chapter 37 diffraction chapter 37 - links to …hep.physics.indiana.edu/~rickv/p222/ch37.pdf ·...

1010 CHAPTER 37 DIFFRACTION

CHAPTER 37

Answer to Checkpoint Questions

1. (a) expand; (b) expand

2. (a) second side maximum; (b) 2:5

3. (a) red; (b) violet

4. diminish

5. (a) increase; (b) same

6. (a) left; (b) less

Answer to Questions

1. (a) contract; (b) contract

2. (a) the m = 5 minimum; (b) (approximately) the maximum between the m = 4 and

m = 5 minima

3. with megaphone (larger opening, less di�raction)

4. (a) a and c tie, then b and d tie; (b) a and b tie , then c and d tie

5. four

6. red

7. (a) larger; (b) red

8. (a) less; (b) greater; (c) greater

9. (a) decrease; (b) same; (c) in place

10. (a) decrease; (b) decrease; (c) to the right

11. (a) A; (b) left; (c) left; (d) right

CHAPTER 37 DIFFRACTION 1011

12. (a) increase; (b) �rst order

Solutions to Exercises & Problems

1E

The condition for a minimum in a single-slit di�raction pattern is given by Eq. 37-3:a sin � = m�, where a is the slit width, � is the wavelength, and m is an integer. Solve for�: � = a sin �=m = (0:022mm)(sin 1:8�)=1 = 6:9� 10�3mm = 690 nm:

2E

(a) � = sin�1(1:50 cm=2:00m) = 0:430�:(b) For the mth di�raction minimum a sin � = m�: Solve for a:

a =m�

sin �=

2(441 nm)

sin 0:430�= 0:118mm :

3E

Use a sin � = m�. The angle � is measured from the forward direction, so for the situationdescribed in the problem it is 0:60�, for m = 1. Thus

a =m�

sin �=

633� 10�9m

sin 0:60�= 6:04� 10�5m :

4E

(a) Use a sin � = m�. For � = �a and m = 1 the angle � is the same as for � = �b andm = 2. Thus �a = 2�b.

(b) Let ma be the integer associated with a minimum in the pattern produced by light withwavelength �a and letmb be the integer associated with a minimum in the pattern producedby light with wavelength �b. A minimum in one pattern coincides with a minimum in theother if they occur at the same angle. This means ma�a = mb�b. Since �a = 2�b, theminima coincide if 2ma = mb. Thus every other minimum of the �b pattern coincides witha minimum of the �a pattern.

5E

(a) Use Eq. 37-3 to calculate the separation between the �rst (m1 = 1) and �fth (m2 = 5)minima:

�y = D�sin � = D�

�m�

a

�=D�

a�m =

D�

a(m2 �m1) :


solve for a:

a =D�(m2 �m1)

�y=

(400mm)(550� 10�6mm)(5� 1)

0:35mm= 2:5mm :

(b) For m = 1

sin � =m�

a=

(1)(550� 10�6mm)

2:5mm= 2:2� 10�4 :

The angle is � = sin�1(2:2� 10�4) = 2:2� 10�4 rad:

6E

From Eq. 37-3a

�=

m

sin �=

1

sin 45:0�= 1:41 :

7E

(a) A plane wave is incident on the lens so it is brought to focus in the focal plane of thelens, a distance of 70 cm from the lens.

(b) Waves leaving the lens at an angle � to the forward direction interfere to produce anintensity minimum if a sin � = m�, where a is the slit width, � is the wavelength, and mis an integer. The distance on the screen from the center of the pattern to the minimumis given by y = D tan �, where D is the distance from the lens to the screen. For theconditions of this problem

sin � =m�

a=

(1)(590� 10�9m)

0:40� 10�3m= 1:475� 10�3 :

This means � = 1:475�10�3 rad and y = (70�10�2m) tan(1:475�10�3 rad) = 1:0�10�3m.

8P

The condition for a minimum of intensity in a single-slit di�raction pattern is a sin � = m�,where a is the slit width, � is the wavelength, and m is an integer. To �nd the angularposition of the �rst minimum to one side of the central maximum set m = 1:

�1 = sin�1��

a

�= sin�1

�589� 10�9m

1:00� 10�3m

�= 5:89� 10�4 rad :

If D is the distance from the slit to the screen, the distance on the screen from the centerof the pattern to the minimum is y1 = D tan �1 = (3:00m) tan(5:89� 10�4 rad) = 1:767�10�3m.


To �nd the second minimum set m = 2:

�2 = sin�1�2(589� 10�9m)

1:00� 10�3m

�= 1:178� 10�3 rad :

The distance from the pattern center to the minimum is y2 = D tan �2 = (3:00m) tan(1:178�10�3 rad) = 3:534 � 10�3m. The separation of the two minima is �y = y2 � y1 =3:534mm� 1:767mm = 1:77mm.

