Wave Interference and Di¤raction

7/31/2019 Wave Interference and Di¤raction

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EP225 Note No. 7

Wave Interference and Di¤raction

7.1 Superposition of Two Waves of Equal Wavelength

When two waves of equal amplitude and wavelength, but with a phase di¤erence are

superposed, E 0 sin(kx !t) + E 0 sin(kx !t + ) (1)

the total amplitude depends on and can vary between zero when = (2m + 1) and 2E 0when = 2m where m is an integer, m = 0;1;2; : The phase di¤erence betweentwo waves occurs through a path di¤erence or can be created arti…cially using a device calledphase shifter. For example, if sound wave is a pipe is split into two branches, the wave alongthe path 2 has to travel additional distance 2d relative to the wave along the path 1, andthe phase di¤erence between the two waves when detected at the exit is

=2

2d rad (2)

When 2d is an integer multiple of the wavelength, 2d = m; the phase di¤erence becomes

= 2m (3)

and the total wave amplitude is doubled. When 2d is equal to (2m + 1)=2; the phasedi¤erence becomes

= (2m + 1) (4)

and the two waves cancel each other. (If the wave e¤ectively vanishes due to interference,where does the wave energy go? Severe wave re‡ection must occur in this case.)

Fig. 7.1. Path di¤erence 2d between the branch 1 and 2 creates a phase di¤erence

=2

2d (rad).

The amplitude of the total wave for arbitrary phase di¤erence falls somewhere between0 and 2E 0; and can be found as a vector sum just like in ac circuit theory,

E = 2E 0 cos

2

(5)

1



Fig. 7.2. Addition of two electric …elds with a phase di¤erence can be done vectorially.

Figure 3 shows wave patterns radiated by two sources of equal frequency (and thus equalwavelength) separated by a distance of 3: Wave intensi…cation occurs along the straightlines on which the distances from the two sources di¤er by 0 (horizontal line), by

(lines

labelled 1); and by 2 (lines labelled 2):

Fig. 7.3. Wave patterns radiated by two wave sources S 1 and S 2 separated by a distance

3: Constructive interference (intensi…cation) occurs along the straight lines on which thepath di¤erences are m; m = 0;1;2:

2



Fig. 7.4. Arrangement of Young’s duoble slit experiment. The slit separation distance d is

exaggerated.

In the early 19th century, Young veri…ed that light was a wave phenomenon throughdemonstration of interference nature of light using a simple arrangement known as Young’sdouble slits. (Actually Young used two pinholes. Light source was sunlight which is domi-nated by green color around at = 550 nm.) When two slits are illuminated by a commonlight source, each slit acts as if it were a new light source as illustrated in Fig. 5. Thisre-radiation process of waves is known by Huygens’ principle which provides a convenientguide to understanding how waves behave when they encounter obstacles.

Fig. 7.5. Re-radiation (or di¤raction) of waves by a small opening on an opaque screenillluminated by a plane wave.

In Fig. 7.4, if the screen distance D is large, the path di¤erence between the light wavesemitted from each slit is approximately

d sin (6)

3



and the corresponding phase di¤erence is

=2

d sin (7)

If is equal to 2m; …eld intensity is doubled and light intensity quadrupled. This construc-

tive interference occurs at given by

sin = m

d

If the distance to the screen D is large, is small and sin can be approximated by

sin ' tan =y

D

The vertical location of intensity maxima is

y = mD

d; m = 0;1;2; (8)

Young was able to determine wavelengths of light sources from the measured values of y(interval of intensity maxima), D and d;

=dy

D(9)

Graphs below show interference patterns (light intensity as functions of the vertical dis-tance y in unit of y = d=D) when the slit opening widths are ideally thin (top) and whenthe slit separation distance d is ten times the opening width a; d = 10a: Intensity modulationis due to di¤raction which we will study in the following section.

cos2(x)

-10 -8 -6 -4 -2 0 2 4 6 8 10

0.2

0.4

0.6

0.8

1.0

x

y

Fig. 7.6 (a). Interference by ideally thin double slits.

cos2(x)

sin(:1x)

:1x

2

4



-20 -15 -10 -5 0 5 10 15 20

0.2

0.4

0.6

0.8

1.0

x

y

Fig. 7.6 (b). When the slit opening is 1/10 of the slit separation a = 0:1d:

Interference pattern can occur only for waves and it is not exaggeration to state what exhibitsinterference should have wave nature. Before the veri…cation by Young, whether light wasa wave phenomenon or particle phenomenon had been a subject of debates. Today, wave-particle dual nature of light (or any moving objects) is well established.

