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浙江大学大学物理甲下 chapter 35
B B0 BM
13
Chapter 35 Magnetic Properties of Materials a long solenoid of circular cross section with a magnetic material BM 0 M B B0 0 M 1 B m B0 (E E0 ) e m is permeability of the material. 0 M ( m 1) B0
U s B
Nobel prize in physics 1952:
Bloch & Purcell for their development of new methods for nuclear magnetic precision methods and discoveries of connection therewith. 10
B dt dL
Δμm is antiparallel with B.
Δμm Δμm
19
Chapter 35 Magnetic Properties of Materials 3. Ferromagnetism The atoms of this material have permanent magnetic dipole moments, and have a strong interaction between neighboring atoms that keeps their dipole moments aligned even when the external magnetic field is removed. κm>>1, B >> B0. κm is not a constant.
浙江大学大学物理甲下 exercise 4
hf h c
hcR (
1 k
2
1 n
2
)
(4) Bohr’s model of the hydrogen atom
E n E n hf
i f
L m vr n
h 2
n
The orbite of Hydrogen atom
1 4
0
e r
2 2
m
v r
2
m v r mr
Solution:
P (x) 2 a sin
2 2
f 1 a
Pdx a
a
2
a 4
sin
2
x
a
dx
0
a
4
(1 cos
2 x a
)d x
a 4 0
0
x
1 a a 2 x ( sin a 4 2 a
)
1 4
1 2
0 . 091
3. Hydrogen atom (1) The Ground State of the Hydrogen Atom
pe h (
0
2 2
)
6.63 10
34
J s(
2.24 10
2
2 2 2
m 2 10
4 4
2
2
2 2
m
2
4.48 10
m
4
)
12
4.44 10
23
kg m s
2. The waves of matter
h p h mv h 2 mK
hcR (
1 k
2
1 n
2
)
(4) Bohr’s model of the hydrogen atom
E n E n hf
i f
L m vr n
h 2
n
The orbite of Hydrogen atom
1 4
0
e r
2 2
m
v r
2
m v r mr
Solution:
P (x) 2 a sin
2 2
f 1 a
Pdx a
a
2
a 4
sin
2
x
a
dx
0
a
4
(1 cos
2 x a
)d x
a 4 0
0
x
1 a a 2 x ( sin a 4 2 a
)
1 4
1 2
0 . 091
3. Hydrogen atom (1) The Ground State of the Hydrogen Atom
pe h (
0
2 2
)
6.63 10
34
J s(
2.24 10
2
2 2 2
m 2 10
4 4
2
2
2 2
m
2
4.48 10
m
4
)
12
4.44 10
23
kg m s
2. The waves of matter
h p h mv h 2 mK
浙江大学大学物理甲下 chapter 30
20
U
q
Chapter 30 Capacitance
Question: Why do we say that the energy is stored in the electric field between the capacitor plates? Take the parallel-plate capacitor as an example.
14
Chapter 30 Capacitance
30-4 Capacitors in Series and Parallel 1. Capacitors connected in Parallel (并联):
C1 a C2
Question: If we identify the above capacitors connected in parallel as a single capacitor,
In general, for capacitors in parallel combination,
Ceq i Ci
2. Capacitors connected in Series (串联): V2 V1 V
a
q q q q C1 C2
b
a
q q Ceq
b
q Ceq V V1 V2 q C1 q C2
Example:
What is the capacitance of the Earth, viewed as an isolated conducting sphere of radius R=6370km?
C 4 0 R 4 3.14 (8.85 10 12 F/m) 6 (6.37 10 m) 7.1 10 4 F 710 μF
U
q
Chapter 30 Capacitance
Question: Why do we say that the energy is stored in the electric field between the capacitor plates? Take the parallel-plate capacitor as an example.
