Question

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

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

Answer 1

Answer: (A)

Pressure = $\frac{I}{C}$
Force = Pressure $\times$ Area = $\frac{I}{C} $ . Area
Momentum transferred = Force .
$\Delta t = \frac{I}{C}$ . Area . $\Delta t$
$ = \frac{25 \times 10^4}{3 \times 10^8} \times 25 \times 10^{-4} \times 40 \times 60$
$ = 5 \times 10^{-3} \; N - s $