9P

Let the �rst mimimum be a distance y from the central axis which is perpendicular to thespeaker. Then sin � = y=(D2 + y2)1=2 = m�=a = �=a (for m = 1). Slove for y:

y =Dp

(a=�)2 � 1=

Dp(af=vs)2 � 1

=100mp

[(0:300m)(3000Hz)=(343m/s)]2 � 1= 41:2m :

10P

From y = m�D=a we get

�y = �

�m�D

a

�=�D

a�m =

(632:8 nm)(2:60)

1:37mm[10� (�10)] = 24:0mm :

11E

From Eq. 37-4

�� =

�2�

�

�(�x sin �) =

�2�

589 nm

��0:10mm

2

�(sin 30�) = 267 rad :

This is equivalent to 267 rad� 84� = 2:79 rad = 160�:

12E

(a) � = sin�1(1:1 cm=3:5m) = 0:18�:

(b) Use Eq. 37-6:

� =

��a

�

�sin � =

�(0:025mm)(sin 0:18�)

538 nm= 0:46 rad :


13P

If you divide the original slit into N strips and represent the light from each strip, when itreaches the screen, by a phasor, then at the central maximum in the di�raction pattern youadd N phasors, all in the same direction and each with the same amplitude. The intensitythere is proportional to N2. If you double the slit width you need 2N phasors if they areeach to have the amplitude of the phasors you used for the narrow slit. The intensity atthe central maximum is proportional to (2N)2 and is therefore 4 times the intensity forthe narrow slit. The energy reaching the screen per unit time, however, is only twice theenergy reaching it per unit time when the narrow slit is in place. The energy is simplyredistributed. For example, the central peak is now half as wide and the integral of theintensity over the peak is only twice the analogous integral for the narrow slit.

14P

Think of the Huygens' explanation of di�raction phenomenon. When A is in place onlythe Huygens' wavelets that pass through the hole get to point P . Suppose they produce aresultant electric �eld EA. When B is in place the light that was blocked by A gets to Pand the light that passed through the hole in A is blocked. Suppose the electric �eld at Pis now EB . The sum EA + EB is the resultant of all waves that get to P when neither Anor B are present. Since P is in the geometric shadow this is zero. Thus EA = �EB andsince the intensity is proportional to the square of the electric �eld, the intensity at P isthe same when A is present as when B is present.

15P

(a) The intensity for a single-slit di�raction pattern is given by

I = Imsin2 �

�2;

where � = (�a=�) sin �, a is the slit width and � is the wavelength. The angle � is measuredfrom the forward direction. You want I = Im=2, so

sin2 � =1

2�2 :

(b) Evaluate sin2 � and �2=2 for � = 1:39 rad and compare the results. To be sure that1:39 rad is closer to the correct value for � than any other value with 3 signi�cant digits,you should also try 1:385 rad and 1:395 rad.

(c) Since � = (�a=�) sin �,

� = sin�1��

�a

�:

Now �=� = 1:39=� = 0:442, so

� = sin�1�0:442�

a

�:

y

α (rad)

y=tan α

y=α

π /2 3π /2π0


The angular separation of the two points of half intensity, one on either side of the centerof the di�raction pattern, is

�� = 2� = 2 sin�1�0:442�

a

�:

(d) For a=� = 1:0,

�� = 2 sin�1(0:442=1:0) = 0:916 rad = 53� ;

for a=� = 5:0,

�� = 2 sin�1(0:442=5:0) = 0:177 rad = 10� ;

and for a=� = 10,

�� = 2 sin�1(0:442=10) = 0:0884 rad = 5:1� :

16P

(a) The intensity for a single-slit di�raction pattern is given by

I = Imsin2 �

�2;

where � = (�a=�) sin �. Here a is the slit width and � is the wavelength. To �nd themaxima and minima, set the derivative of I with respect to � equal to zero and solve for�. The derivative is

dI

d�= 2Im

sin�

�3(� cos�� sin�) :

The derivative vanishes if � 6= 0 but sin� = 0. This yields � = m�, where m is an integer.Except for m = 0 these are the intensity minima: I = 0 for � = m�.

The derivative also vanishes for � cos��sin� = 0. This condition can be written tan� = �.These are the maxima.

(b) The values of � that satisfy tan� = �

can be found by trial and error on a pocketcalculator or computer. Each of them isslightly less than one of the values (m +1

2)� rad, so start with these values. The �rst

few are 0, 4:4934, 7:7252, 10:9041, 14:0662,and 17:2207. They can also be found graph-ically. As in the diagram to the right, ploty = tan� and y = � on the same graph.The intersections of the line with the tan�curves are the solutions. The �rst two solu-tions listed above are shown on the diagram.

φ φ

φ

φω t

E1E1

E2

E3

E


(c) Write � = (m + 1

2)� for the maxima. For the central maximum, � = 0 and m = �1

2.

For the next, � = 4:4934 and m = 0:930. For the next � = 7:7252 and m = 1:959.

17P

Since the slit width is much less than the wavelength of the light, the central peak of thesingle-slit di�raction pattern is spread across the screen and the di�raction envelope canbe ignored. Consider 3 waves, one from each slit. Since the slits are evenly spaced thephase di�erence for waves from the �rst and second slits is the same as the phase di�erencefor waves from the second and third slits. The electric �elds of the waves at the screencan be written E1 = E0 sin(!t), E2 = E0 sin(!t + �), and E3 = E0 sin(!t + 2�), where� = (2�d=�) sin �. Here d is the separation of adjacent slits and � is the wavelength.

