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# Chapter 36–Diffraction Hints

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Chapter 36–Diffraction Hints-5/30/10

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:

 S ingle slit * D ouble 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.. Intensity comparison 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. H ow 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 a are 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 inten­sity I/Io 11I/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 N1 is 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 l and frequency f f 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.

For 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 2πm (m+½)l 2π(m+½) x x [cos(g/2)]2 x 3 ml 2πm (m±1/3)l 2π(m±1/3) (m+½)l 2π(m+½) 1/9[1+2cosg]2 1/9 4 ml 2πm (m±¼)l (m+½)l 2π(m±¼); 2π(m+½) (m±1/3)l 2π(m±1/3) 1/4[cos(3g/2)+ cos(g/2)]2 1/16

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.

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.

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.

a) 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.)

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

o n 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 Eo parallel 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
S ome 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

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 also

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

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