题目内容
单色细光束射到折射率n=
的透明球表面,光束在过球心的平面内,入射角θ1=45°,研究经折射进入球内后又经内表面反射一次,再经球面折射后射出的光线,如图所示(图上已画出入射光线和出射光线).
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241356004638311.png)
(1)在图上大致画出光线在球内的路径和方向;
(2)求入射光线与出射光线之间的夹角α;
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824135600416344.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241356004638311.png)
(1)在图上大致画出光线在球内的路径和方向;
(2)求入射光线与出射光线之间的夹角α;
(1)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241356004946506.png)
(2) α=30°.
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241356004946506.png)
(2) α=30°.
(1)光线在球内的路径和方向如下图所示.
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241356004946506.png)
(2)由折射定律
=n.
得sin θ2=
=
=
,所以θ2=30°
由几何关系及对称性,有
=θ2-(θ1-θ2)=2θ2-θ1,
α=4θ2-2θ1把θ2=30°,θ1=45°代入,得α=30°.
本题考查光的折射定律,光路图如图所示,根据公式
=n.,可算出折射角,然后根据几何对称性,可得入射光线和射出光线的夹角。
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241356004946506.png)
(2)由折射定律
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824135600588616.png)
得sin θ2=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824135600635572.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824135600666546.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824135600713338.png)
由几何关系及对称性,有
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824135600931401.png)
α=4θ2-2θ1把θ2=30°,θ1=45°代入,得α=30°.
本题考查光的折射定律,光路图如图所示,根据公式
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824135600588616.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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