In Young's double-slit interfernce experiment,what is the difference in path length of the light waves from the two slits at the center of the first bright fringe above the central maximum?A.0B.1/4 C.1/2D.1E 3/2注:B,C,D,E项,因为打不出,我漏

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In Young's double-slit interfernce experiment,what is the difference in path length of the light waves from the two slits at the center of the first bright fringe above the central maximum?A.0B.1/4 C.1/2D.1E 3/2注:B,C,D,E项,因为打不出,我漏
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In Young's double-slit interfernce experiment,what is the difference in path length of the light waves from the two slits at the center of the first bright fringe above the central maximum?A.0B.1/4 C.1/2D.1E 3/2注:B,C,D,E项,因为打不出,我漏
In Young's double-slit interfernce experiment,what is the difference in path length of the light waves from the two slits at the center of the first bright fringe above the central maximum?
A.0
B.1/4
C.1/2
D.1
E 3/2
注:B,C,D,E项,因为打不出,我漏了波长的基本单位,中文译音大概是“郎木塔”
说实话,我是一文科生,不太明白这个光的干涉;;还有,回答的时候,能不能注明背景知识///对拉,这是一道单选题

In Young's double-slit interfernce experiment,what is the difference in path length of the light waves from the two slits at the center of the first bright fringe above the central maximum?A.0B.1/4 C.1/2D.1E 3/2注:B,C,D,E项,因为打不出,我漏

额.

这道题问的是光的干涉实验.

不知道你有书没有.就是一道壁上有两个小孔然后光射出小孔发生干涉现象.

然后打到屏幕上的结果就是明暗相间的等距离的平行光纹儿.

题目问的是第一亮纹两个光的光程差是多少.

所有的亮纹的光程差是 N倍的lumbda.n为integer

所有暗纹的光程差是N倍的二分之一lumbda.这里的n是odd.

大概就是这样. 

所以答案为D.

我给你找找图.