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  • A light beam of power and wavelength <img alt="5000 \mathrm{~A}^{\circ}" src="https://entra

A light beam of power 12 \mathrm{mw} and wavelength 5000 \mathrm{~A}^{\circ} falls normally on a partially reflecting plate, which reflects 40 \% of the light falling on it and absorbs the rest. Then the force exerted on the plate is:

Option: 1

11.2 \times 10^{-11} \mathrm{~N}.


Option: 2

5.6 \times 10^{-11} \mathrm{~N}.


Option: 3

3.6 \times 10^{-11} \mathrm{~N}.


Option: 4

2.8 \times 10^{-11} \mathrm{~N}.


Answers (1)

best_answer

No. of photons falling on the plate per sec., n=\frac{P}{(h c / \lambda)}

\begin{aligned} & \text { No. of reflected photon per sec. }=0.4 \mathrm{n} \\\\ & \text { No. of absorbed photons per sec. }=0.6 \mathrm{n} \\\\ & \text { The force exerted by reflected photons }=0.4 \mathrm{n}(2 \mathrm{~h} / \lambda) \\\\ & \text { The force exerted by absorbed photons }=0.6 \mathrm{n}(\mathrm{h} / \lambda) \\\\ & \therefore \quad \text { Net force }=(0.4 \times 2+0.6) \mathrm{n}(\mathrm{h} / \lambda) \\\\ & =1.4\left(\frac{\mathrm{P} \lambda}{\mathrm{hc}}\right) \frac{\mathrm{h}}{\lambda}=1.4 \frac{\mathrm{P}}{\mathrm{c}}=\frac{1.4 \times 12 \times 10^{-3}}{3 \times 10^8}=5.6 \times 10^{-11} \mathrm{~N} . \end{aligned}

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Nehul

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