Diffraction - Deviation of light from rectilinear propagation caused by the obstruction of light waves, .i.e, a physical obstacle . There is no significant physical distinction between interference and diffraction. However, interference generally involves a superposition of only a few waves whereas diffraction involves a large number of waves . Huygens-Fresnel Principle: Every unobstructed point of a wavefront serves as a source of spherical secondary wavelets. The amplitude of the optical field at any point is the superposition of these wavelets . Suppose light strikes a screen containing an aperture. The effects of diffraction can be understood by considering the phasor addition of the electric field for an array of point sources emitting E-M waves that are unobstructed by the aperture . The principles of diffraction can be understood by considering the interference of
Transcript
Diffraction - Deviation of light from rectilinear propagation caused by the obstruction of light waves, .i.e, a physical obstacle.
There is no significant physical distinction between interference and diffraction. However, interference generally involves a superposition of only a few waves whereas diffraction involves a large number of waves.
Huygens-Fresnel Principle: Every unobstructed point of a wavefront serves as a source of spherical secondary wavelets. The amplitude of the optical field at any point is the superposition of these wavelets.
Suppose light strikes a screen containing an aperture. The effects of diffraction can be understood by considering the phasor addition of the electric field for an array of point sources emitting E-M waves that are unobstructed by the aperture.
The principles of diffraction can be understood by considering the interference of water waves whose wavelength is made to vary in comparison to the size of the aperture.
To illustrate diffraction consider water waves in a ripple tank: ABBPAPOPD maxmax
max)( a
max)( b
max)( c
Constructive interference of most phasors for all angles; N small Approx. spherical wave
Destructive interference of most phasors for large angles; N large shadow
Intermediate region; N medium, mixed interference
N=5
Effective number of point sources N for phasor construction
region
Light waves striking an aperture
Fresnel Diffraction (near-field diffraction) when incoming and outgoing waves are both non-planar.
Source S and Screen C are moved to a large distance R, resulting in Fraunhofer diffraction (Far-field diffraction). As a rule-of-thumb: R > a2/ for Fraunhofer diffraction.
Fraunhofer diffraction conditions produced by lenses, leaving source S and screen C in their original position. Note that the lenses act to filter non-plane wave components from striking point P on the observation screen C.
Both incoming and outgoing wave contributions are considered as plane waves, and = const.
R
a
Consider a serious of diffraction measurements on a screen whose distance (R) from the slit varies from close (bottom) to far (top).
The diffraction pattern in the near-field (bottom) is sensitive to variations in R whereas the shape is independent of distance for R large (top).
The sharp structure shown in the near-field pattern is a result of rapidly changing phasor (E-field) orientations whose summations at each point on the screen are sensitive to distance, slit geometry, and angular spread of the waves. As shown, small changes in R in the near-field can cause large changes in the resulting phasor addition and irradiance distribution on the screen. If a >> , the pattern will resemble a sharp geometric shadow for near-field distances.
For large distances (e.g., R6 > a2/), the parallel nature of the plane waves will result in phasor additions yielding smooth distribution in irradiance, whose shape is independent of R.
R1
R2
R3
R4
R5
R6
R6 > R5 > R4 > R3.…
a
As a bridge between interference and diffraction, consider a linear array of equally spaced N coherent point oscillators. We examine the superposition of the fields from each source at a point P sufficiently far such that rays are nearly parallel.
The field of each wave is equal so that E0(r1) = E0(r2) = … =
E0(rN) = E0(r) .
In order to evaluate the field, we employ a phasor sum, as follows :
]...
1[)(
)(...
)()(~
)(
)()()(0
)(0
)(0
)(0
1
13121
21
rrik
rrikrriktkri
tkri
tkritkri
N
N
e
eeerE
erE
erEerEE
Due to a difference in OPL, there is a phase difference between adjacent sources of = k0, where = ndsin, and = kdsin. From the figure, = k(r2-r1), 2 = k(r3-r1),
3 = k(r4-r1)… ,
Therefore, at point P, the phasor sum is obtained using a geometric series (similar to what was done for multiple beam interference) and yields:
2/sin
2/sin
2
~~,
2/sin
2/sin)(
)(1
1)(
...1)(~
2
2
0
*
2/)1(0
2/2/
2/2/
2/
2/
00
)1(320
1
11
1
NI
EEITherefore
NeeerE
ee
ee
e
eeerE
e
eeerE
eeeeeerEE
Niikrti
ii
iNiN
i
iNikrti
i
iNikrti
iNiiiikrti
Note that as 0, I N 2I0, as expected for N coherent sources completely in-phase.
Also note that for N = 2, I = 4I0cos2(/2) using sin(2 ) = 2sin cos and again = kdsin.
