2 Slit Diffraction

▀▄▀▄ Answer: 3 📌📌📌 question ➜ Although we have discussed single-slit diffraction only for a slit, a similar result holds when light bends around a straight, thin object, such as a strand of hair. This work is licensed under a. Recent Comments. Anonymous on Hello world! Proudly powered by WordPress. . As we saw with the two-slit problem, the geometry of the problem is much easier when we go to the limit that the distance to the screen (where the interference pattern is viewed) is much larger than the width of the slit. – Fraunhofer or far-field diffraction. When this isn’t the case, you still get diffraction, it just looks different. 2D FDTD Yee algorithm simulation of an infinite line source. The source is placed 0.25 wavelengths above a 1/20 wavelength wide slit. The slit reproduces a w. Java applet: Diffraction of light by a single slit.

Chapter 36–Diffraction Hints-5/30/10
1. Answer the following questions about diffraction.

  1. How does Huygen’s principle explain diffraction? Under what relationship between opening and wavelength is diffraction most noticeable? The edges of a wave-front act like point sources that spread the wave into the “shadow” regions on either side of an opening. Diffraction is most noticeable when wavelength and opening are of comparable size. When the wave-front is long compared to the opening the edge effects are negligible and the wave appears to go “straight through” the opening.

  2. Although we can hear around corners. We cannot see around corners. How can you explain this in view of the fact that sound and light are both waves? The wavelengths of sound are in the cm->meter range which is comparable to the size of common openings like doors, windows, etc…Wavelengths of light are 400700 nm which makes it hard to detect diffraction of light around ordinary openings.

  3. If a coin is glued to a glass sheet and this arrangement is held in front of a laser beam, the projected shadow of the coin has diffraction rings around its edge and a bright spot in the center (called Poisson’s dot). Explain how this happens. Light that diffracts around the edges of the coin makes rings of constructive and destructive interference in the coin’s “shadow”. On the axis behind the coin’s center there’s constructive interference and a bright spot can be seen. It’s called Poisson’s dot because this famous mathematician was the first to suggest that a wave theory of light would predict such a spot. When this was demonstrated the wave theory of light got a big boost.

  4. Describe the change in width of the central maximum in the single-slit diffraction pattern if you decrease any of the following: (i) the slit width, (ii) the wavelength, (iii) the frequency of the light, (iv) the distance L between the slit and projection screen. Given that sinø=l/a for the first node, and that tanø=y/L: (i) increase; (ii) decrease; (iii) increase; (iv) decrease

  5. The 2-slit diffraction-interference pattern is a superposition of the single slit and double slit pattern. Explain how the 2-slit pattern is modified by the single slit pattern. The second slit doubles the overall intensity of the single slit pattern, but the two separate slits interfere with each other and generate a double-slit pattern that is superimposed on the single-slit pattern. The resultant pattern looks like a series of equally spaced maxima that decrease in intensity as one gets farther from the center.

Single-slit interference pattern and its intensity
2. The analysis of the single-slit pattern has significant differences from the double-slit pattern. With 1-slit we assume that there’s an infinite number of point sources in a finite sized opening, but with 2-slits we assume there’s a finite number of sources (2) and the size of the openings is ignored.

a) Compare and contrast the 1-slit to the 2-slit pattern, in the chart below:




Single slit *

Double slit **

In general

r= a sinø

ß = 2π∆r/l

r = d sinø

g = 2π∆r/l

Maxima or “brights”

Central: ø=0
Secondary:

r = (m+½)l


m=1,2,3..

Central: ß=0
Secondary:
ß = 2π(m+½)
m=1,2,3..

r = ml
m=0,1,2,3..

g = 2πm;
m=0,1,2,3..

Minima or “darks”

r = ml
m=1,2,3..

ß = 2πm;
m=1,2,3..

r = (m+½)l
m=0,1,2,3..

g = 2π(m+½);
m=0,1,2,3..