Example 1: (a) Find the distance between two neighboring maxima y in Young’s ex-periment when d (slit separation) is 0.2 mm, D = 1 m, and = 550 nm. (b) In the samearrangement, what would happen to the interference pattern on the screen if one of the slitsis covered with a plastic …lm having n = 1:4 and thickness t = 10 m?Solution

(a) From Eq. (9), we …nd

y =D

d=

550 109 m 1 m

2 104 m= 2:75 mm

(b) The additional phase di¤erence due to the plastic …lm is

0 =2

(n 1)t

Therefore the entire interference pattern on the screen is shifted vertically by y0 deter-mined from

sin 0 ' y0

D=

(n 1)t

d

y0 =(n 1)Dt

d

= 0:4 10

5

2 104 m

= 2 cm

(Upward if the upper slit is covered with the …lm. Why?)

7.2 Thin Film Interference

Oil …lms on water surface and soap …lms by themselves often appear colored. Thisphenomenon can also be understood in terms of two wave interference. Consider a thin

5



dielectric …lm of index of refraction n and thickness d: Because of impedance mismatch,light falling on the …lm is re‡ected at both surfaces. The path di¤erence between light wavesre‡ected at the top and bottom surfaces is 2d. Since the speed of light in the …lm is c=n; thewavelength in the …lm is shortened by the same factor,

…lm =air

n(10)

and the phase di¤erence due to the path di¤erence alone is

path =2

…lm

2d = n4d

air(11)

The wave re‡ected at the top surface su¤ers change (polarity reversal) because the …lmimpedance is smaller than that in air,

Z …lm =Z 0n

=377

n()

The wave re‡ected at the bottom surface does not su¤er any phase change. Therefore, thetotal phase di¤erence is

= path +

and the condition for constructive interference is = 2m: From

2n

air

2d + = 2m

we …nd

2nd =

m 1

2

air ; m = 1; 2; 3; (12)

It is noted that as the …lm thickness becomes small, re‡ection disappears.. In fact extremelythin soap …lm appears dark because re‡ected waves from both surfaces cancel each other.

Fig. 7.7. Wave re‡ected at the top surface changes its phase by because the …lmimpedance is smaller than air impedance. No phase change occurs for the wave re‡ected at

the lower surface.

6



The case of thin oil …lm on water surface can be analyze in a similar manner. If the indexof refraction of oil is larger than that of water, the condition for constructive interference isthe same,

2noil d =

m 1

2

air ; for constructive interference (13)

If the index of refraction of oil falls between 1 (air) and nwater; the condition is changed as

2noil d = mair ; for constructive interference (14)

because in this case re‡ected waves at both surfaces su¤er phase change of :

Example 2: Newton’s Rings. A plano-convex lens placed on a ‡at glass surface pro-duces interference rings. If the curvature radius of the lens is R; what are the radii of thebright rings?

Thin air …lm between the lens and ‡at glass plate causes interference and createsconcentric Newton’s rings.

Solution The thin air …lm between the lens and ‡at glass causes interference. The intensitymaxima occur at …lm thickness

2d =

m 1

2

air ; for constructive interference

because the index of refraction of air …lm is unity. From

(R + d)

2

= R

2

+ r

2

we …nd the thickness d at radius r is approximately given by

d ' r2

2R

Therefore,

r2 =

m 1

2

Rair

7



r =

s m 1

2

Rair ; m = 1; 2; 3;

determines the radii of bright rings.

7.3 Multiple Slits and Antenna Directivity

Fig. 7.8. Top: Six wave sources equally separated. The phase di¤erence between twoneighboring …leds is = (2=)d sin _: Middle: The total …eld E constracted from 6

vectors. Note that E 0 = 2R sin(=2); E = 2R sin(3 ) where R is the radius of the circle onwhich the vectors lie. Bottom: The maximum total …eld is 6E 0 and the minimum …eld is 0.

For multiple slits with uniform separation distances, the phase di¤erence between electric…elds emitted from two neighboring slits is

=2

d sin : (15)

8



For N slits, the total electric …eld can be found from vector sum of N …elds. In Fig. 8, thecase N = 6 is shown. Since

E 0 = 2R sin

2

E = 2R sin

6

2

we …nd

E = E 0sin62

sin

2

where E 0 is the electric …eld contributed by each slit. Generalization for arbitrary N isstraightforward,

E = E 0sinN2

sin2

(16)

The intensity of the wave is proportional to E 2;

I _ E 20sin2

N2

sin2

2

(17)

9



As N increases, the interference pattern at the intensity maxima, which occur at = 2m;becomes narrower and for large N; it becomes extremely sharp as seen in Fig. 7.9.