The phasor diagram is shown to the right. It yields

E = E0 cos�+E0 +E0 cos� = E0(1 + 2 cos�)

for the amplitude of the resultant wave. Since the in-tensity of a wave is proportional to the square of theelectric �eld, we may write I = AE2

0(1+2 cos�)2, where

A is a constant of proportionality. If Im is the inten-sity at the center of the pattern, for which � = 0, thenIm = 9AE2

0. Take A to be Im=9E

2

0and obtain

I =Im

9(1 + 2 cos�)

2=Im

9

�1 + 4 cos�+ 4 cos2 �

�:

18E

Use Eq. 37-12: sin � = 1:22�=d. In our case � = 2:5�=2 = 1:25�, so

d =1:22�

sin �=

1:22(550 nm)

sin 1:25�= 31�m :

19E

(a) Use the Rayleigh criteria. To resolve two point sources the central maximum of thedi�raction pattern of one must lie at or beyond the �rst minimum of the di�raction patternof the other. This means the angular separation of the sources must be at least �R =1:22�=d, where � is the wavelength and d is the diameter of the aperture. For the headlightsof this problem

�R =1:22(550� 10�9m)

5:0� 10�3m= 1:34� 10�4 rad :

(b) If L is the distance from the headlights to the eye when the headlights are just resolvableand D is the separation of the headlights, then D = L tan �R � L�R, where the small angle


approximation tan �R � �R was made. This is valid if �R is measured in radians. ThusL = D=�R = (1:4m)=(1:34� 10�4 rad) = 1:0� 104m = 10 km.

20E

(a) Use Eq. 37-14:

�R = 1:22�

d=

(1:22)(540� 10�6mm)

5:0mm= 1:3� 10�4 rad :

(b) The linear separation is D = L�R = (160� 103m)(1:3� 10�4 rad) = 21m:

21E

The minimum separation is

Dmin = dem�R = dem

�1:22

�

d

�=

(1:22)(3:82� 108m)(550� 10�9m)

5:1m= 50m :

22E

Lmax =D

�R=

D

1:22�=d=

(5:0� 10�3m)(4:0� 10�3m)

1:22(550� 10�9m)= 30m :

23E

Dmin = L�R = L

�1:22�

d

�=

(1:22)(250mm)(500� 10�3�m)

5:00mm= 30:5�m :

24E

(a) Use Rayleigh's criterion: two objects can be resolved if their angular separation at theobserver is greater than �R = 1:22�=d, where � is the wavelength of the light and d isthe diameter of the aperture (the eye or mirror). If L is the distance from the observerto the objects then the smallest separation D they can have and still be resolvable isD = L tan �R � L�R, where �R is measured in radians. The small angle approximationtan �R � �R was made. Thus

D =1:22L�

d=

1:22(8:0� 1010m)(550� 10�9m)

5:0� 10�3m= 1:1� 107m = 1:1� 104 km :

This distance is greater than the diameter of Mars. One part of the planet's surface cannotbe resolved from another part.


(b) Now d = 5:1m and

D =1:22(8:0� 1010m)(550� 10�9m)

5:1m= 1:1� 104m = 11 km :

25E

Lmax =D

�R=

D

1:22�=d=

(5:0� 10�2m)(4:0� 10�3m)

1:22(0:10� 10�9m)= 1:6� 106m = 1600 km :

26E

Dmin = L�R = L

�1:22�

d

�=

(6:2� 103m)(1:22)(1:6� 10�2m)

2:3m= 53m :

27P

(a) The diameter is

D = L�R = L

�1:22�

d

�=

(2000� 103m)(1:22)(1:40� 10�9m)

0:200� 10�3m= 17:1m :

(b) I=Im = (d=D)2 = (0:200� 10�3m=1:71m)2 = 1:37� 10�10:

28P

(a)

Lmax =D

1:22�=d=

2(50� 10�6m)(1:5� 10�3m)

1:22(650� 10�9m)= 0:19m :

(b) The wavelength of the blue light is shorter so Lmax / ��1 will be larger.

29P

According to Rayleigh's criterion (Eq. 37-14) �R = 1:22�=d. In our case �R � D=L, whereD = 60�m is the size of the object your eyes must resolve, and L is the limiting viewingdistance in question. Also d = 3:00mm is the diameter of your pupil. Solve for L:

L =Dd

1:22�=

(60�m)(3:00mm)

1:22(550 nm)= 27 cm :


30P

(a) Use Eq. 37-12:

� = sin�1�1:22�

d

�= sin�1

�1:22(vs=f)

d

�

= sin�1�

(1:22)(1450m/s)

(25� 103Hz)(0:60m)

�= 6:8� :

(b) Now f = 1:0� 103Hz so

1:22�

d=

(1:22)(1450m/s)

(1:0� 103Hz)(0:60m)= 2:9 > 1 :

Since sin � cannot exceed 1 there is no minimum.

31P

Use 2� = 1:22�=d = D=L:

d =1:22�L

D=

(1:22)(220mi)(1610m/mi)(500� 10�9m)

(30 ft)(0:305m/ft)= 4:7 cm :

32P

From �R = 1:22�=d = D=L we get

d =1:22�L

D=

(1:22)(550� 10�9m)(160� 103m)

0:30m= 0:36m :

33P

(a) The �rst minimum in the di�raction pattern is at an angular position �, measuredfrom the center of the pattern, such that sin � = 1:22�=d, where � is the wavelength andd is the diameter of the antenna. If f is the frequency then the wavelength is � = c=f =(3:00� 108m/s)=(220� 109Hz) = 1:36� 10�3m. Thus

� = sin�1�1:22�

d

�= sin�1

�1:22(1:36� 10�3m)

55:0� 10�2m

�= 3:02� 10�3 rad :

The angular width of the central maximum is twice this, or 6:04� 10�3 rad (0:346�).


