题目内容
(12分)如图4-9所示,一个横截面为直角三角形的三棱镜,ÐA=30°,ÐC=90°.三棱镜材料的折射率是 n=
。一条与BC面成θ=30°角的光线射向BC面,经过AC边一次反射从AB边射出。
(1)求光在三棱镜中的速度
(2)画出此束光线通过棱镜的光路图
(3)求从AB边射出光线与AB边的夹角![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241219419482520.jpg)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824121941932225.gif)
(1)求光在三棱镜中的速度
(2)画出此束光线通过棱镜的光路图
(3)求从AB边射出光线与AB边的夹角
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241219419482520.jpg)
(1)1.73×108m/s
(2)光路图如图所示
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241219419794573.jpg)
(3)30°
(2)光路图如图所示
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241219419794573.jpg)
(3)30°
(1)由折射率公式有 n =
(1分)
代入数据得v=
×108m/s=1.73×108m/s (2分)
(2)光路图如图所示 (4分)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241219419794573.jpg)
(3)由折射定律:
在BC界面:sin(90°-θ)=
sinγ γ=300° ①(1分)
∵sinC=
<60° ②(1分)
∴光线在AC界面发生反射再经AB界面折射(1分)
sin30°=sinγ/ ③ (1分)
γ/=60°
则射出光线与AB面的夹角 β=90°-γ/=30° ④(1分)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824121942010216.gif)
代入数据得v=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824121942213227.gif)
(2)光路图如图所示 (4分)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241219419794573.jpg)
(3)由折射定律:
在BC界面:sin(90°-θ)=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824121941932225.gif)
∵sinC=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824121942275344.gif)
∴光线在AC界面发生反射再经AB界面折射(1分)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824121941932225.gif)
γ/=60°
则射出光线与AB面的夹角 β=90°-γ/=30° ④(1分)
![](http://thumb2018.1010pic.com/images/loading.gif)
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