Ncert Exercises & Solutions

Light with an energy flux of $20~\text{W/cm}^2$ falls on a non-reflecting surface at normal incidence. If the surface has an area of $30~\text{cm}^2$ , the momentum delivered (for complete absorption) during $30$ minutes is:
1. $36\times10^{-5}~\text{kg-m/s}$
2. $36\times10^{-4}~\text{kg-m/s}$
3. $108\times10^{4}~\text{kg-m/s}$
4. $1.08\times10^{7}~\text{kg-m/s}$

(b) Hint: The omentum transferred depends on the energy transferred to the wall.

Step 1: Given, energy flux,

ϕ

$ϕ$ = 20 W/

{cm}^{2}

${cm}^{2}$

Area, A= 30

{cm}^{2}

${cm}^{2}$

Time, t = 30 min = 30 x 60 sec
Now, the total energy falling on the surface in time is, U =

ϕ

$ϕ$ At =20 x 30 x (30 x 60) J

Step 2: The momentum of the incident light =

\frac{U}{c}

$\frac{U}{c}$

= \frac{20 \times 30 \times (30 \times 60)}{3 \times 10^{8}} \Rightarrow 36 \times 10^{- 4} kg - {ms}^{- 1}

$= \frac{20 \times 30 \times (30 \times 60)}{3 \times 10^{8}} \Rightarrow 36 \times 10^{- 4} kg - {ms}^{- 1}$

Step 3: Momentum of the reflected light = 0

$∴$ Momentum delivered to the surface

$= 36 \times 10^{- 4} - 0 = 36 \times 10^{- 4} kg - {ms}^{- 1}$

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