(b) Now � = 1:6 cm and d = 2:3m, so

� = sin�1�1:22(1:6� 10�2m)

2:3m

�= 8:5� 10�3 rad :

The angular width of the central maximum is 1:7� 10�2 rad (0:97�).

34P

(a) The angular separation is

� = �R =1:22�

d=

(1:22)(550� 10�9m)

0:76m= 8:8� 10�7 rad = 0:1800 :

(b) The distance is

D = L�R =(10 ly)(9:46� 1012 km/ly)(0:18)�

(3600)(180)= 8:4� 107 km :

(c) The diameter is

d = 2�RL =2(0:18)(�)(14m)

(3600)(180)= 2:5� 10�5m = 0:025mm :

35P

(a) Since � = 1:22�=d, the larger the wavelength the larger the radius of the �rst minimum(and second maximum, ect). Therefore the white pattern is outlined by red lights (withlonger wavelength than blue lights).(b)

d =1:22�

��

1:22(7� 10�7m)

1:5(0:50�)(�=180�)=2= 1:3� 10�4m :

36P

The energy of the beam of light which is projected onto the moon is concentrated is acircular spot of diameter d1, where d1=dem = 2� = 2(1:22�=d0), with d0 the diameter ofthe mirror on Earth and dem the Earth-Moon separation. The fraction of energy pickedup by the re ector of diameter d2 on the Moon is then �0 = (d2=d1)

2. This re ected light,upon reaching the Earth, has a circular cross section of diameter d3 satisfying d3=dem =2� = 2(1:22�=d2). The fraction of the re ected energy that is picked up by the telescope isthen �00 = (d0=d3)

2: Thus the fraction of the original energy picked up by the detector is

� = �0�00 =

�d0

d3

�2�d2

d1

�2

=

�d0d2

(2:44�dem=d0)(2:44�dem=d2)

�2=

�d0d2

2:44�dem

�4

=

�(2:6m)(0:10m)

2:44(0:69� 10�6m)(3:82� 108m)

�4� 4� 10�13 :


37E

Bright interference fringes occur at angles � given by d sin � = m�, where d is the slitseparation, � is the wavelength, andm is an integer. For the slits of this problem d = 11a=2,so a sin � = 2m�=11. The �rst minimum of the di�raction pattern occurs at the angle �1given by a sin �1 = � and the second occurs at the angle �2 given by a sin �2 = 2�, where ais the slit width. You want to count the values of m for which �1 < � < �2, or what is thesame, the values of m for which sin �1 < sin � < sin �2. This means 1 < (2m=11) < 2. Thevalues are m = 6, 7, 8, 9, and 10. There are �ve bright fringes in all.

38E

The number is 2(d=a)� 1 = 2(2a=a)� 1 = 3:

39E

It is clear from Eq. 37-5 that for a single slit of width 2a the di�raction pattern is given by

I = Im

�sin(2�)

2�

�;

where � = �a sin �=�. Now, if we put d = a in Eq. 37-19, then � = � = �a sin �=�, andEq. 37-19 reduces to

I = Im(cos�)2

�sin�

�

�2

= Im

�2 sin� cos�

2�

�2

= Im

�sin(2�)

2�

�2;

where the trigonometric identity sin(2�) = 2 sin� cos� was used. Thus Eq. 37-19 indeedreduces to the di�raction pattern for a single slit of width 2a.

40P

(a) Let the location of the fourth bright fringe coincide with the �rst minimum of di�ractionpattern: sin � = 4�=d = �=a, or d = 4a.(b) Any bright fringe which happens to be at the same location with a di�raction minimumwill vanish. So let sin � = m1�=d = m2�=a = m1�=4a = m2�=a, or m1 = 4m2 wherem2 = 1; 2; 3; � � � : The fringes missing are thus the 4th, the 8th, the 12th, � � � , i.e., everyfourth fringe is missing.

41P

The angular location of the mth bright fringe is given by d sin � = m� so the linear sepa-ration between two adjacent fringe is

�y = �(D sin �) = �

�Dm�

d

�=D�

d�m =

D�

d:


42P

(a) The angular positions � of the bright interference fringes are given by d sin � = m�,where d is the slit separation, � is the wavelength, and m is an integer. The �rst di�ractionminimum occurs at the angle �1 given by a sin �1 = �, where a is the slit width. Thedi�raction peak extends from ��1 to +�1, so you want to count the number of values ofm for which ��1 < � < +�1, or what is the same, the number of values of m for which� sin �1 < sin � < +sin �1. This means �1=a < m=d < 1=a or �d=a < m < +d=a. Nowd=a = (0:150 � 10�3m)=(30:0 � 10�6m) = 5:00, so the values of m are m = �4, �3, �2,�1, 0, +1, +2, +3, and +4. There are nine fringes.

(b) The intensity at the screen is given by

I = Im�cos2 �

�� sin��

�2

;

where � = (�a=�) sin �, � = (�d=�) sin �, and Im is the intensity at the center of the pattern.For the third bright interference fringe d sin � = 3�, so � = 3� rad and cos2 � = 1. Similarly,� = 3�a=d = 3�=5:00 = 0:600� rad and (sin�)2=�2 = (sin 0:600�)2=(0:600�)2 = 0:255.The intensity ratio is I=Im = 0:255.

