Light with an energy flux of $20W/c{m}^2$ falls on a non-reflecting surface at normal incidence. If the surface has an area of $30c{m}^2,$ the total momentum delivered (for complete absorption) for $30$ minutes is:

1. $1.08\times {10}^{7}kgm/s.$
2. $108\times {10}^{4}kgm/s.$
3. $36\times {10}^{-5}kgm/s.$
4. $36\times {10}^{-4}kgm/s.$

The correct option is D $36\times {10}^{-4}kgm/s.$
Step 1: Find the Energy of incident light.

Formula Used: $I=\frac{E}{At}$

As we know,

the intensity of the light incident is given by :

$I=\frac{E}{At}...\left(i\right)$


Given:

Intensity $\left(I\right)=20W/c{m}^2$

Area of cross section, $A=30c{m}^2,$

Time of absorption,

$\left(t\right)=30mins=30\times 60sec=1800sec$

Therefore, by putting the values in equation $\left(i\right)$ we get,

$20=\frac{E}{30\times 1800}$

$E=20\times 30\times 1800$

$E=1.08\times {10}^{6}J$


Step 2: Find the momentum of incident light.

As we know, $E=\frac{hc}{\lambda }$

And
$\lambda =\frac{h}{p}$

Therefore, $E=p\times c$

$p=\frac{E}{c}$

where, $c=3\times {10}^{8}m/s$

By putting the values in $p,$ we get Momentum will be given by,

$p=\frac{20\times 30\times 1800}{3\times {10}^8}$

$p=36\times {10}^{-4}kgm/s$

Final Answer: Option (b)