Light with an energy flux of 20 W/cm2 falls on a non-reflecting surface at normal incidence. If the surface has an area of 30 cm2, the total momentum delivered (for complete absorption) during 30 minutes is

Question

Light with an energy flux of 20 W/cm2 falls on a non-reflecting surface at normal incidence. If the surface has an area of 30 cm2, the total momentum delivered (for complete absorption) during 30 minutes is (A) 36 × 10-5 kg m/s (B) 36 × 10-4 kg m/s (C) 108 × 104 kg m/s (D) 1.08 × 107 kg m/s

Tardigrade Admin · Accepted Answer

For complete absorption, p = (U/c) where U = Energy carried by light = IAt ∴ p = (IAt/c) Here, I = 20 W/cm2, A = 30 cm2, t = 30 min = 1800 s So, p = (20 × 30 × 1800/3 × 108) = 36 × 10-4 kg m s-1

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

$36 \times 1 0^{- 5} k g m / s$

$36 \times 1 0^{- 4} k g m / s$

$108 \times 1 0^{4} k g m / s$

$1.08 \times 1 0^{7} k g m / s$

For complete absorption,
$p = \frac{U}{c}$
where $U =$ Energy carried by light $= I A t$
$∴ p = \frac{I A t}{c}$
Here, $I = 20 W / c m^{2}$ , $A = 30 c m^{2}$ , $t = 30 min = 1800 s$
So, $p = \frac{20 \times 30 \times 1800}{3 \times 1 0 ^{8}}$
$= 36 \times 1 0^{- 4} k g m s^{- 1}$

Q. Light with an energy flux of 20W/cm2 falls on a non-reflecting surface at normal incidence. If the surface has an area of 30cm2, the total momentum delivered (for complete absorption) during 30 minutes is

36×10−5kgm/s

36×10−4kgm/s

108×104kgm/s

1.08×107kgm/s