题目内容
如图所示,某三棱镜的截面是一直角三角形,棱镜材料的折射率为n,底面BC涂黑,入射光沿平行于底面BC的方向射向AB而,经AB和AC折射后射出.为了使上述入射光线能从AC面射出,求折射率n的取值范围.
![](http://thumb.1010pic.com/pic2/upload/papers/20140610/201406102312052702075.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140610/201406102312052702075.png)
画出光路图如图所示.
因为入射光平行于BC面,i=60°
由折射定律有
=n,解得sinα=
光折到AC面上时,
=n
由几何关系可得:α+β=90°
则sinβ=cosα=
=
sinr=nsinβ=
要使有光线从AC面射出,应有sinr≤1:即
≤1
解得n≤
.
答:折射率n的取值范围1<n≤
.
![](http://thumb.1010pic.com/pic2/upload/papers/20140610/201406102312068628600.png)
因为入射光平行于BC面,i=60°
由折射定律有
sini |
sinα |
| ||
2n |
光折到AC面上时,
sinr |
sinβ |
由几何关系可得:α+β=90°
则sinβ=cosα=
1-sin2α |
| ||
2n |
sinr=nsinβ=
| ||
2 |
要使有光线从AC面射出,应有sinr≤1:即
| ||
2 |
解得n≤
| ||
2 |
答:折射率n的取值范围1<n≤
| ||
2 |
![](http://thumb.1010pic.com/pic2/upload/papers/20140610/201406102312068628600.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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