22
2
0 2/sin
2/sinlim N
N
This is true for 0 and for = 2m, m = 0, 1, 2… ,
For conditions of constructive interference with N sources, = kdsin = 2m (2/)dsinm = 2m which gives dsinm = m and Im = N 2I0, conditions which are identical to two beam interference when there is complete constructive interference.
Consider now a line source of oscillators, where each point emits a spherical wavelet as shown in the figure. The electric field for a spherical wavelet emitted from a point is
krtr
E sin0 where 0 is the source strength for a
point emitter.
Let L = source strength per unit length. Then for each differential segment of source length dy, the contribution of the spherical wavelet at point P is
krtr
dydE L
sin From geometry,
,sin2
90cos222
222
yRRy
yRRyr
from the law of cosines.
y
)90 - (
Y
R
Ysin
Width of line source = D
...sin
...2
sin1
sin21
2
2
2
2
yR
R
y
R
yR
R
y
R
yRr
where R >>D and D > y.
Note that in order to fulfill Fraunhofer (far-field) diffraction conditions, we also require R > D 2/. Thus r R - ysin is the resulting approximation that we will use. We can calculate the total field appearing on the screen at angle , by summing or integrating all contributions of dE on the linear source:
2/
/
2/
2/
2/
2/ 2
)sin(sinsinD
D
LD
D
LD
D
yRktR
dykrtr
dydEE
Note that we are also ignoring the presence of ysin in the denominator since its presence in the phase of the sine in the numerator has a much greater contribution to the integral.
We can further separate terms in the sine by using the following identity:
sinsincossincossinsinsin kykRtkykRtkykRt
Since sin(kysin) is an odd function in y, the last term cancels in the integral.
Therefore, the integral is as follows:
222
2/
0
2/
2/
sin)0(
sin
2
1)(
,sinsin
sin)2/(
sinsin)2/(
sin)2/(sinsin
sin
sin)2/(sin2
sinsincos2
sincossin
IR
DIisirradianceaveragedtimetheand
kRtR
DEkDletSo
kRtkD
kD
R
DkRt
k
kD
R
kRtkydyR
kykRtdyR
E
L
L
LL
DL
D
D
L
We can now solve the single-slit diffraction problem by considering Figs. 10.6 and 10.7 on the next slides where the variable D b and = (kb/2)sin , where b is the slit width (typically a few hundred ) and l is the slit height (typically ~ cm).
Clearly, I() have minima when sin = 0 and = m, m = 1, 2, 3…,
mbm
bmkbNote sinsin
2
2sin)2/( )minima(
Note that the condition for diffraction minima, bsin = m, should not be confused with a similar expression for two-slit interference maxima, asin = m.
Fig. 10.6 (a) Single-slit Fraunhofer diffraction. (b) Diffraction pattern of a single vertical slit under point-source illumination.
Fraunhofer diffraction using lenses so that the source and fringe pattern can both be at
convenient distances from the aperture .
We can determine the conditions for maxima in the diffraction pattern by setting the derivative of the Irradiance to zero:
tansincos)(0sin)(
0sincos
sin02
cos11
sinsin
20
sin0
3
2
2
iiori
I
Id
dI
II
The first condition (i) is just the condition for minima that we have already seen (i.e. = m, m = 1, 2, 3, …)
The second condition (ii) results in a transcendental equation whose graphical solutions can be observed on the left. The solutions can easily be found numerically: 1.43, 2.46, 3.47…which approaches (m+1/2) for large .
Principal or Central Maximum
Subsidiary Maximum (SM)
1st SM 2nd SM 3rd SM
Minima (I = 0)
Phasor constructions of the E-field for (a) the central maximum, (b) a direction slightly displaced from the central maximum, (c) the first minimum, and (d) the first maximum beyond the central maximum for single-slit diffraction. The example corresponds to N = 18 phasors.
For the figure on the far left, the conditions for minima of m = bsin are understood by dividing the phasors of rays in two and four equal segments in (b) and (c), respectively. In (b), for each ray in the top half, a phasor 180 out-of-phase can be found in the bottom half, resulting in zero E-field. Likewise, in (c), each successive quarter contains phasors that are 180 out-of-phase, resulting in zero E-field and zero irradiance.
Figure 10.13 (a) Double-slit geometry. Point P on the screen is essentially infinitely far away. (b) A double-slit diffraction pattern for a =3b.
Consider now the set-up for a double-slit diffraction measurement. The rectangular geometry of the slit is shown such that l >> b, where l and b are the height and width of each slit. The screen on which the diffraction pattern is observed is located a distance R from the slits, where R >> a and R >> b2/. The point P is located on the screen. We will need to calculate the resulting E-field contribution at point P from all points (sources) located in both slits, using essentially the same procedure performed for the single-slit case.
