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

energy is 36 mJ

radiation pressure exerted by the beam is 1.9254 ×
`10^(-6)` Pa

Step-by-step 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