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  • A monochromatic beam of light  incident nor

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

Option: 1

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


Option: 2

3 \times 10^{20} \mathrm{~m}^2 / \mathrm{s}


Option: 3

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


Option: 4

None of the above 


Answers (1)

best_answer

\text{P}=[2(0.25)+0.75] \frac{\mathrm{l}}{\mathrm{c}}=1.25 \frac{\mathrm{l}}{\mathrm{c}} \\\therefore \quad \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 Number of photons incident per unit area per second

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

Posted by

Anam Khan

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