Question

Light is incident normally on a completely absorbing surface with an energy flux of $25{\text{Wcm}}^{-2}$. If the surface has an area of $25{\text{cm}}^2,$ the momentum transferred to the surface in $40\text{min}$ time duration will be:

• A. $1.4\times {10}^{-6}\text{Ns}$
• B. $3.5\times {10}^{-6}\text{Ns}$
• C. $5.0\times {10}^{-3}\text{Ns}$
• D. $6.3\times {10}^{-4}\text{Ns}$

Answer 1

The correct option is C $5.0\times {10}^{-3}\text{Ns}$

Given:
$I=25{\text{Wcm}}^{-2}$.
$A=25{\text{cm}}^2,$
time $\Delta t=40\text{min}$
Pressure , $P=\frac{I}{C}$

$\Rightarrow \frac{F}{A}=\frac{I}{C}$ $(\because P=F/A)$

$\Rightarrow F=\frac{IA}{C}$

Momentum transferred

$\Rightarrow \Delta p=F\Delta t$

$\Rightarrow \Delta p=\frac{I}{C}A\Delta t=\frac{(25\times {10}^4)\times 25\times {10}^{-4}\times 40\times 60}{3\times {10}^8}\text{N-s}$

$\Rightarrow \Delta p=5\times {10}^{-3}\text{Ns}$

Hence option $\left(C\right)$ is correct.