14
Chapter 30 Capacitance
30-4 Capacitors in Series and Parallel 1. Capacitors connected in Parallel (并联):
C1 a C2
Question: If we identify the above capacitors connected in parallel as a single capacitor,
In general, for capacitors in parallel combination,
Ceq i Ci
2. Capacitors connected in Series (串联): V2 V1 V
a
q q q q C1 C2
b
a
q q Ceq
b
q Ceq V V1 V2 q C1 q C2
Example:
What is the capacitance of the Earth, viewed as an isolated conducting sphere of radius R=6370km?
C 4 0 R 4 3.14 (8.85 10 12 F/m) 6 (6.37 10 m) 7.1 10 4 F 710 μF
浙江大学大学物理甲下 chapter 42
Bright fringe at the center m 1,2,3... min ima m 1,2,3... max ima
Secondary maxima
9
N : odd
2019/3/12
Chapter 42
Diffraction
f
Diffraction pattern of single-slit
x0 2 f 1 2 f
Chapter 42
Diffraction
Quick quiz-1 If a classroom door is open slightly, you can hear sounds coming from the hallway. Yet you cannot see what is happening in the hallway. Why is there this difference? (1) light waves do not diffract through the single slit of the open doorway. (2) Sound waves can pass through the walls, but light waves cannot. (3) The open door is a small slit for sound waves, but a large slit for light waves. (4) The open door is a large slit for sound waves, but a small slit for light waves.
Diffraction at a cnterference and diffraction
浙江大学大学物理甲下 chapter 29
Suppose we place a rectangular slab of a conductor in a uniform electric field E0,
6
Chapter 29 The Electrical Properties of Materials The properties of conductor in electrostatic equilibrium 1.The electric field is zero at any point inside a conductor in electrostatic equilibrium. The conductor becomes an equipotential, and its surfaces become equipotential surfaces.
9
Chapter 29 The Electrical Properties of Materials Two new concepts: Electric circuit: the continuous loop of flowing electrons. Electric current: the flow of electrons. The direction of the current i is opposite to the motion of the electrons.
5
Chapter 29 The Electrical Properties of Materials 29-2 A Conductor in an Electric Field ( Static Conditions )
The content in this section is only a review.
6
Chapter 29 The Electrical Properties of Materials The properties of conductor in electrostatic equilibrium 1.The electric field is zero at any point inside a conductor in electrostatic equilibrium. The conductor becomes an equipotential, and its surfaces become equipotential surfaces.
9
Chapter 29 The Electrical Properties of Materials Two new concepts: Electric circuit: the continuous loop of flowing electrons. Electric current: the flow of electrons. The direction of the current i is opposite to the motion of the electrons.
5
Chapter 29 The Electrical Properties of Materials 29-2 A Conductor in an Electric Field ( Static Conditions )
The content in this section is only a review.
浙江大学大学物理甲下 chapter 34
When B is increasing,…
10
Chapter 34 Faraday’s Law of Induction
Sample problem 34-2: r, R (resistance) are known, Bz=-4.0-5.6t+2.2t2. Find induced current in the loop at t=1s and t=2s. Solution: Choose dA into the page, t=1s, Bz=-7.4T; t=2s, Bz=-6.4T, thus in the time interval, Bz<0. z
t 1
“-” means the current is dBz Az 3.9V counterclockwise. dt dBz 5.6 4.4t t 2 3.2T /s, dt t 2
dBz Az 1.03V dt
“+” means the current is clockwise.
12
Chapter 34 Faraday’s Law of Induction
Quick Quiz-1 The Figure shows a graphical representation of the field magnitude versus time for a magnetic field that passes through a fixed loop and is oriented perpendicular to the plane of the loop. The magnitude of the magnetic field at any time is uniform over the area of the loop. Which the magnitude of the emf generated in the loop at the five instants indicated is largest?
