A small object at rest, absorbs a light pulse of power $$20$$ mW and duration $$300$$ ns. Assuming speed of light as $$3 \times 10^8$$ m s$$^{-1}$$. The momentum of the object becomes equal to:

A small object absorbs a light pulse of power $$20$$ mW and duration $$300$$ ns.

First, we calculate the energy of the light pulse.

$$E = P \times t = 20 \times 10^{-3} \times 300 \times 10^{-9} = 6 \times 10^{-9}$$ J

Next, we calculate the momentum transferred.

When light is completely absorbed by an object, the momentum transferred equals:

$$p = \frac{E}{c} = \frac{6 \times 10^{-9}}{3 \times 10^8} = 2 \times 10^{-17}$$ kg m/s

The momentum of the object becomes $$2 \times 10^{-17}$$ kg m s$$^{-1}$$.

The correct answer is Option 2: $$2 \times 10^{-17}$$ kg m s$$^{-1}$$.