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Question: Light with an energy flux of 20W/\(cm^{2}\)fall on a non-reflecting surface at normal incidence. If ...

Light with an energy flux of 20W/cm2cm^{2}fall on a non-reflecting surface at normal incidence. If the surface has an area of 30cm2cm^{2},the total momentum delivered (for complete absorption) during 30 minutes is

A

36×10−5kgm/s36 \times 10^{- 5}kgm/sm/s

B

36×10−4kgm/s36 \times 10^{- 4}kgm/s

C

108×104kgm/s108 \times 10^{4}kgm/s

D

1.08×107kgm/s1.08 \times 10^{7}kgm/s

Answer

36×10−4kgm/s36 \times 10^{- 4}kgm/s

Explanation

Solution

: For complete absorption,P=UcP = \frac{U}{c}

Where U = Energy carried by light = IAt

∴P=IAtc\therefore P = \frac{IAt}{c}

Here, I=20W/cm2,A=30cm2,t=30min⁡=1800sI = 20W/cm^{2},A = 30cm^{2},t = 30\min = 1800s

So, p=20×30×18003×108=36×10−4kgms−1p = \frac{20 \times 30 \times 1800}{3 \times 10^{8}} = 36 \times 10^{- 4}kgms^{- 1}