Thông tin tài liệu
36.1:
m.105.06
m2.00
m)10(7.50m)10(1.35
λ
λ
7
43
1
1
x
ay
a
x
y
36.2:
m.103.21
m1010.2
m)10(5.46m)(0.600
λλ
5
3
7
1
1
y
x
a
a
x
y
36.3:
The angle to the first dark fringe is simply:
θ
arctan
a
λ
= arctan
.0.15
m100.24
m10633
3
9
36.4:
m.105.91
m
10
7.50
m)10(6.33m)2(3.50λ2
2
3
4
7
1
a
x
yD
36.5: The angle to the first minimum is
= arcsin
a
λ
= arcsin
.48.6
cm12.00
cm9.00
So the distance from the central maximum to the first minimum is just
tan
1
xy
cm.45.4)(48.6tancm)(40.0
36.6:
a) According to Eq. 36.2
a
a
m
a
m
θ
λλ
1)0.90(sin
λ
)(sin
Thus,
mm.105.80nm580λ
.
4
a
b) According to Eq. 36.7
.128.0
)(sin
)(sinsin
)(sin
)(sinsin
2
4
4
2
0
λa
λa
I
I
36.7: The diffraction minima are located by
,2,1,sin
maλm
m1.00m;0.2752Hz)(1250)sm(344λ afv
;6.55,3;4.33,2;0.16,1
θmθmm
no solution for
larger
m
36.8: a)
)sin(
max
tωkxEE
Hz105.73
m
10
5.24
sm103.0
λ
λ
m105.24
m1020.1
22
λ
λ
2
14
7
8
7
17
c
fcf
k
k
b)
λsin
a
m101.09
28.6sin
m105.24
sin
λ
6
7
a
c)
,.3,2,1(λsin
mma
)
74
22sin
2
m101.09
m105.24
λ
2
6
7
D
36.9:
aθ λsin
locates the first minimum
m920.0)38.42(sinm)10620(sinλ
38.42andcm)(40.0cm)5.36(tan,tan
9
μθa
xyθθxy
36.10: a)
m.104.17
m103.00
m)10(5.00m)(2.50
λλ
4
3
7
1
1
y
x
a
a
x
y
b)
cm.4.2m104.17
m103.00
m)10(5.00m)(2.50
λ
2
3
5
1
y
x
a
c)
m.104.17
m103.00
m)10(5.00m)(2.50
λ
7
3
10
1
y
x
a
36.11: a)
m.105.43
m
10
3.50
m)10(6.33m)(3.00λ
3
4
7
1
a
x
y
So the width of the
brightest fringe is twice this distance to the first minimum, 0.0109 m.
b) The next dark fringe is at
m0.0109
m
10
3.50
m)106.33(m)2(3.00λ2
4
7
2
a
x
y
.
So the width of the first bright fringe on the side of the central maximum is the distance
from
,yto
12
y
which is
m1043.5
3
.
36.12:
.)m1520(
m)(3.00m)1020.6(
m)1005.4(2
λ
2
sin
λ
2
1
7
4
yy
π
x
yaa
β
a)
.760.0
2
)m1000.1()m1520(
2
:m1000.1
31
3
β
y
0
2
0
2
0
822.0
760.0
)760.0(sin
2
)2(sin
II
β
β
II
b)
.28.2
2
)m1000.3()m1520(
2
:m1000.3
31
3
β
y
.111.0
28.2
)28.2(sin
2
)2(sin
0
2
0
2
0
II
β
β
II
c)
.80.3
2
)m1000.5()m1520(
2
:m1000.5
31
3
β
y
.0259.0
80.3
)80.3(sin
2
)2(sin
0
2
0
2
0
II
β
β
II
36.13: a)
m.10.756
m
10
2.40
m)10(5.40m)(3.00λ
3
4
7
1
a
x
y
b)
.
2
λ
λ
22
λ
2
sin
λ
2
1
π
ax
xπa
x
yπaa
β
.mW1043.2
2
)2(sin
)mW1000.6(
2
)2(sin
2
6
2
26
2
0
π
π
β
β
II
36.14: a)
.00sin
2
:0
o
λ
a
b) At the second minimum from the center
.4
λ2
λ
2
sin
λ
2
π
a
πa
θ
πa
β
c)
1910.7sin
m
10
00
.
6
m)1050.1(2
sin
λ
2
7
4
π
θ
πa
β
rad.
36.15:
m.1036.524.0sin
2
m)1020.3(2
sin
2
λsin
λ
2
6
4
aa
36.16: The total intensity is given by drawing an arc of a circle that has length
0
E
and
finding the length of the cord which connects the starting and ending points of the curve.
