion
Consider two points P1 and P2 on an object. Suppose that ordinary incandescent light has been
sent through a color filter so that the light consists of a narrow band of wavelengths around some
central value . If the object is illuminated with this light and then viewed through a circular
aperture, the two points will each produce a diffraction pattern and, if the points are sufficiently
close together, the corresponding diffraction patterns may overlap so much that the points are
indistinguishable. While there is some subjectivity involved here, a common quantitative
specification is Rayleigh’s criterion, which states that the two points are just barely
distinguishable if the
central diffraction
maximum of one pattern
falls on the first zero of
the diffraction pattern of
the other. Using (7.26),
this criterion implies that
two points cannot be
distinguished if their
angular separations are
less than about 1.2 /a.
(7.40) Rayleigh’s Resolution Criterion:
1.2 /a
Problem: A military spy satellite in orbit at a height of 37,000 km uses a lens to focus an image
that is electronically scanned and radioed back to earth. The mission of the vehicle requires it to
distinguish tanks from trucks on the ground. What minimum linear resolution is required? What
minimum angular resolution is required? If yellow-green light of wavelength 560nm is used,
what is the smallest diameter lens that can be used?
7.10