4.(1)BCE【解析】由题知b光在AC上发生了全反射,恰好发生全反射时,AC右侧亦无b光射出,此时n0由几何关系可知i=37°,得出n0故棱镜对b光的折射率大于等于一,A错误;k光在AC上发生反射因a光的折射率小于b光,二光以相同的入射角射到界面BC上同一点,故a光在b光的右侧,B正确;若k光经M点斜向下入射,当a光射到AC上入射角大于等于其临界角时,将发生全反射,C正确;b光的折射率大于a光,故b光的频率大于a光,b光的波长小于a光,使用a光做双缝干涉实验时,相邻亮条纹间距较大,D错误;由v=一可知,在棱镜中a光的速度大于b光,故由M至N的过程中,a光的传播时间小于b光,E正确(2)(|)6(0.375+2.25k)m/s(k=0,1i ) a=5t-cos 5rI(m/s)【解析】()由图知振幅A=10cm,x=0处质点的振动方程为y=10sin(t +o)(cm) O令t=0,结合波向x轴正方向传播可知6x2=15cm处质点的振动方程为y=1Osin( ot+A·2x/ccm且A>x2,t=0时,y=-10cm解得A=0.45m③x1=0处的质点,t1=0.2s时有+2kx(k=0,1,2,…)A结合v=④解得v=(0.375+2.25k)m/s(k=0,1,2,……)⑤(i)k=0时,T=1.2s,a=-rrad/s⑥x2处质点的振动方程为y=10sin a即y=-10 cos (t(cm)⑦根据瞬时速度的定义有lim△y△根据加速度的定义有lim△5r25联立⑤⑦③②式解得a=1883(m/s)0评分标准:本题共10分。①②③④⑤⑥⑦⑧⑨⑩式各1分。

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