问题 计算题

已知光在真空中的传播速度C=3×108m/s。如图,折射率n=1.5某玻璃立方体放在空气中。光线从立方体的顶面斜射进来,然后投射到它的另一个侧面P点,问:

① 光在玻璃中的传播速度V;

②此光线能否从另一个侧面P点射出?请说明你判断的依据。

答案

①2×108  (m/s)②光线不能在P点射出 

① 光在玻璃中的传播速度V

V="C" / n =3×108 / 1.5 (m/s)= 2×108  (m/s)   (1分)

② 该玻璃发生全反射的临界角C

      故C<450    (1分)

光线在B点发生折射,由折射定律知

  故r<450   (1分)

由几何关系知i1=900-r>450 >C            (1分)

所以光线在P点一定发生全反射,

光线不能在P点射出                   (1分)

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