浙江大学大学物理下总复习word精品文档7页
【电磁学】一、电势：1.静电场是保守场、电场力作功与路径无关，仅与起始、终了位置有关0d =⋅⎰Ll E ϖϖ 环流定律移动电荷q 时电场力做的功为 b a b a ab ab W W U U q W A -=-=∆-=)( 2.电势的定义：⎰⋅=零点pp l E U ϖϖd通常取无限远处（或地）为电势零点U =0 电势差（电压） ⎰⋅=-ba b a l E U U ϖϖd电场强度与电势的关系：重点掌握已知U (x,y,z )求电场强度，注意公式中的负号！ 3. 电势叠加原理：∑=i U U ，记住下面的常用公式： 在真空中：点电荷的电势分布为rq U 04πε=连续带电体rq U 04d d πε=或⎰=rq U 04d πε均匀带电球面或金属带电球体：⎪⎪⎪⎩⎪⎪⎪⎨⎧>=≤=)( 4 )( 400R r r q U R r R q U πεπε球面内是等势体★ . 求电场中任一点的电势可以用电势叠加的方法，也可以用先求电场强度分布，再从定义来分段积分 ★ . 求解电荷非对称分布电场中的电势时，一定用叠加原理，即⎰=r qU 04d πε★ . 有导体存在时，必须先求感应电荷的分布再求电势分布；求感应电荷时必须以对称中心的电势为参考点。
二. 导体和电介质：要先分清是导体还是电介质，如是导体必须判断是否带电或接地等 （1）. 导体：在电场中的导体一定处于静电平衡状态（静电场）计算有导体存在时的电场强度U E ,ϖ分布时★ . 注意导体表面的电荷重新分布，导体接地时是 U =0，两导体相连时是U 1=U 2. ★ . 注意导体附近有点电荷存在时，求感应电荷的方法是以对称中心的电势为参考，叠加各部分电势，通过电势关系求出感应电荷。
（2）. 电介质：在电场中电介质处于极化状态，对各向同性的均匀电介质而言，有：电介质中的高斯定理包围的自由电荷）( d ∑⎰=⋅q S D Sϖϖ，灵活使用补偿或叠加原理。
【精选】浙江大学物理化学(甲)第二章(2)9
H
V
T
U V
T
0
H
p
T
U p
T
0
H H (T )
所以理想气体，单纯pVT变化时，不管恒压与否
H Qp
T2 T1
C
p
dT
T2 T1
nCp,mdT
11
∴ 理想气体的热力学能和焓仅是温度的函数， 而与p、V无关。
Q=0 (绝热) U W (第一定律)
在左侧压力p1下气体膨胀功为：
W1 p1V p1V1 (V =0V1 V1)
气体通过小孔膨胀，功为：
W2 p2V p2V2 (V =V2 0 V2)
24
∴ W W1 W2 p1V1 p2V2
即
U2 U1 p1V1 p2V2
气态物质
(
Vm T
)
p
0
，Cp>CV
理想气体 Cp CV nR
7
2.The Joule experiment, U and H of perfect gas
Joule在1843年做了实验: 实验过程：如右图所示。
实验现象：没有观察到温度变化 （dT=0)
实验结果： (1) 系统（气体）与环境（水浴）
(UT
)p
(UT
)V
(UV
)T
(
V T
)p
CV
TV
6
(3) 讨论：
Cp,m
CV,m
(
Um Vm
)p
浙江大学大学物理甲下 chapter 41
Two different paths
Single source
2012-10-26
Interference possible here
7
Chapter 41
Interference
Superposition of Two Waves Two waves encountered at point P:
y ym1 ym
D
d
Does the distances on the screen between the adjacent maxima are equal all?
