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吴热振波光大物教材参考答案

8.6 假定N 个粒子的速率分布函数为

00v v a

=)(v f

v v >

（1）作出速率分布曲线；（2）由0v 求常数a ；（3）求粒子的平均速率。解：（1）

（2）由归一化条件，有

1)(0

==??

∞

v a d v dv v f

v a =

（3）粒子的平均速率为 0200

121)(0

v a v vadv dv v vf v v ==

∞

8.9 在容积为3

31030m -?的容器中，贮有kg 3

1020-?的气体，其压强为

Pa 3107.50?。试求该气体分子的最概然速率、平均速率及方均根速率。

解：由RT M

pV =

，有

)

(390102010

30107.502223

s m M

pV RT

v p =?????=

--μ

)(44088s m M pV

v ==

ππμ

)(478332s m M

v ==

8.14 温度为c

27时，mol 1氧气具有多少平动动能？多少转动动能？解：气体的平动动能为 )(1074.330031.82

233J RT E t ?=??==

气体的转动动能为 )(1049.230031.82

223J RT E r ?=??==

8.15 在室温K 300时，mol 1氢气的内能是多少？g 1氮气的内能是多少？解：氢气的内能为 )(1023.630031.82

2532J RT E H ?=??==

氮气的内能为 )(22330031.82510281013

J RT M

E N =?????==--μ 7.7 一定质量气体从外界吸收热量J 8.1731，并保持在压强为Pa 5

10013.1?下，体积从L 10膨胀到L 15，问气体对外做功多少？内能增加多少？

解：在等压条件下，气体对外做功

)(50710)1015(10013.1)(3512J V V p A =?-??=-=- 气体的内能增加为

)(8.12065078

.1713J A Q E =-=-=? 7.8 质量为kg 02.0的氦气（R C V 2

），温度由C

17升为C 27，若在升温过程中：（1）体积保持不变；（2）压强保持不变；（3）与外界不交换热量。试分别计算各过程中气体吸收的热量、内能的改变和对外所做的功。

解：已知氦气的摩尔质量mol kg M 3

104-?=，则kg 02.0氦气的摩尔数

m o l 510402

.03

=?=

-ν，内能变化

J T T C E V 623)290300(31.82

5)(12=-???=-=?ν

（1）体积不变时，0=A ，且

J E A E Q 623=?=+?=

（2）压强不变时 J T T C Q p 1040)290300(31.82

5)(12=-???

=-=ν J E Q A 4166231040=-

=?-=

（3）与外界不交换热量，0=Q ，则

J E A 623-=?-=

7.12 mol 10单原子理想气体，在压缩过程中外界对它做功J 209，其温度升高K 1，试求气体吸收的热量与内能的增量，此过程中气体的摩尔热容是多少？

解：内能增量

J T R T C E V 7.124131.82

1023=???=?=?=?ν

ν 由于J A 209-=，则吸热为

J A E Q 3.842097.124-=-=+?= 过程中的摩尔热容为 k m o l J T Q C m ?-=-?=?=

43.81

3.841011ν 7.16 利用过程方程直接证明在绝热线和等温线的交点A 处，绝热线斜率的绝对值比等

温线斜率的绝对值大。

解：绝热过程中，C pV =γ 01=+-pdV V dp V γγγ V

p dV dp Q γ-=)(

等温过程中，C pV =，则 V

p dV dp T -=)(

由于1>γ，所以T Q dV

dV dp )()(

> 7.17 如题图7.17中DC AB ,是绝热线，COA 是等温线。已知系统在COA 中放热

J 100，OAB 的面积是J 30，ODC 的面积是J 70，试问在BOD 过程中系统式吸热还是

放热？热量是多少？

解：整个循环中，0=?E ，

)(403070J S S A O ABO O D CO =-=-= 且有 E A Q ?+=

A J Q Q Q Q BO D CO A BO D ==-=+=)(40100 故BOD 过程中吸热为

)(140)100(40J Q A Q CO A BO D =--=-=

题图7.17 题图7.22

7.22 如题图7.22所示，mol

1单原子理想气体所经历的循环过程，其中ab为等温线，

假定2

V，求循环的效率。

解在a

c-的等体过程中，气体吸热为

)

