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A light beam of intensity $20 \mathrm{~W} / \mathrm{cm}^{2}$ is incident normally on a perfectly reflecting surface of sides $25 \mathrm{~cm} \times 15 \mathrm{~cm}$. The momentum imparted to the surface by the light per second is
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The correct answer is:
$5 \times 10^{-5} \mathrm{~kg}-\mathrm{ms}^{-1}$
Given, intensity, $I=20 \mathrm{~W} / \mathrm{cm}^{2}$
Area of surface, $A=(25 \times 15) \mathrm{cm}^{2}$
and time, $t=1 \mathrm{~s}$
The intensity of light of energy $E$ is given by
$\begin{aligned} I &=\frac{E}{A t} \Rightarrow 20=\frac{E}{25 \times 15 \times 1} \\ \Rightarrow \quad E &=20 \times 25 \times 15 \mathrm{~J} \end{aligned}$
The momentum imparted to a perfectly reflecting surface is given by
$p=\frac{2 E}{c}=\frac{2 \times 20 \times 25 \times 15}{3 \times 10^{8}}=5 \times 10^{-5} \mathrm{~kg} \mathrm{~ms}^{-1}$
Area of surface, $A=(25 \times 15) \mathrm{cm}^{2}$
and time, $t=1 \mathrm{~s}$
The intensity of light of energy $E$ is given by
$\begin{aligned} I &=\frac{E}{A t} \Rightarrow 20=\frac{E}{25 \times 15 \times 1} \\ \Rightarrow \quad E &=20 \times 25 \times 15 \mathrm{~J} \end{aligned}$
The momentum imparted to a perfectly reflecting surface is given by
$p=\frac{2 E}{c}=\frac{2 \times 20 \times 25 \times 15}{3 \times 10^{8}}=5 \times 10^{-5} \mathrm{~kg} \mathrm{~ms}^{-1}$
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