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Physics Question on Electromagnetic waves

A light beam of intensity 20 W/cm220\, W/cm^2 is incident normally on a perfectly reflecting surface of sides 25 cm×15 cm25 \,cm \times 15 \,cm. The momentum imparted to the surface by the light per second is

A

2×10−5kg ms−12\times10^{-5} kg \,ms^{-1}

B

1×10−5kg ms−11\times10^{-5} kg\, ms^{-1}

C

5×10−5kg ms−15\times10^{-5} kg \,ms^{-1}

D

1.2×10−5kg ms−11.2\times10^{-5} kg\, ms^{-1}

Answer

5×10−5kg ms−15\times10^{-5} kg \,ms^{-1}

Explanation

Solution

Intensity I=E/AI = E / A
where EE is energy of radiation and AA is incident area
⇒E=IA\Rightarrow E = IA
Momentum of radiation is given by
P=2E/c=2IA/cP =2 E / c =2 IA / c
Where cc is speed of light.
P=2×20×104Wm2×375×10−4m23×108m/sP=\frac{2 \times 20 \times 10^{4} \frac{ W }{ m ^{2}} \times 375 \times 10^{-4} m ^{2}}{3 \times 10^{8} m / s }
=5×10−5kg ms−1=5 \times 10^{-5} kg\,ms ^{-1}