(

)

(

)

在b

a-等温过程中，气体吸热为

在c

b-等压过程中，气体放热为

)

(

)

(

)

(

整个循环中A

?,0

吸

放

吸

)

(

)

(

已知

=，所以

)

7.26 一卡诺热机工作于温度为K

1000与K

300的两个热源之间，如果（1）将高温热源的温度提高K

100；（2）将低温热源的温度降低K

100，试问理论上热机的效率各增加多少？解：卡诺热机工作在K

1000与K

300之间时，其效率为

1000

300

（1）若把高温热源的温度提高K

100时，其效率为

%731100

300

11'

21=-

=T T η 即效率提高了%31=-ηη

（2）若把低温热源的温度降低K 100时，其效率为

%801000

200

111'

22=-=-=T T η 即效率提高了%102=-ηη

6.3 一物体沿x 轴作谐振动，振幅为cm 0.10周期为s 0.2，在0=t 时，坐标为cm 0.5，

且向x 轴负向运动，求在cm x 0.6-=处，沿x 轴负方向运动时，物体的速度和加速度以及它从这个位置回到平衡位置所需要的最短时间。

解；已知 s T 2=，所以 ππ

ω==

2 设振动方程为 cm t x )cos(10?π+=

则速度为 )s i n (

10?ππ+-=t v 加速度为 )c o s (102?ππ+-=t a

0=t 时，cm x 5=，0

?=，

所以 cm t x )3

cos(10π

π+

设在时刻'

t ，振子位于cm x 6-=处，且向x 轴负方向运动，对应于旋转矢量图，

则有 53)3cos('

-=+

πt ，所以 5

4)3s i n ('=+ππt 所以 s cm t v 1.25)3

sin(10'

-=+

-=π

ππ

92.59)3

cos(10s cm t a =+

-=π

ππ

设弹簧振子回到平衡位置的时刻为'

't ，对应旋转矢量图可知 2

'π

π=

t 故从上述位置回到平衡位置所需时间为 s t t 8.0/]}3

)53[arccos()323{(

'=----=-ππ

ππ

6.6 喇叭膜片做谐振动，频率为Hz 440，其最大位移为mm 75.0，试求：（1）角频率；（2）最大速率；（3）最大加速度。

解：设膜片的振动方程为 )c o s

(?ω+=t A x （1）πππνω88044022=?==

（2）s m A v m 07.21075.08803=??==-πω

（3）233221073.51075.0)880(s m A a m ?=??==-πω

6.14 一质量为g 10的物体做简谐振动，其振幅为cm 24，周期为s 0.4。当0=t 时，位移为cm 24。试求（1）s t 5.0=时，物体所在的位置；（2）s t 5.0=时，物体所受力的大小和方向；（3）由起始位置运动到m x 12.0=处所需的最少时间；（4）在m x 12.0=处，物体的速度、动能、势能和总能量。