So graphically we can find the electric field at a point by examining the geometry as
shown below for three cases.
a)
.
2
λ
λ
2
sin
λ
2
a
aa
From the diagram,
.
2
2
00
EEE
E
p
p
So the intensity is just:
.
42
2
0
0
2
I
II
This agrees with Eq. (36.5).
b)
.2
λ
λ
2
sin
λ
2
a
aa
From the diagram, it is clear that the total amplitude
is zero, as is the intensity. This also agrees with Eq. (36.5).
c)
.3
2
λ3
λ
2
sin
λ
2
a
aa
From the diagram,
.
3
2
2
3
00
EEE
E
p
p
So
the intensity is just:
.
9
4
3
2
0
2
0
2
III
This agrees with Eq. (36.5).
36.17: a)
25.3sin
rad0.56
m)1005.1(2
sin
2
λsin
λ
2
4
aa
.m1068.6
7
b)
.)1036.9(
20.56
)20.56(sin
2
)2sin(
0
5
2
0
2
0
IIII
36.18:
a) Ignoring diffraction, the first five maxima will occur as given by:
.5,4,3,2,1for,
4
λ
arcsin
λ
arcsinλsin
m
a
m
d
m
md
b)
.2
λ
λ
2
sin
λ
2
and,
2
λ
λ
2
sin
λ
2
m
d
mddm
d
maa
So including diffraction, the intensity:
.
4/
)4/(sin
4/
)4/(sin
2
2
cos
2
)2sin(
2
cos
2
0
2
2
0
2
2
0
m
m
I
m
mm
III
So for
;405.0
2
)2(sin
:2;811.0
4
)4(sin
:1
00
2
200
2
1
IIImIIIm
0
)(sin
:4;0901.0
43
)43(sin
:3
0
2
400
2
3
IImIIIm
.0324.0
45
)45(sin
:5
00
2
5
IIIm
36.19: a)
,3If
a
d
then there are five fringes:
.2,1,0
m
b) The
6
m
interference fringe coincides with the second diffraction minimum, so
there are two fringes
)5,4(
mm
within the first diffraction maximum on one side
of the central maximum.
36.20:
By examining the diagram, we see that every fourth slit cancels each other.
36.21:
a) If the slits are very narrow, then the first maximum is at
.1sin
λ
1
d
.0627.0
m1030.5
m1080.5
arcsin
λ
arcsin
4
7
1
d
Also, the second maximum is at
2sin
λ
2
d
.125.0
m105.30
m)102(5.80
arcsin
λ2
arcsin
4
7
2
d
b)
,1
2
cosbut
2
2)sin(
2
cos
2
2
0
II
since we are at the 2 slit maximum. So
2
0
2
1
1
01
)sin(
λsin
)
λsinsin(
da
da
I
θa
θa
II
.249.0
m)10(5.30m)1020.3(
m))10(5.30m)1020.3(sin(
0
2
44
44
01
III
And
.0256.0
m)1030.5(m)1020.3(2
))m10(5.30m)10(3.20(2sin
2
)sin(2
λsin
)
λsinsin(
0
2
44
44
02
2
0
2
2
2
02
III
da
da
I
a
a
II
36.22: We will use
2
2
0
2
2)sin(
2
cos
II
, and must calculate the phases
and
.
Using
,sin
λ
2
and,sin
λ
2
da
we have:
a)
.757.0)1.01(57.9and),0.177(10.12:rad1025.1
0
4
II
b)
.268.0)2.03(116.3and),0.355(20.32:rad1050.2
0
4
II
c)
.)114.0()2.43(139.2and),0.426(24.12:rad1000.3
0
4
II
36.23:
With four slits there must be four vectors in each phasor diagram, with the
orientation of each successive one determined by the relative phase shifts. So:
We see that destructive interference occurs from adjacent slits in case (ii) and from
alternate slits in cases (i) and (iii).
36.24: Diffraction dark fringes occur for
,
λ
sin
a
m
d
and interference maxima occur
for
d
m
i
λ
sin
. Setting them equal to each other yields a missing bright spot whenever
the destructive interference matches the bright spots. That is:
.3
λλ
ddi
di
mm
a
d
m
a
m
d
m
That is, the missing parts of the pattern occur for
integers.for,3 9,6,3 mmm
i
36.25:
a) Interference maxima: Diffraction minima:
mλθd
i
sin
and
.λsin nθa
d
If the
thm
interference maximum corresponds to the
thn
diffraction minimum then
.
di
θθ
or
n
m
a
d
so
mm.0.280mm)840.0(
3
1
d
m
n
a
b) The diffraction minima will squelch the interference maxima for all
3
n
m
up to
the highest seen order. For
nm,630λ
the largest value of
m
will be when
.90
θ
.1333
m1030.6
m1040.8
λ
7
4
max
d
m
.444
m)1030.6(3
m1040.8
λ
7
4
max
a
n
So after
444.,.,.3,2,1for1332.,.,.9,6,3
nmm
will also be missing.
c) By changing
λ
we only change the highest order seen.
