Question

A laser emits light at power 6.01 mW and wavelength 633 nm. The laser beam is focused (narrowed) until its diameter matches the 1060 nm diameter of a sphere placed in its path. The sphere is perfectly absorbing and has density 5.00 × 10³ kg/m³. What are (a) the beam intensity at the sphere's location, (b) the radiation pressure on the sphere, (c) the magnitude of the corresponding force, and (d) the magnitude of the acceleration that force alone would give the sphere?

Answer 1

(a) We note that the cross section area of the beam is πd²/4, where d is the diameter of the spot (d= 1060 nm). The beam intensity is

I = P/(πd²/4) = 0.00601 x 4/(3.14 x (1060 x 10-9)²) = 6.81 x 10⁹ W/m²

(b) The radiation pressure is

p = I/c = 68.1 x 10⁸/(3 x 10⁸) = 22.7 Pa.

(c) In computing the corresponding force, we can use the power and intensity to eliminate the area (mentioned in part (a)). We obtain

F = Pp/I = 0.00601 x 22.7/6.81 x 10⁹ = 2.00345 x 10^-11 N

(d) The acceleration of the sphere is

a= F/m = F/(ρπd³/6) = 6 x 2.00345 x 10^-11/(5 x 10³ x π x (1060 x 10^-9)³) = 6.4253 x 10⁹ m/s²