14
2012-10-26
Chapter 41
Interference
Young’s double-slit experiment (1801)
Chapter 41
Interference
Content of this Chapter
Two-source interference Coherence Intensity in double-slit interference
Chapterinterference 41 Interference Double-slit
relation)
6
2012-10-26
Chapter 4coherent light from separate sources. (f 1014 Hz)
• Need two waves from single source taking two different paths – One source, two slits – Reflection (thin films) – Diffraction*
Single source
2012-10-26
Interference possible here
7
Chapter 41
Interference
Superposition of Two Waves Two waves encountered at point P:
y ym1 ym
D
d
Does the distances on the screen between the adjacent maxima are equal all?
14
2012-10-26
Chapter 41
Interference
Young’s double-slit experiment (1801)
Chapter 41
Interference
Content of this Chapter
Two-source interference Coherence Intensity in double-slit interference
Chapterinterference 41 Interference Double-slit
relation)
6
2012-10-26
Chapter 4coherent light from separate sources. (f 1014 Hz)
• Need two waves from single source taking two different paths – One source, two slits – Reflection (thin films) – Diffraction*
浙江大学物理化学甲第九章原电池
浙江大学物理化学甲第九章原电池
3
如： (1)
环保型能源的开发——燃料电池
反应: H2 + O2→H2O 若通过直接燃烧→电能，则能源利用率<25%
若设计成氢氧燃料电池 能源利用率可达 72%
同时生成的水对环境没有污染。
(2) 蓄电池的开发 蓄电池作为贮能器，要求容量大，可多次充放电。
(3) 人造器官的电源的开发，如：用于人造心脏的电池 (4) 无记忆的锂电池的开发
负极： H2 (p1) 2H+(aH+) +2e 正极： Cl2(p2) + 2e 2Cl- (aCl-) 电池反应：H2 (p1)+ Cl2(p2) 2H+(aH+) + 2Cl- (aCl-)
浙江大学物理化学甲第九章原电池
19
根据化学反应等温式：
rGmrGm RTlnaaH H 2 2aaC C 222ll
得待测电池的浙电江动大学势物理：化E学x甲第九A章C原电EA池S.HC.
15
2.Weston标准电池
在测量原电池的电动势时，需要一个电动势值为已知 的标准电池，常用有Weston标准电池，它是一个高度可逆电 池，特点：
(1) 电动势相当稳定
(2) E 温度系数很小，-4×10-5V/℃，无需恒温。
从上面分析可知，电池反应为可逆：
Zn(s) + 2AgCl(s)放电 Zn2+ + 2Ag(s) + 2Cl充电
并且满足：I0， 上述电池为可逆电池。 研究可逆电池的电动势，可知最大非体积功和化学反应过程中 的热力学函数的变化。
浙江大n和Cu棒插入H2SO4溶液中构成的电池。
(2)当ΔrGm>0，这个反应是热力学上非自发反应，则E<0
浙大《大学物理》第五章
五、简谐振动的能量 势能
线性回复力是保守力
1 2 E p x kxdx kx 2
0
1 2 1 2 1 2 E p kx kA cos (t ) m 2 A 2 cos 2 (t ) 2 2 2
动能
与弹性势能有相同的形式，但是两个不同的量
1 1 1 2 2 2 2 2 E k m v m A sin (t ) kA sin 2 (t ) 2 2 2
总机械能
1 1 2 1 2 1 2 E E k E p m v kx kA m 2 A 2 2 2 2 2
例
x
o
5-2 简谐振动的动力学方程
21 2011-4-14
解
（1） 角频率
k m
0.72N m 1 6.0s 1 0.02kg
x0=0.05， v0=0
由初始条件确定常数A和
2 A x0
2 v0 2
x0 0.05m
v0 tan 0 x0
0 或 π
T 2π
取 0
A
A
)
o
o
A
x
xt图
T
t
t
v
vt
图
v A sin( t ) A cos( t
2
2
T
A
a A cos(t ) A cos(t )
2
A 2
a
a t图
o
T
t
8 2011-4-14
d 2θ mglθ J 2 dt
A c o s ( t )
浙江大学大学物理甲下 chapter 49
Is it soft or hard for a solid ? Can it be hammered into a thin sheet or drawn into a fine wire ? Is it transparent ? What kind of waves travel through it and at what speeds ? Does it conduct heat ? What are its magnetic properties ? What is its crystal structure?
