如图所示,高度为H=1m圆柱形容器中盛满折射率n=2的某种透明液体,容器底部安装一块平面镜,容器直径L=2H 如图所示,一不透明的圆柱形容器内装满折射率n= ...

\u9ad8\u4e2d\u7269\u7406 \u6298\u5c04\u7387

\u8bbe\u5bb9\u5668\u9ad8\u4e3aH\uff0e\u4f5c\u56fe\u627e\u51fa\u53d1\u5149\u70b9S\u5728\u5e73\u9762\u955c\u4e2d\u7684\u50cf\u70b9S\u2032\uff0c\u8fde\u63a5SM\u5ef6\u957f\u4ea4\u76f4\u5c3a\u4e8eH\u70b9\uff0e \u7531\u9898\uff1a\u5149\u7ebfSM\u7684\u5165\u5c04\u89d2\u4e3a45\u00b0\uff0c\u6839\u636e\u53cd\u5c04\u5b9a\u5f8b\u53ef\u77e5\u53cd\u5c04\u89d2\u7b49\u4e8e45\u00b0\uff0c\u5219MH\u6cbf\u7ad6\u76f4\u65b9\u5411\uff0e \u8fde\u63a5S\u2032P\uff0c\u6839\u636e\u5bf9\u79f0\u6027\u5f97 S\u2032M=OM=1m\uff0c\u5219\u5728Rt\u25b3PR S\u2032\u4e2d\uff0cR S\u2032= 3 dm\uff0cPR=H+S\u2032M=3dm \u7531\u51e0\u4f55\u77e5\u8bc6\u5f97\u2220r=30\u00b0\uff0e \u6839\u636e\u6298\u5c04\u5b9a\u5f8b\u53ef\u5f97\uff1an= sini sinr \u89e3\u5f97sini= 2 2 \uff0c\u2220i=45\u00b0\uff0c \u6545\u523b\u5ea6\u5c3a\u4e0a\u88ab\u7167\u4eae\u7684\u8303\u56f4QH=1dm+ 3 dm=\uff081+ 3 \uff09dm\u7b54\uff1a\u523b\u5ea6\u5c3a\u4e0a\u88ab\u7167\u4eae\u7684\u8303\u56f4QH=1dm+ 3 dm=\uff081+ 3 \uff09dm\uff0e

解答:解:(1)由n=
c
v
得,点光源S发出的光在水中传播的速度:v=
c
n
=
3×108
2
=1.5×108m/s;
(2)点光源S通过平面镜所成像为S′,如图所示,如果反射光线能照亮全部液面则入射角应满足i≤C,C为全反射临界角.
由 sinC=
1
n
,C=30°
tani=
L
2(H+h)
≤tanC
,L=2H
解得:h≥(


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