Given that the power of the incident light is $60 \, \text{W}$, the energy per second is $60 \, \text{J}$. The speed of light is $3 \times 10^8 \, \text{m/s}$. Therefore, the momentum of the incident light is
\[p_1 = \frac{60}{3 \times 10^8} = 2 \times 10^{-7} \,
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