A) 6.63 ´ 10?27 kg-m/sec
B) 2 ´ 10?27 kg-m/sec
C) 10?27 kg-m/sec
D) None of these
Correct Answer: B
Solution :
The momentum of the incident radiation is given as \[p=\frac{h}{\lambda }\]. When the light is totally reflected normal to the surface the direction of the ray is reversed. That means it reverses the direction of it?s momentum without changing it?s magnitude \ \[\Delta p=2p=\frac{2h}{\lambda }=\frac{2\times 6.6\times {{10}^{-34}}}{6630\times {{10}^{-10}}}\]= 2 ´ 10?27 kg-m/sec.
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