题目内容
(选做题,选修3-4)
如图所示,△ABC为一直角三棱镜的截面,其顶角θ=30°,P为垂直于直线BCD的光屏,现一宽度等于AB的单色平行光束垂直射向AB面,在屏P上形成一条宽度等于
的光带,试作出光路图并求棱镜的折射率。
如图所示,△ABC为一直角三棱镜的截面,其顶角θ=30°,P为垂直于直线BCD的光屏,现一宽度等于AB的单色平行光束垂直射向AB面,在屏P上形成一条宽度等于
![](http://thumb.1010pic.com/pic1/upload/papers/g04/20120324/20120324113527068942.gif)
![](http://thumb.1010pic.com/pic1/upload/papers/g04/20120324/201203241135270742435.gif)
解:
平行光束经棱镜折射后的出射光束仍是平行光束,如图所示。图中θ1、θ2为AC面上入射角和折射角,根据折射定律,有nsin θ1=sin θ2
![](http://thumb.1010pic.com/pic1/upload/papers/g04/20120905/201209051357164844718.png)
设出射光线与水平方向成 α角,则θ2=θ1+α
由于![](http://thumb.1010pic.com/pic1/upload/papers/g04/20120324/201203241135270881089.gif)
所以![](http://thumb.1010pic.com/pic1/upload/papers/g04/20120324/201203241135270931034.gif)
而![](http://thumb.1010pic.com/pic1/upload/papers/g04/20120324/201203241135270981079.gif)
所以tan α=![](http://thumb.1010pic.com/pic1/upload/papers/g04/20120324/201203241135271031069.gif)
可得α=30°,θ2=60°,所以
。
平行光束经棱镜折射后的出射光束仍是平行光束,如图所示。图中θ1、θ2为AC面上入射角和折射角,根据折射定律,有nsin θ1=sin θ2
![](http://thumb.1010pic.com/pic1/upload/papers/g04/20120905/201209051357164844718.png)
设出射光线与水平方向成 α角,则θ2=θ1+α
由于
![](http://thumb.1010pic.com/pic1/upload/papers/g04/20120324/201203241135270881089.gif)
所以
![](http://thumb.1010pic.com/pic1/upload/papers/g04/20120324/201203241135270931034.gif)
而
![](http://thumb.1010pic.com/pic1/upload/papers/g04/20120324/201203241135270981079.gif)
所以tan α=
![](http://thumb.1010pic.com/pic1/upload/papers/g04/20120324/201203241135271031069.gif)
可得α=30°,θ2=60°,所以
![](http://thumb.1010pic.com/pic1/upload/papers/g04/20120324/201203241135271071101.gif)
![](http://thumb2018.1010pic.com/images/loading.gif)
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