Intensitycomparison

I/Io=[sin(ß/2)/ ß/2]2;
Also used to compare 2-slit maxs

Within one antinode: I/Io=[cos(g/2)]2
Antinode/Antinode: Im/Io Use 1-slit formula

*The ∆r and ß in the single-slit pattern compare the edges of a single wave-front.
** The ∆r and g in the double-slit pattern compare two separate wave-fronts.

b) What is ß when ø=0 (the central antinode) in the single-slit pattern? Why isn’t the intensity of the central antinode zero? When ø=0, ß=0, but this doesn’t mean the intensity of the central antinode is zero, on the contrary, it is the most intense (Io). As ß0, sin (ß/2) (ß/2), so at Ao , I/Io =[(ß/2)/(ß/2)]2= 1


3. Consider two slits side by side, each slit’s size a is 2l and the slit separation d (center-to-center) is 6l.

  1. How many double-slit antinodes would you be able to see within the central maxima of the single slit pattern? Make a rough sketch of the pattern. N1 (ø=30º) in the 1-slit pattern
    corresponds to m=3 (A
    3)in the 2-slit pattern. So m=0±2, or 5 maxima, would
    be visible (m=±3, would not be visible because it would coincide with the N
    1 node).

  2. Repeat (a) assuming that the separation d is 7l. Here N1 in the single-slit pattern corresponds to m=3.5 in the double-slit pattern, which corresponds to N3 in the double-slit pattern. So m=0±3, or 7 maxima would be visible, with m=±3 very faint.

  3. In general, if two slits with openings aare a distance d apart, how many 2-slit antinodes would fit into the major 1-slit antinode? It’s not possible to come up with a simple formula that works all the time, it depends on whether N1 in the single-slit pattern corresponds to a max or min in the double-slit pattern, and whether you count the faint maxima near N1, but roughly…2(d/a) or [2(d/a)-1] works.

  4. Each slit produces a single slit pattern of its own, how do the two single-slit patterns affect each other, irrespective of the double slit pattern? For example, how is the major single-slit antinode affected by the presence of the other single-slit so close by? A second slit doubles the overall intensity of the single-slit pattern without shifting it much (recall the slits separation is in the order of 100s of nanometers whereas the interference pattern spreads over centimeters).

4. Light of wavelength 589 nm illuminates a single slit 0.75 mm in width.



  1. At what distance from the slit should a screen be located if the first minimum in the diffraction pattern is to be 0.80 mm from the center of the screen? L=ya/l =1.02 m

  2. What is the linear and angular width of the central maximum? Measuring from N1 to N-1, ∆y=1.6 mm and ∆ø=1.57 x 10-3 rad (0.090º).

  3. What’s the linear distance between the first and second minima in the pattern?... between the first and third minima? ∆y­12=0.80 mm; ∆y­13=1.6 mm.

  4. Compare the width of the central antinode to the other antinodes in the single-slit pattern. The central max is 2x larger than the secondary maxima (only within the small angle region). This is different from the double slit pattern where all the maxima are of equal width, including the central one.

  5. If a second slit of the same width were added 5 µm from the first, how many orders of the double slit pattern would be visible within the large central antinode? d/a=8.49, this means 8 orders plus m=0 (17 maxima visible).

5.[]JJ A single-slit diffraction pattern is formed on a screen 6.0 m away from a 0.35-mm-wide slit. Monochromatic 546-nm light is used.



  1. What is the angular position (ø) of a point on the screen 4.8 mm from the principal maximum? sinø=4.8/6000=0.80 x 10-3 rad (0.046º)~ø.

  2. What is the phase difference (ß) between the edges of the beam at this point? Make sure you understand the difference between ø and ß. Why is it important that ß be measured in radians? ß=2πasinø/l=3.2 rad. Angle ø is the angular position (a real angle in space) of the point in question relative to Ao; angle ß is the phase difference (an angle in a phase diagram) between the edges of the beam that interfere at that point.
    It is important that ß be measured in radians because the intensity formula was derived using the radian definition of ß.


  3. Calculate the fractional inten­sityI/Io11I/I at this point on the screen. Formula plug: I/Io = 0.38511I/I

  4. Where is the nearest minimum to this point located on the screen? The angular location of the point is less than the angular location of the first node (sinøN1=l/a=1.56 x 10-3 rad), so N1is the nearest minimum.

  5. Repeat this problem assuming that the screen is only 2.0 m away from the slits. Answers: sinø=2.4 x 10-3 rad~ø; ß=3.1π rad; I/Io = 0.042; this point is between N1 and N2.

6. Compare the intensity of the first two secondary antinodes to the central antinode intensity in the single-slit pattern. That is, find IA1/Io and IA2/Io. In reality our simplified analysis underestimates the intensities of these antinodes.




r

ß/2

sinß/2

I/Io

A1

3l/2

3/2

-1

0.045 (4.5%)

A2

5l/2

5/2

+1

0.016 (1.6%)
Note how much less intense the
secondary maxima are.