.666
3
2000
λ
.2000
m1020.4
m1040.8
λ
max
7
4
max
a
n
d
m
So
.666.,.,.3,2,1for1999.,.,.9,6,3
nm
36.26:
The third bright band is missing because the first order single slit minimum
occurs at the same angle as the third order double slit maximum.
91
.
1
tan
cm90
cm3
θ
θ
Single-slit dark spot:
λsin
θa
nm(width)1050.1
91.1sin
nm500
sin
λ
4
θ
a
Double-slit bright fringe:
λ3sin
d
ion)nm(separat1050.4
91.1sin
nm)500(3
sin
λ3
4
θ
d
36.27: a) Find
λsin: mθdd
cm10086.2sinλso,nm681λand3for4.78
4
mdm
The number of slits per cm is
cmslits47901 d
b) 1st order:
1.19and)m10086.2()m10681(λsinso,1
69
dm
2nd order:
40.8and/2λsinso,2 θdm
c) For
dm /4sin,4
is greater than 1.00, so there is no 4th-order bright band.
36.28: First-order:
λsin
1
θd
Fourth-order:
λ4sin
1
θd
4.38
94.8sin4sin4sin
λ
λ
4
sin
sin
4
14
1
4
θ
θθ
θ
d
θd
36.29: a)
m10111.1cm
5
900
1
d
For
.106.3λnm,700λ
4
d
The first-order lines are located at
θd sin;λsin
is small enough for
θθ
sin
to be an excellent approximation.
b)
m.2.50where,λ xdxy
The distance on the screen between 1st order bright bands for two different
wavelengths is
xyddyxy )(λso,)(
nm13.3m)(2.50m)103.00()m10111.1(
35
36.30:
a)
slits.1820
m)106.5627m102(6.5645
m105645.6
λ
λ
λ
λ
77
7
m
NNmR
b)
d
m
d
λ
sin)mslits000,500()mmslits500(
111
.
0137
.
0
41.0160500,000)m)105627.6()2((sin
41.0297500,000)m)105645.6()2((sin
71
2
71
1
θ
36.31:
))396.0((arcsin
m101.60
m)10m(6.328
arcsin
λ
arcsin
6
7
m
d
m
.3.52:2;3.23:1
1
mm
All other m-values lead to angles greater than
.90
o
36.32:
m.1000.2
m1000.5
1
cmslits5000
6
15
d
a)
m.1067.4
1
13.5sin m)1000.2(sin
λλsin
7
6
m
d
md
b)
.8.27
m102.00
m)1067.4(2
arcsinarcsin:2
6
7
o
d
m
λ
m
36.33:
:Thenm.1086.2
m1050.3
1
mmslits350
6
15
d
.1.33:3;3.21:2;5.10:1
))182.0((arcsin
m102.86
m)1020.5(
arcsin
λ
arcsinλsin
6
7
mmm
m
m
d
m
md
[...]... 1.22(0. 036 m) sin 1 D W 2.8 10 4 m D 1.88 m 36. 41: sin 1 1.22 36. 42: sin 1 1.22 D sin 1 D1 (8.00 10 6 m)(1.00 10 8 ) λ λ D 1.22 1.22 1.22 λ 0.0656 m 6.56 cm 6.20 10 7 m λ 1.22 0.102 The screen is 4.5 m away, so the 7.4 10 6 m D diameter of the Airy ring is given by trigonometry: D 2 y 2 x tan θ 2 x sin θ 2(4.5 m)(0.102) 91.8 cm 36. 43: sin 1 1.22 36. 44:... : 1 11.2; m 2 : 22.9; so, a) 1 4.00, b) 2 8.77 36. 36: 587.8002 nm λ 587.8002 Nm N λ mλ (587.9782 nm 587.8002 nm) 0.178 N 3302 slits 3302 slits N 2752 1.20 cm 1.20 cm cm 36. 37: For x-ray diffraction, 2d sin mλ d 2(8.50 10 11 m) mλ d 2.32 10 10 m 2 sin 2 sin 21.5 36. 38: For the first order maximum in Bragg reflection: 2d sin 2(4.40... width a is increased ii) 36. 52: If the apparatus of Exercise 36. 4 is placed in water, then all that changes is the λ 2 xλ 2 xλ D 5.91 10 3 m wavelength λ λ So : D 2 y1 n a an n 1.33 4.44 10 3 m 4.44 mm 36. 53: sin θ λ a locates the first dark band λ liquid λ sin air air ; sin liquid a a sin liquid λ liquid λ air sin 0.4 836 air λ λ air n... s 2 fy 2(0.180 m)(8.00 10 3 m) D 2 | y | y 1.15 10 4 m 0.115 mm s 25.0 m s 36. 