解：已知m A 24.0=，s T 4=，kg m 01.0=

当0=t ，A m x ==24.0，因而该谐振动的初相为0=?，所以，谐振动的振动方程为 t t x 2c o s 24.042cos 24.0ππ

== （1）当s t 5.0=时，物体所在的位置为 m x 17.04

cos

24.01==π

（2）由运动方程可得 t a 2

c o s )2(24.02

π-=

所以，s t 5.0=时，物体所受的合力大小为 )(1019.44

cos

(24.001.032

N a m F -?=???==π

其方向为x 轴负方向，指向平衡位置。

（3）由旋转矢量法可知，m x 12.0=时，其相位为

为整数n n t ,3

ππ

因此，由起始位置运动到m x 12.0=处所需的最少时间为

)(667.03

s t ==

（4）在m x 12.0=处，物体的速度为

s m v 326.0)3

2sin(

(24.0-=?-=π

π 物体的动能为 J mv E k 4221033.5)326.0(01.02

21-?=??==

在0=x 处，物体所具有的动能即为总机械能，所以

J A m E 42221011.7)2

()24.0(01.021)(21-?=???==

ω 在m x 12.0=处，物体的势能为

J E E E k p 441078.110)33.511.7(--?=?-=-=

6.16 一物体悬挂于弹簧下端并做谐振动，当物体位移为振幅的一半时，这个振动系统的动能占总能量的多大部分？势能占多大部分？又位移多大时，动能、势能各占总能量的一半？解：当物体的位移为振幅的一半时，系统的势能为

22281)2(2121kA A k kx E p === %254

1218122===kA kA E E p

这时动能占总能量的部分为

%754

411)(==-

=-=E E E E E p k 动能势能各占总能量一半时，有

2222

141)21(2121kx kA kA E E p ====

解得，这时位移大小为 2

A x =

12.8 已知某一维平面简谐波的周期s T 3105.2-?=，振幅m A 2

100.1-?=，波长

m 0.1=λ，沿x 轴正向传播。试写出此一维平面简谐波的波函数（设0=t 时，0=x 处质

点在正的最大位移处）。

解：0=t 时，在0=x 处，质点恰好处于正的最大位移处，其振动的初相为0，振动方程为 )2cos()02cos(0T

t A T t A y ππ

=+= 在x 轴上任取一点P ，如图，坐标为x ，P 点相位落后于原点，相位差为λ

π?x

2=?，

其振动方程为

)(2c o s )2

c o s (λ

π?πx T t A T t A y P -=?-= P 点是任选的一点，所以波函数为 )()400(2cos 100.1)(2cos 2m x t x

T t A y -?=-=-πλ

12.10 波源的振动方程为)(5

cos

0.62

m t y π

-?=，它所激起的波以s m 0.2的速度

在一直线上传播，求：

（1）距波源m 0.6处一点的振动方程；（2）该点与波源的相位差。解：（1）m 0.6处质点的相位落后于波源，相位差为

ππωω?5

265=?=??

=??=?u x t 所以，该质点的振动方程为

)()5

cos(100.6)5

cos(

100.622m t t y π

?π

?=?-?=-- （2）相位差 π?5

? 12.12 一横波沿绳子传播时的波函数为 )410cos(

05.0x t y ππ-=，式中y x ,以米计，t 以秒计。

（1）求次波的波长和波速；

（2）求m x 2.0=处的质点，在s t 1=时的相位，它是原点处质点在哪一时刻的相位？

解：（1）把)410cos(

05.0x t y ππ-=与波函数的标准形式)

22cos(?λ

πν+-=x t A y 对比，对应项相等，即

ππν102=，

πλ

42=

得 Hz 5=ν， m 5.0=λ 则 s m u 5.25.05=?==νλ （2）s t 1=时，m x 2.0=处质点的相位为

πππππ?2.92.04110410'=?-?=-=x t

原点在t 时刻相位为

t t π?10)(0=

若)(0't ??=，有 ππ2.910=t

则 s t 92.0=，即原点在s t 92.0=时的相位等于所求相位。

12.14 一平面简谐波，沿x 轴正向传播，波速为s m 4，已知位于坐标原点处的波源的振动曲线如图12.14（a ）所示。（1）写出此波的波函数；（2）试画出s t 3=时刻的波形图。