Chapter 49 Electrical Conduction in Solids
According to these assumptions we have derived the Ohm’s law: 2 j nevd ne E m
ห้องสมุดไป่ตู้
1 ne m
2
v m 8kT 1
2. Quantum-mechanical free electron model (QMFE) The electrons are free to move through the solid also. Aside from collision with the ions, the electrostatic interaction between the electrons and the lattice ions is ignored, and the interaction between the electrons is also ignored. But: The electrons must be treated quantum mechanically. This will quantize the energy spectrum of the electron gas.
Chapter 49 Electrical Conduction in Solids
According to these assumptions we have derived the Ohm’s law: 2 j nevd ne E m
ห้องสมุดไป่ตู้
1 ne m
2
v m 8kT 1
2. Quantum-mechanical free electron model (QMFE) The electrons are free to move through the solid also. Aside from collision with the ions, the electrostatic interaction between the electrons and the lattice ions is ignored, and the interaction between the electrons is also ignored. But: The electrons must be treated quantum mechanically. This will quantize the energy spectrum of the electron gas.
浙江大学大学物理甲下chapter
l(n 1) 2 r2 r1 0
r2
r1
[l(n
1)
2
]
7
2
r2
r1
d D
x
x
D d
(r2
r1)
D d
7
2
2.5mm
r1
x
r2
Example 2: There is an oil film (n2=1.20) on flat grass plate (n1=1.52). A light ray (λ=600nm) illuminates normally and is observed from above by reflected light. Find the thickness of the film at 5th bright fringe from the edge of oil film.
(1) SS2 S2O l nl (SS1 S1O)
(2m 1) , m 3
SS2
2 SS1 l(n 1)
7
2
D
l(n 1) 7 l 4μm
22
(2) SS2 r2 l nl (SS1 r1) 0
d sin m, 2400 sin 30 m 400 m 3
d 3, a 800nm a
Example 9: E43-11
Solution: If the second order spectra overlaps the third order, it is because the 700nm second order line is at a larger angle than the 400nm third order line.
r2
r1
[l(n
1)
2
]
7
2
r2
r1
d D
x
x
D d
(r2
r1)
D d
7
2
2.5mm
r1
x
r2
Example 2: There is an oil film (n2=1.20) on flat grass plate (n1=1.52). A light ray (λ=600nm) illuminates normally and is observed from above by reflected light. Find the thickness of the film at 5th bright fringe from the edge of oil film.
(1) SS2 S2O l nl (SS1 S1O)
(2m 1) , m 3
SS2
2 SS1 l(n 1)
7
2
D
l(n 1) 7 l 4μm
22
(2) SS2 r2 l nl (SS1 r1) 0
d sin m, 2400 sin 30 m 400 m 3
d 3, a 800nm a
Example 9: E43-11
Solution: If the second order spectra overlaps the third order, it is because the 700nm second order line is at a larger angle than the 400nm third order line.
浙江大学大学物理甲下
A1 A2 , Imax Amin A1 A2 , I min
Φ 2 n2r2 n1r1
,
m 0,1,2...Imax m 0,1,2...Imin
(1) Double-slit interference
nor1 nor2 d sin m
n=1.4 n=1.5
Solution: for maximum , 2nd m,
at the edge of film , d 0, is bright.