43P

(a) The �rst minimum of the di�raction pattern is at 5:00� so a = �= sin � = 0:440�m=sin 5:00� = 5:05�m.(b) Since the fourth bright fringe is missing d = 4a = 4(5:05�m) = 20:2�m:(c) For the m = 1 bright fringe

� =�a sin �

�=�(5:05�m) sin 1:25�

0:440�m= 0:787 rad ;

so the intensity of the m = 1 fringe is

I = Im

�sin�

�

�2

= (7:0mW/cm2)

�sin 0:787 rad

0:787

�2

= 5:7mW/cm2 ;

which agrees with what Fig. 37-39 indicates. Similarly for m = 2 I = 2:9mW/cm2, alsoin agreement with Fig. 37-39.

44P

As the phase di�erence �� is varied from zero to �, both the intensity pro�le of the di�rac-tion and the location of the interference maximum change. At �� = �, the original centraldi�raction envelop is now a minimum, and the two maxima of the di�raction intensitypro�le are now centered where the �rst minima were. The locations of the intensity max-ima/minima due to interference exchange, with the original locations of the maxima nowthose of minima, and vice versa.


As �� is further varied from � to 2�, the intensity pattern is gradually changed back,resuming the original pattern at �� = 2�.

45E

(a) d = 20:0mm=6000 = 0:00333mm = 3:33�m:

(b) Let d sin � = m� (m = 0; �1; �2; � � � ), we �nd � = 0 for m = 0, � = sin�1(��=d) =sin�1(�0:589�m= 3:30�m) = �10:2� form = �1, and similarly�20:7� form = �2; �32:2�

for m = �3; �45� for m = �4, and �62:2� for m = �5. Since jmj�=d > 1 for jmj � 6 theseare all the maxima.

46E

(a) Let d sin � = m� and solve for �:

� =d sin �

m=

(1:0mm=200)(sin 30�)

m=

2500 nm

m;

where m = 1; 2; 3 � � � . In the visible light range m can assume the following values:m1 = 4; m2 = 5 and m3 = 6. The corresponding wavelengths are �1 = 2500 nm=4 =625 nm; �2 = 2500 nm=5 = 500 nm, and �3 = 2500 nm=6 = 416 nm.

(b) The colors are orange (for �1 = 625 nm), blue-green (for �2 = 500 nm), and violet (for�3 = 416 nm).

47E

The angular location of the mth order di�raction maximum is given by m� = d sin �. Tobe able to observe the �fth-order one we must let sin �jm=5 = 5�=d < 1, or

� <d

5=

1:00 nm=315

5= 635 nm :

So all wavelengths shorter than 635 nm can be used.

48E

The ruling separation is d = 1=(400mm�1) = 2:5 � 10�3mm. Di�raction lines occur atangles � such that d sin � = m�, where � is the wavelength and m is an integer. Notice thatfor a given order the line associated with a long wavelength is produced at a greater anglethan the line associated with a shorter wavelength. Take � to be the longest wavelengthin the visible spectrum (700 nm) and �nd the greatest integer value of m such that � isless than 90�. That is, �nd the greatest integer value of m for which m� < d. Sinced=� = (2:5 � 10�6m)=(700 � 10�9m) = 3:57 that value is m = 3. There are 3 completeorders on each side of the m = 0 order. The second and third orders overlap (see 59P).


49E

Let the total number of lines on the grating be N , then d = L=N where L = 3:00 cm. Forthe second order di�raction maximum d sin � = (L=N) sin � = m� = 2�, so

N =L sin �

2�=

(3:00� 10�2m)(sin 33�)

2(600� 10�9m)= 13; 600 :

50E

Use Eq. 37-25 for di�raction maxima: d sin � = m�. In our case since the angle between them = 1 and m = �1 maxima is 26� the angle � corresponding to m = 1 is � = 26�=2 = 13�.Solve for d:

d =m�

sin �=

(1)(550 nm)

sin 13�= 2:4�m :

51P

Let d sin � = (L=N) sin � = m�, we get

� =(L=N) sin �

m=

(1:0� 107 nm)(sin 30�)

(1)(10; 000)= 500 nm :

52P

(a) Maxima of a two-slit interference pattern occur at angles � given by d sin � = m�,where d is the slit separation, � is the wavelength, and m is an integer. The two linesare adjacent so their order numbers di�er by unity. Let m be the order number for theline with sin � = 0:2 and m + 1 be the order number for the line with sin � = 0:3. Then0:2d = m� and 0:3d = (m + 1)�. Subtract the �rst equation from the second to obtain0:1d = �, or d = �=0:1 = (600� 10�9m)=0:1 = 6:0� 10�6m.

(b) Minima of the single-slit di�raction pattern occur at angles � given by a sin � = m�,where a is the slit width. Since the fourth order interference maximum is missing it mustfall at one of these angles. If a is the smallest slit width for which this order is missingthe angle must be given by a sin � = �. It is also given by d sin � = 4�, so a = d=4 =(6:0� 10�6m)=4 = 1:5� 10�6m.