N = 2;

sin2x

2sin x

2

-2.5 -2.0 -1.5 -1.0 -0.5 0.0 0.5 1.0 1.5 2.0 2.5

0.5

1.0

x

I

N = 6;

sin6x

6sin x

2

-2.5 -2.0 -1.5 -1.0 -0.5 0.0 0.5 1.0 1.5 2.0 2.5

0.5

1.0

x

I

N = 20;

sin20x

20sin x

2

-2.5 -2.0 -1.5 -1.0 -0.5 0.0 0.5 1.0 1.5 2.0 2.5

0.5

1.0

x

I

N = 50; sin50x

50sin x2

-2.5 -2.0 -1.5 -1.0 -0.5 0.0 0.5 1.0 1.5 2.0 2.5

0.5

1.0

x

I

10



N = 50 (detail),

sin50x

50sin x

2

-0.10 -0.08 -0.06 -0.04 -0.02 0.00 0.02 0.04 0.06 0.08 0.10

0.5

1.0

x

I

Fig. 7.9. From top, N = 2; 6; 20 and 50: The last …gure shows detail of the case N = 50around x = sin =d = 0:

Compare the last …gure with the di¤raction pattern in Fig. 12. They are essentially identical.In fact, we will analyze di¤raction as a limiting case of large N in the section to follow.

A device called optical spectrometer is equipped with a grating having many grooves(often more than 10,000) which acts as multiple slits. It can resolve colors (wavelength andintensity) of a given light source.

Example 3: A grating in a spectrometer has 50,000 grooves over a width 9 cm. Find theangular locations of intensity peaks for blue (400 nm) and red (700 nm) colors.

Optical spectrometer. The grating has many grooves which act as multiple slits. The oneshown is transmission type. Re‡ection type gratings are commonly used. Only the …rst

order angular locations are shown. Other colors fall between the blue and red.

Solution The spacing between two neighboring grooves is

d =9 cm

50000= 1:8 m

Then intensity maxima occur at

d

sin = 0;1 (order 1);2 (order 2);3 (order 3),

11



For blue color,sin B = 0;0:22;0:44;0:66;0:88

B = 0;12:7;26:1;41:4;61:6

For red color,sin R = 0;0:39;0:78

R = 0;23:0;51:6

All colors fall at = 0 and this angle is not useful for spectroscopy. In the …gure, order1 angular locations of blue and red colors are shown. The resolution of a spectrometer(sharpness of the lines) is limited by di¤raction and each line is subject to an angular spreadof order

'

a

where a is the width of ‡at surfaces of the grooves. .

Example 4: Four antennas excited by a common generator form an array. The distancebetween two neighboring antennas is one half wavelength (d = =2) and the phase also

between two neighboring antennas can be controlled electronically. Sketch the angular dis-tribution of radiation power for various :

Antanna array spaced at =2: The phase between two neighboring antennas can becontrolled electroncially.

Solution In Eq. (17), N = 4 and

=2

2sin + = sin +

Therefore, the angular dependence of the Poynting ‡ux is given by

f (; ) =

sin(2 )

sin2

!2

=

sin(2 sin + 2)

sin sin +

2

!2

12



The function f (; ) is plotted below for various phase angle : As can be seen, the directivityof the antenna array can be varied without physically rotating the array.

sin(2 sin + 2 0)

sin

sin +0

2

!2

-16 -14 -12 -10 -8 -6 -4 -2 2 4 6 8 10 12 14 16-2

2

x

y

= =6;

0@sin(2 sin + 2 =6)

sin sin +=6

2

1A

2

-14 -12 -10 -8 -6 -4 -2 2 4 6 8 10 12 14 16

-4

-2 xy

= 2=3;

0@sin(2 sin + 2 =3)

sin sin +=3

2

1A

2

-14 -12 -10 -8 -6 -4 -2 2 4 6 8 10 12 14

-6

-4

-2 x

y

= =2;

0@sin(2 sin + 2 =2)

sin sin +=2

2

1A

2

-14 -12 -10 -8 -6 -4 -2 2 4 6 8 10 12 14

-8

-6

-4

-2

x

y

= 2=3;

