题目内容
(1)一列简谐横波在某一时刻的波形图如图甲所示,图中P、Q两质点的横坐标分别为x=1.5m和x=4.5m.P点的振动图象如图乙所示.![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131028202033370685101/SYS201310282020333706851013_ST/images0.png)
在下列四幅图中,Q点的振动图象可能是______
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131028202033370685101/SYS201310282020333706851013_ST/images1.png)
(2)用氦氖激光器进行双缝干涉实验,已知所用双缝间的距离为d=0.1mm,双缝到屏的距离为l=6.0m,测得屏上干涉条纹中相邻明条纹的间距是3.8cm,氦氖激光器发出的红光的波长是多少?假如把整个干涉装置放入折射率为
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131028202033370685101/SYS201310282020333706851013_ST/0.png)
【答案】分析:(1)根据PQ横坐标之间的距离为3m,知PQ间的距离是波长的
倍,波若向右传播,P点在平衡位置向上振动,则Q点处于波峰;若波向左传播,P点在平衡位置向下振动,则Q点处于波谷.
(2)根据
,可得红光的波长.进入水中,红光的波长发生改变,则明条纹的间距发生变化.
解答:解:(1)PQ横坐标之间的距离为3m,是波长的
倍.波若向右传播,P点在平衡位置向上振动,则Q点处于波峰;若波向左传播,P点在平衡位置向下振动,则Q点处于波谷.故B、C正确,A、D错误.
故选BC.
(2)根据
,红光的波长为:
=
.
因为
,![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131028202033370685101/SYS201310282020333706851013_DA/7.png)
则红光进入水中的波长![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131028202033370685101/SYS201310282020333706851013_DA/8.png)
所以
′=
=2.85×10-2m
故屏上的明条纹间距是2.85×10-2m.
点评:解决本题的关键掌握波动和振动的关系,以及条纹间距公式
.
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131028202033370685101/SYS201310282020333706851013_DA/0.png)
(2)根据
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131028202033370685101/SYS201310282020333706851013_DA/1.png)
解答:解:(1)PQ横坐标之间的距离为3m,是波长的
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131028202033370685101/SYS201310282020333706851013_DA/2.png)
故选BC.
(2)根据
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131028202033370685101/SYS201310282020333706851013_DA/3.png)
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131028202033370685101/SYS201310282020333706851013_DA/4.png)
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131028202033370685101/SYS201310282020333706851013_DA/5.png)
因为
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131028202033370685101/SYS201310282020333706851013_DA/6.png)
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131028202033370685101/SYS201310282020333706851013_DA/7.png)
则红光进入水中的波长
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131028202033370685101/SYS201310282020333706851013_DA/8.png)
所以
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131028202033370685101/SYS201310282020333706851013_DA/9.png)
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131028202033370685101/SYS201310282020333706851013_DA/10.png)
故屏上的明条纹间距是2.85×10-2m.
点评:解决本题的关键掌握波动和振动的关系,以及条纹间距公式
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131028202033370685101/SYS201310282020333706851013_DA/11.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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