45: sin 1 1.22 1.22 λ λ R D 1.22λ D sin 1 W 5.93 1011 m 1.22(5.0 10 m) 1.45 m 2.50 10 5 m 7 36. 46: sin 1 1.22 (4.00 10 3 m)(0.0720 m) λ y yD s 429 m D s 1.22(5.50 10 7 m) 1.22λ 36. 47: Let y be the separation between the two points being resolved and let s be their λ y distance... adjacent slits means that the total amplitude is 6E , and the intensity is 36I b) If the phase difference is 2 , then we have the same phasor diagram as above, and equal amplitude, 6E , and intensity, 36I c) There is an interference minimum whenever the phasor diagrams close on themselves, such as in the five cases below 36. 61: a) For the maxima to occur for N slits, the sum of all the phase differences... m 1 36. 39: The best resolution is 0.3 arcseconds, which is about 8.33 10 5 1.22λ 1.22(5.5 10 7 m) a) D 0.46 m 0.5 m sin 1 sin(8.33 10 5) b) The Keck telescopes are able to gather more light than the Hale telescope, and hence they can detect fainter objects However, their larger size does not allow them to have greater resolutionatmospheric conditions limit the resolution 36. 40:... locates the first dark band λ liquid λ sin air air ; sin liquid a a sin liquid λ liquid λ air sin 0.4 836 air λ λ air n (Eq.33.5), so n λ air λ liquid 1 0.4 836 2.07 1 36. 54: For bright spots, N sin λ 1 Red: N sin θR 700 nm 1 Violet: N sin θV 400 nm sin R 7 sin V 4 R V 15 R V 15 sin(V 15) 7 sin V 4 sin θV cos 15 cos θ... (1.20 m)(5.40 10 7 m) 1.80 10 3 m 4 3.60 10 m a 1 sin(a sin λ ) a sin θ b) 1.39 a sin λ λ 2 36. 55: a) y1 (1.39)(5.40 10 7 m) sin 6.64 10 4 4 (3.60 10 m) y x tan x sin (1.20 m)(6.64 10 4 ) 7.97 10 4 m 0.797 mm 2 sin γ 36. 56: a) I I 0 γ The maximum intensity occurs when the derivative of the intensity function with respect... be N 2 Therefore total phase shifts of these minima are 2m N Hence the angle at which they are found, and the angular width, will be: 36. 57: The phase shift for adjacent slits is λ 2 mλ λ 2λ θ 2m 2d N d dN dN 2 2 2 36. 58: a) E p E p x E p y So, from the diagram at right, we have: 2 Ep E 2 0 (1 cos cos 2 ) 2 (sin sin 2 ) 2 (2 cos 2 ... in Section 35.5 d) Below are phasor diagrams for specific phase shifts 36. 59: a) For eight slits, the phasor diagrams must have eight vectors: 3 5 7 , , and , totally destructive interference occurs between 4 4 4 3 , totally destructive interference occurs with every second slit slits four apart For 2 b) For 36. 60: For six slits, the phasor diagrams must have six vectors a) Zero .
tan
1
xy
cm.45.4)(48.6tancm)(40.0
36. 6:
a) According to Eq. 36. 2
a
a
m
a
m
θ
λλ
1)0.90(sin
λ
)(sin
Thus,
mm.105.80nm580λ
.
4
a
b) According to Eq. 36. 7
.128.0
)(sin
)(sinsin
)(sin
)(sinsin
2
4
4
2
0
λa
λa
I
I
36. 7:.
74
22sin
2
m101.09
m105.24
λ
2
6
7
D
36. 9:
aθ λsin
locates the first minimum
m920.0)38.42(sinm)10620(sinλ
38.42andcm)(40.0cm)5 .36( tan,tan
9
μθa
xyθθxy
36. 10:
Ngày đăng: 17/01/2014, 04:20
Xem thêm: Tài liệu Physics exercises_solution: Chapter 36 ppt
