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第24章 光的偏振
一、选择题
1.三个偏振片Pl、P2与P3堆叠在一起,Pl与P2的偏振化方向间的夹角为30°,Pl与P3的偏振化方向相互垂直。强度为的自然光垂直入射于偏振片 Pl,并依次通过偏振片Pl、P2与P3,则通过三个偏振片后的光强为( )。[北京邮电大学2010研]
A.I0/4
B.3I0/8
C.3I0/32
D.I0/16
【答案】C
【解析】强度为的自然光垂直入射于偏振片 Pl后强度变为
,则由马吕斯定律可知
得
![](https://epubservercos.yuewen.com/98CD16/15436378304486206/epubprivate/OEBPS/Images/image634.png?sign=1738852340-UawozBn5V3rEdtx7dcsp7Jys9Ez4nWIa-0-0109a39d123933a77e6ea0549478149e)
2.光线从折射率为n1的介质入射到折射率为n2的介质界面上反射时,反射光是完全偏振光,则入射角io满足的方程是( )。[电子科技大学2006研]
A.
B.
C.
D.
【答案】C
二、填空题
1.光强为I0的自然光依次通过两个偏振片P1和P2。若P1和P2的偏振化方向的夹角α=30°,则透射偏振光的强度I是______。[华南理工大学2010研]
A.I0/4
B.
C.
D.I0/8
E.3I0/8
【答案】E
【解析】自然光通过偏振片后,光强变为
,且偏振方向沿偏振片
的偏振方向;然后通过偏振片
时,光强则满足:
。
2.两个偏振片堆叠在一起,其偏振化方向相互垂直,若一束强度为IO的线偏振光入射,其光矢量振动方向与第一偏振片偏振化方向夹角为π/4,则穿过第一偏振片后的光强为______,穿过两个偏振片后的光强为______。[华南理工大学2009研]
【答案】;0
【解析】由马吕斯定律可知
![](https://epubservercos.yuewen.com/98CD16/15436378304486206/epubprivate/OEBPS/Images/image647.png?sign=1738852340-CZMcMlOKarkeGu1iwn0KNDsAlYnMQ3vX-0-b49373c52d7cb487876d735e1c0292f8)
![](https://epubservercos.yuewen.com/98CD16/15436378304486206/epubprivate/OEBPS/Images/image648.png?sign=1738852340-8qVfthR7XgixAidzGUY49z6elepVOVvx-0-5d34f1d049c893a3d4373b0459f2b9a7)
3.光强为I的一束自然光,经过三个偏振片后,透射光强为I/16,已知第一个偏振片和第三个偏振片的偏振化方向相互垂直,则第一个偏振片和第二个偏振片的偏振化方向夹角为______。[南京航空航天大学2007研]
【答案】或
4.一束平行的自然光,以角入射到平玻璃表面上,若反射光束是完全偏振的,则透射光束的折射角是______;玻璃的折射率为______。[南京航空航天大学2006研]
【答案】;1.73
三、计算题
1(1)一强度为I0的自然光先后垂直穿过偏振化方向夹角为60°的两偏振片P1和P2,求P2出射光的强度。
(2)若在P1和P2之间插入一个1/4波片,其光轴与P1和P2的偏振化方向夹角均为30°,则P2的出射光强度变为多少? [厦门大学2011研]
解:(1)自然光经过P1偏振片后,光强减半,并且变成沿P1透振方向的线偏光。再经过P2偏振片,设出射光强为I2,由马吕斯定律,
![](https://epubservercos.yuewen.com/98CD16/15436378304486206/epubprivate/OEBPS/Images/image652.png?sign=1738852340-m229FiSNFLs6MwmxzXvToIZoiMlyxihz-0-2d7013d91933a78391b3d8eda4ab3aca)
(2)设1/4波片光轴方向沿y轴,P1偏振片出射光波幅为A1。经过波片后线偏光变为椭圆偏振光,如图24-1可得,
![](https://epubservercos.yuewen.com/98CD16/15436378304486206/epubprivate/OEBPS/Images/image653.png?sign=1738852340-EgTxOSSK7msJAKpD1iaxWaUtZBq07rCl-0-09f087f2b566cf7335578278de0aa18f)
![](https://epubservercos.yuewen.com/98CD16/15436378304486206/epubprivate/OEBPS/Images/image654.png?sign=1738852340-dDoCgYbVdbzSnIwXyXRfJU4RB0QsaZ0w-0-f24429e0dc0c85335295b7bd979d9294)
经过波片后o光和e光相位相差90°,故出射光电场分量可表示为
![](https://epubservercos.yuewen.com/98CD16/15436378304486206/epubprivate/OEBPS/Images/image655.png?sign=1738852340-POW7homAIFkzE2zbRI8Ed2B2KhrTVJ03-0-c9dfd6d5b1d539c19a2d8a8821aa8f77)
再经过P2偏振片,椭圆偏光变成沿P2透振方向的线偏光,将在P2方向做投影可以得到出射光的电场分量,
![](https://epubservercos.yuewen.com/98CD16/15436378304486206/epubprivate/OEBPS/Images/image657.png?sign=1738852340-RfMoMBhS1ObtB23E9s0Hhju44pcwvbCN-0-c81b032f074aa17ec909a6f58c237fa1)
可进一步约化为:
![](https://epubservercos.yuewen.com/98CD16/15436378304486206/epubprivate/OEBPS/Images/image658.png?sign=1738852340-GETc0iSe89okkn1qO5Dk507ul2AkkeOY-0-22949c49ae998302b689a42bbaf6a9ab)
所以出射光的波幅为:
![](https://epubservercos.yuewen.com/98CD16/15436378304486206/epubprivate/OEBPS/Images/image659.png?sign=1738852340-KFjJXb9gmUlFs5HpgZwciklRLMaxBn9a-0-9faf60ef086455fa74594fd08eb201f8)
