Average force exerted on a non-reflecting surface at normal incidence is $2.4 \times 10^{-4} \mathrm{~N}$. If $360 \mathrm{~W} / \mathrm{cm}^{2}$ is the light energy flux during span of 1 hour 30 minutes. Then the area of the surface is:

JEE Main · 2024 · 08 Apr 2024 · Shift 1 · EM Waves

Average force exerted on a non-reflecting surface at normal incidence is

2.4 \times 1 0^{- 4} N

. If

360 W / cm^{2}

is the light energy flux during span of 1 hour 30 minutes. Then the area of the surface is:

(A) $0.2 m^{2}$
(B) $0.02 m^{2}$ ✓ Correct
(C) $20 m^{2}$
(D) $0.1 m^{2}$

Answer: (B)

Solution

For a perfectly absorbing surface at normal incidence the radiation pressure is

\frac{I}{c} = \frac{F}{A}

, so

A = \frac{F c}{I}

.
With

I = 360 W/cm^{2} = 3.6 \times 1 0^{6} W/m^{2}

A = \frac{( 2.4 \times 1 0 ^{- 4} ) ( 3 \times 1 0 ^{8} )}{3.6 \times 1 0 ^{6}} = 0.02 m^{2}

Option (B). (The 1 h 30 min exposure is irrelevant to the instantaneous force.)

JEE Main · 2024 · 08 Apr 2024 · Shift 1 · Single correct option

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