Resolution of two objects:
7. Diffraction makes it hard to visually distinguish (or “resolve”) two light sources or objects that are close together.

a) Explain how the Rayleigh criterion is derived and how it is used. The Rayleigh criterion states that the limit of resolution of two distant objects occurs when the angular separation between them equals the angular size of the first node produced by the light from the objects passing through an opening. That is, sinøR=l/a. When the “slit” is circular the Rayleigh criterion becomes sinøR=1.22l/D, where D is the diameter of the aperture. Proof of this later statement is beyond the scope of this course. The smaller the resolution angle, the greater the ability (or power) to resolve to objects; that is, “resolving power” is inversely proportional to øR.

b) You use a lens of diameter D, and light of wavelength land frequency ff to form an image of two closely spaced and distant objects. Which of the following will increase the resolving power? (i) a smaller diameter lens, (ii) higher frequency light, (iii) longer wavelength light? Justify your answers. Decreasing øR increases the resolving power of the lens, so: (i) decreases; (ii) increases( f=c/l); (iii) decreases.(a) Use a lens with a smaller diameter; (b) use light of higher fre­quency; (c) use light of longer wavelength. In each case justify your answer. Q36.4. Light of wavelengthand frequencypasses through a single slit of width a. The diffraction pattern is observed on a screen a distance x from the slit. Which of the following will decrease the width of the central maximum? (a) Decrease the slit

Use a lens with a smaller diameter; (b) use light of higher fre­quency; (c) use light of longer wavelength. In each case justify your answer. (a) Use a lens with a smaller diameter; (b) use light of higher fre­quency; (c) use light of longer wavelength. In each case justify your answer


8. A helium-neon laser emits light that has a wavelength of 632.8 nm. The circular aperture through which the beam emerges has a diameter of 0.50 cm. Esti­mate the diameter of the beam 1.0 km from the laser.
Using sinøR=1.22l/D=1.22(0.6328 µm/0.5 cm)=1.54 x10-4 rad~øR.
Diameter of the beam at 1 km is approx. equal to an arclength of radius 1km, D~ s=RøR=0.154 m=15 cm
9. The Moon is approximately 4 x 105 km from the Earth. Can a telescope on the Earth resolve two lunar craters 50 km apart, if the telescope mirror has a diameter of l5 cm? Can craters 1.0 km apart be resolved? Take the wavelength to be 700 nm, and jus­tify tify your answers with approximate calculations.
Rayleigh criterion: sinøR=1.22(.7 µm/15 cm)=5.7 x10-6 rad~øR.
The angular separation between the two craters is approx. ø =50/4 x 105 =1.25 x10-4 rad, which is larger than øR, so they will be resolved.
DiffractionFor a 1 km separation, ø=1/4 x 105 =2.5 x10-6 rad, which is smaller than øR, so they will not be resolved.
10. Suppose you are standing on a straight highway watching a car moving away from you at 30 m/s. The air is perfectly clear, and after 10 min you see only one tail-light. If the diameter of your pupil is 7.0 mm and the index of refraction of your eye is 1.33, estimate the approximate width of the car from the tail-light separation. Remember red light is ~700 nm in wavelength.
At the moment the taillights blend into one, your eye has reached the Rayleigh limit. To apply the Rayleigh criterion to the eye, you need the wavelength in the eye (l =700 nm/1.33) then sinøR=1.22(7nm/1.33x7 mm)=92 x10-6 rad ~øR. The separation between the tail-lights is approx. an arc with a radius equal to the distance to the car. So s=R øR =92 x10-6 (30x600)=1.6 m
Multiple-slit interference
11. Describe how the interference pattern changes when more slits of the same size and the same separation are added. How do we explain the narrowing of the antinodes? In the chart below compare some features of the 2-slit, the 3-slit, and (if you have the stamina) the 4-slit interference pattern.
The antinodes (maxima) of the interference pattern become narrower. The “nodal” regions (minima) become wider and have faint “secondary maxima” within them that fade to zero intensity as the number of slits increases. The reason for the broader “nodal regions” is that, with more slits, there are more “destructive interference opportunities” than “constructive interference opportunities”. Thus maxima become narrower and more intense while the minima widen.
* Given the time constraints at this point in the course I will not be covering this material in any formal way. But you should have a general idea of what happens as more slits are brought into play.
Below is a chart summarizing comparing 2, 3, and 4 slits. You will NOT be tested on the fine points of the theory.

