Light is incident normally on a completely absorbing surface with an energy flux of $25 \mathrm{~W} \mathrm{~cm}^{-2}$. If the surface has an area of $25 \mathrm{~cm}^{2}$, the maximum transferred to the surface in 40 min time duration will be

(C) $5.0 \times 10^{-3} \mathrm{Ns}$

(A) $6.3 \times 10^{-4} \mathrm{Ns}$

(B) $3.5 \times 10^{-6} \mathrm{Ns}$

(D) $1.4 \times 10^{-6} \mathrm{Ns}$

JEE Main · 2019 · 10 Apr 2019 · Shift 2 · EM Waves

Light is incident normally on a completely absorbing surface with an energy flux of

25 W cm^{- 2}

. If the surface has an area of

25 cm^{2}

, the maximum transferred to the surface in 40 min time duration will be

(A) $6.3 \times 1 0^{- 4} Ns$
(B) $3.5 \times 1 0^{- 6} Ns$
(C) $5.0 \times 1 0^{- 3} Ns$ ✓ Correct
(D) $1.4 \times 1 0^{- 6} Ns$

Answer: (C)

Solution

The whole beam lands on the surface, so the incident power is

P = 25 W cm^{- 2} \times 25 cm^{2} = 625 W

. Over

40

minutes the energy delivered is

E = P t = 625 \times 40 \times 60 = 1.5 \times 1 0^{6} J

. For a perfectly absorbing surface each unit of energy carries momentum

E / c

, so the momentum transferred is

\frac{1.5 \times 1 0 ^{6}}{3 \times 1 0 ^{8}} = 5.0 \times 1 0^{- 3} Ns

JEE Main · 2019 · 10 Apr 2019 · Shift 2 · Single correct option

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