(c) First set � = 90� and �nd the largest value of m for which m� < d sin �. This is thehighest order that is di�racted toward the screen. The condition is the same as m < d=�

and since d=� = (6:0� 10�6m)=(600� 10�9m) = 10:0, the highest order seen is the m = 9order. The fourth and eighth orders are missing so the observable orders are m = 0, 1, 2,3, 5, 6, 7, and 9.

53P

(a) For the maximum with the greatest value of m (= M) we have M� = a sin � < d,so M < d=� = 900 nm=600 nm = 1:5, or M = 1. Thus three maxima can be seen, withm = 0; �1:


(b) From Eq. 37-28

��hw =�

Nd cos �=

d sin �

Nd cos �=

tan �

N=

1

N

�sin�1

��

d

��

=1

1000tan

�sin�1

�600 nm

900 nm

��= 0:051� :

54P

The angular positions of the �rst-order di�raction lines are given by d sin � = �, where d isthe slit separation and � is the wavelength. Let �1 be the shorter wavelength (430 nm) and� be the angular position of the line associated with it. Let �2 be the longer wavelength(680 nm) and let � + �� be the angular position of the line associated with it. Here�� = 20�. Then d sin � = �1 and d sin(� + ��) = �2. Use a trigonometric identity toreplace sin(�+��) with sin � cos��+cos � sin��, then use the equation for the �rst line to

replace sin � with �1=d and cos � withp1� �2

1=d2. After multiplying by d you should obtain

�1 cos�� +pd2 � �2

1sin�� = �2. Rearrange to get

pd2 � �2

1sin�� = �2 � �1 cos��.

Square both sides and solve for d. You should get

d =

r(�2 � �1 cos��)2 + (�1 sin��)2

sin2��

=

s[(680 nm)� (430 nm) cos 20�]

2+ [(430 nm) sin 20�]

2

sin2 20�

= 914 nm = 9:14� 10�4mm :

There are 1=d = 1=(9:14� 10�4mm) = 1090 rulings per mm.

55P

Use Eq. 37-25: m� = d sin �. For m = �1

� =d sin �

m=

(1:73�m) sin(�17:6�)

�1= 523 nm ;

and for m = �2

� =(1:73�m) sin(�37:3�)

�2= 524 nm :

Similarly we may compute the values of � corresponding to the angles for m = �3 . Theaverage value of these �0s is 523 nm.

1

2ψ θ

θψ

d

B

A C

θ1ψψ

incident wavediffracted wave

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.60.4

0.6

0.8

1

δ (rad)

ψ (rad)


56P

The di�erence in path lengths between the two ad-jacent light rays shown to the right is �x = jABj+jBCj = d sin + d sin �. The condition for brightfringes to occur is thus

�x = d (sin + sin �) = m� ;

where m = 0; 1; 2; � � � :

57P

From the �gure to the right we see that the angulardeviation of the �rst-order maximum from the inci-dent direction is � = + �1. Here sin �1 = �=d �sin = (600 nm=1:50�m) � sin = 0:400 � sin :Thus

� = + �1 = + sin�1��

d� sin

�= + sin�1(0:400� sin ) :

The plot is as follows. The angles are given in radians.

58P

From Eq. 37-25 we get

�(sin �) = �

�m�

d

�=m��

d;


but for small ��; �sin � � (d sin �=d�)�� = cos �� so

�� m��

d cos �=

m��

dp1� sin2 �

=m��

dp1� (m�=d)2

=��p

(d=m)2 � �2:

59P

From d sin � = m� we see that if two spectral lines (labeled 1 and 2, respectively) overlap,meaning they share the same value of �, then m1�1 = m2�2. For m1 = 2 and m2 = 3 thisbecomes

�1

�2=m2

m1

=3

2:

Since 400 nm < �2 < �1 < 700 nm we can always �nd suitable values of �1 and �2 whichsatisfy this condition. For example �1 = 600 nm, and �2 = 400 nm, or �1 = 660 nm and�2 = 440 nm, etc. So these two spectra always overlap, regardless of the value of d.

60P

In this case a = d=2, and the formula for the locations of the mth di�raction minimum,m� = a sin � = (d=2) sin �, may be re-written as (2m)� = d sin �, which we recognize asthe formula for the location of the (2m)-th maximum of interference. Thus all the (2m)-th(i.e., even) orders of maxima will be eliminated (exept m = 0).

61P

At the location of the hole sin � � 50mm=30 cm = 0:164, and from m� = d sin � we �nd

sin � � 5:0 cm=p(30 cm)2 + (5:0 cm)2 = 0:164; so

m =d sin �

�=

(1:00� 106 nm=350)(0:164)

�=

470 nm

�:

Since for white light � > 400 nm the only integer m allowed here is m = 1. At one edge ofthe hole � = 477 nm, while at the other edge

� = d sin �0 =

�1:00� 106 nm

350

�"50mm+ 10mmp(30 cm)2 + (6:0 cm)2

#= 560 nm :

So the range of wavelength is from 470 to 560 nm.

