JEE Main · 2021 · 17 Mar 2021 · Shift 1 · EM Waves

2.5 \times 1 0^{- 6} N

average force is exerted by a light wave on a non- reflecting surface of

30 cm^{2}

area during 40 minutes of time span, the energy flux of light just before it falls on the surface is ______

W / cm^{2}

. (Round off to the Nearest Integer)
(Assume complete absorption and normal incidence conditions are there)

Answer: 25

Solution

For complete absorption at normal incidence the radiation pressure is

P = \frac{I}{c}

, where

I

is the energy flux. The force on the surface is

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

I = \frac{F c}{A} = \frac{( 2.5 \times 1 0 ^{- 6} ) ( 3 \times 1 0 ^{8} )}{30} = 25 W c m^{- 2}

with the area left in

c m^{2}

because the flux is asked for per

c m^{2}

. The quoted force is already a time average, so the

40

minute span does not enter. The energy flux of the light is

25 W c m^{- 2}

JEE Main · 2021 · 17 Mar 2021 · Shift 1 · Integer answer