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Q.

A collimated beam of light of flux density $30 k W m^{- 2}$ is incident normally on a $100 m m^{2}$ completely absorbing screen. If $P$ is the pressure exerted on the scruen and $Δ p$ is the momentum transferred to the screen during a $1000 s$ interval then,

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a

$P = 10^{- 3} N m^{- 2}$

b

$P = 10^{- 4} N m^{- 2}$

c

$4 p = 10^{- 1} k g m s^{- 1}$

d

$Δ p = 10^{- 5} k g m s^{- 1}$

answer is B.

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Detailed Solution

$Pressure P = \frac{I}{c} = \frac{30 \times 10^{3}}{3 \times 10^{8}} = 10^{- 4} {Nm}^{- 2} ∵ P = \frac{F}{A} = \frac{1}{A} \frac{Δ p}{Δ t} Momentum transferred Δ p = P A Δ t = 10^{- 4} \times 100 \times 10^{- 6} \times 1000 = 10^{- 5} {kgms}^{- 1}$

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