故出射光强为:
![](https://epubservercos.yuewen.com/98CD16/15436378304486206/epubprivate/OEBPS/Images/image660.png?sign=1738852340-oFZoH2IiYep8Xn5maZV0jw1VnHnb7Ew0-0-2f938c1215d34ce239ed77d6fbbb7789)
图24-1
2.一束自然光以起偏角自某透明液体入射到玻璃表面上,若玻璃的折射率为1.56,求:
(1)该液体的折射率。
(2)折射角。[宁波大学2009研]
解:(1)设该液体的折射率为,由布儒斯特定律得:
![](https://epubservercos.yuewen.com/98CD16/15436378304486206/epubprivate/OEBPS/Images/image664.png?sign=1738852340-j1AOREW44Admxpxg7EeIa3UamwJaK2Ma-0-5dc7a2a519db4b42d90fa2f42a6a585d)
可得:
![](https://epubservercos.yuewen.com/98CD16/15436378304486206/epubprivate/OEBPS/Images/image665.png?sign=1738852340-rBdf3ZQj71MxUJBmzmMR76BaKqiScaIn-0-afb3b7bbe41d5430bb52910e0a2421bd)
(2)折射角:
![](https://epubservercos.yuewen.com/98CD16/15436378304486206/epubprivate/OEBPS/Images/image666.png?sign=1738852340-qXHF0sS29RseKLHF0INly55TD1GuCVbk-0-b24ac5b776a2d89a64dcc914bc1f7a3b)
3.当一束自然光以入射角57°由空气投射到一块平板玻璃表面上时,反射光为完全偏振光,则此时光线的折射角为多大?[浙江大学2008研]
解:由可知,折射角为
![](https://epubservercos.yuewen.com/98CD16/15436378304486206/epubprivate/OEBPS/Images/image668.png?sign=1738852340-uypKZbY99vE18EUzeZJYs7ql8GZyWFpX-0-8298eb9825716f531b6c48a7cdc5eae2)
4.如图24-2所示,杨氏双孔干涉装置,当透振方向夹角为的理想偏振片
、
移动时,屏上最大光强为
,干涉条纹衬比度为1,试求:插入
、
后屏上最大光强和干涉条纹衬比度各是多少?[山东大学研]
![](https://epubservercos.yuewen.com/98CD16/15436378304486206/epubprivate/OEBPS/Images/image672.jpg?sign=1738852340-1yfJCTQWTg5pohVhjKfHgkxQRl6P2JiX-0-956ab497ce721e9e96c62b6a2c9f0c45)
图24-2
解:合成光强:
![](https://epubservercos.yuewen.com/98CD16/15436378304486206/epubprivate/OEBPS/Images/image673.png?sign=1738852340-IqAL28NXBMfA9scD6ChHdUsdiP5n3ye2-0-09e66dbd2e362434631b2d82294f293c)
![](https://epubservercos.yuewen.com/98CD16/15436378304486206/epubprivate/OEBPS/Images/image674.png?sign=1738852340-cOhPxJHUMtW22hRKjwCKbRp8SqYcQP1H-0-c6c610500cc0fac56f9b16f680c2aeb7)
最大光强:
![](https://epubservercos.yuewen.com/98CD16/15436378304486206/epubprivate/OEBPS/Images/image675.png?sign=1738852340-pxHRY1Duf2j6JnNBw0y6H9cehHBrHcRx-0-dbacbcadaeb627b8f9ee2343ccfb580b)
最小光强:
![](https://epubservercos.yuewen.com/98CD16/15436378304486206/epubprivate/OEBPS/Images/image676.png?sign=1738852340-Y9dgLgnMz7TD2wO660jBCw6XM1J0BtCe-0-12e445299e57c4d71f5f270825a83c6d)
干涉条纹衬比度:
![](https://epubservercos.yuewen.com/98CD16/15436378304486206/epubprivate/OEBPS/Images/image677.png?sign=1738852340-OVVVSBqNCcdJPrSKuXZO5O2tyjUL0doi-0-294283f92cfa09ce2fa1563c6f2954c5)
5.如图24-3所示,强度为的单色自然光照明扬氏双孔装置,P是理想偏振片(道振方向为P),求屏幕Π上的光强分布和干涉条纹的衬比度。[山东大学研]
![](https://epubservercos.yuewen.com/98CD16/15436378304486206/epubprivate/OEBPS/Images/image678.png?sign=1738852340-v5R67568g1Yrvwiu5RlwC7Z8HeSSz9Yu-0-b436d2515daa61f446e6c04b16ae76f5)
图24-3
解:由光强合成公式,得:
![](https://epubservercos.yuewen.com/98CD16/15436378304486206/epubprivate/OEBPS/Images/image679.png?sign=1738852340-jiMtOede3RJqotbgGuFW6QKfIT5dWY0a-0-ed3e6693f1ad1c50d084520bb7ce0093)
衬比度:
![](https://epubservercos.yuewen.com/98CD16/15436378304486206/epubprivate/OEBPS/Images/image680.png?sign=1738852340-9tNJglqFoR2SARVOFCnNOsqDurHaFY8y-0-e7f311dc67b4982472ec2e60c26567b5)