Problem : Compute the energy density of a 2.0 mW laser beam with a diameter of 2mm.The energy outputted by the laser per unit time (second) is 0.002 J/s. The area of the beam is Πr2 = Π×(0.001)2 = 3.14×10-6 m2. We want the energy per volume, so just by dimensional analysis we can say J/m2 = J/s×1/m2×s/m, so the energy density is given by u = power×1/area×1/c = 0.002× = 2.12×10-6 J/m3.
Problem : The power per unit area incident on the earth due to sunlight is about 1.4 kW/m2. What is the approximate magnitude of the electric field vector on sunlight near the earth (μ0âÉá4Π×10-7 N.s2/C2)?Since the electric and magnetic field vectors are perpendicular, we can write the expression for the incident power per unit area as S = . We can assume the fields are of equal magnitude, so E2E = 4.19×10-2 V/m.
Problem : Write an expression (in SI units) for the magnetic field of a 10 MHz light wave traveling in the negative z direction and an electric field amplitude of .First let us consider the direction of the wave. We know that must yield the direction and is given as in the positive x direction. Thus must point in the negative y direction by the right hand rule. Thus we want an equation of the form . The amplitude must be given by B0 = E0/c = 0.1/3.0×108 = 3.33×10-10. σ = 2Πν = 2Π×107 and k = 2Π/λ = 2Πν/c = 2Π×0.033. So the expression is:
Problem : Consider the following electric and magnetic waves:
Problem : Consider a 60W light bulb. Assume that all radiation given off is in the form of light of wavelength 650 nm. Calculate the number of photons given off per second.Each second the total energy given off is 60 Joules. Each photon has an energy given by E = hν where ν = c/λ = 3.0×108/(650×10-9) = 4.62×1014 Hz. So E = 6.626×10-34×4.62×1014 = 3.06×10-19 Joules per photon. Thus the total number of photons must be 60/(3.06×10-19) 1.96×1020 photons emitted every second!
Take a Study Break!