N

Primary maxima

Minima

Secondary maxima

I/Io

Isec/Ipri

r=dsinø

g

r=dsinø

g

r=dsinø

g

2

ml

m

(m+½)l

(m+½)

x

x

[cos(g/2)]2

x

3

ml

m

(m±1/3)l

(m±1/3)

(m+½)l

(m+½)

1/9[1+2cosg]2

1/9

4

ml

m

(m±¼)l
(m+½)l

(m±¼);

(m+½)



(m±1/3)l

(m±1/3)

1/4[cos(3g/2)+ cos(g/2)]2

1/16

The diffraction grating
12. Answer the following questions about the diffraction grating.

  1. How is the diffraction grating different from a small number of slits? What advantages does it have? The maxima are extremely narrow and secondary maxima are completely negligible. The sharpness of the maxima make it possible to resolve different wavelengths in a spectrum of light.

  2. What is the diffraction grating primarily used for? To separate the wavelengths (colors) in a light spectrum to they can be analyzed…

  3. The minimum “resolving power” R needed to distinguish two close wavelengths (where l~l) is defined by R=(l/∆lIt can also be proven that a grating’s resolving power is given by Rgrating=Nm, where N is the number of lines and m is the order. Explain why this last expression for R makes conceptual sense. The larger the no. of slits (N), the narrower the maxima which makes them more distinct. The larger the order (m), the bigger the separation between different wavelength maxima. Both these factors make it easier to resolve wavelengths with a grating.

  4. What is the difference between a “continuous spectrum” and a “line spectrum”? Which one is generated by an incandescent light-bulb?... a fluorescent light-bulb? A “continuous spectrum” contains the full range of colors in white light. A “line-spectrum” contains a few separate “lines” of color. The glowing wire of an incandescent bulb generates a “continuous spectrum”. The fluorescent bulb uses mercury gas which generates a “line-spectrum”. If fact, all glowing elemental gases (helium, hydrogen, sodium, mercury etc…) generate line spectra unique to each element, a sort of elemental “fingerprint”.

  5. How is the diffraction grating similar to a prism? Both disperse light, creating a “rainbow effect”.

13. A grating with 250 lines/mm is used with an incan­descent light source. Assume the visible spectrum ranges in wavelength from 400 to 700 nm. In how many orders (m’s) can one see (a) the entire visible spec­trum and (b) the short-wavelength region only? The separation d= 1mm/250= 4 µm.


Max order for red: mred=d/=4/.7=5.7; max order for violet: mviol=d/=4/.4=10The entire spectrum exists in 5 orders, the short-wavelength region in ~9 orders.
14.!)7,ILight of wavelength 500 nm is incident normally on a diffraction grating. The third-order (m=3) maxima of the diffraction pattern is observed at an angle of 37°, determine the following

a) The line separation and the number of rulings per centimeter for the grating. Since sinø = ml/d sin37º=3(500 nm)/d d=2.5 µm # lines/cm=1/d=4000/cm

b) The total number of primary maxima that can be observed in this situation. The largest possible “m” is given by: sin90º= ml/d m=2.5/.5=5 So observable m=0±5 (11 maxima).
15. A grating is illumi­nated by a sodium discharge tube. The lines in in the grating are uniformly spaced at 775 nm. The grating has enough lines to resolve the two wavelengths forming the sodium doublet (l1 589.0 nm and l2 589.6 nm).


  1. Calculate the angular separation in the first-order spectrum between the doublet wavelengths.

ø1=arcsin(589/775)=49.464º and ø2=arcsin(589.6/775)=49.532º∆ø=0.068º =0.0012 rad

  1. What is the angular separation in the second order spectrum?

This is a “trick question”; no second order spectra is possible.

  1. What minimum resolving power R is required here? R=(l/∆l 

  2. What minimum number of slits do you need for the required resolving power R in (a) and in (b)?
    R=Nm At m=1, N=R=982 At m=2, N=R/2=491, but there is no m=2 possible in this problem.