for minimum, 2nd (2m 1) , m 0,1, 2
2
If the 1st minimum correspond to m=0, the center point is
2d
n22
n12sin
2i
2
,
m
m 1,2,3......max ima
(2m
1)
2
m 0,1,2......min ima
If i =0, 2nd
2
( 2d n ,
n
2
in the film)
Example 1: Two slips interference, S1S2=0.7mm, D=100cm, λ=500nm. Now the single slip is shifted and SS1-SS2= λ/2, an optical medium plates with thickness of l and index of
1)
2
]
7
2
r2
r1
d D
x
x
D d
(r2
r1)
D d
7
2
2.5mm
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m0,1,2,3,..(.maximum)
dsin (m1)
2 m0, 1, 2, 3,...(minimum)
dsindtandym
D
ymmdD m0,1,2,3,.(..maxima)
yym1ymdD
(2) Interference from thin film
2d
n22
n12sin2i
2
Exercise Lesson
Wave optics
1. Interference Phase difference and optical path difference
Φ
2
n2 r2
2
2
n1r1
1
( Let 1 2 )
2
(n2r2
n1r1)
2m , Amax (2m 1) ,
refraction of 1.5 placed at the back of one slits. The point o is 4th minimum fringe. Find (1) l, (2) the distance the central
maximum to the point o.
Solution:
(4) missing order
asinm1 dsinm2
dam m12 m2dam1 ismissing
(5) Bragg’s law
2dsin m
m 1,2,3,...
The angle θ is the angle between the incident x-ray and the plane.
Example 5: (E43-12) Assume that light is incident on a grating at an angle φ as shown. Show that the condition for a diffraction maximum is d(sin φ ±sinθ) =mλ.
Solution:
fom r ax,im 2 un2m dm
m0,1,2
At the edge of film, d=0 (m=0), is bright.
Thus the 5th bright fringe is m=4. dm1.0106m
2n2
Example 3: As shown, λ=560nm, center point is dark, there are 20 dark rings in the outer region. Find the thickness of the film at center.
,
m
m1,2,3.....m. axima
(2m1) 2 m0,1,2.....m. inima
If i =0, 2nd
(2dn, in
2
n
2
tfhilem
Example 1: Two slips interference, S1S2=0.7mm, D=100cm, λ=500nm. Now the single slip is shifted and SS1-SS2= λ/2, an optical medium plates with thickness of l and index of
(1)S2SS2Oln l(S1SS1O)
(2m1) , m3
SS2
2 SS1 l(n
1)
7 2
D
l(n1) 7 l 4μm
22
(2)S2Sr2ln l(S1Sr1)0
l(n1)2r2r10
r2
r1
[l(n
1)
2
]
Hale Waihona Puke 72r2r1
d D
x
x
D d
(r2
r1)
D d
7
2
2.5mm
r1
x
r2
Example 2: There is an oil film (n2=1.20) on flat grass plate (n1=1.52). A light ray (λ=600nm) illuminates normally and is observed from above by reflected light. Find the thickness of the film at 5th bright fringe from the edge of oil film.
A1 Amin
A2 , I max A1 A2 , I min
Φ 2 n2r2 n1r1 2
m,
n2 r2
n1r1
(2m 1)
2
,
m 0,1,2...Imax m 0,1,2...Imin
(1) Double-slit interference
nor1nor2 dsinm
a
(2) Grating diffraction
d si n mm 0 , 1 , 2 ,.P.ri.nciple maxima
(3) Resolving power
Airy spot:
R min1.22d, R1/R resolutiaobnility
Grating:
D dcmos
R mN
m=20.
2nd (4 01), d41n0m 0
2
If the 1st minimum correspond to m=1, the center point is
m=21.
