题目内容
一半径为R的1/4球体放置在水平面上,球体由折射率为
的透明材料制成。现有一束位于过球心O的竖直平面内的光线,平行于桌面射到球体表面上,折射入球体后再从竖直表面射出,如图所示。已知入射光线与桌面的距离为
。求出射角。
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250004403142840.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000440096337.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000440221554.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250004403142840.png)
θ=60°
试题分析:设入射光线与1/4球体的交点为C,连接OC,OC即为入射点的法线。因此,图中的角α为入射角。过C点作球体水平表面的垂线,垂足为B。依题意,
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000440330581.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000440361493.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000440377721.png)
设光线在C点的折射角为β,由折射定律得
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000440392807.png)
由①②式得
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000440502524.png)
由几何关系知,光线在球体的竖直表面上的入射角γ(见图)为30°
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250004405172848.png)
由折射定律得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000440533762.png)
因此
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000440564696.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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