16. It is possible to prove that, for two close wavelength l and (l+∆l where ∆l << l the angular separation between the mth-order spectra is given by the formula ∆ø=∆l[(d/m)2- l2]-1/2 where d is the slit separation and m is the order number.

a) Apply this formula to the problem above and show that you get pretty much the same answer for ∆ø.
Just plug-in and note that answer is in radians: ∆ø= (0.6)[(775/1)2-589.32]=0.0012 rad

b) Derive the expression given in the stem of this problem for ∆ø.


Since l =dsinø/m Take the derivative , ∂lø =dcosø/m=(d/m)(1-sin2ø)1/2∂lø ~∆l∆ø =(d/m)[1-(ml/d)2]1/2=[(d/m)2- l2]1/2…etc
17. Suppose that there is a single slit 6.0 cm wide in front of a microwave source operating at 7.5 GHz (109 Hz). (a) Calculate the angle subtended by the first minimum in the diffraction pattern. (b) What is the relative intensity I/Io at ø= = 15°? (c) Consider the case when there are two such sources, separated laterally by 20 cm, behind the slit. What is the maximum distance between the plane of the sources and the slit to be able to resolve the diffraction patterns? (Note that in this case, the approximation sin~(J = tan (J is not valid because of the relatively small value of a/ l.
2 slit diffraction experimenta) Wavelength here is l=c/f=(3 x 108)/(7.5 x 109)=0.04 m; and sinø=l/a=4/6=0.67ø=42º.A.)
b) The intensity depends on ß/2=π asinø/l =π(6)sin 15º/4=1.22 rads I/Io=[sin1.22rad/1.22]2=0.59
c) Applying the Rayleigh criterion, øR =arcsin (l/a) =42º.
d) Since we can’t use the small angle approximation, we apply
trig to right triangle in diagram, 10/L=tan 21ºL=26 cm
Review of polarization and scattering of light (this material was covered in Ch. 33)
18. Explain what polarization is and how Polaroid filters work. Explain how reflection and scattering partially polarize light. Discussed in class and in the textbook.

  1. How does polarization prove that light is a transverse wave and not longitudinal? Only transverse waves can be polarized. Since light can be polarized, it must be a transverse wave.

  2. Certain sunglasses use a polarizing material to reduce the intensity of light reflected from shiny surfaces. Why does this work? What ori­entation of polarization should the material have to be most effective? Reflection partially polarizes light, increasing its electric field parallel to the reflecting surface and reducing the other component. Most of the annoying reflections that reach our eyes come from horizontal surfaces; so Polaroid filters with a vertical axis will reduce a lot of the reflected light.

  3. Is light from the sky polarized? Why is it that clouds seen through Polaroid glasses stand out in bold contrast to the sky? Sky-light must be partially polarized since Polaroids filter out some of the bright sky-light and make clouds stand out.

  4. With a more advanced analysis it can be shown that the amount of light scattering around us is proportional to the frequency of the light to the 4th power. How does this fact help explain the blueness of the sky, the whiteness of the clouds, and the color of the sunset? Blue light has a higher frequency so it is more effectively scattered by atmospheric particles, making the sky appear blue. When looking at the sun, the light you see is more direct and has a deficit of blue-frequencies from scattering, so it appears yellow or even red at sunset. Light going through clouds encounter a great number of water molecules and most wavelengths, not just the blue, end up scattered, so the clouds appear white. The textbook has a good discussion of this.

19. What is the polarizing angle? Also called the “Brewster angle”, it’s the angle of incidence that generates totally polarized reflected light. The reflected is polarized parallel to the reflecting surface. At this angle the reflected and refracted rays are perpendicular to each other, so that Snell’s law gives n1 sinøp=n2 sin (90º- øp )= n2 cosøp tan øp=n2 /n1



  1. If a light beam is incident on heavy flint glass ((n = 1.65) at the polarizing angle. Calculate the angle of refraction for the transmitted ray. øp=arctan(1.65)= 59.8º Refraction angle=90º-59.8º=31.2º

  2. For a particular transparent medium surrounded by air, show that the critical angle for internal reflection and the polarizing angle are related by cot øp =()p = sinøc. Recall sin øc= n1 /n2

