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IV. Numerical Problems · Q3

Q.A pulse of light of duration 10−610^{-6} s is absorbed completely by a small object initially at rest. If the power of the pulse is 60×10−360\times10^{-3} W, calculate the final momentum of the object.

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Step 1. The pulse's total energy is its power times its duration: U=P t=(60×10−3 W)(10−6 s)=60×10−9=6×10−8U=P\,t=(60\times10^{-3}\ \text{W})(10^{-6}\ \text{s})=60\times10^{-9}=6\times10^{-8} J.

Step 2. For electromagnetic radiation completely absorbed by an object, the momentum delivered equals the energy absorbed divided by the speed of light: p=U/cp=U/c.

Step 3. Substituting: p=6×10−83×108=2×10−16p=\dfrac{6\times10^{-8}}{3\times10^8}=2\times10^{-16} kg m s−1^{-1}. …

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