解：波速s m u 4=，由图可知，周期s T 4=，所以波长为

)(1644m uT =?==λ

由图可知，s t 0=时刻，原点处波源处于正的最大位移处，所以波源初相0=?，其振动方程为 t A t A y πνπν2cos )02cos(=+=

所以波函数为 )16

4(2cos 4)22cos(x

t x t A y -=-

=πλπ

πν （2）把s t 3=代入波函数，可得波形曲线方程为 )16

43(

2cos 4x

y -=π 波形曲线如图12.14（b ）所示。

图12.14（a ）图12.14（b ） 12.16已知一平面简谐波沿x 轴正向传播，如图所示，周期为s T 5.0=，振幅m A 1.0=，当0=t 时，波源振动的位移恰好为正的最大值，若波源取做坐标原点，求：（1）沿波的传播方向距离波源为

2λ

处质点的振动方程；（2）当2T t =时，4

λ=x 处质点的振动速度。

解：（1）0=t 时，波源达到正的最大位移，所以其初相为0，振动方程为

)()4c o s (1.0)5

.02c o s (1.0)02

c o s (1.00m t t

T t y πππ==+= 在2λ

处的质点，其相位落后于波源，相位差πλλπ?=?=

?22，此质点振动方程为 ))(4cos(1.0)4cos(1.0m t t y ππ?π-=?-= （2）与上述过程同理，4

=x 处质点的振动方程为

))(2

4cos(1.0m t y π

π-=

质点的速度 t t s dt dy v πππ

ππ4cos 4.0)2

4(sin 41.0=-?-==

当s T

t 25.02

时，速度为)(4.0)25.04cos(4.0s m v πππ-=?= 12.20 B A 、

为两个同振幅、同相位的相干波源，它们在同一介质中相距2

3λ

，P 为B A 、连线延长线上的任意点，如图所示。求：（1）自B A 、

两波源发出的波在P 点引起的两个振动的相位差；（2）P 点的合振动的振幅。

解：（1）把B A 、

两波源的相位用)(t ?表示。波源A 在P 点引起振动，其相位落后于A 点，相位差PA λ

?2=

?，此振动的相位为 PA t t P λ

??2)()(-

同理，波源B 在P 点引起振动，其相位为 PB t t P λ

??2)()('

两振动的相位差为

λλπ

λπλ

?λ

????32

22)

2)(()2)(()()('

-=?-=-=-

--=-=?AB PB t PA t t t P P

（2）两振动反相，所以P 点合振动振幅等于0.

12.22如图所示，B A 、两点为同一介质中的两相干平面波波源，其振幅皆为m 05.0，频率为Hz 100，但当A 点为波峰时，B 恰为波谷，设在介质中的波速为s m 10，试写出由B A 、发出的两列波传到P 点时的干涉结果。

解：设B A 、两波源至P 点的距离分别为1r 和2r ，如图所示。

m r 151= m r 252015222=+=

两波的波长为m u

1.0100

λ 则两波在P 点激起的两振动的相位差为

ππ

πλ

???2011

.015

252)(21212-=---=--

-=?r r 所以两波在P 点干涉相消。

13.6 汞弧灯发出的光通过一绿色滤光片后射到相距mm 60.0的双缝上，在距双缝m 5.2处的屏幕上出现干涉条纹。测得两相邻明纹中心的距离为mm 27.2，试计算入射光的波长。