2nd (2m1) , m12,
2
Example 4 (E41-40):
2d(n1)60, n1601.00030
2d
2. Diffraction (1) Single-slit diffraction
n=1.4 n=1.5
Solution: fom r axi,m 2nudm ,
at tehdegoefil,m d0,isbrig . ht
fo m r i,n 2 n i m d (2 m 1 u ),m m 0 ,1 2 , 2
If the 1st minimum correspond to m=0, the center point is
0
Brighftringaet thceenter
asin 2m2 m
m1,2,3...minima
(2m1)2
m1,2,3..m . axima
asinatanaaxmm
f
x02f12f a
w oifd tc h th e enm traax l im
xf,
d tow ho f ithfreirn ges
dsin (m1)
2 m0, 1, 2, 3,...(minimum)
dsindtandym
D
ymmdD m0,1,2,3,.(..maxima)
yym1ymdD
(2) Interference from thin film
2d
n22
n12sin2i
2
Exercise Lesson
Wave optics
1. Interference Phase difference and optical path difference
Φ
2
n2 r2
2
2
n1r1
1
( Let 1 2 )
2
(n2r2
n1r1)
2m , Amax (2m 1) ,
refraction of 1.5 placed at the back of one slits. The point o is 4th minimum fringe. Find (1) l, (2) the distance the central
maximum to the point o.
Solution:
(4) missing order
asinm1 dsinm2
dam m12 m2dam1 ismissing
(5) Bragg’s law
2dsin m
m 1,2,3,...
The angle θ is the angle between the incident x-ray and the plane.
Example 5: (E43-12) Assume that light is incident on a grating at an angle φ as shown. Show that the condition for a diffraction maximum is d(sin φ ±sinθ) =mλ.
Solution:
fom r ax,im 2 un2m dm
m0,1,2
At the edge of film, d=0 (m=0), is bright.
Thus the 5th bright fringe is m=4. dm1.0106m
2n2
Example 3: As shown, λ=560nm, center point is dark, there are 20 dark rings in the outer region. Find the thickness of the film at center.
,
m
m1,2,3.....m. axima
(2m1) 2 m0,1,2.....m. inima
If i =0, 2nd
(2dn, in
2
n
2
tfhilem
Example 1: Two slips interference, S1S2=0.7mm, D=100cm, λ=500nm. Now the single slip is shifted and SS1-SS2= λ/2, an optical medium plates with thickness of l and index of
(1)S2SS2Oln l(S1SS1O)
(2m1) , m3
SS2
2 SS1 l(n
1)
7 2
D
l(n1) 7 l 4μm
22
(2)S2Sr2ln l(S1Sr1)0
l(n1)2r2r10
r2
r1
[l(n
1)
2
]
Hale Waihona Puke 72r2r1
d D
x
x
D d
(r2
r1)
D d
7
2
2.5mm
r1
x
r2
Example 2: There is an oil film (n2=1.20) on flat grass plate (n1=1.52). A light ray (λ=600nm) illuminates normally and is observed from above by reflected light. Find the thickness of the film at 5th bright fringe from the edge of oil film.
A1 Amin
A2 , I max A1 A2 , I min
Φ 2 n2r2 n1r1 2
m,
n2 r2
n1r1
(2m 1)
2
,
m 0,1,2...Imax m 0,1,2...Imin
(1) Double-slit interference
nor1nor2 dsinm
a
(2) Grating diffraction
d si n mm 0 , 1 , 2 ,.P.ri.nciple maxima
(3) Resolving power
Airy spot:
R min1.22d, R1/R resolutiaobnility
Grating:
D dcmos
R mN
m=20.
2nd (4 01), d41n0m 0
2
If the 1st minimum correspond to m=1, the center point is
m=21.
2nd (2m1) , m12,
2
Example 4 (E41-40):
2d(n1)60, n1601.00030
2d
2. Diffraction (1) Single-slit diffraction
n=1.4 n=1.5
Solution: fom r axi,m 2nudm ,
at tehdegoefil,m d0,isbrig . ht
fo m r i,n 2 n i m d (2 m 1 u ),m m 0 ,1 2 , 2
If the 1st minimum correspond to m=0, the center point is
0
Brighftringaet thceenter
asin 2m2 m
m1,2,3...minima
(2m1)2
m1,2,3..m . axima
asinatanaaxmm
f
x02f12f a
w oifd tc h th e enm traax l im
xf,
d tow ho f ithfreirn ges