62P

The derivation is similar to that used to obtain Eq. 37-27. At the �rst minimum beyondthe mth principal maximum two waves from adjacent slits have a phase di�erence of �� =2�m + (2�=N), where N is the number of slits. This implies a di�erence in path length


of �L = (��=2�)� = m� + (�=N). If �m is the angular position of the mth maximumthen the di�erence in path length is also given by �L = d sin(�m +��). Thus d sin(�m +��hw) = m�+ (�=N). Use the trigonometric identity sin(�m +��hw) = sin �m cos��hw +cos �m sin��hw. Since ��hw is small, we may approximate sin��hw by ��hw in radiansand cos��hw by unity. Thus d sin �m + d��hw cos �m = m� + (�=N). Use the conditiond sin �m = m� to obtain d��hw cos �m = �=N and

��hw =�

Nd cos �m:

63E

Let R = �=�� = Nm and solve for N :

N =�

m��=

(589:6 nm + 589:0 nm)=2

2(589:6 nm� 589:0 nm)= 491 :

64E

(a) Solve �� from R = �=�� = Nm:

�� =�

Nm=

500 nm

(600=mm)(5:0mm)(3)= 0:056 nm56 pm :

(b) Since sin � = mmax�=d < 1,

mmax <d

�=

1

(600=mm)(500� 10�6mm)= 3:3 ;

thus mmax = 3. No higher orders of maxima can be seen.

65E

If a grating just resolves two wavelengths whose mean is � and whose separation is ��then its resolving power is de�ned by R = �=��. The text shows this is Nm, where N isthe number of rulings in the grating and m is the order of the lines. Thus �=�� = Nm

and

N =�

m��=

656:3 nm

(1)(0:18 nm)= 3650 rulings :

66E

(a) From R = �=�� = Nm we �nd

N =�

m��=

(415:496 nm + 415:487 nm)=2

2(415:96 nm� 415:487 nm)= 23; 100 :


(b) The maxima are found at

� = sin�1�m�

d

�= sin�1

�(2)(415:5 nm)

4:0� 107 nm=23; 100

�= 28:7� :

67E

(a) From d sin � = m� we �nd

d =m�

sin �=

3(589:3 nm)

sin 10�= 1:0� 104 nm = 10�m :

(b) The total width of the ruling is

L = Nd =Rd

m=

Nd

m��

=(589:3 nm)(10�m)

3(589:59 nm� 589:00 nm)= 3:3� 103�m = 3:3mm :

68E

The dispersion of a grating is given by D = d�=d�, where � is the angular position of aline associated with wavelength �. The angular position and wavelength are related byd sin � = m�, where d is the slit separation and m is an integer. Di�erentiate this withrespect to � to obtain (d�=d�) d cos � = m or

D =d�

d�=

m

d cos �:

Now m = (d=�) sin �, so

D =d sin �

d� cos �=

tan �

�:

The trigonometric identity tan � = sin �= cos � was used.

69E

(a) For the �rst order maxima � = d sin �, which gives

� = sin�1��

d

�= sin�1

�589 nm

76� 106 nm=40; 000

�= 18� ;

so from 68E D = tan �=� = tan 18�=589 nm = 0:032�=nm: Similarly for m = 2 and m = 3we have � = 38� and 68�, and the corresponding values of dispersion are 0:076�=nm and0:24�=nm, respectively.


(b) R = mN = 40000m = 40; 000 for (m = 1); 80; 000 (for m = 2); and 120; 000 (form = 3).

70P

(a) We require that sin � = m�1;2=d � sin 30�, where m = 1; 2 and �1 = 500 nm. This gives

d �2�s

sin 30�=

2(600 nm)

sin 30�= 2400 nm :

For a grating of given totla width L we have N = L=d / d�1, so we need to minimize d tomaximize R = mN / d�1. Thus we choose d = 2400 nm.(b) Let the third-order maximum for �2 = 600 nm be the �rst minimum for the single-slitdi�raction pro�le. This requires that d sin � = 3�2 = a sin �, or a = d=3 = 2400 nm=3 =800 nm:(c) Let sin � = mmax�2=d � 1 to obtain

mmax �d

�2=

2400 nm

800 nm= 3 :

Since the third order is missing the only maxima present are the ones with m = 0; 1 and2.

71P

(a) From the expression for the half-width ��hw (given by Eq. 37-28) and that for theresolving power R (given by Eq. 37-32), we �nd the product of ��hw and R to be

��hwR =

��

Nd cos �

�Nm =

m�

d cos �=d sin �

d cos �= tan � ;

where we used m� = d sin � (see Eq. 37-25).(b) For �rst order m = 1, so the corresponding angle �1 satis�es d sin �1 = m� = �. Thusthe product in question is given by

tan �1 =sin �1

cos �1=

sin �1p1� sin2 �1

=1p

(1= sin �1)2 � 1

=1p

(d=�)2 � 1=

1p(900 nm=600 nm)2 � 1

= 0:89 :

72P

(a) Since the resolving power of a grating is given by R = �=�� and by Nm, the range ofwavelengths that can just be resolved in order m is �� = �=Nm. Here N is the number


of rulings in the grating. The frequency f is related to the wavelength by f� = c, wherec is the speed of light. This means f �� + ��f = 0, so �� = �(�=f)�f = �(�2=c)�f ,where f = c=� was used. The negative sign means simply that an increase in frequencycorresponds to a decrease in wavelength. If we interpret �f as the range of frequenciesthat can be resolved we may take it to be positive. Then

�2

c�f =

�

Nm

and

�f =c

Nm�:

(b) The di�erence in travel time for waves traveling along the two extreme rays is �t =�L=c, where �L is the di�erence in path length. The waves originate at slits that areseparated by (N � 1)d, where d is the slit separation and N is the number of slits, so thepath di�erence is �L = (N � 1)d sin � and the time di�erence is

�t =(N � 1)d sin �

c:

If N is large this may be approximated by �t = (Nd=c) sin �. The lens does not a�ect thetravel time.