解：由双缝干涉的条纹间距公式可得

)(547)(1045.55

.21060.01027.273

3nm m D d x =?=???=??=---λ 13.10 用折射率58.1=n 的很薄的云母片覆盖在双缝实验中的一条缝上，这时屏上的第

七级亮条纹移动到原来的零级亮条纹的位置上。如果入射光的波长为nm 550，试问次云母片的厚度为多少？

解：设云母片的厚度为d ，无云母片时，零级亮条纹在屏上P 点，则到达P 点的光程差为

012=-=r r δ

加上云母片后，到达P 点的两光束的光程差为

λδ7)1()(12'=-=-+-=d n r nd d r

所以有 )(6.61

58.1105507179

m n d μλ=-??=-=

- 13.11 利用等厚干涉可以测量微小的角度。如图所示，折射率4.1=n 的劈尖状板，在

某单色光的垂直照射下，量出两相邻明条纹间距cm l 25.0=，已知单色光在空气中的波长nm 700=λ，求劈尖顶角θ。

解：由劈尖干涉相邻明条纹间距公式 θ

n l 2=

可得 )(100.110

5.24.12107002439

rad nl ---?=????==λ

θ 13.13 如图所示，用紫色光观察牛顿环时，测得第k 级暗环的半径mm r k 4=，第5+k 级暗环的半径mm r k 65=+，所用平凸透镜的曲率半径m R 10=，求紫光的波长和级数k 。

解：牛顿环暗环半径为 λkR r k = （1） λR k r k )5(5+=+ （2）由式（1）、（2）得

)(10410

510)46(576

22225m R r r k k --+?=??-=-=

λ 44

64552

2252=-?=-=+k k k r r r k 13.15 一平面单色光波垂直照射在厚度均匀的薄油膜上，油膜覆盖在玻璃板上，所用光

源波长可连续变化，观察到nm 500和nm 700这两个波长的光在反射中消失。油的折射率为30.1，玻璃的折射率为50.1，试求油膜的厚度。

解：由于在油膜的上下表面反射都有半波损失，暗纹条件为

)

12(21

λ+=k nd （1）

]1)1(2[22

λ++=k nd （2）

由式（1）和（2）解得 )(1073.6)

500700(30.12700

500)(222121nm n d ?=-???=-=

λλλλ

13.16白光垂直照射在空气中的厚度为m μ40.0的玻璃片上，玻璃片的折射率为50.1。试问在可见光范围内（nm 700~400=λ），哪些波长的光在反射中加强？哪些波长的光在透射中加强？

解：反射加强的条件为

),3,2,1(22 ==+

k k nd λλ

即 1

24-=k nd

仅当3=k 时，λ为可见光波长，因此求得

)(48.01

3240

.050.14m μλ=-???=

透射光加强的条件即反射减弱的条件，即 ),3,2,1(2

)

12(2

2 =+=+

k k nd λ

由此得 k

24=

λ 当2=k 时，)(60.02

240

.050.14m μλ=???=

当3=k 时，)(40.03

240

.050.14m μλ=???=

13.18 波长nm 500=λ的平行光，垂直地入射到一宽度为mm a 0.1=的单逢上，若在

缝的后面有一焦距为cm f 100=的凸透镜，使光线聚焦于屏上，试问从衍射图样的中心到下列各点的距离如何？（1）第一级极小，（2）第一级亮条纹的极大处，（3）第三级极小。

解：（1）由单缝衍射暗纹公式 λ?=1s i n a ，得

9110510

110500sin ---?=??==a λ

? 从中心到第一级极小处的距离

)(105105100sin tan )(24111cm f f x --?=??=≈=??暗

(2)由亮纹公式2

)

12(sin λ

?+=k a k 得

)

112(sin 1λ

?+?=a

1105.710

1210500323sin ---?=????==a λ? 从中心到第一级极大处的距离为

)(105.7105.7100sin tan )(24111cm f f x --?=??=≈=??明

（3）同理 3

393105.110

11050033sin ---?=???==a λ? 所以从中心到第三级极小处的距离为

)(15.010

5.1100sin tan )(3

333cm f f x =??=≈=-??暗

13.20 有一单缝，宽mm 1.0，在缝后放一焦距为cm 50的凸透镜，用波长nm 546=λ的

平行绿光垂直照射单缝，求位于透镜焦平面处的屏上的中央亮条纹的宽度。如果把此装置浸入水中，中央亮条纹宽度如何变化？

解：中央明纹的宽度由相邻中央明纹两侧的暗纹（1±=k ）位置决定，根据公式

λ?=1sin a 有

911046.5101.010546sin ---?=??==a λ

中央明纹的宽度为

)(546.01046.5502sin 2310cm f x =???=≈-?