20. Three polarizing disks whose planes are parallel are centered

on a common axis. The direction of the transmission axis in each

case is shown in the illustration relative to the common vertical

direction. An unpolarized beam of light with intensity Io == 10 units

(arbi­trary) is incident from the left on the first disk. Calculate the

transmitted intensity after going through three Polaroid filters, If when

a) ø1= 20º, ø2= 40º, and ø3= 60º. I=Io(½)cos2(40º-20º) cos2(60º-40º)=0.39Io

b) ø1= 0º, ø2= 30º, and ø3= 90º. I=Io(½)cos2(30º-0º) cos2(90º-30º)=0.094Io

c) How would the final intensity change if the middle filter is removed? Ia=Io(½)cos2(60º-20º)=0. 29Io; Ib=Io(½)cos2(90º-0º)=0

d) Repeat the problem assuming that the incoming beam is initially plane-polarized with Eoparallel to the vertical reference direction. Ia =Io cos2(20º-0º) cos2(40º-20º) cos2(60º-40º)=0.69Io; Ib =Io cos2(30º-0º) cos2(90º-30º)=0.19Io
Some Challenge problems
21. If the light strikes the single slit at an angle of ∂ from the perpendicular

direction, show that the condition for destructive in­terference, must be modified

to read sinø = m(l/a) ± sin∂.
The proof requires a careful drawing of the light going though the slit. Generally ml=asinø, but here the central antinode is along the ∂ direction (not the 0º direction), so ml=asinø ± asin∂. The ± sign in the expression depends on which side of the ∂ direction of the beam you’re looking at in the pattern.
22.The left and right po­larizing disks in problem 18 have their transmission axes fixed perpen­dicular to each other (ie. ø1= 0º and ø3= 90º). Assume that the center disk is rotating on the common axis with an angular speed w.


  1. Show that, if unpolarized light is incident on the left disk with an intensity Io, the intensity of the beam emerging from the right disk is I=Io(1-cos 4wt)/16. Hint: You will need the trigo­nometric identity, sin2ø =(1- cos2ø) /2, and recall that ø== wt.
    The intensity of the emerging is given by I =I
    o(½)cos2(wt)cos2(90º-wt)= Io(½)cos2(wt)sin2(wt)= Io(½)[½sin(2wt)]2


  2. I'?:;, I

2 Slit Diffraction



  1. How many times a rotating period will the intensity of the emerging beam be 0? The solution indicates that the intensity frequency is 4 times the rotating filter frequency, so 4 times.

  2. If the middle filter is rotating at a rate of 100 Hz, find the pulsing frequency of the emerging light. 400Hz (see b)

23. Show that the resolving power of a diffraction grating, defined as R=ll is also equal to Nm, where m is spectrum order and N the number of lines in the diffraction grating. This proof is pretty sophisticated and if you’re interested, the Young & Friedman text has a detailed derivation in page 1383. But be aware that the textbook uses ø for the phase angle, while I’ve been using g.


24.Light strikes a water surface at the polarizing angle. The part of the

beam refracted into the water strikes a submerged glass slab (n=1.5) as

shown. A minimum amount of light is reflected from the upper surface


of the slab, find the angle between water surface and glass slab.
The fact that a minimum of light is reflected from the glass slab suggests that the ray alsoSlit
strikes the glass at the polarizing angle in the water-glass transition. The reasoning is that most of the parallel field component has already been extracted from the beam in the first polarizing reflection between the air and the water and there is little left for the second polarizing reflection. The polarizing
angles are: ø1=arctan (1.33/1)=53º, and ø2=arctan (1.50/1.33)=42º. Geometry gives ø=11º

Diffraction

Light is a transverse electromagnetic wave. Diffraction, and interference are phenomena observed with all waves.

Diffraction is the tendency of a wave emitted from a finite source or passing through a finite aperture to spread out as it propagates. Diffraction results from the interference of an infinite number of waves emitted by a continuous distribution of source points. According to Huygens' principle every point on a wave front of light can be considered to be a secondary source of spherical wavelets. These waveletspropagate outward with the characteristic speed of the wave. The wavelets emitted by all points on the wave front interfere with each other to produce the traveling wave. Huygens' principle also holds for electromagnetic waves. When studying the propagation of light, we can replace any wave front by a collection of sources distributed uniformly over the wave front, radiating in phase.
When light passes through a small opening, comparable in size to the wavelength λ of the light, in an otherwise opaque obstacle, the wave front on the other side of the opening resembles the wave front shown on the right..

The light spreads around the edges of the obstacle. This is the phenomenon of diffraction. Diffraction is a wave phenomenon and is also observed with water waves in a ripple tank.

