If \[2.5 \times {10^{ - 6}}N\] average force is exerted by a light wave on a non-reflecting surface of \[30c{m^2}\] area during \[40\]minutes of time span, the energy flux of light just before it falls on the surface is _______ \[W/c{m^2}\].
(Round off to the nearest integer) (Assume complete absorption and normal incidence conditions are there)

Hint: Whenever these forms of questions are given we have to understand that the concept here used is to solve the question for finding out the energy flux intensity of the light incident on the surface which is non-reflecting. Non-reflecting here means this the photons in the light energy are totally absorbed in the surface.

Formula used:
\[F = \dfrac{{IA}}{C}\]

Complete answer:
Given,
\[F = 2.5 \times {10^{ - 6}}~N\]
\[Area = A = 30~c{m^2}\],
\[time,T = 40~{minutes}\]

Here, we have to draw a diagram showing the phenomenon occurring according to the question.

Then let us use a formula for the force exerted on the surface by the light wave as below:
\[F = \dfrac{{IA}}{C}\]
\[ \ldots {\rm{ }}(I - {\rm{ }}intensity{\rm{ }}of{\rm{ }}light{\rm{ }}wave,C - {\rm{ }}speed{\rm{ }}of{\rm{ }}light)\]

Now, placing all the values in the equation above, we get
\[I = \dfrac{{FC}}{A}\]
\[I = \dfrac{{2.5 \times {{10}^6} \times 3 \times {{10}^8}}}{{30 \times {{10}^{ - 4}}}}\]
\[I = 2.5 \times {10^{ - 6}} \times {10^{11}}\]
\[I = 2.5 \times {10^5}\]
\[I = 25 \times {10^4}watt/{m^2}\]

But it has been asked in centimetre unit, so we have to convert meter to the centimetre, so that
\[I = 25 \times \dfrac{{{{10}^4}}}{{{{10}^4}}}watt/c{m^2}\]
\[I = 25W/c{m^2}\]

So, the intensity of the light wave is \[25W/c{m^2}\].

Note: Let us understand the concept used here, light is totally observed since, it is mentioned in the question that the surface is non-reflecting that means there will be no reflection considered. Hence, we just have to apply the formula and do the proper calculation. Must remember that give answer in the asked unit only.