0@sin(2 sin + 2 2=3)

sin sin +2=3

2

1A

2

13



-12 -10 -8 -6 -4 -2 2 4 6 8 10 12

-10

-8

-6

-4

-2

2

x

y

= ;

sin(2 sin + 2 )

sin sin +

2

!2

-5 5

-16

-14

-12-10

-8

-6

-4

-2

2

4

6

8

10

12

14

16

x

y

= 5=4;

sin(2 sin + 2 1:25)

sin sin +1:25

2

!2

14



-10 -8 -6 -4 -2 2 4 6 8 10

-6

-4

-2

2

4

6

8

10

12

x

y

= 3=2;sin(2 sin + 2

1:5)

sin sin +1:5

2

!2

-14 -12 -10 -8 -6 -4 -2 2 4 6 8 10 12 14

2

4

6

8

x

y

= 7=4;

sin(2 sin + 2 1:75)

sin sin +1:75

2

!2

-14 -12 -10 -8 -6 -4 -2 2 4 6 8 10 12 14

2

4

x

y

Fig. 7.10. Directivity of the four-antenna array as the phase is varied. Note that thedirection of maximum radiation can be controlled by without physically rotating the

array.

7.4 Single Slit Di¤raction, Resolving Power of Optical Devices

15



Fig. 7.11. Di¤raction by a single slit can be analyzed by assuming a large number of wavesources (Huygens sources) at the opening.

A pinhole or slit illuminated by a light source reradiates light wave as if it were a new

light source, according to the Huygens’ principle. (See Fig. 5.) A widely spread di¤ractionpattern is formed on a screen behind a slit with little resemblance to the shape of the slit.Di¤raction can be analyzed using the formula Eq. (16) for multiple slit. For this purpose,we divide the opening of the slit a into N sub-slits and eventually let N go to in…nity. Thephase for N slits is

=2

a

N sin : (18)

Therefore,

E = E 0sinN2

sin 2

= N E 0sina sin

a

sin

= N E 0sin

; (19)

where =

a

sin : (20)

Since E 0 is the …eld due to each slit, N E 0 is the electric …eld at the slit opening E 0 and we…nd

E = E 0sin

: (21)

Then Poynting ‡ux is proportional to the following function,

f () =

sin

2

(22)

which is plotted below as a function of x = a sin =;sin x

x

2

16



-3 -2 -1 0 1 2 3

0.5

1.0

x

y

Fig. 7.12. Di¤raction pattern as a function of a sin =: Compare this with the last …gurein Fig. 7 (multiple slit interference pattern near a peak with N = 50):

The light intensity is maximum at = 0 but the intensity pattern is spread about the peak.The …rst minima occur at x = 1; or

sin =

a(23)

The smaller the slit opening is, the larger is the di¤raction spread. If light were not of wavenature, such di¤raction would not occur. Also, di¤raction is more pronounced for longerwavelength..

For circular opening of radius a, similar di¤raction e¤ect occurs and imposes a theoreticallimit on the minimum angular resolution of optical instruments such as telescopes. Imagesof distant objects such as stars are actually di¤raction patterns formed by telescope mirrorsand each image has angular spread of order

' 1:22

D

where D = 2a is the diameter of the mirror. The numerical factor 1.22 stems from the …rstroot of the …rst order Bessel function J 1(x) = 0; x = 3:81;

2

a sin = 3:8

sin = 3:81

2a= 1:22

D

Since light waves from two stars are uncorrelated, intensities of each image are additive.Note that resolution improves with the diameter of the mirror. A larger telescope has higherresolving power.

Example 5: (a) What should the diameter of a telescope be if it is to resolve the angulardiameter of a star which is known to be 0:0500? Assume = 550 nm. (b) Find the spot sizeof a ruby laser ( = 690 nm) beam shot at the moon if its original diameter on the earth is1 cm. The earth-moon distance is 3:8 108 m.

Solution

(a) From the resolution limit

= 1:22

D

17



we …nd

D =1:22

=1:22 550 109 m

0:05=(60)2 =180 rad= 2:77 m

(b) The angular spread due to di¤raction is

= 1:22

D

1:22 690 109 m

0:01 m= 8:4 105 rad

Therefore, the radius of the “spot” at the moon is

r = 3:8 108 m 8:4 105 rad= 31:9 km

Beam spreading is due to di¤raction and thus unavoidable as long as laser light is of wave nature.

Note Babinet’s Principle. Di¤raction due to an opening on an opaque screen isidentical to that due to an obstacle with the same shape. This is known as Babinet’sprinciple and can be proved as follows. In calculating the electric …eld due to an opening,we add all contributions from the Huygens sources on the opening. In mathematical form,this is given by an integral

E _

Z opening

=

Z entire screen

Z

entire screenopening

(24)

The …rst integral in RHS simply yields uniform illumination without di¤raction which is of no interest. The last integral is the contribution from the shadow having the same shape asthe opening. The di¤racted electric …eld due to the shadow is equal to negative of that due toan opening. Since the light intensity is proportional to square of the …led, the sign di¤erenceis immaterial as long as di¤racted power is concerned. This proves Babinet’s principle. Inthe above example, water particles cause shadow di¤raction which is identical to di¤ractiondue to openings on an opaque screen with the same diameter.