Water waves in a ripple tank

A single large slit:

A single small slit:

The single slit

When light passes through a single slit whose width w is on the order of the wavelength of the light, then we can observe a single slit diffraction pattern on a screen that is a distance L >> w away from the slit. The intensity is a function of angle. Huygens' principle tells us that each part of the slit can be thought of as an emitter of waves. All these waves interfere to produce the diffraction pattern. Where crest meets crest we have constructive interference and where crest meets trough we have destructive interference.
Very far from a point source the wave fronts are essentially plane waves. This is called the Fraunhofer regime, and the diffraction pattern is called Fraunhofer diffraction. The positions of all maxima and minima in the Fraunhofer diffraction pattern from a single slit can be found from the following simple arguments.
  • Consider a slit of width w, as shown in the diagram on the right. A plane wave is incident from the bottom and all points oscillate in phase inside the slit.
  • For light leaving the slit in a particular direction defined by the angle θ, we may have destructive interference between the ray at the right edge (ray 1) and the middle ray (ray 7). To arrive at a distant screen perpendicular to the direction of propagation of the rays, the rays coming from different points inside the slit have to travel different distances. They have a different optical path length. If ray 7 has to travel an extra distance if one-half wavelength (λ/2) compared to ray 1, then ray 1 and ray 7 destructively interfere. Crest meets trough.
If the optical path length of two rays differs by λ/2, the two rays interfere destructively. For ray 1 and ray 7 to be half a wavelength out of phase we need

(w/2)sinθ = λ/2 or wsinθ = λ.

But from geometry, if these two rays interfere destructively, so do rays 2 and 8, 3 and 9, and 4 and 10, 5 and 11, and 6 and 12.

In effect, light from one half of the opening interferes destructively and cancels out light from the other half.

Destructive interference produces the dark fringes. Dark fringes in the diffraction pattern of a single slit are found at angles θ for which

w sinθ = mλ,

where m is an integer, m = 1, 2, 3, ... . For the first dark fringe we have w sinθ = λ.

When w is smaller than λ , the equation wsinθ = λ has no solution and no dark fringes are produced.

If the interference pattern is viewed on a screen a distance L from the slits, then the wavelength can be found from the spacing of the fringes. We have λ = w sinθ/m and sinθ = z/(L2 + z2)1/2), or

λ = zw/(m(L2 + z2)1/2),

where z is the distance from the center of the interference pattern to the mth dark line in the pattern.

If L >> z then (L2 + z2)1/2 ~ L and we can write

λ = zw/(mL).

Problem:

Slit

When a monochromatic light source shines through a 0.2 mm wide slit onto a screen 3.5 m away, the first dark band in the pattern appears 9.1 mm from the center of the bright band. What is the wavelength of the light?

  • Solution:
    z = 9.1 mm = 9.1*10-3 m
    L = 3.5 m
    w = 0.2 mm = 2*10-4 m
    L >> z, therefore λ = zw/(mL)
    λ = (9.1*10-3 m)(2*10-4 m)/(3.5 m)
    λ = 5.2*10-7 m = 520 nm

Link: Single slit diffraction II

Diffraction, and interference are phenomena observed with all waves.

  • Observe single and double slit diffraction with water waves.

Interference

The double slit

If light is incident onto an obstacle which contains two very small slits a distance d apart, then the wavelets emanating from each slit will interfere behind the obstacle. Waves passing through each slit are diffracted and spread out. At angles where the single slit diffraction pattern produces nonzero intensity, the waves from the two slits can now constructively or destructively interfere.

If we let the light fall onto a screen behind the obstacle, we will observe a pattern of bright and dark stripes on the screen, in the region where with a single slit we only observe a diffraction maximum. This pattern of bright and dark lines is known as an interference fringe pattern. The bright lines indicate constructive interference and the dark lines indicate destructive interference.

The bright fringe in the middle of the diagram on the right is caused by constructive interference of the light from the two slits traveling the same distance to the screen. It is known as the zero-order fringe. Crest meets crest and trough meets trough. The dark fringes on either side of the zero-order fringe are caused by destructive interference. Light from one slit travels a distance that is ½ wavelength longer than the distance traveled by light from the other slit. Crests meet troughs at these locations. T he dark fringes are followed by the first-order fringes, one on each side of the zero-order fringe. Light from one slit travels a distance that is one wavelength longer than the distance traveled by light from the other slit to reach these positions. Crest again meets crest.