在水中，光的波长n

λ=

，所以装置浸入水中（33.1=n ）时，有

11011.410

1.033.110546sin ---?=???===a n a λ

? 则)(411.01011.4502sin 2)(310cm f x =???=≈-‘

水?

13.21在单缝夫琅禾费衍射实验中，波长为λ的单色光第三极亮条纹与nm 630'

=λ的单色光的第二级亮条纹恰好重合，试计算λ的数值。解：根据题意有

)132(s i n

3λ

?+?=a 2

)

122(sin '

2λ?+?=a

且 32s i n s i n ??= 联立解得 nm 4507

630

575'=?==

λλ 13.25 为了测定一光栅的光栅常数，用波长为nm 8.632=λ的氦氖激光器的激光垂直照射光栅，做光栅的衍射光谱实验，已知第一级亮条纹出现在

30的方向上，问这光栅的光栅常数是多大？这光栅的1厘米内有多少条缝？第二级亮条纹是否可能出现？为什么？

解：光栅常数为

)(10266.130

sin 108.632sin 4

71cm b a --?=?==+

?λ 每厘米内的缝数为条）(109.710

266.11

134

?=?=+=

-b a N 当2=k 时，有 130sin 2sin 22sin 22=?==+=

?λ

a 所以第二级亮条纹出现在无穷远处，不会出现在接收屏上。

13.26波长nm 600=λ的单色光垂直入射在一光栅上，第2级、第3级光谱线分别出现在衍射角32??、满足下式的方向上，即30.0sin ,20.0sin 32==??，第4级缺级，试问：（1）光栅常数等于多少？（2）光栅上狭缝宽度有多大？（3）在屏上可能出现的全部光谱线的级数？

解：（1）根据光栅方程有

)(0.6)(100.620

.0106002sin 269

2m m b a μ?λ=?=??==+--

（2）根据缺级公式

a b a =+，当第四级缺级时，4=k ，则41'<≤k ，所以有当1'=k 时，第四级缺级，符合题意，把4=k ，1'

=k 代入缺级公式，得m a μ5.1=

当2'

=k 时，所缺谱线级数为2、4，根据已知，第二级不缺级，不符合题意，故舍去。

当3'=k 时，第四级缺级，符合题意，把4=k ，3'

=k 代入缺级公式，得m a μ5.4=

所以，满足题目条件的狭缝宽度有两个：m a μ5.1=或m a μ5.4= （3）根据光栅方程有 1010

6000

.6sin )(3

=?=

+≤

-λ

a b a k 所以，在屏上出现谱线的最大级数为9。

光谱缺级级数为4，8，12，…，则屏上出现全部谱线的级数为 0，±1，±2，±3，±5，±6，±7，±9

13.28一束自然光和线偏振光的混合光，当它通过一偏振片时，发现光强取决于偏振片的取向，透射的最大光强是最小光强的5倍。求入射光束中两种光的光强各占总入射光强的几分之几

解：设自然光、偏振光的光强分别为偏自、I I ，总入射光光强偏自总I I I +=，当偏振片的偏振化方向与偏振光的光振动方向平行时，可得最大光强的透射光。偏自I I I +=

max 当偏振片的偏振化方向与偏振光光振动方向垂直时，可得最小光强的透射光。自I I 2

min = 根据题意有

5min

max

=I I 所以可得

=总偏I I 3

1=总自I I 13.30 利用布儒斯特定律，可以测定不透明电介质的折射率，今测到某一电介质的起偏角为

57，试求这一电介质的折射率。

解：54.157tan tan ===

b i n

13.32 已知某一物质的全反射临界角是

45，它的起偏角是多大？解：全反射时有 90sin sin 122n i n = 即 2

2sin n n i =

由布儒斯特定律得 245

sin 1sin 1tan 212==== i n n i b 所以 '4254 =b i

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