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Light with an energy flux of \mathrm{18 \mathrm{~W} / \mathrm{cm}^2} falls on a non-reflecting surface at normal incidence. If the surface has an area of \mathrm{20 \mathrm{~cm}^2}, the average force exerted on the surface during a span 30 minutes

Option: 1

1.2 \times 10^{-6} \mathrm{~N} \\


Option: 2

10^{-3} \mathrm{~N} \\


Option: 3

4 \times 10^{-7} \mathrm{~N} \\


Option: 4

5 \times 10^{-4} \mathrm{~N}


Answers (1)

best_answer

Total energy falling on the surface

                       \begin{aligned} & \text{U}=\left(18 \mathrm{~W} / \mathrm{cm}^2\right)(30 \times 60 \mathrm{~s})\left(20 \mathrm{~cm}^2\right) \\ \\& \text{U}=6.48 \times 10^5 \mathrm{~J} \end{aligned}

Total momentum, 

                       \mathrm{p=\frac{U}{c}=\frac{6.48 \times 10^5}{3 \times 10^8} \mathrm{~kg} \mathrm{~m} / \mathrm{s}}

\mathrm{\Rightarrow \quad \quad \quad\rho=2.16 \times 10^{-3} \mathrm{~kg} \mathrm{~m} / \mathrm{s}}

\therefore     Average force exerted on the surface is 

                    \mathrm{F=\frac{p}{t}=\frac{2.16 \times 10^{-3}}{30 \times 60}=1.2 \times 10^{-6} \mathrm{~N}}

Posted by

Rishabh

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