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A monochromatic beam of light (\lambda=4900 \AA) incident normally upon a surface produces a pressure of 5 \times 10^{-7} \mathrm{~N} / \mathrm{m}^2 on it. Assuming that 25 \% of the light incident is reflected and the rest absorbed, what is the number of photons falling per second on a unit area of thin surface.

Option: 1

3 \times 10^{20} m^{-2} s^{-1}


Option: 2

2 \times 10^{20} \mathrm{~m}^{-2} \mathrm{~s}^{-1}


Option: 3

10^{20} \mathrm{~m}^{-2} \mathrm{~s}^{-1}


Option: 4

4 \times 10^{20} \mathrm{~m}^{-2} \mathrm{~s}^{-1}


Answers (1)

best_answer

Momentum of light falling on the surface per second

P=[2(0.25)+0.75] \frac{\mathrm{I}}{\mathrm{c}}=1.25 \frac{\mathrm{I}}{\mathrm{c}}

\therefore \text { Intensity of light, } \mathrm{I}=\frac{\mathrm{cP}}{1.25}=\frac{\left(3 \times 10^8\right)\left(5 \times 10^{-7}\right)}{1.25}=120 \mathrm{Wm}^{-2}

\text { Energy of photon, } \mathrm{E}=\frac{\mathrm{hc}}{\lambda}=\frac{\left(6.63 \times 10^{-34}\right)\left(3 \times 10^8\right)}{0.49 \times 10^{-6}}=4 \times 10^{-19} \mathrm{~J}

\therefore \quad \text { Number of photons incident per unit area per second }

\mathrm{n}=\frac{I}{E}=\frac{120}{4 \times 10^{-19}}=3 \times 10^{20} \mathrm{~m}^{-2} \mathrm{~s}^{-1}

Posted by

Devendra Khairwa

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