Question: Q. 2. Light with an energy flux of $20 \mathrm{~W} / \mathrm{cm}^{2}$ falls on a non-reflecting surface at normal incidence. If the surface has an area of $30 \mathrm{~cm}^{2}$, the total momentum delivered (for complete absorption) during 30 minutes is (a) $36 \times 10^{-5} \mathrm{~kg} \mathrm{~m} / \mathrm{s}$. (c) $108 \times 10^{4} \mathbf{~ k g ~ m} / \mathrm{s}$. (b) $36 \times 10^{-4} \mathrm{~kg} \mathrm{~m} / \mathrm{s}$. (d) $1.08 \times 10^{7} \mathrm{~kg} \mathrm{~m} / \mathrm{s}$.
[NCERT Exemplar]
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Solution:
Ans. Correct option : (b)
Explanation: Energy flux $=\phi=20 \mathrm{~W} / \mathrm{cm}^{2}$
Area $A=30 \mathrm{~cm}^{2}$, time $t=30 \times 60 \mathrm{sec}$
$U=$ Total energy falling in $t \mathrm{sec}$
$=$ Energy flux $\times$ Area $\times$ time $=\phi A t$
$U=20 \times 30 \times 30 \times 60 \mathrm{~J}$
Momentum of the incident light
$=\frac{U}{c}=\frac{20 \times 30 \times 30 \times 60}{3 \times 10^{8}}=36 \times 10^{-4} \mathrm{~kg}-\mathrm{ms}^{-1}$
As no reflection from the surface and for complete absorption so momentum of reflected radiation is zero.
Momentum delivered to surface
$$ \begin{aligned} & =\text { Change in momentum } \ & =p_{f}-p_{i}=0-36 \times 10^{-4} \mathrm{kgm} / \mathrm{s} \ & =-36 \times 10^{-4} \mathrm{~kg} \mathrm{~m} / \mathrm{s} \end{aligned} $$
(-) sign shows the direction of momentum.
