A) \[\frac{E}{c}\]
B) \[\frac{2E}{c}\]
C) \[\frac{2E}{{{c}^{2}}}\]
D) \[\frac{E}{{{c}^{2}}}\]
Correct Answer: B
Solution :
The radiation energy is given by \[E=\frac{hc}{\lambda }\] Initial momentum of the radiation is \[{{\mathbf{P}}_{i}}=\frac{h}{\lambda }=\frac{E}{c}\] The reflected momentum is \[{{\mathbf{P}}_{r}}=-\frac{h}{\lambda }=-\frac{E}{c}\] So, the change in momentum of light is \[\Delta {{\mathbf{P}}_{light}}={{\mathbf{P}}_{r}}-{{\mathbf{P}}_{i}}=-\frac{2E}{c}\] Thus, the momentum transferred to the surface is \[\Delta {{\mathbf{P}}_{light}}=\frac{2E}{c}\]
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