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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