Example 6: Show that the interference-di¤raction pattern due to double slits with sepa-ration distance d and slit openings a is given by

I () = I 0 cos2

2

sin

2

where

=2

d sin ; =

a sin

18



Double-slit with a …nite opening widths a:

Solution The electric …eld due to interference by ideally thin slits is proportional to

cos

2

Di¤raction e¤ect modulates the …eld as

cos

2sin

Therefore, the light intensity is given by

I () = I 0 cos2

2

sin

2

The case d = 10a is shown in Fig. 7.6 (b) (lower …gure).

Near Field (Fresnel) Di¤raction

Fig. 7.13. Di¤raction near a sharp knife edge.

The di¤raction phenomenon discussed in the preceding section is based on the assumptionthat the observing point is su¢ciently far away. In this case, the path di¤erence betweensources separated by a distance h can be approximated by

h sin ;

19



independent of the screen distance D: As D decreases, the path di¤erence should be calcu-lated more accurately from p

D2 + (h + h0)2 p

D2 + h2

=hh0

D+

h2

2D

' h sin + h2

2D(25)

where use is made of the expansion

p 1 + x ' 1 +

1

2x for small x

In Fraunhofer (far …eld) di¤raction, only the …rst term is retained. In Fresnel (near …eld)di¤raction, the second term is dominant provided the angle is small. The phase angle

=2

h2

2D=

h2

D(26)

is now quadratic in h in contrast to the case of Fraunhofer di¤raction.

Fig. 7.14. When the distance to the screen D is not large, the path di¤erence between r1

and r2 is not simply h sin and should be calculated more accurately.

20



Fig. 7.15. In Fraunnhofer di¤raction (left), the phase di¤erence is proportional to the

distance between sources h(_

l): In Fresnel di¤raction (right), is proportional to h

2

andthe vector diagram becomes a spiral.

Figure 7.15 shows the di¤erence between Fraunhofer di¤raction (left) and Fresnel di¤rac-tion (right). In Fraunhofer di¤raction, the phase is linear with h (the distance betweensources), and the electric …eld vector lies on a circle. However, in Fresnel di¤raction, isquadratic in h; and more rapidly increases with h resulting in formation of a spiral. Thespiral (known as Cornu) is described by the following parametric expressions for (x; y), andshown in Fig. 7.16.

For near …eld di¤raction by a sharp knife edge, the unperturbed electric …eld, which isobserved at a position well above the edge, corresponds to the distance between the centers

of the two spirals. The …eld at the same height as the edge corresponds to the distancebetween the origin and one of the spiral center and is one half the unperturbed …eld. Thelight intensity is thus one quarter of the unperturbed intensity. The …eld below the edgegradually decreases and asymptotically goes to zero. The …eld above the edge exhibitsoscillation and approaches the unperturbed …led far above the edge. Di¤raction is morepronounced as the wavelength increases because variation in the phase,

=

h2

D

decreases as increases and it requires a longer vertical distance for the same phase changeto occur.

Example 6: What is the maximum intensity in di¤raction by a knife edge?

Solution From the Cornu diagram, the maximum electric …eld can be found to be

E max ' 1:18E 0

where E 0 is the unperturbed …eld. Therefore, the intensity maximum is

I max ' 1:4I 0

21



- 0 . 6

- 0 . 4

- 0 . 2

0

0 . 2

0 . 4

0 . 6

- 0 . 8 - 0 . 6 - 0 . 4 - 0 . 2 0 . 2 0 . 4 0 . 6 0 . 8

Figure 1: Fig. 7.16. The unperturbed electric …eld is the distance between the centers of two spirals. The …eld on the plane y = 0 is one half the unperturbed …eld.

which occurs at a height

h 'r 34D

Note that di¤raction e¤ect is proportional to the square root of the wavelength in the limitof Fresnel di¤raction. In Fraunhofer di¤raction, it is proportional to the wavelength.

22



420-2-4

1.2

1

0.8

0.6

0.4

0.2

0

Figure 2: Fig. 7.17. Light intensity as a function of y (height) near a knife edge. y isnormalized by

p D=2: The unperturbed intensity at y = 1 is chosen unity. The region

y < 0 is behind the knife edge.

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Wave Interference and Di¤raction

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