题目内容
由于激光是亮度高、平行度好、单色好的相干光,所以光导纤维中用激光作为信息高速传输的载体.要使射到粗细均匀的圆形光导纤维一个端面上的激光束都能从另一个端面射出,而不会从侧壁“泄漏”出来,光导纤维所用材料的折射率至少应为多大?
n≥![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120031945225.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120031945225.gif)
设激光束在光导纤维端面的入射角为i,折射角为α.折射光线射向侧面时的入射角为β,如图13-8-2所示.
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241200319611466.jpg)
图13-8-2
由折射定律:n=![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120031976459.gif)
由几何关系:α+β=90°
sinα=cosβ
由全反射临界角的公式:sinβ=![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120031976232.gif)
cosβ=![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120031992360.gif)
要保证从端面射入的任何光线都能发生全反射,应有i=90°,sini=1.故
n=
=
=
=![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120032163344.gif)
解得n=![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120031945225.gif)
光导纤维的折射率应为n≥
.
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241200319611466.jpg)
图13-8-2
由折射定律:n=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120031976459.gif)
由几何关系:α+β=90°
sinα=cosβ
由全反射临界角的公式:sinβ=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120031976232.gif)
cosβ=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120031992360.gif)
要保证从端面射入的任何光线都能发生全反射,应有i=90°,sini=1.故
n=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120031976459.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120032132473.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120032148405.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120032163344.gif)
解得n=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120031945225.gif)
光导纤维的折射率应为n≥
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120031945225.gif)
![](http://thumb2018.1010pic.com/images/loading.gif)
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