Question

Light with energy flux 1.2 W/m^2 falls on a non-reflecting surface at normal incidence. The pressure on the plate is
(1) 3 * 10^-8 N/m^2
(2) 2 * 10^-8 N/m^2
(3) 4 * 10^-8 N/m^2
(4) 2 * 10^-9 N/m^2
Please give a proper explanation

Answer 1

Given info : Light with energy flux 1.2 W/m² falls on a non - reflecting surface at normal incidence.

To find : The pressure on the plate is ...

solution : we know, pressure = F/A

and force , F = dp/dt , where p denotes linear momentum.

momentum of light particle, p = E/c

[ ∵ E = mc² and mc²/c = mc = momentum ]

now, dp/dt = d(E/c)/dt = 1/c dE/dt

now pressure = 1/c dE/dt/A = 1/c [dE/A dt ]

1/c [energy flux ]

[ energy flux = rate of change of energy per unit area per unit time ]

here energy flux = 1.2 W/m² , c = 3 × 10^8 m/s

so pressure = 1.2/3 × 10^8 = 0.4 × 10^-8 N/m² = 4 × 10^-9 N/m²

Therefore the pressure on the plate is 4 × 10^-9 N/m²