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A perfectly reflecting mirror has an area of 1 \mathrm{~cm}^2 Light energy is allowed to fall on it for 1 \mathrm{~h} at the rate of \mathrm{10 \mathrm{Wcm}^{-2}.} The force that acts on the mirror is

Option: 1

3.35 \times 10^{-8} \mathrm{~N}


Option: 2

6.7 \times 10^{-8} \mathrm{~N}


Option: 3

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


Option: 4

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


Answers (1)

best_answer

Let E= Energy falling on the surface per second = \mathrm{10 \mathrm{~J} }
Momentum Of photons
\mathrm{\begin{aligned} p & =\frac{h}{\lambda}=\frac{h}{\left(c_1 v\right)} \\ & =\frac{h v}{c}=\frac{E}{c} \end{aligned} }
On reflection,
Change in momentum per second = \mathrm{2 p=\frac{2 E}{c} }
We know that,
Change in momentum per second = force
\mathrm{\begin{aligned} F & =\frac{2 E}{c}=\frac{2 \times 10}{3 \times 10^8} \\ & =6.7 \times 10^{-8} \mathrm{~N} \end{aligned} }

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Deependra Verma

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