A helium-neon laser of the type often found in physicslaboratories has a beam po

Question

A helium-neon laser of the type often found in physicslaboratories has a beam power of 5.50 mW at a wavelength of 633 nm.The beam is focused by a lens to a circular spot whose effectivediameter may be taken to be equal to 2.00 wavelengths.
Calculate the intensity of the focused beam.
Calculate the radiation pressure exerted ona tiny perfectly absorbing sphere whose diameter is that of thefocal spot.

Calculate the force exerted on thissphere.

Calculate the magnitude of the accelerationimparted to it. Assume a sphere density of3.80×103kg/m3
A helium-neon laser of the type often found in physicslaboratories has a beam power of 5.50 mW at a wavelength of 633 nm.The beam is focused by a lens to a circular spot whose effectivediameter may be taken to be equal to 2.00 wavelengths.
Calculate the intensity of the focused beam.
Calculate the radiation pressure exerted ona tiny perfectly absorbing sphere whose diameter is that of thefocal spot.

Calculate the force exerted on thissphere.

Calculate the magnitude of the accelerationimparted to it. Assume a sphere density of3.80×103kg/m3

Explanation / Answer

The beam power of the helium-neon laser is P = 5.50 mW = 5.50* 10-3 W The beam power is at a wavelength of = 633 nm = 633 *10-9 m The effective diameter of the circular spot is d =2 The intensity of the focussed beam is I = (P/A) ----------------(1) Here,A = (d2/4) = ( *(2)2/4) = ( * 2) Substituting the above value of A in equation (1),we get I = (P/( * 2)) Substituting the values in the above equation,we get I = (5.50 * 10-3/(3.14 * (633 *10-9)2) or I = 4.37 * 109 W/m2 The radiation pressure exerted on a tiny perfectly absorbingsphere whose diameter is that of the focal spot is p = (I/c) -----------------(2) Here,c = 3 * 108 m/s Substituting the values in equation (2),we get p = (4.37 * 109/3 * 108) or p = 14.57 Wm-3s-1 Let the force exerted on this sphere be F,therefore,weget p = (F/A) or F = p * A = p * 4(d2/4) = p *d2 or F = 14.57 * 3.14 * (2 * 633 *10-9)2 or F = 0.073 N Let the magnitude of the acceleration imparted to the spherebe a,therefore,we get F = m * a or a = (F/m) ------------------(3) Here,m = * V = * (4/3)(d/2)3 = * (4/3)(2/2)3 =(4/3)3 Here, = 3.80 * 103kg/m3 or m = (4/3) * 3.80 * 103 * 3.14 * (633 *10-9)3 or m = 0.4 * 10-14 kg Substituting the above values in equation (3),we get a = (0.073/0.4 * 10-14) = 1825 * 1010m/s2 Here, = 3.80 * 103kg/m3 or m = (4/3) * 3.80 * 103 * 3.14 * (633 *10-9)3 or m = 0.4 * 10-14 kg Substituting the above values in equation (3),we get a = (0.073/0.4 * 10-14) = 1825 * 1010m/s2 or m = 0.4 * 10-14 kg Substituting the above values in equation (3),we get a = (0.073/0.4 * 10-14) = 1825 * 1010m/s2

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