The E-field is obtained, as before, by integration over all sources in each slit:
sinsin)()()(
2/
2/
2/
2/
zRktzFandR
CwheredzzFCdzzFCE Lba
ba
b
b
Integration procedure is similar to that of the single slit and results in:
sin2/sin2/2sinsinsin
kbandkawherekRtkRtbCE
Note that the definition for is precisely as before for single-slit diffraction. Using the identity: sin x + sin (x + 2y) = 2cos y sin(x + y), we can express E as
kRtbCE sincos
sin2
After squaring and time-averaging, we obtain the irradiance as:
22
2
0 cossin
4
II
in which = = 0 represents = 0 and I(0) = 4I0 since E0 = 2bC and twice that relative to a single slit. Note that I0= (bC)2 = (E0/2)2 is the contribution from one slit.
Examine some limits:
If kb << 1, (sin / ) 1 and we get I() = 4I0 cos2 = 4I0 cos2(/2), which yields the expected result for Young’s two slit interference when the slit width b is very small.
If a = 0, = 0, then I() =4I0 (sin / )2 = I0’ (sin / )2 , which is just the situation of two slits coalescing into a single slit.
Note that the expression I() =4I0 (sin / )2 cos2 is that of an interference term modulated by a diffraction effect.
Note that in general for two slits, minima occur when = m, m = 1, 2, 3, … and = (2m+1)/2, m = 0, 1, 2, 3… ,
As a general rule-of-thumb, for a = mb, we observe 2m bright fringes within the central diffraction peak, including fractional fringes. So, in Fig. 10.13, for a =3b, we have 5 + 2(1/2) = 6 bright fringes within the central
maximum .
When an interference maximum overlaps with a diffraction minimum (as shown in Fig. 10.13c) it is often referred to as a missing order.
Single- and double-slit Fraunhofer patterns. (a) Photographs taken with monochromatic light. The faint cross-hatching arises entirely in the printing process. (b) When the slit spacing equals b, the two slit coalesce into one (of width 2b) and the single-slit pattern appears (i.e., that’s the first curve closest to you). The farthest curve corresponds to the two slits separated by a =10b. Notice that the two-slit patterns all have their first diffraction minimum at a distance from the central maximum of Zo. Note how the curves gradually match Fig. 10.13(b) as the slit width b gets smaller in comparison to the separation a.
)a(
)b(
Figure 10.14
For diffraction from many slits (i.e., the N-slit problem), we generalize the 2-slit problem:
sinsin)(
)(....)()()(2/
2/
2/2
2/2
2/)1(
2/)1(
2/
2/
zRktzFandR
Cwhere
dzzFCdzzFCdzzFCdzzFCE
L
ba
ba
ba
ba
baN
baN
b
b
j
j
N
j
kRjkR
kajaRRwhere
jkRtbCE
2
sin2
,sin
2sinsin1
0
The integrals can be evaluated as before for the single- and double-slit cases using the same identities. The result is
)see figure for Rj(
The summation can be evaluated most easily using phasors, as before:
1
0
21
0
Imsin
2sinsin N
j
jikRtiN
j
eebCjkRtbCE
Again, the phasor sum is evaluated as a geometric series, as was done for the case of multiple beam interference. Therefore, we can write
sin2
sin2
,)1(sinsin
sinsin kband
kaNkRt
NbCE
Therefore, the time-averaged irradiance can be expressed as
22
0 sin
sinsin)(
N
II
Note that I0 is the irradiance in for = 0 emitted by one slit and I(0) = N2I0, which is seen by taking the limit for = 0 of the above expression. Also note that if kb << 1 (i.e. very narrow slits), we get the same expression for a linear coherent array of oscillators.
22
2
22
0 sin
sinsin)0(
sin
sinsin)(,
N
N
INIITherefore
The principal maxima are given by the following positions: = 0, , 2, 3, … = m (m = 0, 1, 2, 3 ..) and = (ka/2)sin a sinm = m and (sinN/sin)2 = N2 at these positions.
Minima are determined by (sinN/sin)2 = 0 = /N, 2/N, 3/N, …, (N-1)/N, (N+1)/N,… Therefore, between consecutive principle maxima we have (N – 1) minima (see figures on next slide).
Subsidiary maxima: Between consecutive principle maxima we have (N-2) subsidiary maxima when |sinN| 1. This occurs for 3/2N, 5/2N, 7/2N, …For large N, sin2 2 and the intensity of the firs subsidiary peak is approximately
222sin
22
)0(
3
2sin)0()(
I
II
Diffraction term multiple-slit interference term
Figure 10.17 multiple-slit pattern with a = 4b, N = 6
Finally, note that for N = 2 slits, we observe that