(c) Substitute the expressions you derived for �t and �f to obtain

�f �t =� c

Nm�

� �Nd sin �c

�=d sin �

m�= 1 :

The condition d sin � = m� for a di�raction line was used to obtain the last result.

73E

Bragg's law gives the condition for a di�raction maximum:

2d sin � = m� ;

where d is the spacing of the crystal planes and � is the wavelength. The angle � is measuredfrom the normal to the planes. For a second order re ection m = 2, so

d =m�

2 sin �=

2(0:12� 10�9m)

2 sin 28�= 2:6� 10�10m = 0:26 nm :

74E

Use Eq. 37-34. For smallest value of � let m = 1:

�min = sin�1�m�

2d

�= sin�1

�(1)(30 pm)

2(0:30� 103 pm)

�= 2:9� :


75E

For �rst order re ection 2d sin �1 = � and for the second order one 2d sin �2 = 2�. Solvefor �2:

�2 = sin�1(2 sin �1) = sin�1(2 sin 3:4�) = 6:8� :

76E

Use Eq. 37-34. From the peak on the left at angle �1 2d sin �1 = �1, or �1 = 2d sin �1 =2(0:94 nm) sin(0:75�) = 0:025 nm = 25pm. From the next peak �2 = 2d sin �2 = 2(0:94 nm)sin(1:15�) = 0:038 nm = 38pm. You can check that the third peak from left is just thesecond-order one for �1.

77E

For the �rst beam 2d sin �1 = �A and for the second one 2d sin �2 = 3�B . The values of dand �A can then be solved.(a)

d =3�B

2 sin �2=

3(97 pm)

2 sin 60�= 1:7� 102 pm :

(b)

�A = 2d sin �1 = 2(1:7� 102 pm)(sin 23�) = 1:3� 102 pm :

78E

� = 2d sin � = 2(39:8 pm)(sin 30:0�) = 39:8 pm :

79P

There are two unknowns, the X-ray wavelength � and the plane separation d, so data forscattering at two angles from the same planes should su�ce. The observations obey Bragg'slaw, so

2d sin �1 = m1�

and2d sin �2 = m2� :

However, these cannot be solved for the unknowns. For example, use �rst equation toeliminate � from the second. You obtain m2 sin �1 = m1 sin �2, an equation that does notcontain either of the unknowns.

80P

The wavelengths satisfy m� = 2d sin � = 2(275 pm)(sin 45�) = 389 pm. In the range ofwavelengths given, the allowed values of m are m = 3; 4, with corresponding wavelengthsbeing 389 pm=3 = 130 pm and 389 pm=4 = 97:2 pm, respectively.

(i) (ii)

(iii ) (iv) (v)

na0

ma0

θ

φ


81P

The angle of incidence on the re ection planes is � = 63:8� � 45:0� = 18:8�, and theplane-plane separation is d = a0=

p2. Thus from 2d sin � = � we get

a0 =p2d =

p2�

2 sin �=

0:260 nmp2 sin 18:8�

= 0:570 nm :

82P

(a) The sets of planes with the next �ve smallerinterplanar spacings (after a0) are shown in the di-agram to the right. In terms of a0 the spacings are:

(i): a0=p2 = 0:7071a0 ;

(ii): a0=p5 = 0:4472a0 ;

(iii): a0=p10 = 0:3162a0 ;

(iv): a0=p13 = 0:2774a0 ;

(v): a0=p17 = 0:2425a0 :

(b) Since any crystal plane passes through latticepoints its slope can be written as the ratio oftwo integers. Consider a set planes with slopem=n, as shown in the diagram to the right. The�rst and last planes shown pass through adjacentlattice points along a horizontal line and thereare m� 1 planes between. If h is the separationof the �rst and last planes, then the interplanarspacing is d = h=m. If the planes make the angle� with the horizontal, then the normal to planes(shown dotted) makes the angle � = 90� � �.The distance h is given by h = a0 cos� and theinterplanar spacing is d = h=m = (a0=m) cos�.Since tan � = m=n, tan� = n=m and cos� =

1=p1 + tan2 � = m=

pn2 +m2. Thus

d =h

m=a0 cos�

m=

a0pn2 +m2

:


83P

The angles of incidence which correspond to intensity maxima in re ected beam of lightsatisfy 2d sin � = m�, or

sin � =m�

2d=m(0:125 nm)

2(0:252 nm)=

m

4:032:

Since j sin �j < 1 the allowed values form arem = 1; 2; 3; 4: Correspondingly the values of �are � = 14:4�; 29:7�; 48:1�; and 82:8�. Therefore the crystal should be rotated counterclock-wise by 48:1�� 45:0� = 3:1� or 82:8�� 45:0� = 37:8�, or clockwise by 45:0�� 14:4� = 30:6�

or 45:0� � 29:7� = 15:3�:

84

� = 0:143 rad, I=Im = 4:72� 10�2;� = 0:247 rad, I=Im = 1:65� 10�2;� = 0:353 rad, I=Im = 8:35� 10�3

chapter 37 diffraction chapter 37 - links to …hep.physics.indiana.edu/~rickv/p222/ch37.pdf ·...

Documents