The diagram on the right shows the geometry for the fringe pattern. If light with wavelength λ passes through two slits separated by a distance d, we will observe constructively interference at certain angles. These angles are found by applying the condition for constructive interference, which is

d sinθ = mλ, m = 0, 1, 2, … .

The distances from the two slits to the screen differ by an integer number of wavelengths. Crest meets crest.

The angles at which dark fringes occur can be found be applying the condition for destructive interference, which is

d sinθ = (m+½)λ, m = 0, 1, 2, … .

The distances from the two slits to the screen differ by an integer number of wavelengths + ½ wavelength. Crest meets trough.

If the interference pattern is viewed on a screen a distance L from the slits, then the wavelength can be found from the spacing of the fringes.
We have sinθ = z/(L2 + z2)1/2 and λ = zd/(m(L2 + z2)1/2), where z is the distance from the center of the interference pattern to the mth bright line in the pattern.
If L >> z then (L2 + z2)1/2 ~ L and we can write

λ = zd/(mL).

Link: Physics 2000: Wave Interference

Diffraction grating

We have seen that diffraction patterns can be produced by a single slit or by two slits. When light encounters an entire array of identical, equally-spaced slits, called a diffraction grating, the bright fringes, which come from constructive interference of the light waves from different slits, are found at the same angles they are found if there are only two slits. But the pattern is much sharper.

The figure on the right shows the interference pattern for various numbers of slits. The width of all slits is 50 micrometers and the spacing between all slits is 150 micrometers. The location of the maxima for two slits is also the location of the maxima for multiple slits. The single slit pattern acts as an envelope for the multiple slit patterns.

Diffraction gratings contain a large number of parallel, closely spaced slits or grooves. They produce interference maxima at angles θ given by d sinθ = mλ. Because the spacing between the slits is generally very small, the angles θ are generally quite large. We cannot use the small angle approximation for relating wavelength and the position of the maxima on a screen for gratings, but have to use

sinθ = z/(L2 + z2)1/2.

Diffraction gratings disperse white light into its component colors because different wavelengths produce bright fringes at different angles.

d sinθ = mλ,
for a given m, bigger wavelength <> bigger angle

The spectral pattern is repeated on either side of the main pattern. These repetitions are called 'higher order spectra'. There are often many of them, each one fainter than the previous one. If the distance between slits is d, and the angle to a bright fringe of a particular color is θ, the wavelength of the light can be calculated.

Problem:

The first order bright line appears 0.25 cm from the center bright line when a double slit grating is used. The distance between the slits is 0.5 mm and the screen is 2.7 m from the grating. Find the wavelength.

2 Slit Diffraction

  • Solution:
    z = 0.25 cm = 2.5*10-3 m
    L = 2.7 m
    d = 0.5 mm = 5*10-4 m
    L >> z, therefore λ = zd/(mL)
    λ = (2.5*10-3 m)(5*10-4 m)/(2.7 m)
    λ = 4.63*10-7 m = 463 nm

Problem:

A diffraction grating has 420 lines per mm. The grating is used to observe normally incident light with a wavelength of 440 nm. The grating is placed 1.3 m from a screen. Where on the screen will the first order bright line appear?

  • Solution:
    d sinθ = λ
    d=(1/420) mm=2.38*10-3 mm=2.38*10-6 m
    λ=440 nm=4.40*10-7 m
    L=1.3 m = distance to the screen.
    sinθ = λ/d = 10.65o.
    z= L tanq =24.5 cm = distance from the central maximum.

Diffraction patterns can be observed when light passes through a set of regularly spaced slits. For a diffraction to produce an observable pattern, the spacing of the slits must be comparable to the wavelength of the radiation. Visible light has a wavelength range from ~400 nm to ~700 nm. A typical diffraction grating for visible light with 300 grooves per mm has a slit spacing of (1/300)mm = 3 mm = 3000 nm. This spacing is 4 to 8 times larger than the wavelengths of visible light and produces an easily observable pattern.

2 Slit Diffraction Experiment

The wavelengths of x-rays lie in the 1 nm to 1 pm range. A typical diffraction grating will not produce an observable pattern. But the wavelengths of x-rays are comparable to the spacing of atoms in common crystals, and material with a regularly spaced grid of atoms can diffract x-rays and produce diffraction patterns that can be captured on photographic film.