Question

A laser emits a cylindrical beam of light 2.3 mm in diameter. The average power of the laser is 2.4 mW . The laser shines its light on a perfectly absorbing surface. Part A How much energy does the surface receive in 15 s ?
Part B What is the radiation pressure exerted by the beam?

Answer 1

Answer:

energy is 36 mJ

radiation pressure exerted by the beam is 1.9254 × $10^{-6}$ Pa

Explanation:

Given data

diameter = 2.3 mm = 2.3 × $10^{-3}$ m

power  = 2.4 mW = 2.4 × $10^{-3}$ J/s

time = 15 s

to find out

energy and radiation pressure exerted

solution

we know for energy

that is energy = power × time

energy =  2.4 × $10^{-3}$ × 15

energy = 36 × $10^{-3}$ J

so energy is 36 mJ

and

now first we find cross section area that i s

cross section Area A = πd²/4

A = π(2.3 × $10^{-3}$)²/4

A = 4.155 × $10^{-6}$ m²

and intensity of radiation will be power / area

so intensity of radiation = 2.4 × $10^{-3}$ /  4.155 × $10^{-6}$

intensity of radiation = 577.62 W/m²

so that now for radiation pressure for beam that is

pressure = intensity of radiation / speed of light

we know speed of light is 3 × $10^{8}$ m/s

so

pressure = 577.62 / 3 × $10^{8}$

pressure = 1.9254 × $10^{-6}$ Pa

so radiation pressure exerted by the beam is 1.9254 × $10^{-6}$ Pa