ps10soln

# ps10soln - ELEC 262 Problem set 10 solutions 1(a The...

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ELEC 262 – Problem set 10 solutions 1(a) The coherence time is the inverse of the spectral width, and the coherence length is c 0 divided by the coherence time, L c = c 0 / ∆ν . In this case, we compute ∆ν according to: () 210 2 1 11 c ∆ν = ν − ν = λ − λ , so 1 8 1 632.799 1 632.801 2 10 nm 0.2 m =− = × = c L . (b) We have seen that the spacing between longitudinal modes in a laser is given by the inverse of the cavity round-trip time. Since the round-trip length is 0.6 m, we can find the mode spacing by 8 0 15 1 0 H z δν = τ = = × RT RT cL . Divide this into ∆ν to find the number of modes: 3 ∆ν δν = . Only three distinct modes can oscillate in this laser. 2. The de Broglie wavelength of the electrons is related to their momentum by λ= = hp hm v . The energy of each electron is 30 keV, which is the same as 4.8 × 10 -15 Joules. We can compute the velocity v because the energy of the wave is 2 1 2 mv . Thus we have the velocity as 8 21 0 m / s == vE m , which means that the wavelength is 7.3 × 10 -12 meters. The diffraction- limited spot size for a focused wave is 2 λ (f/#). In this case, that works out to 60 λ , or 0.44 nanometers. A value of 1.22 λ (f/#) = 0.27 nanometers is also an acceptable answer here, since the problem does not specify whether the electron beam is to be treated as a Gaussian beam or a plane wave.

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ps10soln - ELEC 262 Problem set 10 solutions 1(a The...

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