Saved Bookmarks
| 1. |
A radiation of energy `E` falls normally on a perfctly refelecting surface . The momentum transferred to the surface isA. `E//c`B. `2E//c`C. `Ec`D. `E//c^(2)` |
|
Answer» Correct Answer - B Initial momentum of surface `P_(i) = (E)/(C)` where c = velocity of light (constant). Since, the surface is perfectly reflecting so the same momentum will be reflected completely Final momentum `P_(i)=(E)/(C)`(negative value) `:.` Change in momentum `Delta_(p) = p_(f)-P_(i) = -(E)/(C)-(E)/(C) = -(2E)/(C)` Thus, momentum transferred to the surface is `Delta_(p)=|Delta_(p)|=(2E)/(C)`. |
|